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48\makeindex
49\begin{document}
50\frontmatter
51\pagestyle{empty}
52\title{Multi--Precision Math}
53\author{\mbox{
54%\begin{small}
55\begin{tabular}{c}
56Tom St Denis \\
57Algonquin College \\
58\\
59Mads Rasmussen \\
60Open Communications Security \\
61\\
62Greg Rose \\
63QUALCOMM Australia \\
64\end{tabular}
65%\end{small}
66}
67}
68\maketitle
69This text has been placed in the public domain.  This text corresponds to the v0.39 release of the 
70LibTomMath project.
71
72\begin{alltt}
73Tom St Denis
74111 Banning Rd
75Ottawa, Ontario
76K2L 1C3
77Canada
78
79Phone: 1-613-836-3160
80Email: tomstdenis@gmail.com
81\end{alltt}
82
83This text is formatted to the international B5 paper size of 176mm wide by 250mm tall using the \LaTeX{} 
84{\em book} macro package and the Perl {\em booker} package.
85
86\tableofcontents
87\listoffigures
88\chapter*{Prefaces}
89When I tell people about my LibTom projects and that I release them as public domain they are often puzzled.  
90They ask why I did it and especially why I continue to work on them for free.  The best I can explain it is ``Because I can.''  
91Which seems odd and perhaps too terse for adult conversation. I often qualify it with ``I am able, I am willing.'' which 
92perhaps explains it better.  I am the first to admit there is not anything that special with what I have done.  Perhaps
93others can see that too and then we would have a society to be proud of.  My LibTom projects are what I am doing to give 
94back to society in the form of tools and knowledge that can help others in their endeavours.
95
96I started writing this book because it was the most logical task to further my goal of open academia.  The LibTomMath source
97code itself was written to be easy to follow and learn from.  There are times, however, where pure C source code does not
98explain the algorithms properly.  Hence this book.  The book literally starts with the foundation of the library and works
99itself outwards to the more complicated algorithms.  The use of both pseudo--code and verbatim source code provides a duality
100of ``theory'' and ``practice'' that the computer science students of the world shall appreciate.  I never deviate too far
101from relatively straightforward algebra and I hope that this book can be a valuable learning asset.
102
103This book and indeed much of the LibTom projects would not exist in their current form if it was not for a plethora
104of kind people donating their time, resources and kind words to help support my work.  Writing a text of significant
105length (along with the source code) is a tiresome and lengthy process.  Currently the LibTom project is four years old,
106comprises of literally thousands of users and over 100,000 lines of source code, TeX and other material.  People like Mads and Greg 
107were there at the beginning to encourage me to work well.  It is amazing how timely validation from others can boost morale to 
108continue the project. Definitely my parents were there for me by providing room and board during the many months of work in 2003.  
109
110To my many friends whom I have met through the years I thank you for the good times and the words of encouragement.  I hope I
111honour your kind gestures with this project.
112
113Open Source.  Open Academia.  Open Minds.
114
115\begin{flushright} Tom St Denis \end{flushright}
116
117\newpage
118I found the opportunity to work with Tom appealing for several reasons, not only could I broaden my own horizons, but also 
119contribute to educate others facing the problem of having to handle big number mathematical calculations.
120
121This book is Tom's child and he has been caring and fostering the project ever since the beginning with a clear mind of 
122how he wanted the project to turn out. I have helped by proofreading the text and we have had several discussions about 
123the layout and language used.
124
125I hold a masters degree in cryptography from the University of Southern Denmark and have always been interested in the 
126practical aspects of cryptography. 
127
128Having worked in the security consultancy business for several years in S\~{a}o Paulo, Brazil, I have been in touch with a 
129great deal of work in which multiple precision mathematics was needed. Understanding the possibilities for speeding up 
130multiple precision calculations is often very important since we deal with outdated machine architecture where modular 
131reductions, for example, become painfully slow.
132
133This text is for people who stop and wonder when first examining algorithms such as RSA for the first time and asks 
134themselves, ``You tell me this is only secure for large numbers, fine; but how do you implement these numbers?''
135
136\begin{flushright}
137Mads Rasmussen
138
139S\~{a}o Paulo - SP
140
141Brazil
142\end{flushright}
143
144\newpage
145It's all because I broke my leg. That just happened to be at about the same time that Tom asked for someone to review the section of the book about 
146Karatsuba multiplication. I was laid up, alone and immobile, and thought ``Why not?'' I vaguely knew what Karatsuba multiplication was, but not 
147really, so I thought I could help, learn, and stop myself from watching daytime cable TV, all at once.
148
149At the time of writing this, I've still not met Tom or Mads in meatspace. I've been following Tom's progress since his first splash on the 
150sci.crypt Usenet news group. I watched him go from a clueless newbie, to the cryptographic equivalent of a reformed smoker, to a real
151contributor to the field, over a period of about two years. I've been impressed with his obvious intelligence, and astounded by his productivity. 
152Of course, he's young enough to be my own child, so he doesn't have my problems with staying awake.
153
154When I reviewed that single section of the book, in its very earliest form, I was very pleasantly surprised. So I decided to collaborate more fully, 
155and at least review all of it, and perhaps write some bits too. There's still a long way to go with it, and I have watched a number of close 
156friends go through the mill of publication, so I think that the way to go is longer than Tom thinks it is. Nevertheless, it's a good effort, 
157and I'm pleased to be involved with it.
158
159\begin{flushright}
160Greg Rose, Sydney, Australia, June 2003. 
161\end{flushright}
162
163\mainmatter
164\pagestyle{headings}
165\chapter{Introduction}
166\section{Multiple Precision Arithmetic}
167
168\subsection{What is Multiple Precision Arithmetic?}
169When we think of long-hand arithmetic such as addition or multiplication we rarely consider the fact that we instinctively
170raise or lower the precision of the numbers we are dealing with.  For example, in decimal we almost immediate can 
171reason that $7$ times $6$ is $42$.  However, $42$ has two digits of precision as opposed to one digit we started with.  
172Further multiplications of say $3$ result in a larger precision result $126$.  In these few examples we have multiple 
173precisions for the numbers we are working with.  Despite the various levels of precision a single subset\footnote{With the occasional optimization.}
174 of algorithms can be designed to accomodate them.  
175
176By way of comparison a fixed or single precision operation would lose precision on various operations.  For example, in
177the decimal system with fixed precision $6 \cdot 7 = 2$.
178
179Essentially at the heart of computer based multiple precision arithmetic are the same long-hand algorithms taught in
180schools to manually add, subtract, multiply and divide.  
181
182\subsection{The Need for Multiple Precision Arithmetic}
183The most prevalent need for multiple precision arithmetic, often referred to as ``bignum'' math, is within the implementation
184of public-key cryptography algorithms.   Algorithms such as RSA \cite{RSAREF} and Diffie-Hellman \cite{DHREF} require 
185integers of significant magnitude to resist known cryptanalytic attacks.  For example, at the time of this writing a 
186typical RSA modulus would be at least greater than $10^{309}$.  However, modern programming languages such as ISO C \cite{ISOC} and 
187Java \cite{JAVA} only provide instrinsic support for integers which are relatively small and single precision.
188
189\begin{figure}[!here]
190\begin{center}
191\begin{tabular}{|r|c|}
192\hline \textbf{Data Type} & \textbf{Range} \\
193\hline char  & $-128 \ldots 127$ \\
194\hline short & $-32768 \ldots 32767$ \\
195\hline long  & $-2147483648 \ldots 2147483647$ \\
196\hline long long & $-9223372036854775808 \ldots 9223372036854775807$ \\
197\hline
198\end{tabular}
199\end{center}
200\caption{Typical Data Types for the C Programming Language}
201\label{fig:ISOC}
202\end{figure}
203
204The largest data type guaranteed to be provided by the ISO C programming 
205language\footnote{As per the ISO C standard.  However, each compiler vendor is allowed to augment the precision as they 
206see fit.}  can only represent values up to $10^{19}$ as shown in figure \ref{fig:ISOC}. On its own the C language is 
207insufficient to accomodate the magnitude required for the problem at hand.  An RSA modulus of magnitude $10^{19}$ could be 
208trivially factored\footnote{A Pollard-Rho factoring would take only $2^{16}$ time.} on the average desktop computer, 
209rendering any protocol based on the algorithm insecure.  Multiple precision algorithms solve this very problem by 
210extending the range of representable integers while using single precision data types.
211
212Most advancements in fast multiple precision arithmetic stem from the need for faster and more efficient cryptographic 
213primitives.  Faster modular reduction and exponentiation algorithms such as Barrett's algorithm, which have appeared in 
214various cryptographic journals, can render algorithms such as RSA and Diffie-Hellman more efficient.  In fact, several 
215major companies such as RSA Security, Certicom and Entrust have built entire product lines on the implementation and 
216deployment of efficient algorithms.
217
218However, cryptography is not the only field of study that can benefit from fast multiple precision integer routines.  
219Another auxiliary use of multiple precision integers is high precision floating point data types.  
220The basic IEEE \cite{IEEE} standard floating point type is made up of an integer mantissa $q$, an exponent $e$ and a sign bit $s$.  
221Numbers are given in the form $n = q \cdot b^e \cdot -1^s$ where $b = 2$ is the most common base for IEEE.  Since IEEE 
222floating point is meant to be implemented in hardware the precision of the mantissa is often fairly small 
223(\textit{23, 48 and 64 bits}).  The mantissa is merely an integer and a multiple precision integer could be used to create
224a mantissa of much larger precision than hardware alone can efficiently support.  This approach could be useful where 
225scientific applications must minimize the total output error over long calculations.
226
227Yet another use for large integers is within arithmetic on polynomials of large characteristic (i.e. $GF(p)[x]$ for large $p$).
228In fact the library discussed within this text has already been used to form a polynomial basis library\footnote{See \url{http://poly.libtomcrypt.org} for more details.}.
229
230\subsection{Benefits of Multiple Precision Arithmetic}
231\index{precision}
232The benefit of multiple precision representations over single or fixed precision representations is that 
233no precision is lost while representing the result of an operation which requires excess precision.  For example, 
234the product of two $n$-bit integers requires at least $2n$ bits of precision to be represented faithfully.  A multiple 
235precision algorithm would augment the precision of the destination to accomodate the result while a single precision system 
236would truncate excess bits to maintain a fixed level of precision.
237
238It is possible to implement algorithms which require large integers with fixed precision algorithms.  For example, elliptic
239curve cryptography (\textit{ECC}) is often implemented on smartcards by fixing the precision of the integers to the maximum 
240size the system will ever need.  Such an approach can lead to vastly simpler algorithms which can accomodate the 
241integers required even if the host platform cannot natively accomodate them\footnote{For example, the average smartcard 
242processor has an 8 bit accumulator.}.  However, as efficient as such an approach may be, the resulting source code is not
243normally very flexible.  It cannot, at runtime, accomodate inputs of higher magnitude than the designer anticipated.
244
245Multiple precision algorithms have the most overhead of any style of arithmetic.  For the the most part the 
246overhead can be kept to a minimum with careful planning, but overall, it is not well suited for most memory starved
247platforms.  However, multiple precision algorithms do offer the most flexibility in terms of the magnitude of the 
248inputs.  That is, the same algorithms based on multiple precision integers can accomodate any reasonable size input 
249without the designer's explicit forethought.  This leads to lower cost of ownership for the code as it only has to 
250be written and tested once.
251
252\section{Purpose of This Text}
253The purpose of this text is to instruct the reader regarding how to implement efficient multiple precision algorithms.  
254That is to not only explain a limited subset of the core theory behind the algorithms but also the various ``house keeping'' 
255elements that are neglected by authors of other texts on the subject.  Several well reknowned texts \cite{TAOCPV2,HAC} 
256give considerably detailed explanations of the theoretical aspects of algorithms and often very little information 
257regarding the practical implementation aspects.  
258
259In most cases how an algorithm is explained and how it is actually implemented are two very different concepts.  For 
260example, the Handbook of Applied Cryptography (\textit{HAC}), algorithm 14.7 on page 594, gives a relatively simple 
261algorithm for performing multiple precision integer addition.  However, the description lacks any discussion concerning 
262the fact that the two integer inputs may be of differing magnitudes.  As a result the implementation is not as simple
263as the text would lead people to believe.  Similarly the division routine (\textit{algorithm 14.20, pp. 598}) does not 
264discuss how to handle sign or handle the dividend's decreasing magnitude in the main loop (\textit{step \#3}).
265
266Both texts also do not discuss several key optimal algorithms required such as ``Comba'' and Karatsuba multipliers 
267and fast modular inversion, which we consider practical oversights.  These optimal algorithms are vital to achieve 
268any form of useful performance in non-trivial applications.  
269
270To solve this problem the focus of this text is on the practical aspects of implementing a multiple precision integer
271package.  As a case study the ``LibTomMath''\footnote{Available at \url{http://math.libtomcrypt.com}} package is used 
272to demonstrate algorithms with real implementations\footnote{In the ISO C programming language.} that have been field 
273tested and work very well.  The LibTomMath library is freely available on the Internet for all uses and this text 
274discusses a very large portion of the inner workings of the library.
275
276The algorithms that are presented will always include at least one ``pseudo-code'' description followed 
277by the actual C source code that implements the algorithm.  The pseudo-code can be used to implement the same 
278algorithm in other programming languages as the reader sees fit.  
279
280This text shall also serve as a walkthrough of the creation of multiple precision algorithms from scratch.  Showing
281the reader how the algorithms fit together as well as where to start on various taskings.  
282
283\section{Discussion and Notation}
284\subsection{Notation}
285A multiple precision integer of $n$-digits shall be denoted as $x = (x_{n-1}, \ldots, x_1, x_0)_{ \beta }$ and represent
286the integer $x \equiv \sum_{i=0}^{n-1} x_i\beta^i$.  The elements of the array $x$ are said to be the radix $\beta$ digits 
287of the integer.  For example, $x = (1,2,3)_{10}$ would represent the integer 
288$1\cdot 10^2 + 2\cdot10^1 + 3\cdot10^0 = 123$.  
289
290\index{mp\_int}
291The term ``mp\_int'' shall refer to a composite structure which contains the digits of the integer it represents, as well 
292as auxilary data required to manipulate the data.  These additional members are discussed further in section 
293\ref{sec:MPINT}.  For the purposes of this text a ``multiple precision integer'' and an ``mp\_int'' are assumed to be 
294synonymous.  When an algorithm is specified to accept an mp\_int variable it is assumed the various auxliary data members 
295are present as well.  An expression of the type \textit{variablename.item} implies that it should evaluate to the 
296member named ``item'' of the variable.  For example, a string of characters may have a member ``length'' which would 
297evaluate to the number of characters in the string.  If the string $a$ equals ``hello'' then it follows that 
298$a.length = 5$.  
299
300For certain discussions more generic algorithms are presented to help the reader understand the final algorithm used
301to solve a given problem.  When an algorithm is described as accepting an integer input it is assumed the input is 
302a plain integer with no additional multiple-precision members.  That is, algorithms that use integers as opposed to 
303mp\_ints as inputs do not concern themselves with the housekeeping operations required such as memory management.  These 
304algorithms will be used to establish the relevant theory which will subsequently be used to describe a multiple
305precision algorithm to solve the same problem.  
306
307\subsection{Precision Notation}
308The variable $\beta$ represents the radix of a single digit of a multiple precision integer and 
309must be of the form $q^p$ for $q, p \in \Z^+$.  A single precision variable must be able to represent integers in 
310the range $0 \le x < q \beta$ while a double precision variable must be able to represent integers in the range 
311$0 \le x < q \beta^2$.  The extra radix-$q$ factor allows additions and subtractions to proceed without truncation of the 
312carry.  Since all modern computers are binary, it is assumed that $q$ is two.
313
314\index{mp\_digit} \index{mp\_word}
315Within the source code that will be presented for each algorithm, the data type \textbf{mp\_digit} will represent 
316a single precision integer type, while, the data type \textbf{mp\_word} will represent a double precision integer type.  In 
317several algorithms (notably the Comba routines) temporary results will be stored in arrays of double precision mp\_words.  
318For the purposes of this text $x_j$ will refer to the $j$'th digit of a single precision array and $\hat x_j$ will refer to 
319the $j$'th digit of a double precision array.  Whenever an expression is to be assigned to a double precision
320variable it is assumed that all single precision variables are promoted to double precision during the evaluation.  
321Expressions that are assigned to a single precision variable are truncated to fit within the precision of a single
322precision data type.
323
324For example, if $\beta = 10^2$ a single precision data type may represent a value in the 
325range $0 \le x < 10^3$, while a double precision data type may represent a value in the range $0 \le x < 10^5$.  Let
326$a = 23$ and $b = 49$ represent two single precision variables.  The single precision product shall be written
327as $c \leftarrow a \cdot b$ while the double precision product shall be written as $\hat c \leftarrow a \cdot b$.
328In this particular case, $\hat c = 1127$ and $c = 127$.  The most significant digit of the product would not fit 
329in a single precision data type and as a result $c \ne \hat c$.  
330
331\subsection{Algorithm Inputs and Outputs}
332Within the algorithm descriptions all variables are assumed to be scalars of either single or double precision
333as indicated.  The only exception to this rule is when variables have been indicated to be of type mp\_int.  This 
334distinction is important as scalars are often used as array indicies and various other counters.  
335
336\subsection{Mathematical Expressions}
337The $\lfloor \mbox{ } \rfloor$ brackets imply an expression truncated to an integer not greater than the expression 
338itself.  For example, $\lfloor 5.7 \rfloor = 5$.  Similarly the $\lceil \mbox{ } \rceil$ brackets imply an expression
339rounded to an integer not less than the expression itself.  For example, $\lceil 5.1 \rceil = 6$.  Typically when 
340the $/$ division symbol is used the intention is to perform an integer division with truncation.  For example, 
341$5/2 = 2$ which will often be written as $\lfloor 5/2 \rfloor = 2$ for clarity.  When an expression is written as a 
342fraction a real value division is implied, for example ${5 \over 2} = 2.5$.  
343
344The norm of a multiple precision integer, for example $\vert \vert x \vert \vert$, will be used to represent the number of digits in the representation
345of the integer.  For example, $\vert \vert 123 \vert \vert = 3$ and $\vert \vert 79452 \vert \vert = 5$.  
346
347\subsection{Work Effort}
348\index{big-Oh}
349To measure the efficiency of the specified algorithms, a modified big-Oh notation is used.  In this system all 
350single precision operations are considered to have the same cost\footnote{Except where explicitly noted.}.  
351That is a single precision addition, multiplication and division are assumed to take the same time to 
352complete.  While this is generally not true in practice, it will simplify the discussions considerably.
353
354Some algorithms have slight advantages over others which is why some constants will not be removed in 
355the notation.  For example, a normal baseline multiplication (section \ref{sec:basemult}) requires $O(n^2)$ work while a 
356baseline squaring (section \ref{sec:basesquare}) requires $O({{n^2 + n}\over 2})$ work.  In standard big-Oh notation these 
357would both be said to be equivalent to $O(n^2)$.  However, 
358in the context of the this text this is not the case as the magnitude of the inputs will typically be rather small.  As a 
359result small constant factors in the work effort will make an observable difference in algorithm efficiency.
360
361All of the algorithms presented in this text have a polynomial time work level.  That is, of the form 
362$O(n^k)$ for $n, k \in \Z^{+}$.  This will help make useful comparisons in terms of the speed of the algorithms and how 
363various optimizations will help pay off in the long run.
364
365\section{Exercises}
366Within the more advanced chapters a section will be set aside to give the reader some challenging exercises related to
367the discussion at hand.  These exercises are not designed to be prize winning problems, but instead to be thought 
368provoking.  Wherever possible the problems are forward minded, stating problems that will be answered in subsequent 
369chapters.  The reader is encouraged to finish the exercises as they appear to get a better understanding of the 
370subject material.  
371
372That being said, the problems are designed to affirm knowledge of a particular subject matter.  Students in particular
373are encouraged to verify they can answer the problems correctly before moving on.
374
375Similar to the exercises of \cite[pp. ix]{TAOCPV2} these exercises are given a scoring system based on the difficulty of
376the problem.  However, unlike \cite{TAOCPV2} the problems do not get nearly as hard.  The scoring of these 
377exercises ranges from one (the easiest) to five (the hardest).  The following table sumarizes the 
378scoring system used.
379
380\begin{figure}[here]
381\begin{center}
382\begin{small}
383\begin{tabular}{|c|l|}
384\hline $\left [ 1 \right ]$ & An easy problem that should only take the reader a manner of \\
385                            & minutes to solve.  Usually does not involve much computer time \\
386                            & to solve. \\
387\hline $\left [ 2 \right ]$ & An easy problem that involves a marginal amount of computer \\
388                     & time usage.  Usually requires a program to be written to \\
389                     & solve the problem. \\
390\hline $\left [ 3 \right ]$ & A moderately hard problem that requires a non-trivial amount \\
391                     & of work.  Usually involves trivial research and development of \\
392                     & new theory from the perspective of a student. \\
393\hline $\left [ 4 \right ]$ & A moderately hard problem that involves a non-trivial amount \\
394                     & of work and research, the solution to which will demonstrate \\
395                     & a higher mastery of the subject matter. \\
396\hline $\left [ 5 \right ]$ & A hard problem that involves concepts that are difficult for a \\
397                     & novice to solve.  Solutions to these problems will demonstrate a \\
398                     & complete mastery of the given subject. \\
399\hline
400\end{tabular}
401\end{small}
402\end{center}
403\caption{Exercise Scoring System}
404\end{figure}
405
406Problems at the first level are meant to be simple questions that the reader can answer quickly without programming a solution or
407devising new theory.  These problems are quick tests to see if the material is understood.  Problems at the second level 
408are also designed to be easy but will require a program or algorithm to be implemented to arrive at the answer.  These
409two levels are essentially entry level questions.  
410
411Problems at the third level are meant to be a bit more difficult than the first two levels.  The answer is often 
412fairly obvious but arriving at an exacting solution requires some thought and skill.  These problems will almost always 
413involve devising a new algorithm or implementing a variation of another algorithm previously presented.  Readers who can
414answer these questions will feel comfortable with the concepts behind the topic at hand.
415
416Problems at the fourth level are meant to be similar to those of the level three questions except they will require 
417additional research to be completed.  The reader will most likely not know the answer right away, nor will the text provide 
418the exact details of the answer until a subsequent chapter.  
419
420Problems at the fifth level are meant to be the hardest 
421problems relative to all the other problems in the chapter.  People who can correctly answer fifth level problems have a 
422mastery of the subject matter at hand.
423
424Often problems will be tied together.  The purpose of this is to start a chain of thought that will be discussed in future chapters.  The reader
425is encouraged to answer the follow-up problems and try to draw the relevance of problems.
426
427\section{Introduction to LibTomMath}
428
429\subsection{What is LibTomMath?}
430LibTomMath is a free and open source multiple precision integer library written entirely in portable ISO C.  By portable it 
431is meant that the library does not contain any code that is computer platform dependent or otherwise problematic to use on 
432any given platform.  
433
434The library has been successfully tested under numerous operating systems including Unix\footnote{All of these
435trademarks belong to their respective rightful owners.}, MacOS, Windows, Linux, PalmOS and on standalone hardware such 
436as the Gameboy Advance.  The library is designed to contain enough functionality to be able to develop applications such 
437as public key cryptosystems and still maintain a relatively small footprint.
438
439\subsection{Goals of LibTomMath}
440
441Libraries which obtain the most efficiency are rarely written in a high level programming language such as C.  However, 
442even though this library is written entirely in ISO C, considerable care has been taken to optimize the algorithm implementations within the 
443library.  Specifically the code has been written to work well with the GNU C Compiler (\textit{GCC}) on both x86 and ARM 
444processors.  Wherever possible, highly efficient algorithms, such as Karatsuba multiplication, sliding window 
445exponentiation and Montgomery reduction have been provided to make the library more efficient.  
446
447Even with the nearly optimal and specialized algorithms that have been included the Application Programing Interface 
448(\textit{API}) has been kept as simple as possible.  Often generic place holder routines will make use of specialized 
449algorithms automatically without the developer's specific attention.  One such example is the generic multiplication 
450algorithm \textbf{mp\_mul()} which will automatically use Toom--Cook, Karatsuba, Comba or baseline multiplication 
451based on the magnitude of the inputs and the configuration of the library.  
452
453Making LibTomMath as efficient as possible is not the only goal of the LibTomMath project.  Ideally the library should 
454be source compatible with another popular library which makes it more attractive for developers to use.  In this case the
455MPI library was used as a API template for all the basic functions.  MPI was chosen because it is another library that fits 
456in the same niche as LibTomMath.  Even though LibTomMath uses MPI as the template for the function names and argument 
457passing conventions, it has been written from scratch by Tom St Denis.
458
459The project is also meant to act as a learning tool for students, the logic being that no easy-to-follow ``bignum'' 
460library exists which can be used to teach computer science students how to perform fast and reliable multiple precision 
461integer arithmetic.  To this end the source code has been given quite a few comments and algorithm discussion points.  
462
463\section{Choice of LibTomMath}
464LibTomMath was chosen as the case study of this text not only because the author of both projects is one and the same but
465for more worthy reasons.  Other libraries such as GMP \cite{GMP}, MPI \cite{MPI}, LIP \cite{LIP} and OpenSSL 
466\cite{OPENSSL} have multiple precision integer arithmetic routines but would not be ideal for this text for 
467reasons that will be explained in the following sub-sections.
468
469\subsection{Code Base}
470The LibTomMath code base is all portable ISO C source code.  This means that there are no platform dependent conditional
471segments of code littered throughout the source.  This clean and uncluttered approach to the library means that a
472developer can more readily discern the true intent of a given section of source code without trying to keep track of
473what conditional code will be used.
474
475The code base of LibTomMath is well organized.  Each function is in its own separate source code file 
476which allows the reader to find a given function very quickly.  On average there are $76$ lines of code per source
477file which makes the source very easily to follow.  By comparison MPI and LIP are single file projects making code tracing
478very hard.  GMP has many conditional code segments which also hinder tracing.  
479
480When compiled with GCC for the x86 processor and optimized for speed the entire library is approximately $100$KiB\footnote{The notation ``KiB'' means $2^{10}$ octets, similarly ``MiB'' means $2^{20}$ octets.}
481 which is fairly small compared to GMP (over $250$KiB).  LibTomMath is slightly larger than MPI (which compiles to about 
482$50$KiB) but LibTomMath is also much faster and more complete than MPI.
483
484\subsection{API Simplicity}
485LibTomMath is designed after the MPI library and shares the API design.  Quite often programs that use MPI will build 
486with LibTomMath without change. The function names correlate directly to the action they perform.  Almost all of the 
487functions share the same parameter passing convention.  The learning curve is fairly shallow with the API provided 
488which is an extremely valuable benefit for the student and developer alike.  
489
490The LIP library is an example of a library with an API that is awkward to work with.  LIP uses function names that are often ``compressed'' to 
491illegible short hand.  LibTomMath does not share this characteristic.  
492
493The GMP library also does not return error codes.  Instead it uses a POSIX.1 \cite{POSIX1} signal system where errors
494are signaled to the host application.  This happens to be the fastest approach but definitely not the most versatile.  In
495effect a math error (i.e. invalid input, heap error, etc) can cause a program to stop functioning which is definitely 
496undersireable in many situations.
497
498\subsection{Optimizations}
499While LibTomMath is certainly not the fastest library (GMP often beats LibTomMath by a factor of two) it does
500feature a set of optimal algorithms for tasks such as modular reduction, exponentiation, multiplication and squaring.  GMP 
501and LIP also feature such optimizations while MPI only uses baseline algorithms with no optimizations.  GMP lacks a few
502of the additional modular reduction optimizations that LibTomMath features\footnote{At the time of this writing GMP
503only had Barrett and Montgomery modular reduction algorithms.}.  
504
505LibTomMath is almost always an order of magnitude faster than the MPI library at computationally expensive tasks such as modular
506exponentiation.  In the grand scheme of ``bignum'' libraries LibTomMath is faster than the average library and usually  
507slower than the best libraries such as GMP and OpenSSL by only a small factor.
508
509\subsection{Portability and Stability}
510LibTomMath will build ``out of the box'' on any platform equipped with a modern version of the GNU C Compiler 
511(\textit{GCC}).  This means that without changes the library will build without configuration or setting up any 
512variables.  LIP and MPI will build ``out of the box'' as well but have numerous known bugs.  Most notably the author of 
513MPI has recently stopped working on his library and LIP has long since been discontinued.  
514
515GMP requires a configuration script to run and will not build out of the box.   GMP and LibTomMath are still in active
516development and are very stable across a variety of platforms.
517
518\subsection{Choice}
519LibTomMath is a relatively compact, well documented, highly optimized and portable library which seems only natural for
520the case study of this text.  Various source files from the LibTomMath project will be included within the text.  However, 
521the reader is encouraged to download their own copy of the library to actually be able to work with the library.  
522
523\chapter{Getting Started}
524\section{Library Basics}
525The trick to writing any useful library of source code is to build a solid foundation and work outwards from it.  First, 
526a problem along with allowable solution parameters should be identified and analyzed.  In this particular case the 
527inability to accomodate multiple precision integers is the problem.  Futhermore, the solution must be written
528as portable source code that is reasonably efficient across several different computer platforms.
529
530After a foundation is formed the remainder of the library can be designed and implemented in a hierarchical fashion.  
531That is, to implement the lowest level dependencies first and work towards the most abstract functions last.  For example, 
532before implementing a modular exponentiation algorithm one would implement a modular reduction algorithm.
533By building outwards from a base foundation instead of using a parallel design methodology the resulting project is 
534highly modular.  Being highly modular is a desirable property of any project as it often means the resulting product
535has a small footprint and updates are easy to perform.  
536
537Usually when I start a project I will begin with the header files.  I define the data types I think I will need and 
538prototype the initial functions that are not dependent on other functions (within the library).  After I 
539implement these base functions I prototype more dependent functions and implement them.   The process repeats until
540I implement all of the functions I require.  For example, in the case of LibTomMath I implemented functions such as 
541mp\_init() well before I implemented mp\_mul() and even further before I implemented mp\_exptmod().  As an example as to 
542why this design works note that the Karatsuba and Toom-Cook multipliers were written \textit{after} the 
543dependent function mp\_exptmod() was written.  Adding the new multiplication algorithms did not require changes to the 
544mp\_exptmod() function itself and lowered the total cost of ownership (\textit{so to speak}) and of development 
545for new algorithms.  This methodology allows new algorithms to be tested in a complete framework with relative ease.
546
547FIGU,design_process,Design Flow of the First Few Original LibTomMath Functions.
548
549Only after the majority of the functions were in place did I pursue a less hierarchical approach to auditing and optimizing
550the source code.  For example, one day I may audit the multipliers and the next day the polynomial basis functions.  
551
552It only makes sense to begin the text with the preliminary data types and support algorithms required as well.  
553This chapter discusses the core algorithms of the library which are the dependents for every other algorithm.
554
555\section{What is a Multiple Precision Integer?}
556Recall that most programming languages, in particular ISO C \cite{ISOC}, only have fixed precision data types that on their own cannot 
557be used to represent values larger than their precision will allow. The purpose of multiple precision algorithms is 
558to use fixed precision data types to create and manipulate multiple precision integers which may represent values 
559that are very large.  
560
561As a well known analogy, school children are taught how to form numbers larger than nine by prepending more radix ten digits.  In the decimal system
562the largest single digit value is $9$.  However, by concatenating digits together larger numbers may be represented.  Newly prepended digits 
563(\textit{to the left}) are said to be in a different power of ten column.  That is, the number $123$ can be described as having a $1$ in the hundreds 
564column, $2$ in the tens column and $3$ in the ones column.  Or more formally $123 = 1 \cdot 10^2 + 2 \cdot 10^1 + 3 \cdot 10^0$.  Computer based 
565multiple precision arithmetic is essentially the same concept.  Larger integers are represented by adjoining fixed 
566precision computer words with the exception that a different radix is used.
567
568What most people probably do not think about explicitly are the various other attributes that describe a multiple precision 
569integer.  For example, the integer $154_{10}$ has two immediately obvious properties.  First, the integer is positive, 
570that is the sign of this particular integer is positive as opposed to negative.  Second, the integer has three digits in 
571its representation.  There is an additional property that the integer posesses that does not concern pencil-and-paper 
572arithmetic.  The third property is how many digits placeholders are available to hold the integer.  
573
574The human analogy of this third property is ensuring there is enough space on the paper to write the integer.  For example,
575if one starts writing a large number too far to the right on a piece of paper they will have to erase it and move left.  
576Similarly, computer algorithms must maintain strict control over memory usage to ensure that the digits of an integer
577will not exceed the allowed boundaries.  These three properties make up what is known as a multiple precision 
578integer or mp\_int for short.  
579
580\subsection{The mp\_int Structure}
581\label{sec:MPINT}
582The mp\_int structure is the ISO C based manifestation of what represents a multiple precision integer.  The ISO C standard does not provide for 
583any such data type but it does provide for making composite data types known as structures.  The following is the structure definition 
584used within LibTomMath.
585
586\index{mp\_int}
587\begin{figure}[here]
588\begin{center}
589\begin{small}
590%\begin{verbatim}
591\begin{tabular}{|l|}
592\hline
593typedef struct \{ \\
594\hspace{3mm}int used, alloc, sign;\\
595\hspace{3mm}mp\_digit *dp;\\
596\} \textbf{mp\_int}; \\
597\hline
598\end{tabular}
599%\end{verbatim}
600\end{small}
601\caption{The mp\_int Structure}
602\label{fig:mpint}
603\end{center}
604\end{figure}
605
606The mp\_int structure (fig. \ref{fig:mpint}) can be broken down as follows.
607
608\begin{enumerate}
609\item The \textbf{used} parameter denotes how many digits of the array \textbf{dp} contain the digits used to represent
610a given integer.  The \textbf{used} count must be positive (or zero) and may not exceed the \textbf{alloc} count.  
611
612\item The \textbf{alloc} parameter denotes how 
613many digits are available in the array to use by functions before it has to increase in size.  When the \textbf{used} count 
614of a result would exceed the \textbf{alloc} count all of the algorithms will automatically increase the size of the 
615array to accommodate the precision of the result.  
616
617\item The pointer \textbf{dp} points to a dynamically allocated array of digits that represent the given multiple 
618precision integer.  It is padded with $(\textbf{alloc} - \textbf{used})$ zero digits.  The array is maintained in a least 
619significant digit order.  As a pencil and paper analogy the array is organized such that the right most digits are stored
620first starting at the location indexed by zero\footnote{In C all arrays begin at zero.} in the array.  For example, 
621if \textbf{dp} contains $\lbrace a, b, c, \ldots \rbrace$ where \textbf{dp}$_0 = a$, \textbf{dp}$_1 = b$, \textbf{dp}$_2 = c$, $\ldots$ then 
622it would represent the integer $a + b\beta + c\beta^2 + \ldots$  
623
624\index{MP\_ZPOS} \index{MP\_NEG}
625\item The \textbf{sign} parameter denotes the sign as either zero/positive (\textbf{MP\_ZPOS}) or negative (\textbf{MP\_NEG}).  
626\end{enumerate}
627
628\subsubsection{Valid mp\_int Structures}
629Several rules are placed on the state of an mp\_int structure and are assumed to be followed for reasons of efficiency.  
630The only exceptions are when the structure is passed to initialization functions such as mp\_init() and mp\_init\_copy().
631
632\begin{enumerate}
633\item The value of \textbf{alloc} may not be less than one.  That is \textbf{dp} always points to a previously allocated
634array of digits.
635\item The value of \textbf{used} may not exceed \textbf{alloc} and must be greater than or equal to zero.
636\item The value of \textbf{used} implies the digit at index $(used - 1)$ of the \textbf{dp} array is non-zero.  That is, 
637leading zero digits in the most significant positions must be trimmed.
638   \begin{enumerate}
639   \item Digits in the \textbf{dp} array at and above the \textbf{used} location must be zero.
640   \end{enumerate}
641\item The value of \textbf{sign} must be \textbf{MP\_ZPOS} if \textbf{used} is zero; 
642this represents the mp\_int value of zero.
643\end{enumerate}
644
645\section{Argument Passing}
646A convention of argument passing must be adopted early on in the development of any library.  Making the function 
647prototypes consistent will help eliminate many headaches in the future as the library grows to significant complexity.  
648In LibTomMath the multiple precision integer functions accept parameters from left to right as pointers to mp\_int 
649structures.  That means that the source (input) operands are placed on the left and the destination (output) on the right.   
650Consider the following examples.
651
652\begin{verbatim}
653   mp_mul(&a, &b, &c);   /* c = a * b */
654   mp_add(&a, &b, &a);   /* a = a + b */
655   mp_sqr(&a, &b);       /* b = a * a */
656\end{verbatim}
657
658The left to right order is a fairly natural way to implement the functions since it lets the developer read aloud the
659functions and make sense of them.  For example, the first function would read ``multiply a and b and store in c''.
660
661Certain libraries (\textit{LIP by Lenstra for instance}) accept parameters the other way around, to mimic the order
662of assignment expressions.  That is, the destination (output) is on the left and arguments (inputs) are on the right.  In 
663truth, it is entirely a matter of preference.  In the case of LibTomMath the convention from the MPI library has been 
664adopted.  
665
666Another very useful design consideration, provided for in LibTomMath, is whether to allow argument sources to also be a 
667destination.  For example, the second example (\textit{mp\_add}) adds $a$ to $b$ and stores in $a$.  This is an important 
668feature to implement since it allows the calling functions to cut down on the number of variables it must maintain.  
669However, to implement this feature specific care has to be given to ensure the destination is not modified before the 
670source is fully read.
671
672\section{Return Values}
673A well implemented application, no matter what its purpose, should trap as many runtime errors as possible and return them 
674to the caller.  By catching runtime errors a library can be guaranteed to prevent undefined behaviour.  However, the end 
675developer can still manage to cause a library to crash.  For example, by passing an invalid pointer an application may
676fault by dereferencing memory not owned by the application.
677
678In the case of LibTomMath the only errors that are checked for are related to inappropriate inputs (division by zero for 
679instance) and memory allocation errors.  It will not check that the mp\_int passed to any function is valid nor 
680will it check pointers for validity.  Any function that can cause a runtime error will return an error code as an 
681\textbf{int} data type with one of the following values (fig \ref{fig:errcodes}).
682
683\index{MP\_OKAY} \index{MP\_VAL} \index{MP\_MEM}
684\begin{figure}[here]
685\begin{center}
686\begin{tabular}{|l|l|}
687\hline \textbf{Value} & \textbf{Meaning} \\
688\hline \textbf{MP\_OKAY} & The function was successful \\
689\hline \textbf{MP\_VAL}  & One of the input value(s) was invalid \\
690\hline \textbf{MP\_MEM}  & The function ran out of heap memory \\
691\hline
692\end{tabular}
693\end{center}
694\caption{LibTomMath Error Codes}
695\label{fig:errcodes}
696\end{figure}
697
698When an error is detected within a function it should free any memory it allocated, often during the initialization of
699temporary mp\_ints, and return as soon as possible.  The goal is to leave the system in the same state it was when the 
700function was called.  Error checking with this style of API is fairly simple.
701
702\begin{verbatim}
703   int err;
704   if ((err = mp_add(&a, &b, &c)) != MP_OKAY) {
705      printf("Error: %s\n", mp_error_to_string(err));
706      exit(EXIT_FAILURE);
707   }
708\end{verbatim}
709
710The GMP \cite{GMP} library uses C style \textit{signals} to flag errors which is of questionable use.  Not all errors are fatal 
711and it was not deemed ideal by the author of LibTomMath to force developers to have signal handlers for such cases.
712
713\section{Initialization and Clearing}
714The logical starting point when actually writing multiple precision integer functions is the initialization and 
715clearing of the mp\_int structures.  These two algorithms will be used by the majority of the higher level algorithms.
716
717Given the basic mp\_int structure an initialization routine must first allocate memory to hold the digits of
718the integer.  Often it is optimal to allocate a sufficiently large pre-set number of digits even though
719the initial integer will represent zero.  If only a single digit were allocated quite a few subsequent re-allocations
720would occur when operations are performed on the integers.  There is a tradeoff between how many default digits to allocate
721and how many re-allocations are tolerable.  Obviously allocating an excessive amount of digits initially will waste 
722memory and become unmanageable.  
723
724If the memory for the digits has been successfully allocated then the rest of the members of the structure must
725be initialized.  Since the initial state of an mp\_int is to represent the zero integer, the allocated digits must be set
726to zero.  The \textbf{used} count set to zero and \textbf{sign} set to \textbf{MP\_ZPOS}.
727
728\subsection{Initializing an mp\_int}
729An mp\_int is said to be initialized if it is set to a valid, preferably default, state such that all of the members of the
730structure are set to valid values.  The mp\_init algorithm will perform such an action.
731
732\index{mp\_init}
733\begin{figure}[here]
734\begin{center}
735\begin{tabular}{l}
736\hline Algorithm \textbf{mp\_init}. \\
737\textbf{Input}.   An mp\_int $a$ \\
738\textbf{Output}.  Allocate memory and initialize $a$ to a known valid mp\_int state.  \\
739\hline \\
7401.  Allocate memory for \textbf{MP\_PREC} digits. \\
7412.  If the allocation failed return(\textit{MP\_MEM}) \\
7423.  for $n$ from $0$ to $MP\_PREC - 1$ do  \\
743\hspace{3mm}3.1  $a_n \leftarrow 0$\\
7444.  $a.sign \leftarrow MP\_ZPOS$\\
7455.  $a.used \leftarrow 0$\\
7466.  $a.alloc \leftarrow MP\_PREC$\\
7477.  Return(\textit{MP\_OKAY})\\
748\hline
749\end{tabular}
750\end{center}
751\caption{Algorithm mp\_init}
752\end{figure}
753
754\textbf{Algorithm mp\_init.}
755The purpose of this function is to initialize an mp\_int structure so that the rest of the library can properly
756manipulte it.  It is assumed that the input may not have had any of its members previously initialized which is certainly
757a valid assumption if the input resides on the stack.  
758
759Before any of the members such as \textbf{sign}, \textbf{used} or \textbf{alloc} are initialized the memory for
760the digits is allocated.  If this fails the function returns before setting any of the other members.  The \textbf{MP\_PREC} 
761name represents a constant\footnote{Defined in the ``tommath.h'' header file within LibTomMath.} 
762used to dictate the minimum precision of newly initialized mp\_int integers.  Ideally, it is at least equal to the smallest
763precision number you'll be working with.
764
765Allocating a block of digits at first instead of a single digit has the benefit of lowering the number of usually slow
766heap operations later functions will have to perform in the future.  If \textbf{MP\_PREC} is set correctly the slack 
767memory and the number of heap operations will be trivial.
768
769Once the allocation has been made the digits have to be set to zero as well as the \textbf{used}, \textbf{sign} and
770\textbf{alloc} members initialized.  This ensures that the mp\_int will always represent the default state of zero regardless
771of the original condition of the input.
772
773\textbf{Remark.}
774This function introduces the idiosyncrasy that all iterative loops, commonly initiated with the ``for'' keyword, iterate incrementally
775when the ``to'' keyword is placed between two expressions.  For example, ``for $a$ from $b$ to $c$ do'' means that
776a subsequent expression (or body of expressions) are to be evaluated upto $c - b$ times so long as $b \le c$.  In each
777iteration the variable $a$ is substituted for a new integer that lies inclusively between $b$ and $c$.  If $b > c$ occured
778the loop would not iterate.  By contrast if the ``downto'' keyword were used in place of ``to'' the loop would iterate 
779decrementally.
780
781EXAM,bn_mp_init.c
782
783One immediate observation of this initializtion function is that it does not return a pointer to a mp\_int structure.  It 
784is assumed that the caller has already allocated memory for the mp\_int structure, typically on the application stack.  The 
785call to mp\_init() is used only to initialize the members of the structure to a known default state.  
786
787Here we see (line @23,XMALLOC@) the memory allocation is performed first.  This allows us to exit cleanly and quickly
788if there is an error.  If the allocation fails the routine will return \textbf{MP\_MEM} to the caller to indicate there
789was a memory error.  The function XMALLOC is what actually allocates the memory.  Technically XMALLOC is not a function
790but a macro defined in ``tommath.h``.  By default, XMALLOC will evaluate to malloc() which is the C library's built--in
791memory allocation routine.
792
793In order to assure the mp\_int is in a known state the digits must be set to zero.  On most platforms this could have been
794accomplished by using calloc() instead of malloc().  However,  to correctly initialize a integer type to a given value in a 
795portable fashion you have to actually assign the value.  The for loop (line @28,for@) performs this required
796operation.
797
798After the memory has been successfully initialized the remainder of the members are initialized 
799(lines @29,used@ through @31,sign@) to their respective default states.  At this point the algorithm has succeeded and
800a success code is returned to the calling function.  If this function returns \textbf{MP\_OKAY} it is safe to assume the 
801mp\_int structure has been properly initialized and is safe to use with other functions within the library.  
802
803\subsection{Clearing an mp\_int}
804When an mp\_int is no longer required by the application, the memory that has been allocated for its digits must be 
805returned to the application's memory pool with the mp\_clear algorithm.
806
807\begin{figure}[here]
808\begin{center}
809\begin{tabular}{l}
810\hline Algorithm \textbf{mp\_clear}. \\
811\textbf{Input}.   An mp\_int $a$ \\
812\textbf{Output}.  The memory for $a$ shall be deallocated.  \\
813\hline \\
8141.  If $a$ has been previously freed then return(\textit{MP\_OKAY}). \\
8152.  for $n$ from 0 to $a.used - 1$ do \\
816\hspace{3mm}2.1  $a_n \leftarrow 0$ \\
8173.  Free the memory allocated for the digits of $a$. \\
8184.  $a.used \leftarrow 0$ \\
8195.  $a.alloc \leftarrow 0$ \\
8206.  $a.sign \leftarrow MP\_ZPOS$ \\
8217.  Return(\textit{MP\_OKAY}). \\
822\hline
823\end{tabular}
824\end{center}
825\caption{Algorithm mp\_clear}
826\end{figure}
827
828\textbf{Algorithm mp\_clear.}
829This algorithm accomplishes two goals.  First, it clears the digits and the other mp\_int members.  This ensures that 
830if a developer accidentally re-uses a cleared structure it is less likely to cause problems.  The second goal
831is to free the allocated memory.
832
833The logic behind the algorithm is extended by marking cleared mp\_int structures so that subsequent calls to this
834algorithm will not try to free the memory multiple times.  Cleared mp\_ints are detectable by having a pre-defined invalid 
835digit pointer \textbf{dp} setting.
836
837Once an mp\_int has been cleared the mp\_int structure is no longer in a valid state for any other algorithm
838with the exception of algorithms mp\_init, mp\_init\_copy, mp\_init\_size and mp\_clear.
839
840EXAM,bn_mp_clear.c
841
842The algorithm only operates on the mp\_int if it hasn't been previously cleared.  The if statement (line @23,a->dp != NULL@)
843checks to see if the \textbf{dp} member is not \textbf{NULL}.  If the mp\_int is a valid mp\_int then \textbf{dp} cannot be
844\textbf{NULL} in which case the if statement will evaluate to true.
845
846The digits of the mp\_int are cleared by the for loop (line @25,for@) which assigns a zero to every digit.  Similar to mp\_init()
847the digits are assigned zero instead of using block memory operations (such as memset()) since this is more portable.  
848
849The digits are deallocated off the heap via the XFREE macro.  Similar to XMALLOC the XFREE macro actually evaluates to
850a standard C library function.  In this case the free() function.  Since free() only deallocates the memory the pointer
851still has to be reset to \textbf{NULL} manually (line @33,NULL@).  
852
853Now that the digits have been cleared and deallocated the other members are set to their final values (lines @34,= 0@ and @35,ZPOS@).
854
855\section{Maintenance Algorithms}
856
857The previous sections describes how to initialize and clear an mp\_int structure.  To further support operations
858that are to be performed on mp\_int structures (such as addition and multiplication) the dependent algorithms must be
859able to augment the precision of an mp\_int and 
860initialize mp\_ints with differing initial conditions.  
861
862These algorithms complete the set of low level algorithms required to work with mp\_int structures in the higher level
863algorithms such as addition, multiplication and modular exponentiation.
864
865\subsection{Augmenting an mp\_int's Precision}
866When storing a value in an mp\_int structure, a sufficient number of digits must be available to accomodate the entire 
867result of an operation without loss of precision.  Quite often the size of the array given by the \textbf{alloc} member 
868is large enough to simply increase the \textbf{used} digit count.  However, when the size of the array is too small it 
869must be re-sized appropriately to accomodate the result.  The mp\_grow algorithm will provide this functionality.
870
871\newpage\begin{figure}[here]
872\begin{center}
873\begin{tabular}{l}
874\hline Algorithm \textbf{mp\_grow}. \\
875\textbf{Input}.   An mp\_int $a$ and an integer $b$. \\
876\textbf{Output}.  $a$ is expanded to accomodate $b$ digits. \\
877\hline \\
8781.  if $a.alloc \ge b$ then return(\textit{MP\_OKAY}) \\
8792.  $u \leftarrow b\mbox{ (mod }MP\_PREC\mbox{)}$ \\
8803.  $v \leftarrow b + 2 \cdot MP\_PREC - u$ \\
8814.  Re-allocate the array of digits $a$ to size $v$ \\
8825.  If the allocation failed then return(\textit{MP\_MEM}). \\
8836.  for n from a.alloc to $v - 1$ do  \\
884\hspace{+3mm}6.1  $a_n \leftarrow 0$ \\
8857.  $a.alloc \leftarrow v$ \\
8868.  Return(\textit{MP\_OKAY}) \\
887\hline
888\end{tabular}
889\end{center}
890\caption{Algorithm mp\_grow}
891\end{figure}
892
893\textbf{Algorithm mp\_grow.}
894It is ideal to prevent re-allocations from being performed if they are not required (step one).  This is useful to 
895prevent mp\_ints from growing excessively in code that erroneously calls mp\_grow.  
896
897The requested digit count is padded up to next multiple of \textbf{MP\_PREC} plus an additional \textbf{MP\_PREC} (steps two and three).  
898This helps prevent many trivial reallocations that would grow an mp\_int by trivially small values.  
899
900It is assumed that the reallocation (step four) leaves the lower $a.alloc$ digits of the mp\_int intact.  This is much 
901akin to how the \textit{realloc} function from the standard C library works.  Since the newly allocated digits are 
902assumed to contain undefined values they are initially set to zero.
903
904EXAM,bn_mp_grow.c
905
906A quick optimization is to first determine if a memory re-allocation is required at all.  The if statement (line @24,alloc@) checks
907if the \textbf{alloc} member of the mp\_int is smaller than the requested digit count.  If the count is not larger than \textbf{alloc}
908the function skips the re-allocation part thus saving time.
909
910When a re-allocation is performed it is turned into an optimal request to save time in the future.  The requested digit count is
911padded upwards to 2nd multiple of \textbf{MP\_PREC} larger than \textbf{alloc} (line @25, size@).  The XREALLOC function is used
912to re-allocate the memory.  As per the other functions XREALLOC is actually a macro which evaluates to realloc by default.  The realloc
913function leaves the base of the allocation intact which means the first \textbf{alloc} digits of the mp\_int are the same as before
914the re-allocation.  All	that is left is to clear the newly allocated digits and return.
915
916Note that the re-allocation result is actually stored in a temporary pointer $tmp$.  This is to allow this function to return
917an error with a valid pointer.  Earlier releases of the library stored the result of XREALLOC into the mp\_int $a$.  That would
918result in a memory leak if XREALLOC ever failed.  
919
920\subsection{Initializing Variable Precision mp\_ints}
921Occasionally the number of digits required will be known in advance of an initialization, based on, for example, the size 
922of input mp\_ints to a given algorithm.  The purpose of algorithm mp\_init\_size is similar to mp\_init except that it 
923will allocate \textit{at least} a specified number of digits.  
924
925\begin{figure}[here]
926\begin{small}
927\begin{center}
928\begin{tabular}{l}
929\hline Algorithm \textbf{mp\_init\_size}. \\
930\textbf{Input}.   An mp\_int $a$ and the requested number of digits $b$. \\
931\textbf{Output}.  $a$ is initialized to hold at least $b$ digits. \\
932\hline \\
9331.  $u \leftarrow b \mbox{ (mod }MP\_PREC\mbox{)}$ \\
9342.  $v \leftarrow b + 2 \cdot MP\_PREC - u$ \\
9353.  Allocate $v$ digits. \\
9364.  for $n$ from $0$ to $v - 1$ do \\
937\hspace{3mm}4.1  $a_n \leftarrow 0$ \\
9385.  $a.sign \leftarrow MP\_ZPOS$\\
9396.  $a.used \leftarrow 0$\\
9407.  $a.alloc \leftarrow v$\\
9418.  Return(\textit{MP\_OKAY})\\
942\hline
943\end{tabular}
944\end{center}
945\end{small}
946\caption{Algorithm mp\_init\_size}
947\end{figure}
948
949\textbf{Algorithm mp\_init\_size.}
950This algorithm will initialize an mp\_int structure $a$ like algorithm mp\_init with the exception that the number of 
951digits allocated can be controlled by the second input argument $b$.  The input size is padded upwards so it is a 
952multiple of \textbf{MP\_PREC} plus an additional \textbf{MP\_PREC} digits.  This padding is used to prevent trivial 
953allocations from becoming a bottleneck in the rest of the algorithms.
954
955Like algorithm mp\_init, the mp\_int structure is initialized to a default state representing the integer zero.  This 
956particular algorithm is useful if it is known ahead of time the approximate size of the input.  If the approximation is
957correct no further memory re-allocations are required to work with the mp\_int.
958
959EXAM,bn_mp_init_size.c
960
961The number of digits $b$ requested is padded (line @22,MP_PREC@) by first augmenting it to the next multiple of 
962\textbf{MP\_PREC} and then adding \textbf{MP\_PREC} to the result.  If the memory can be successfully allocated the 
963mp\_int is placed in a default state representing the integer zero.  Otherwise, the error code \textbf{MP\_MEM} will be 
964returned (line @27,return@).  
965
966The digits are allocated and set to zero at the same time with the calloc() function (line @25,XCALLOC@).  The 
967\textbf{used} count is set to zero, the \textbf{alloc} count set to the padded digit count and the \textbf{sign} flag set 
968to \textbf{MP\_ZPOS} to achieve a default valid mp\_int state (lines @29,used@, @30,alloc@ and @31,sign@).  If the function 
969returns succesfully then it is correct to assume that the mp\_int structure is in a valid state for the remainder of the 
970functions to work with.
971
972\subsection{Multiple Integer Initializations and Clearings}
973Occasionally a function will require a series of mp\_int data types to be made available simultaneously.  
974The purpose of algorithm mp\_init\_multi is to initialize a variable length array of mp\_int structures in a single
975statement.  It is essentially a shortcut to multiple initializations.
976
977\newpage\begin{figure}[here]
978\begin{center}
979\begin{tabular}{l}
980\hline Algorithm \textbf{mp\_init\_multi}. \\
981\textbf{Input}.   Variable length array $V_k$ of mp\_int variables of length $k$. \\
982\textbf{Output}.  The array is initialized such that each mp\_int of $V_k$ is ready to use. \\
983\hline \\
9841.  for $n$ from 0 to $k - 1$ do \\
985\hspace{+3mm}1.1.  Initialize the mp\_int $V_n$ (\textit{mp\_init}) \\
986\hspace{+3mm}1.2.  If initialization failed then do \\
987\hspace{+6mm}1.2.1.  for $j$ from $0$ to $n$ do \\
988\hspace{+9mm}1.2.1.1.  Free the mp\_int $V_j$ (\textit{mp\_clear}) \\
989\hspace{+6mm}1.2.2.   Return(\textit{MP\_MEM}) \\
9902.  Return(\textit{MP\_OKAY}) \\
991\hline
992\end{tabular}
993\end{center}
994\caption{Algorithm mp\_init\_multi}
995\end{figure}
996
997\textbf{Algorithm mp\_init\_multi.}
998The algorithm will initialize the array of mp\_int variables one at a time.  If a runtime error has been detected 
999(\textit{step 1.2}) all of the previously initialized variables are cleared.  The goal is an ``all or nothing'' 
1000initialization which allows for quick recovery from runtime errors.
1001
1002EXAM,bn_mp_init_multi.c
1003
1004This function intializes a variable length list of mp\_int structure pointers.  However, instead of having the mp\_int
1005structures in an actual C array they are simply passed as arguments to the function.  This function makes use of the 
1006``...'' argument syntax of the C programming language.  The list is terminated with a final \textbf{NULL} argument 
1007appended on the right.  
1008
1009The function uses the ``stdarg.h'' \textit{va} functions to step portably through the arguments to the function.  A count
1010$n$ of succesfully initialized mp\_int structures is maintained (line @47,n++@) such that if a failure does occur,
1011the algorithm can backtrack and free the previously initialized structures (lines @27,if@ to @46,}@).  
1012
1013
1014\subsection{Clamping Excess Digits}
1015When a function anticipates a result will be $n$ digits it is simpler to assume this is true within the body of 
1016the function instead of checking during the computation.  For example, a multiplication of a $i$ digit number by a 
1017$j$ digit produces a result of at most $i + j$ digits.  It is entirely possible that the result is $i + j - 1$ 
1018though, with no final carry into the last position.  However, suppose the destination had to be first expanded 
1019(\textit{via mp\_grow}) to accomodate $i + j - 1$ digits than further expanded to accomodate the final carry.  
1020That would be a considerable waste of time since heap operations are relatively slow.
1021
1022The ideal solution is to always assume the result is $i + j$ and fix up the \textbf{used} count after the function
1023terminates.  This way a single heap operation (\textit{at most}) is required.  However, if the result was not checked
1024there would be an excess high order zero digit.  
1025
1026For example, suppose the product of two integers was $x_n = (0x_{n-1}x_{n-2}...x_0)_{\beta}$.  The leading zero digit 
1027will not contribute to the precision of the result.  In fact, through subsequent operations more leading zero digits would
1028accumulate to the point the size of the integer would be prohibitive.  As a result even though the precision is very 
1029low the representation is excessively large.  
1030
1031The mp\_clamp algorithm is designed to solve this very problem.  It will trim high-order zeros by decrementing the 
1032\textbf{used} count until a non-zero most significant digit is found.  Also in this system, zero is considered to be a 
1033positive number which means that if the \textbf{used} count is decremented to zero, the sign must be set to 
1034\textbf{MP\_ZPOS}.
1035
1036\begin{figure}[here]
1037\begin{center}
1038\begin{tabular}{l}
1039\hline Algorithm \textbf{mp\_clamp}. \\
1040\textbf{Input}.   An mp\_int $a$ \\
1041\textbf{Output}.  Any excess leading zero digits of $a$ are removed \\
1042\hline \\
10431.  while $a.used > 0$ and $a_{a.used - 1} = 0$ do \\
1044\hspace{+3mm}1.1  $a.used \leftarrow a.used - 1$ \\
10452.  if $a.used = 0$ then do \\
1046\hspace{+3mm}2.1  $a.sign \leftarrow MP\_ZPOS$ \\
1047\hline \\
1048\end{tabular}
1049\end{center}
1050\caption{Algorithm mp\_clamp}
1051\end{figure}
1052
1053\textbf{Algorithm mp\_clamp.}
1054As can be expected this algorithm is very simple.  The loop on step one is expected to iterate only once or twice at
1055the most.  For example, this will happen in cases where there is not a carry to fill the last position.  Step two fixes the sign for 
1056when all of the digits are zero to ensure that the mp\_int is valid at all times.
1057
1058EXAM,bn_mp_clamp.c
1059
1060Note on line @27,while@ how to test for the \textbf{used} count is made on the left of the \&\& operator.  In the C programming
1061language the terms to \&\& are evaluated left to right with a boolean short-circuit if any condition fails.  This is 
1062important since if the \textbf{used} is zero the test on the right would fetch below the array.  That is obviously 
1063undesirable.  The parenthesis on line @28,a->used@ is used to make sure the \textbf{used} count is decremented and not
1064the pointer ``a''.  
1065
1066\section*{Exercises}
1067\begin{tabular}{cl}
1068$\left [ 1 \right ]$ & Discuss the relevance of the \textbf{used} member of the mp\_int structure. \\
1069                     & \\
1070$\left [ 1 \right ]$ & Discuss the consequences of not using padding when performing allocations.  \\
1071                     & \\
1072$\left [ 2 \right ]$ & Estimate an ideal value for \textbf{MP\_PREC} when performing 1024-bit RSA \\
1073                     & encryption when $\beta = 2^{28}$.  \\
1074                     & \\
1075$\left [ 1 \right ]$ & Discuss the relevance of the algorithm mp\_clamp.  What does it prevent? \\
1076                     & \\
1077$\left [ 1 \right ]$ & Give an example of when the algorithm  mp\_init\_copy might be useful. \\
1078                     & \\
1079\end{tabular}
1080
1081
1082%%%
1083% CHAPTER FOUR
1084%%%
1085
1086\chapter{Basic Operations}
1087
1088\section{Introduction}
1089In the previous chapter a series of low level algorithms were established that dealt with initializing and maintaining
1090mp\_int structures.  This chapter will discuss another set of seemingly non-algebraic algorithms which will form the low 
1091level basis of the entire library.  While these algorithm are relatively trivial it is important to understand how they
1092work before proceeding since these algorithms will be used almost intrinsically in the following chapters.
1093
1094The algorithms in this chapter deal primarily with more ``programmer'' related tasks such as creating copies of
1095mp\_int structures, assigning small values to mp\_int structures and comparisons of the values mp\_int structures
1096represent.   
1097
1098\section{Assigning Values to mp\_int Structures}
1099\subsection{Copying an mp\_int}
1100Assigning the value that a given mp\_int structure represents to another mp\_int structure shall be known as making
1101a copy for the purposes of this text.  The copy of the mp\_int will be a separate entity that represents the same
1102value as the mp\_int it was copied from.  The mp\_copy algorithm provides this functionality. 
1103
1104\newpage\begin{figure}[here]
1105\begin{center}
1106\begin{tabular}{l}
1107\hline Algorithm \textbf{mp\_copy}. \\
1108\textbf{Input}.  An mp\_int $a$ and $b$. \\
1109\textbf{Output}.  Store a copy of $a$ in $b$. \\
1110\hline \\
11111.  If $b.alloc < a.used$ then grow $b$ to $a.used$ digits.  (\textit{mp\_grow}) \\
11122.  for $n$ from 0 to $a.used - 1$ do \\
1113\hspace{3mm}2.1  $b_{n} \leftarrow a_{n}$ \\
11143.  for $n$ from $a.used$ to $b.used - 1$ do \\
1115\hspace{3mm}3.1  $b_{n} \leftarrow 0$ \\
11164.  $b.used \leftarrow a.used$ \\
11175.  $b.sign \leftarrow a.sign$ \\
11186.  return(\textit{MP\_OKAY}) \\
1119\hline
1120\end{tabular}
1121\end{center}
1122\caption{Algorithm mp\_copy}
1123\end{figure}
1124
1125\textbf{Algorithm mp\_copy.}
1126This algorithm copies the mp\_int $a$ such that upon succesful termination of the algorithm the mp\_int $b$ will
1127represent the same integer as the mp\_int $a$.  The mp\_int $b$ shall be a complete and distinct copy of the 
1128mp\_int $a$ meaing that the mp\_int $a$ can be modified and it shall not affect the value of the mp\_int $b$.
1129
1130If $b$ does not have enough room for the digits of $a$ it must first have its precision augmented via the mp\_grow 
1131algorithm.  The digits of $a$ are copied over the digits of $b$ and any excess digits of $b$ are set to zero (step two
1132and three).  The \textbf{used} and \textbf{sign} members of $a$ are finally copied over the respective members of
1133$b$.
1134
1135\textbf{Remark.}  This algorithm also introduces a new idiosyncrasy that will be used throughout the rest of the
1136text.  The error return codes of other algorithms are not explicitly checked in the pseudo-code presented.  For example, in 
1137step one of the mp\_copy algorithm the return of mp\_grow is not explicitly checked to ensure it succeeded.  Text space is 
1138limited so it is assumed that if a algorithm fails it will clear all temporarily allocated mp\_ints and return
1139the error code itself.  However, the C code presented will demonstrate all of the error handling logic required to 
1140implement the pseudo-code.
1141
1142EXAM,bn_mp_copy.c
1143
1144Occasionally a dependent algorithm may copy an mp\_int effectively into itself such as when the input and output
1145mp\_int structures passed to a function are one and the same.  For this case it is optimal to return immediately without 
1146copying digits (line @24,a == b@).  
1147
1148The mp\_int $b$ must have enough digits to accomodate the used digits of the mp\_int $a$.  If $b.alloc$ is less than
1149$a.used$ the algorithm mp\_grow is used to augment the precision of $b$ (lines @29,alloc@ to @33,}@).  In order to
1150simplify the inner loop that copies the digits from $a$ to $b$, two aliases $tmpa$ and $tmpb$ point directly at the digits
1151of the mp\_ints $a$ and $b$ respectively.  These aliases (lines @42,tmpa@ and @45,tmpb@) allow the compiler to access the digits without first dereferencing the
1152mp\_int pointers and then subsequently the pointer to the digits.  
1153
1154After the aliases are established the digits from $a$ are copied into $b$ (lines @48,for@ to @50,}@) and then the excess 
1155digits of $b$ are set to zero (lines @53,for@ to @55,}@).  Both ``for'' loops make use of the pointer aliases and in 
1156fact the alias for $b$ is carried through into the second ``for'' loop to clear the excess digits.  This optimization 
1157allows the alias to stay in a machine register fairly easy between the two loops.
1158
1159\textbf{Remarks.}  The use of pointer aliases is an implementation methodology first introduced in this function that will
1160be used considerably in other functions.  Technically, a pointer alias is simply a short hand alias used to lower the 
1161number of pointer dereferencing operations required to access data.  For example, a for loop may resemble
1162
1163\begin{alltt}
1164for (x = 0; x < 100; x++) \{
1165    a->num[4]->dp[x] = 0;
1166\}
1167\end{alltt}
1168
1169This could be re-written using aliases as 
1170
1171\begin{alltt}
1172mp_digit *tmpa;
1173a = a->num[4]->dp;
1174for (x = 0; x < 100; x++) \{
1175    *a++ = 0;
1176\}
1177\end{alltt}
1178
1179In this case an alias is used to access the 
1180array of digits within an mp\_int structure directly.  It may seem that a pointer alias is strictly not required 
1181as a compiler may optimize out the redundant pointer operations.  However, there are two dominant reasons to use aliases.
1182
1183The first reason is that most compilers will not effectively optimize pointer arithmetic.  For example, some optimizations 
1184may work for the Microsoft Visual C++ compiler (MSVC) and not for the GNU C Compiler (GCC).  Also some optimizations may 
1185work for GCC and not MSVC.  As such it is ideal to find a common ground for as many compilers as possible.  Pointer 
1186aliases optimize the code considerably before the compiler even reads the source code which means the end compiled code 
1187stands a better chance of being faster.
1188
1189The second reason is that pointer aliases often can make an algorithm simpler to read.  Consider the first ``for'' 
1190loop of the function mp\_copy() re-written to not use pointer aliases.
1191
1192\begin{alltt}
1193    /* copy all the digits */
1194    for (n = 0; n < a->used; n++) \{
1195      b->dp[n] = a->dp[n];
1196    \}
1197\end{alltt}
1198
1199Whether this code is harder to read depends strongly on the individual.  However, it is quantifiably slightly more 
1200complicated as there are four variables within the statement instead of just two.
1201
1202\subsubsection{Nested Statements}
1203Another commonly used technique in the source routines is that certain sections of code are nested.  This is used in
1204particular with the pointer aliases to highlight code phases.  For example, a Comba multiplier (discussed in chapter six)
1205will typically have three different phases.  First the temporaries are initialized, then the columns calculated and 
1206finally the carries are propagated.  In this example the middle column production phase will typically be nested as it
1207uses temporary variables and aliases the most.
1208
1209The nesting also simplies the source code as variables that are nested are only valid for their scope.  As a result
1210the various temporary variables required do not propagate into other sections of code.
1211
1212
1213\subsection{Creating a Clone}
1214Another common operation is to make a local temporary copy of an mp\_int argument.  To initialize an mp\_int 
1215and then copy another existing mp\_int into the newly intialized mp\_int will be known as creating a clone.  This is 
1216useful within functions that need to modify an argument but do not wish to actually modify the original copy.  The 
1217mp\_init\_copy algorithm has been designed to help perform this task.
1218
1219\begin{figure}[here]
1220\begin{center}
1221\begin{tabular}{l}
1222\hline Algorithm \textbf{mp\_init\_copy}. \\
1223\textbf{Input}.   An mp\_int $a$ and $b$\\
1224\textbf{Output}.  $a$ is initialized to be a copy of $b$. \\
1225\hline \\
12261.  Init $a$.  (\textit{mp\_init}) \\
12272.  Copy $b$ to $a$.  (\textit{mp\_copy}) \\
12283.  Return the status of the copy operation. \\
1229\hline
1230\end{tabular}
1231\end{center}
1232\caption{Algorithm mp\_init\_copy}
1233\end{figure}
1234
1235\textbf{Algorithm mp\_init\_copy.}
1236This algorithm will initialize an mp\_int variable and copy another previously initialized mp\_int variable into it.  As 
1237such this algorithm will perform two operations in one step.  
1238
1239EXAM,bn_mp_init_copy.c
1240
1241This will initialize \textbf{a} and make it a verbatim copy of the contents of \textbf{b}.  Note that 
1242\textbf{a} will have its own memory allocated which means that \textbf{b} may be cleared after the call
1243and \textbf{a} will be left intact.  
1244
1245\section{Zeroing an Integer}
1246Reseting an mp\_int to the default state is a common step in many algorithms.  The mp\_zero algorithm will be the algorithm used to
1247perform this task.
1248
1249\begin{figure}[here]
1250\begin{center}
1251\begin{tabular}{l}
1252\hline Algorithm \textbf{mp\_zero}. \\
1253\textbf{Input}.   An mp\_int $a$ \\
1254\textbf{Output}.  Zero the contents of $a$ \\
1255\hline \\
12561.  $a.used \leftarrow 0$ \\
12572.  $a.sign \leftarrow$ MP\_ZPOS \\
12583.  for $n$ from 0 to $a.alloc - 1$ do \\
1259\hspace{3mm}3.1  $a_n \leftarrow 0$ \\
1260\hline
1261\end{tabular}
1262\end{center}
1263\caption{Algorithm mp\_zero}
1264\end{figure}
1265
1266\textbf{Algorithm mp\_zero.}
1267This algorithm simply resets a mp\_int to the default state.  
1268
1269EXAM,bn_mp_zero.c
1270
1271After the function is completed, all of the digits are zeroed, the \textbf{used} count is zeroed and the 
1272\textbf{sign} variable is set to \textbf{MP\_ZPOS}.
1273
1274\section{Sign Manipulation}
1275\subsection{Absolute Value}
1276With the mp\_int representation of an integer, calculating the absolute value is trivial.  The mp\_abs algorithm will compute
1277the absolute value of an mp\_int.
1278
1279\begin{figure}[here]
1280\begin{center}
1281\begin{tabular}{l}
1282\hline Algorithm \textbf{mp\_abs}. \\
1283\textbf{Input}.   An mp\_int $a$ \\
1284\textbf{Output}.  Computes $b = \vert a \vert$ \\
1285\hline \\
12861.  Copy $a$ to $b$.  (\textit{mp\_copy}) \\
12872.  If the copy failed return(\textit{MP\_MEM}). \\
12883.  $b.sign \leftarrow MP\_ZPOS$ \\
12894.  Return(\textit{MP\_OKAY}) \\
1290\hline
1291\end{tabular}
1292\end{center}
1293\caption{Algorithm mp\_abs}
1294\end{figure}
1295
1296\textbf{Algorithm mp\_abs.}
1297This algorithm computes the absolute of an mp\_int input.  First it copies $a$ over $b$.  This is an example of an
1298algorithm where the check in mp\_copy that determines if the source and destination are equal proves useful.  This allows,
1299for instance, the developer to pass the same mp\_int as the source and destination to this function without addition 
1300logic to handle it.
1301
1302EXAM,bn_mp_abs.c
1303
1304This fairly trivial algorithm first eliminates non--required duplications (line @27,a != b@) and then sets the
1305\textbf{sign} flag to \textbf{MP\_ZPOS}.
1306
1307\subsection{Integer Negation}
1308With the mp\_int representation of an integer, calculating the negation is also trivial.  The mp\_neg algorithm will compute
1309the negative of an mp\_int input.
1310
1311\begin{figure}[here]
1312\begin{center}
1313\begin{tabular}{l}
1314\hline Algorithm \textbf{mp\_neg}. \\
1315\textbf{Input}.   An mp\_int $a$ \\
1316\textbf{Output}.  Computes $b = -a$ \\
1317\hline \\
13181.  Copy $a$ to $b$.  (\textit{mp\_copy}) \\
13192.  If the copy failed return(\textit{MP\_MEM}). \\
13203.  If $a.used = 0$ then return(\textit{MP\_OKAY}). \\
13214.  If $a.sign = MP\_ZPOS$ then do \\
1322\hspace{3mm}4.1  $b.sign = MP\_NEG$. \\
13235.  else do \\
1324\hspace{3mm}5.1  $b.sign = MP\_ZPOS$. \\
13256.  Return(\textit{MP\_OKAY}) \\
1326\hline
1327\end{tabular}
1328\end{center}
1329\caption{Algorithm mp\_neg}
1330\end{figure}
1331
1332\textbf{Algorithm mp\_neg.}
1333This algorithm computes the negation of an input.  First it copies $a$ over $b$.  If $a$ has no used digits then
1334the algorithm returns immediately.  Otherwise it flips the sign flag and stores the result in $b$.  Note that if 
1335$a$ had no digits then it must be positive by definition.  Had step three been omitted then the algorithm would return
1336zero as negative.
1337
1338EXAM,bn_mp_neg.c
1339
1340Like mp\_abs() this function avoids non--required duplications (line @21,a != b@) and then sets the sign.  We
1341have to make sure that only non--zero values get a \textbf{sign} of \textbf{MP\_NEG}.  If the mp\_int is zero
1342than the \textbf{sign} is hard--coded to \textbf{MP\_ZPOS}.
1343
1344\section{Small Constants}
1345\subsection{Setting Small Constants}
1346Often a mp\_int must be set to a relatively small value such as $1$ or $2$.  For these cases the mp\_set algorithm is useful.
1347
1348\newpage\begin{figure}[here]
1349\begin{center}
1350\begin{tabular}{l}
1351\hline Algorithm \textbf{mp\_set}. \\
1352\textbf{Input}.   An mp\_int $a$ and a digit $b$ \\
1353\textbf{Output}.  Make $a$ equivalent to $b$ \\
1354\hline \\
13551.  Zero $a$ (\textit{mp\_zero}). \\
13562.  $a_0 \leftarrow b \mbox{ (mod }\beta\mbox{)}$ \\
13573.  $a.used \leftarrow  \left \lbrace \begin{array}{ll}
1358                              1 &  \mbox{if }a_0 > 0 \\
1359                              0 &  \mbox{if }a_0 = 0 
1360                              \end{array} \right .$ \\
1361\hline                              
1362\end{tabular}
1363\end{center}
1364\caption{Algorithm mp\_set}
1365\end{figure}
1366
1367\textbf{Algorithm mp\_set.}
1368This algorithm sets a mp\_int to a small single digit value.  Step number 1 ensures that the integer is reset to the default state.  The
1369single digit is set (\textit{modulo $\beta$}) and the \textbf{used} count is adjusted accordingly.
1370
1371EXAM,bn_mp_set.c
1372
1373First we zero (line @21,mp_zero@) the mp\_int to make sure that the other members are initialized for a 
1374small positive constant.  mp\_zero() ensures that the \textbf{sign} is positive and the \textbf{used} count
1375is zero.  Next we set the digit and reduce it modulo $\beta$ (line @22,MP_MASK@).  After this step we have to 
1376check if the resulting digit is zero or not.  If it is not then we set the \textbf{used} count to one, otherwise
1377to zero.
1378
1379We can quickly reduce modulo $\beta$ since it is of the form $2^k$ and a quick binary AND operation with 
1380$2^k - 1$ will perform the same operation.
1381
1382One important limitation of this function is that it will only set one digit.  The size of a digit is not fixed, meaning source that uses 
1383this function should take that into account.  Only trivially small constants can be set using this function.
1384
1385\subsection{Setting Large Constants}
1386To overcome the limitations of the mp\_set algorithm the mp\_set\_int algorithm is ideal.  It accepts a ``long''
1387data type as input and will always treat it as a 32-bit integer.
1388
1389\begin{figure}[here]
1390\begin{center}
1391\begin{tabular}{l}
1392\hline Algorithm \textbf{mp\_set\_int}. \\
1393\textbf{Input}.   An mp\_int $a$ and a ``long'' integer $b$ \\
1394\textbf{Output}.  Make $a$ equivalent to $b$ \\
1395\hline \\
13961.  Zero $a$ (\textit{mp\_zero}) \\
13972.  for $n$ from 0 to 7 do \\
1398\hspace{3mm}2.1  $a \leftarrow a \cdot 16$ (\textit{mp\_mul2d}) \\
1399\hspace{3mm}2.2  $u \leftarrow \lfloor b / 2^{4(7 - n)} \rfloor \mbox{ (mod }16\mbox{)}$\\
1400\hspace{3mm}2.3  $a_0 \leftarrow a_0 + u$ \\
1401\hspace{3mm}2.4  $a.used \leftarrow a.used + 1$ \\
14023.  Clamp excess used digits (\textit{mp\_clamp}) \\
1403\hline
1404\end{tabular}
1405\end{center}
1406\caption{Algorithm mp\_set\_int}
1407\end{figure}
1408
1409\textbf{Algorithm mp\_set\_int.}
1410The algorithm performs eight iterations of a simple loop where in each iteration four bits from the source are added to the 
1411mp\_int.  Step 2.1 will multiply the current result by sixteen making room for four more bits in the less significant positions.  In step 2.2 the
1412next four bits from the source are extracted and are added to the mp\_int. The \textbf{used} digit count is 
1413incremented to reflect the addition.  The \textbf{used} digit counter is incremented since if any of the leading digits were zero the mp\_int would have
1414zero digits used and the newly added four bits would be ignored.
1415
1416Excess zero digits are trimmed in steps 2.1 and 3 by using higher level algorithms mp\_mul2d and mp\_clamp.
1417
1418EXAM,bn_mp_set_int.c
1419
1420This function sets four bits of the number at a time to handle all practical \textbf{DIGIT\_BIT} sizes.  The weird
1421addition on line @38,a->used@ ensures that the newly added in bits are added to the number of digits.  While it may not 
1422seem obvious as to why the digit counter does not grow exceedingly large it is because of the shift on line @27,mp_mul_2d@ 
1423as well as the  call to mp\_clamp() on line @40,mp_clamp@.  Both functions will clamp excess leading digits which keeps 
1424the number of used digits low.
1425
1426\section{Comparisons}
1427\subsection{Unsigned Comparisions}
1428Comparing a multiple precision integer is performed with the exact same algorithm used to compare two decimal numbers.  For example,
1429to compare $1,234$ to $1,264$ the digits are extracted by their positions.  That is we compare $1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10^1 + 4 \cdot 10^0$
1430to $1 \cdot 10^3 + 2 \cdot 10^2 + 6 \cdot 10^1 + 4 \cdot 10^0$ by comparing single digits at a time starting with the highest magnitude 
1431positions.  If any leading digit of one integer is greater than a digit in the same position of another integer then obviously it must be greater.  
1432
1433The first comparision routine that will be developed is the unsigned magnitude compare which will perform a comparison based on the digits of two
1434mp\_int variables alone.  It will ignore the sign of the two inputs.  Such a function is useful when an absolute comparison is required or if the 
1435signs are known to agree in advance.
1436
1437To facilitate working with the results of the comparison functions three constants are required.  
1438
1439\begin{figure}[here]
1440\begin{center}
1441\begin{tabular}{|r|l|}
1442\hline \textbf{Constant} & \textbf{Meaning} \\
1443\hline \textbf{MP\_GT} & Greater Than \\
1444\hline \textbf{MP\_EQ} & Equal To \\
1445\hline \textbf{MP\_LT} & Less Than \\
1446\hline
1447\end{tabular}
1448\end{center}
1449\caption{Comparison Return Codes}
1450\end{figure}
1451
1452\begin{figure}[here]
1453\begin{center}
1454\begin{tabular}{l}
1455\hline Algorithm \textbf{mp\_cmp\_mag}. \\
1456\textbf{Input}.   Two mp\_ints $a$ and $b$.  \\
1457\textbf{Output}.  Unsigned comparison results ($a$ to the left of $b$). \\
1458\hline \\
14591.  If $a.used > b.used$ then return(\textit{MP\_GT}) \\
14602.  If $a.used < b.used$ then return(\textit{MP\_LT}) \\
14613.  for n from $a.used - 1$ to 0 do \\
1462\hspace{+3mm}3.1  if $a_n > b_n$ then return(\textit{MP\_GT}) \\
1463\hspace{+3mm}3.2  if $a_n < b_n$ then return(\textit{MP\_LT}) \\
14644.  Return(\textit{MP\_EQ}) \\
1465\hline
1466\end{tabular}
1467\end{center}
1468\caption{Algorithm mp\_cmp\_mag}
1469\end{figure}
1470
1471\textbf{Algorithm mp\_cmp\_mag.}
1472By saying ``$a$ to the left of $b$'' it is meant that the comparison is with respect to $a$, that is if $a$ is greater than $b$ it will return
1473\textbf{MP\_GT} and similar with respect to when $a = b$ and $a < b$.  The first two steps compare the number of digits used in both $a$ and $b$.  
1474Obviously if the digit counts differ there would be an imaginary zero digit in the smaller number where the leading digit of the larger number is.  
1475If both have the same number of digits than the actual digits themselves must be compared starting at the leading digit.  
1476
1477By step three both inputs must have the same number of digits so its safe to start from either $a.used - 1$ or $b.used - 1$ and count down to
1478the zero'th digit.  If after all of the digits have been compared, no difference is found, the algorithm returns \textbf{MP\_EQ}.
1479
1480EXAM,bn_mp_cmp_mag.c
1481
1482The two if statements (lines @24,if@ and @28,if@) compare the number of digits in the two inputs.  These two are 
1483performed before all of the digits are compared since it is a very cheap test to perform and can potentially save 
1484considerable time.  The implementation given is also not valid without those two statements.  $b.alloc$ may be 
1485smaller than $a.used$, meaning that undefined values will be read from $b$ past the end of the array of digits.
1486
1487
1488
1489\subsection{Signed Comparisons}
1490Comparing with sign considerations is also fairly critical in several routines (\textit{division for example}).  Based on an unsigned magnitude 
1491comparison a trivial signed comparison algorithm can be written.
1492
1493\begin{figure}[here]
1494\begin{center}
1495\begin{tabular}{l}
1496\hline Algorithm \textbf{mp\_cmp}. \\
1497\textbf{Input}.   Two mp\_ints $a$ and $b$ \\
1498\textbf{Output}.  Signed Comparison Results ($a$ to the left of $b$) \\
1499\hline \\
15001.  if $a.sign = MP\_NEG$ and $b.sign = MP\_ZPOS$ then return(\textit{MP\_LT}) \\
15012.  if $a.sign = MP\_ZPOS$ and $b.sign = MP\_NEG$ then return(\textit{MP\_GT}) \\
15023.  if $a.sign = MP\_NEG$ then \\
1503\hspace{+3mm}3.1  Return the unsigned comparison of $b$ and $a$ (\textit{mp\_cmp\_mag}) \\
15044   Otherwise \\
1505\hspace{+3mm}4.1  Return the unsigned comparison of $a$ and $b$ \\
1506\hline
1507\end{tabular}
1508\end{center}
1509\caption{Algorithm mp\_cmp}
1510\end{figure}
1511
1512\textbf{Algorithm mp\_cmp.}
1513The first two steps compare the signs of the two inputs.  If the signs do not agree then it can return right away with the appropriate 
1514comparison code.  When the signs are equal the digits of the inputs must be compared to determine the correct result.  In step 
1515three the unsigned comparision flips the order of the arguments since they are both negative.  For instance, if $-a > -b$ then 
1516$\vert a \vert < \vert b \vert$.  Step number four will compare the two when they are both positive.
1517
1518EXAM,bn_mp_cmp.c
1519
1520The two if statements (lines @22,if@ and @26,if@) perform the initial sign comparison.  If the signs are not the equal then which ever
1521has the positive sign is larger.   The inputs are compared (line @30,if@) based on magnitudes.  If the signs were both 
1522negative then the unsigned comparison is performed in the opposite direction (line @31,mp_cmp_mag@).  Otherwise, the signs are assumed to 
1523be both positive and a forward direction unsigned comparison is performed.
1524
1525\section*{Exercises}
1526\begin{tabular}{cl}
1527$\left [ 2 \right ]$ & Modify algorithm mp\_set\_int to accept as input a variable length array of bits. \\
1528                     & \\
1529$\left [ 3 \right ]$ & Give the probability that algorithm mp\_cmp\_mag will have to compare $k$ digits  \\
1530                     & of two random digits (of equal magnitude) before a difference is found. \\
1531                     & \\
1532$\left [ 1 \right ]$ & Suggest a simple method to speed up the implementation of mp\_cmp\_mag based  \\
1533                     & on the observations made in the previous problem. \\
1534                     &
1535\end{tabular}
1536
1537\chapter{Basic Arithmetic}
1538\section{Introduction}
1539At this point algorithms for initialization, clearing, zeroing, copying, comparing and setting small constants have been 
1540established.  The next logical set of algorithms to develop are addition, subtraction and digit shifting algorithms.  These 
1541algorithms make use of the lower level algorithms and are the cruicial building block for the multiplication algorithms.  It is very important 
1542that these algorithms are highly optimized.  On their own they are simple $O(n)$ algorithms but they can be called from higher level algorithms 
1543which easily places them at $O(n^2)$ or even $O(n^3)$ work levels.  
1544
1545MARK,SHIFTS
1546All of the algorithms within this chapter make use of the logical bit shift operations denoted by $<<$ and $>>$ for left and right 
1547logical shifts respectively.  A logical shift is analogous to sliding the decimal point of radix-10 representations.  For example, the real 
1548number $0.9345$ is equivalent to $93.45\%$ which is found by sliding the the decimal two places to the right (\textit{multiplying by $\beta^2 = 10^2$}).  
1549Algebraically a binary logical shift is equivalent to a division or multiplication by a power of two.  
1550For example, $a << k = a \cdot 2^k$ while $a >> k = \lfloor a/2^k \rfloor$.
1551
1552One significant difference between a logical shift and the way decimals are shifted is that digits below the zero'th position are removed
1553from the number.  For example, consider $1101_2 >> 1$ using decimal notation this would produce $110.1_2$.  However, with a logical shift the 
1554result is $110_2$.  
1555
1556\section{Addition and Subtraction}
1557In common twos complement fixed precision arithmetic negative numbers are easily represented by subtraction from the modulus.  For example, with 32-bit integers
1558$a - b\mbox{ (mod }2^{32}\mbox{)}$ is the same as $a + (2^{32} - b) \mbox{ (mod }2^{32}\mbox{)}$  since $2^{32} \equiv 0 \mbox{ (mod }2^{32}\mbox{)}$.  
1559As a result subtraction can be performed with a trivial series of logical operations and an addition.
1560
1561However, in multiple precision arithmetic negative numbers are not represented in the same way.  Instead a sign flag is used to keep track of the
1562sign of the integer.  As a result signed addition and subtraction are actually implemented as conditional usage of lower level addition or 
1563subtraction algorithms with the sign fixed up appropriately.
1564
1565The lower level algorithms will add or subtract integers without regard to the sign flag.  That is they will add or subtract the magnitude of
1566the integers respectively.
1567
1568\subsection{Low Level Addition}
1569An unsigned addition of multiple precision integers is performed with the same long-hand algorithm used to add decimal numbers.  That is to add the 
1570trailing digits first and propagate the resulting carry upwards.  Since this is a lower level algorithm the name will have a ``s\_'' prefix.  
1571Historically that convention stems from the MPI library where ``s\_'' stood for static functions that were hidden from the developer entirely.
1572
1573\newpage
1574\begin{figure}[!here]
1575\begin{center}
1576\begin{small}
1577\begin{tabular}{l}
1578\hline Algorithm \textbf{s\_mp\_add}. \\
1579\textbf{Input}.   Two mp\_ints $a$ and $b$ \\
1580\textbf{Output}.  The unsigned addition $c = \vert a \vert + \vert b \vert$. \\
1581\hline \\
15821.  if $a.used > b.used$ then \\
1583\hspace{+3mm}1.1  $min \leftarrow b.used$ \\
1584\hspace{+3mm}1.2  $max \leftarrow a.used$ \\
1585\hspace{+3mm}1.3  $x   \leftarrow a$ \\
15862.  else  \\
1587\hspace{+3mm}2.1  $min \leftarrow a.used$ \\
1588\hspace{+3mm}2.2  $max \leftarrow b.used$ \\
1589\hspace{+3mm}2.3  $x   \leftarrow b$ \\
15903.  If $c.alloc < max + 1$ then grow $c$ to hold at least $max + 1$ digits (\textit{mp\_grow}) \\
15914.  $oldused \leftarrow c.used$ \\
15925.  $c.used \leftarrow max + 1$ \\
15936.  $u \leftarrow 0$ \\
15947.  for $n$ from $0$ to $min - 1$ do \\
1595\hspace{+3mm}7.1  $c_n \leftarrow a_n + b_n + u$ \\
1596\hspace{+3mm}7.2  $u \leftarrow c_n >> lg(\beta)$ \\
1597\hspace{+3mm}7.3  $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\
15988.  if $min \ne max$ then do \\
1599\hspace{+3mm}8.1  for $n$ from $min$ to $max - 1$ do \\
1600\hspace{+6mm}8.1.1  $c_n \leftarrow x_n + u$ \\
1601\hspace{+6mm}8.1.2  $u \leftarrow c_n >> lg(\beta)$ \\
1602\hspace{+6mm}8.1.3  $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\
16039.  $c_{max} \leftarrow u$ \\
160410.  if $olduse > max$ then \\
1605\hspace{+3mm}10.1  for $n$ from $max + 1$ to $oldused - 1$ do \\
1606\hspace{+6mm}10.1.1  $c_n \leftarrow 0$ \\
160711.  Clamp excess digits in $c$.  (\textit{mp\_clamp}) \\
160812.  Return(\textit{MP\_OKAY}) \\
1609\hline
1610\end{tabular}
1611\end{small}
1612\end{center}
1613\caption{Algorithm s\_mp\_add}
1614\end{figure}
1615
1616\textbf{Algorithm s\_mp\_add.}
1617This algorithm is loosely based on algorithm 14.7 of HAC \cite[pp. 594]{HAC} but has been extended to allow the inputs to have different magnitudes.  
1618Coincidentally the description of algorithm A in Knuth \cite[pp. 266]{TAOCPV2} shares the same deficiency as the algorithm from \cite{HAC}.  Even the 
1619MIX pseudo  machine code presented by Knuth \cite[pp. 266-267]{TAOCPV2} is incapable of handling inputs which are of different magnitudes.
1620
1621The first thing that has to be accomplished is to sort out which of the two inputs is the largest.  The addition logic
1622will simply add all of the smallest input to the largest input and store that first part of the result in the
1623destination.  Then it will apply a simpler addition loop to excess digits of the larger input.
1624
1625The first two steps will handle sorting the inputs such that $min$ and $max$ hold the digit counts of the two 
1626inputs.  The variable $x$ will be an mp\_int alias for the largest input or the second input $b$ if they have the
1627same number of digits.  After the inputs are sorted the destination $c$ is grown as required to accomodate the sum 
1628of the two inputs.  The original \textbf{used} count of $c$ is copied and set to the new used count.  
1629
1630At this point the first addition loop will go through as many digit positions that both inputs have.  The carry
1631variable $\mu$ is set to zero outside the loop.  Inside the loop an ``addition'' step requires three statements to produce
1632one digit of the summand.  First
1633two digits from $a$ and $b$ are added together along with the carry $\mu$.  The carry of this step is extracted and stored
1634in $\mu$ and finally the digit of the result $c_n$ is truncated within the range $0 \le c_n < \beta$.
1635
1636Now all of the digit positions that both inputs have in common have been exhausted.  If $min \ne max$ then $x$ is an alias
1637for one of the inputs that has more digits.  A simplified addition loop is then used to essentially copy the remaining digits
1638and the carry to the destination.
1639
1640The final carry is stored in $c_{max}$ and digits above $max$ upto $oldused$ are zeroed which completes the addition.
1641
1642
1643EXAM,bn_s_mp_add.c
1644
1645We first sort (lines @27,if@ to @35,}@) the inputs based on magnitude and determine the $min$ and $max$ variables.
1646Note that $x$ is a pointer to an mp\_int assigned to the largest input, in effect it is a local alias.  Next we
1647grow the destination (@37,init@ to @42,}@) ensure that it can accomodate the result of the addition. 
1648
1649Similar to the implementation of mp\_copy this function uses the braced code and local aliases coding style.  The three aliases that are on 
1650lines @56,tmpa@, @59,tmpb@ and @62,tmpc@ represent the two inputs and destination variables respectively.  These aliases are used to ensure the
1651compiler does not have to dereference $a$, $b$ or $c$ (respectively) to access the digits of the respective mp\_int.
1652
1653The initial carry $u$ will be cleared (line @65,u = 0@), note that $u$ is of type mp\_digit which ensures type 
1654compatibility within the implementation.  The initial addition (line @66,for@ to @75,}@) adds digits from
1655both inputs until the smallest input runs out of digits.  Similarly the conditional addition loop
1656(line @81,for@ to @90,}@) adds the remaining digits from the larger of the two inputs.  The addition is finished 
1657with the final carry being stored in $tmpc$ (line @94,tmpc++@).  Note the ``++'' operator within the same expression.
1658After line @94,tmpc++@, $tmpc$ will point to the $c.used$'th digit of the mp\_int $c$.  This is useful
1659for the next loop (line @97,for@ to @99,}@) which set any old upper digits to zero.
1660
1661\subsection{Low Level Subtraction}
1662The low level unsigned subtraction algorithm is very similar to the low level unsigned addition algorithm.  The principle difference is that the
1663unsigned subtraction algorithm requires the result to be positive.  That is when computing $a - b$ the condition $\vert a \vert \ge \vert b\vert$ must 
1664be met for this algorithm to function properly.  Keep in mind this low level algorithm is not meant to be used in higher level algorithms directly.  
1665This algorithm as will be shown can be used to create functional signed addition and subtraction algorithms.
1666
1667MARK,GAMMA
1668
1669For this algorithm a new variable is required to make the description simpler.  Recall from section 1.3.1 that a mp\_digit must be able to represent
1670the range $0 \le x < 2\beta$ for the algorithms to work correctly.  However, it is allowable that a mp\_digit represent a larger range of values.  For 
1671this algorithm we will assume that the variable $\gamma$ represents the number of bits available in a 
1672mp\_digit (\textit{this implies $2^{\gamma} > \beta$}).  
1673
1674For example, the default for LibTomMath is to use a ``unsigned long'' for the mp\_digit ``type'' while $\beta = 2^{28}$.  In ISO C an ``unsigned long''
1675data type must be able to represent $0 \le x < 2^{32}$ meaning that in this case $\gamma \ge 32$.
1676
1677\newpage\begin{figure}[!here]
1678\begin{center}
1679\begin{small}
1680\begin{tabular}{l}
1681\hline Algorithm \textbf{s\_mp\_sub}. \\
1682\textbf{Input}.   Two mp\_ints $a$ and $b$ ($\vert a \vert \ge \vert b \vert$) \\
1683\textbf{Output}.  The unsigned subtraction $c = \vert a \vert - \vert b \vert$. \\
1684\hline \\
16851.  $min \leftarrow b.used$ \\
16862.  $max \leftarrow a.used$ \\
16873.  If $c.alloc < max$ then grow $c$ to hold at least $max$ digits.  (\textit{mp\_grow}) \\
16884.  $oldused \leftarrow c.used$ \\ 
16895.  $c.used \leftarrow max$ \\
16906.  $u \leftarrow 0$ \\
16917.  for $n$ from $0$ to $min - 1$ do \\
1692\hspace{3mm}7.1  $c_n \leftarrow a_n - b_n - u$ \\
1693\hspace{3mm}7.2  $u   \leftarrow c_n >> (\gamma - 1)$ \\
1694\hspace{3mm}7.3  $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\
16958.  if $min < max$ then do \\
1696\hspace{3mm}8.1  for $n$ from $min$ to $max - 1$ do \\
1697\hspace{6mm}8.1.1  $c_n \leftarrow a_n - u$ \\
1698\hspace{6mm}8.1.2  $u   \leftarrow c_n >> (\gamma - 1)$ \\
1699\hspace{6mm}8.1.3  $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\
17009. if $oldused > max$ then do \\
1701\hspace{3mm}9.1  for $n$ from $max$ to $oldused - 1$ do \\
1702\hspace{6mm}9.1.1  $c_n \leftarrow 0$ \\
170310. Clamp excess digits of $c$.  (\textit{mp\_clamp}). \\
170411. Return(\textit{MP\_OKAY}). \\
1705\hline
1706\end{tabular}
1707\end{small}
1708\end{center}
1709\caption{Algorithm s\_mp\_sub}
1710\end{figure}
1711
1712\textbf{Algorithm s\_mp\_sub.}
1713This algorithm performs the unsigned subtraction of two mp\_int variables under the restriction that the result must be positive.  That is when
1714passing variables $a$ and $b$ the condition that $\vert a \vert \ge \vert b \vert$ must be met for the algorithm to function correctly.  This
1715algorithm is loosely based on algorithm 14.9 \cite[pp. 595]{HAC} and is similar to algorithm S in \cite[pp. 267]{TAOCPV2} as well.  As was the case
1716of the algorithm s\_mp\_add both other references lack discussion concerning various practical details such as when the inputs differ in magnitude.
1717
1718The initial sorting of the inputs is trivial in this algorithm since $a$ is guaranteed to have at least the same magnitude of $b$.  Steps 1 and 2 
1719set the $min$ and $max$ variables.  Unlike the addition routine there is guaranteed to be no carry which means that the final result can be at 
1720most $max$ digits in length as opposed to $max + 1$.  Similar to the addition algorithm the \textbf{used} count of $c$ is copied locally and 
1721set to the maximal count for the operation.
1722
1723The subtraction loop that begins on step seven is essentially the same as the addition loop of algorithm s\_mp\_add except single precision 
1724subtraction is used instead.  Note the use of the $\gamma$ variable to extract the carry (\textit{also known as the borrow}) within the subtraction 
1725loops.  Under the assumption that two's complement single precision arithmetic is used this will successfully extract the desired carry.  
1726
1727For example, consider subtracting $0101_2$ from $0100_2$ where $\gamma = 4$ and $\beta = 2$.  The least significant bit will force a carry upwards to 
1728the third bit which will be set to zero after the borrow.  After the very first bit has been subtracted $4 - 1 \equiv 0011_2$ will remain,  When the 
1729third bit of $0101_2$ is subtracted from the result it will cause another carry.  In this case though the carry will be forced to propagate all the 
1730way to the most significant bit.  
1731
1732Recall that $\beta < 2^{\gamma}$.  This means that if a carry does occur just before the $lg(\beta)$'th bit it will propagate all the way to the most 
1733significant bit.  Thus, the high order bits of the mp\_digit that are not part of the actual digit will either be all zero, or all one. All that
1734is needed is a single zero or one bit for the carry.  Therefore a single logical shift right by $\gamma - 1$ positions is sufficient to extract the 
1735carry.  This method of carry extraction may seem awkward but the reason for it becomes apparent when the implementation is discussed.  
1736
1737If $b$ has a smaller magnitude than $a$ then step 9 will force the carry and copy operation to propagate through the larger input $a$ into $c$.  Step
173810 will ensure that any leading digits of $c$ above the $max$'th position are zeroed.
1739
1740EXAM,bn_s_mp_sub.c
1741
1742Like low level addition we ``sort'' the inputs.  Except in this case the sorting is hardcoded 
1743(lines @24,min@ and @25,max@).  In reality the $min$ and $max$ variables are only aliases and are only 
1744used to make the source code easier to read.  Again the pointer alias optimization is used 
1745within this algorithm.  The aliases $tmpa$, $tmpb$ and $tmpc$ are initialized
1746(lines @42,tmpa@, @43,tmpb@ and @44,tmpc@) for $a$, $b$ and $c$ respectively.
1747
1748The first subtraction loop (lines @47,u = 0@ through @61,}@) subtract digits from both inputs until the smaller of
1749the two inputs has been exhausted.  As remarked earlier there is an implementation reason for using the ``awkward'' 
1750method of extracting the carry (line @57, >>@).  The traditional method for extracting the carry would be to shift 
1751by $lg(\beta)$ positions and logically AND the least significant bit.  The AND operation is required because all of 
1752the bits above the $\lg(\beta)$'th bit will be set to one after a carry occurs from subtraction.  This carry 
1753extraction requires two relatively cheap operations to extract the carry.  The other method is to simply shift the 
1754most significant bit to the least significant bit thus extracting the carry with a single cheap operation.  This 
1755optimization only works on twos compliment machines which is a safe assumption to make.
1756
1757If $a$ has a larger magnitude than $b$ an additional loop (lines @64,for@ through @73,}@) is required to propagate 
1758the carry through $a$ and copy the result to $c$.  
1759
1760\subsection{High Level Addition}
1761Now that both lower level addition and subtraction algorithms have been established an effective high level signed addition algorithm can be
1762established.  This high level addition algorithm will be what other algorithms and developers will use to perform addition of mp\_int data 
1763types.  
1764
1765Recall from section 5.2 that an mp\_int represents an integer with an unsigned mantissa (\textit{the array of digits}) and a \textbf{sign} 
1766flag.  A high level addition is actually performed as a series of eight separate cases which can be optimized down to three unique cases.
1767
1768\begin{figure}[!here]
1769\begin{center}
1770\begin{tabular}{l}
1771\hline Algorithm \textbf{mp\_add}. \\
1772\textbf{Input}.   Two mp\_ints $a$ and $b$  \\
1773\textbf{Output}.  The signed addition $c = a + b$. \\
1774\hline \\
17751.  if $a.sign = b.sign$ then do \\
1776\hspace{3mm}1.1  $c.sign \leftarrow a.sign$  \\
1777\hspace{3mm}1.2  $c \leftarrow \vert a \vert + \vert b \vert$ (\textit{s\_mp\_add})\\
17782.  else do \\
1779\hspace{3mm}2.1  if $\vert a \vert < \vert b \vert$ then do (\textit{mp\_cmp\_mag})  \\
1780\hspace{6mm}2.1.1  $c.sign \leftarrow b.sign$ \\
1781\hspace{6mm}2.1.2  $c \leftarrow \vert b \vert - \vert a \vert$ (\textit{s\_mp\_sub}) \\
1782\hspace{3mm}2.2  else do \\
1783\hspace{6mm}2.2.1  $c.sign \leftarrow a.sign$ \\
1784\hspace{6mm}2.2.2  $c \leftarrow \vert a \vert - \vert b \vert$ \\
17853.  Return(\textit{MP\_OKAY}). \\
1786\hline
1787\end{tabular}
1788\end{center}
1789\caption{Algorithm mp\_add}
1790\end{figure}
1791
1792\textbf{Algorithm mp\_add.}
1793This algorithm performs the signed addition of two mp\_int variables.  There is no reference algorithm to draw upon from 
1794either \cite{TAOCPV2} or \cite{HAC} since they both only provide unsigned operations.  The algorithm is fairly 
1795straightforward but restricted since subtraction can only produce positive results.
1796
1797\begin{figure}[here]
1798\begin{small}
1799\begin{center}
1800\begin{tabular}{|c|c|c|c|c|}
1801\hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert > \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\
1802\hline $+$ & $+$ & Yes & $c = a + b$ & $a.sign$ \\
1803\hline $+$ & $+$ & No  & $c = a + b$ & $a.sign$ \\
1804\hline $-$ & $-$ & Yes & $c = a + b$ & $a.sign$ \\
1805\hline $-$ & $-$ & No  & $c = a + b$ & $a.sign$ \\
1806\hline &&&&\\
1807
1808\hline $+$ & $-$ & No  & $c = b - a$ & $b.sign$ \\
1809\hline $-$ & $+$ & No  & $c = b - a$ & $b.sign$ \\
1810
1811\hline &&&&\\
1812
1813\hline $+$ & $-$ & Yes & $c = a - b$ & $a.sign$ \\
1814\hline $-$ & $+$ & Yes & $c = a - b$ & $a.sign$ \\
1815
1816\hline
1817\end{tabular}
1818\end{center}
1819\end{small}
1820\caption{Addition Guide Chart}
1821\label{fig:AddChart}
1822\end{figure}
1823
1824Figure~\ref{fig:AddChart} lists all of the eight possible input combinations and is sorted to show that only three 
1825specific cases need to be handled.  The return code of the unsigned operations at step 1.2, 2.1.2 and 2.2.2 are 
1826forwarded to step three to check for errors.  This simplifies the description of the algorithm considerably and best 
1827follows how the implementation actually was achieved.
1828
1829Also note how the \textbf{sign} is set before the unsigned addition or subtraction is performed.  Recall from the descriptions of algorithms
1830s\_mp\_add and s\_mp\_sub that the mp\_clamp function is used at the end to trim excess digits.  The mp\_clamp algorithm will set the \textbf{sign}
1831to \textbf{MP\_ZPOS} when the \textbf{used} digit count reaches zero.
1832
1833For example, consider performing $-a + a$ with algorithm mp\_add.  By the description of the algorithm the sign is set to \textbf{MP\_NEG} which would
1834produce a result of $-0$.  However, since the sign is set first then the unsigned addition is performed the subsequent usage of algorithm mp\_clamp 
1835within algorithm s\_mp\_add will force $-0$ to become $0$.  
1836
1837EXAM,bn_mp_add.c
1838
1839The source code follows the algorithm fairly closely.  The most notable new source code addition is the usage of the $res$ integer variable which
1840is used to pass result of the unsigned operations forward.  Unlike in the algorithm, the variable $res$ is merely returned as is without
1841explicitly checking it and returning the constant \textbf{MP\_OKAY}.  The observation is this algorithm will succeed or fail only if the lower
1842level functions do so.  Returning their return code is sufficient.
1843
1844\subsection{High Level Subtraction}
1845The high level signed subtraction algorithm is essentially the same as the high level signed addition algorithm.  
1846
1847\newpage\begin{figure}[!here]
1848\begin{center}
1849\begin{tabular}{l}
1850\hline Algorithm \textbf{mp\_sub}. \\
1851\textbf{Input}.   Two mp\_ints $a$ and $b$  \\
1852\textbf{Output}.  The signed subtraction $c = a - b$. \\
1853\hline \\
18541.  if $a.sign \ne b.sign$ then do \\
1855\hspace{3mm}1.1  $c.sign \leftarrow a.sign$ \\
1856\hspace{3mm}1.2  $c \leftarrow \vert a \vert + \vert b \vert$ (\textit{s\_mp\_add}) \\
18572.  else do \\
1858\hspace{3mm}2.1  if $\vert a \vert \ge \vert b \vert$ then do (\textit{mp\_cmp\_mag}) \\
1859\hspace{6mm}2.1.1  $c.sign \leftarrow a.sign$ \\
1860\hspace{6mm}2.1.2  $c \leftarrow \vert a \vert  - \vert b \vert$ (\textit{s\_mp\_sub}) \\
1861\hspace{3mm}2.2  else do \\
1862\hspace{6mm}2.2.1  $c.sign \leftarrow  \left \lbrace \begin{array}{ll}
1863                              MP\_ZPOS &  \mbox{if }a.sign = MP\_NEG \\
1864                              MP\_NEG  &  \mbox{otherwise} \\
1865                              \end{array} \right .$ \\
1866\hspace{6mm}2.2.2  $c \leftarrow \vert b \vert  - \vert a \vert$ \\
18673.  Return(\textit{MP\_OKAY}). \\
1868\hline
1869\end{tabular}
1870\end{center}
1871\caption{Algorithm mp\_sub}
1872\end{figure}
1873
1874\textbf{Algorithm mp\_sub.}
1875This algorithm performs the signed subtraction of two inputs.  Similar to algorithm mp\_add there is no reference in either \cite{TAOCPV2} or 
1876\cite{HAC}.  Also this algorithm is restricted by algorithm s\_mp\_sub.  Chart \ref{fig:SubChart} lists the eight possible inputs and
1877the operations required.
1878
1879\begin{figure}[!here]
1880\begin{small}
1881\begin{center}
1882\begin{tabular}{|c|c|c|c|c|}
1883\hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert \ge \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\
1884\hline $+$ & $-$ & Yes & $c = a + b$ & $a.sign$ \\
1885\hline $+$ & $-$ & No  & $c = a + b$ & $a.sign$ \\
1886\hline $-$ & $+$ & Yes & $c = a + b$ & $a.sign$ \\
1887\hline $-$ & $+$ & No  & $c = a + b$ & $a.sign$ \\
1888\hline &&&& \\
1889\hline $+$ & $+$ & Yes & $c = a - b$ & $a.sign$ \\
1890\hline $-$ & $-$ & Yes & $c = a - b$ & $a.sign$ \\
1891\hline &&&& \\
1892\hline $+$ & $+$ & No  & $c = b - a$ & $\mbox{opposite of }a.sign$ \\
1893\hline $-$ & $-$ & No  & $c = b - a$ & $\mbox{opposite of }a.sign$ \\
1894\hline
1895\end{tabular}
1896\end{center}
1897\end{small}
1898\caption{Subtraction Guide Chart}
1899\label{fig:SubChart}
1900\end{figure}
1901
1902Similar to the case of algorithm mp\_add the \textbf{sign} is set first before the unsigned addition or subtraction.  That is to prevent the 
1903algorithm from producing $-a - -a = -0$ as a result.  
1904
1905EXAM,bn_mp_sub.c
1906
1907Much like the implementation of algorithm mp\_add the variable $res$ is used to catch the return code of the unsigned addition or subtraction operations
1908and forward it to the end of the function.  On line @38, != MP_LT@ the ``not equal to'' \textbf{MP\_LT} expression is used to emulate a 
1909``greater than or equal to'' comparison.  
1910
1911\section{Bit and Digit Shifting}
1912MARK,POLY
1913It is quite common to think of a multiple precision integer as a polynomial in $x$, that is $y = f(\beta)$ where $f(x) = \sum_{i=0}^{n-1} a_i x^i$.  
1914This notation arises within discussion of Montgomery and Diminished Radix Reduction as well as Karatsuba multiplication and squaring.  
1915
1916In order to facilitate operations on polynomials in $x$ as above a series of simple ``digit'' algorithms have to be established.  That is to shift
1917the digits left or right as well to shift individual bits of the digits left and right.  It is important to note that not all ``shift'' operations
1918are on radix-$\beta$ digits.  
1919
1920\subsection{Multiplication by Two}
1921
1922In a binary system where the radix is a power of two multiplication by two not only arises often in other algorithms it is a fairly efficient 
1923operation to perform.  A single precision logical shift left is sufficient to multiply a single digit by two.  
1924
1925\newpage\begin{figure}[!here]
1926\begin{small}
1927\begin{center}
1928\begin{tabular}{l}
1929\hline Algorithm \textbf{mp\_mul\_2}. \\
1930\textbf{Input}.   One mp\_int $a$ \\
1931\textbf{Output}.  $b = 2a$. \\
1932\hline \\
19331.  If $b.alloc < a.used + 1$ then grow $b$ to hold $a.used + 1$ digits.  (\textit{mp\_grow}) \\
19342.  $oldused \leftarrow b.used$ \\
19353.  $b.used \leftarrow a.used$ \\
19364.  $r \leftarrow 0$ \\
19375.  for $n$ from 0 to $a.used - 1$ do \\
1938\hspace{3mm}5.1  $rr \leftarrow a_n >> (lg(\beta) - 1)$ \\
1939\hspace{3mm}5.2  $b_n \leftarrow (a_n << 1) + r \mbox{ (mod }\beta\mbox{)}$ \\
1940\hspace{3mm}5.3  $r \leftarrow rr$ \\
19416.  If $r \ne 0$ then do \\
1942\hspace{3mm}6.1  $b_{n + 1} \leftarrow r$ \\
1943\hspace{3mm}6.2  $b.used \leftarrow b.used + 1$ \\
19447.  If $b.used < oldused - 1$ then do \\
1945\hspace{3mm}7.1  for $n$ from $b.used$ to $oldused - 1$ do \\
1946\hspace{6mm}7.1.1  $b_n \leftarrow 0$ \\
19478.  $b.sign \leftarrow a.sign$ \\
19489.  Return(\textit{MP\_OKAY}).\\
1949\hline
1950\end{tabular}
1951\end{center}
1952\end{small}
1953\caption{Algorithm mp\_mul\_2}
1954\end{figure}
1955
1956\textbf{Algorithm mp\_mul\_2.}
1957This algorithm will quickly multiply a mp\_int by two provided $\beta$ is a power of two.  Neither \cite{TAOCPV2} nor \cite{HAC} describe such 
1958an algorithm despite the fact it arises often in other algorithms.  The algorithm is setup much like the lower level algorithm s\_mp\_add since 
1959it is for all intents and purposes equivalent to the operation $b = \vert a \vert + \vert a \vert$.  
1960
1961Step 1 and 2 grow the input as required to accomodate the maximum number of \textbf{used} digits in the result.  The initial \textbf{used} count
1962is set to $a.used$ at step 4.  Only if there is a final carry will the \textbf{used} count require adjustment.
1963
1964Step 6 is an optimization implementation of the addition loop for this specific case.  That is since the two values being added together 
1965are the same there is no need to perform two reads from the digits of $a$.  Step 6.1 performs a single precision shift on the current digit $a_n$ to
1966obtain what will be the carry for the next iteration.  Step 6.2 calculates the $n$'th digit of the result as single precision shift of $a_n$ plus
1967the previous carry.  Recall from ~SHIFTS~ that $a_n << 1$ is equivalent to $a_n \cdot 2$.  An iteration of the addition loop is finished with 
1968forwarding the carry to the next iteration.
1969
1970Step 7 takes care of any final carry by setting the $a.used$'th digit of the result to the carry and augmenting the \textbf{used} count of $b$.  
1971Step 8 clears any leading digits of $b$ in case it originally had a larger magnitude than $a$.
1972
1973EXAM,bn_mp_mul_2.c
1974
1975This implementation is essentially an optimized implementation of s\_mp\_add for the case of doubling an input.  The only noteworthy difference
1976is the use of the logical shift operator on line @52,<<@ to perform a single precision doubling.  
1977
1978\subsection{Division by Two}
1979A division by two can just as easily be accomplished with a logical shift right as multiplication by two can be with a logical shift left.
1980
1981\newpage\begin{figure}[!here]
1982\begin{small}
1983\begin{center}
1984\begin{tabular}{l}
1985\hline Algorithm \textbf{mp\_div\_2}. \\
1986\textbf{Input}.   One mp\_int $a$ \\
1987\textbf{Output}.  $b = a/2$. \\
1988\hline \\
19891.  If $b.alloc < a.used$ then grow $b$ to hold $a.used$ digits.  (\textit{mp\_grow}) \\
19902.  If the reallocation failed return(\textit{MP\_MEM}). \\
19913.  $oldused \leftarrow b.used$ \\
19924.  $b.used \leftarrow a.used$ \\
19935.  $r \leftarrow 0$ \\
19946.  for $n$ from $b.used - 1$ to $0$ do \\
1995\hspace{3mm}6.1  $rr \leftarrow a_n \mbox{ (mod }2\mbox{)}$\\
1996\hspace{3mm}6.2  $b_n \leftarrow (a_n >> 1) + (r << (lg(\beta) - 1)) \mbox{ (mod }\beta\mbox{)}$ \\
1997\hspace{3mm}6.3  $r \leftarrow rr$ \\
19987.  If $b.used < oldused - 1$ then do \\
1999\hspace{3mm}7.1  for $n$ from $b.used$ to $oldused - 1$ do \\
2000\hspace{6mm}7.1.1  $b_n \leftarrow 0$ \\
20018.  $b.sign \leftarrow a.sign$ \\
20029.  Clamp excess digits of $b$.  (\textit{mp\_clamp}) \\
200310.  Return(\textit{MP\_OKAY}).\\
2004\hline
2005\end{tabular}
2006\end{center}
2007\end{small}
2008\caption{Algorithm mp\_div\_2}
2009\end{figure}
2010
2011\textbf{Algorithm mp\_div\_2.}
2012This algorithm will divide an mp\_int by two using logical shifts to the right.  Like mp\_mul\_2 it uses a modified low level addition
2013core as the basis of the algorithm.  Unlike mp\_mul\_2 the shift operations work from the leading digit to the trailing digit.  The algorithm
2014could be written to work from the trailing digit to the leading digit however, it would have to stop one short of $a.used - 1$ digits to prevent
2015reading past the end of the array of digits.
2016
2017Essentially the loop at step 6 is similar to that of mp\_mul\_2 except the logical shifts go in the opposite direction and the carry is at the 
2018least significant bit not the most significant bit.  
2019
2020EXAM,bn_mp_div_2.c
2021
2022\section{Polynomial Basis Operations}
2023Recall from ~POLY~ that any integer can be represented as a polynomial in $x$ as $y = f(\beta)$.  Such a representation is also known as
2024the polynomial basis \cite[pp. 48]{ROSE}. Given such a notation a multiplication or division by $x$ amounts to shifting whole digits a single 
2025place.  The need for such operations arises in several other higher level algorithms such as Barrett and Montgomery reduction, integer
2026division and Karatsuba multiplication.  
2027
2028Converting from an array of digits to polynomial basis is very simple.  Consider the integer $y \equiv (a_2, a_1, a_0)_{\beta}$ and recall that
2029$y = \sum_{i=0}^{2} a_i \beta^i$.  Simply replace $\beta$ with $x$ and the expression is in polynomial basis.  For example, $f(x) = 8x + 9$ is the
2030polynomial basis representation for $89$ using radix ten.  That is, $f(10) = 8(10) + 9 = 89$.  
2031
2032\subsection{Multiplication by $x$}
2033
2034Given a polynomial in $x$ such as $f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_0$ multiplying by $x$ amounts to shifting the coefficients up one 
2035degree.  In this case $f(x) \cdot x = a_n x^{n+1} + a_{n-1} x^n + ... + a_0 x$.  From a scalar basis point of view multiplying by $x$ is equivalent to
2036multiplying by the integer $\beta$.  
2037
2038\newpage\begin{figure}[!here]
2039\begin{small}
2040\begin{center}
2041\begin{tabular}{l}
2042\hline Algorithm \textbf{mp\_lshd}. \\
2043\textbf{Input}.   One mp\_int $a$ and an integer $b$ \\
2044\textbf{Output}.  $a \leftarrow a \cdot \beta^b$ (equivalent to multiplication by $x^b$). \\
2045\hline \\
20461.  If $b \le 0$ then return(\textit{MP\_OKAY}). \\
20472.  If $a.alloc < a.used + b$ then grow $a$ to at least $a.used + b$ digits.  (\textit{mp\_grow}). \\
20483.  If the reallocation failed return(\textit{MP\_MEM}). \\
20494.  $a.used \leftarrow a.used + b$ \\
20505.  $i \leftarrow a.used - 1$ \\
20516.  $j \leftarrow a.used - 1 - b$ \\
20527.  for $n$ from $a.used - 1$ to $b$ do \\
2053\hspace{3mm}7.1  $a_{i} \leftarrow a_{j}$ \\
2054\hspace{3mm}7.2  $i \leftarrow i - 1$ \\
2055\hspace{3mm}7.3  $j \leftarrow j - 1$ \\
20568.  for $n$ from 0 to $b - 1$ do \\
2057\hspace{3mm}8.1  $a_n \leftarrow 0$ \\
20589.  Return(\textit{MP\_OKAY}). \\
2059\hline
2060\end{tabular}
2061\end{center}
2062\end{small}
2063\caption{Algorithm mp\_lshd}
2064\end{figure}
2065
2066\textbf{Algorithm mp\_lshd.}
2067This algorithm multiplies an mp\_int by the $b$'th power of $x$.  This is equivalent to multiplying by $\beta^b$.  The algorithm differs 
2068from the other algorithms presented so far as it performs the operation in place instead storing the result in a separate location.  The
2069motivation behind this change is due to the way this function is typically used.  Algorithms such as mp\_add store the result in an optionally
2070different third mp\_int because the original inputs are often still required.  Algorithm mp\_lshd (\textit{and similarly algorithm mp\_rshd}) is
2071typically used on values where the original value is no longer required.  The algorithm will return success immediately if 
2072$b \le 0$ since the rest of algorithm is only valid when $b > 0$.  
2073
2074First the destination $a$ is grown as required to accomodate the result.  The counters $i$ and $j$ are used to form a \textit{sliding window} over
2075the digits of $a$ of length $b$.  The head of the sliding window is at $i$ (\textit{the leading digit}) and the tail at $j$ (\textit{the trailing digit}).  
2076The loop on step 7 copies the digit from the tail to the head.  In each iteration the window is moved down one digit.   The last loop on 
2077step 8 sets the lower $b$ digits to zero.
2078
2079\newpage
2080FIGU,sliding_window,Sliding Window Movement
2081
2082EXAM,bn_mp_lshd.c
2083
2084The if statement (line @24,if@) ensures that the $b$ variable is greater than zero since we do not interpret negative
2085shift counts properly.  The \textbf{used} count is incremented by $b$ before the copy loop begins.  This elminates 
2086the need for an additional variable in the for loop.  The variable $top$ (line @42,top@) is an alias
2087for the leading digit while $bottom$ (line @45,bottom@) is an alias for the trailing edge.  The aliases form a 
2088window of exactly $b$ digits over the input.  
2089
2090\subsection{Division by $x$}
2091
2092Division by powers of $x$ is easily achieved by shifting the digits right and removing any that will end up to the right of the zero'th digit.  
2093
2094\newpage\begin{figure}[!here]
2095\begin{small}
2096\begin{center}
2097\begin{tabular}{l}
2098\hline Algorithm \textbf{mp\_rshd}. \\
2099\textbf{Input}.   One mp\_int $a$ and an integer $b$ \\
2100\textbf{Output}.  $a \leftarrow a / \beta^b$ (Divide by $x^b$). \\
2101\hline \\
21021.  If $b \le 0$ then return. \\
21032.  If $a.used \le b$ then do \\
2104\hspace{3mm}2.1  Zero $a$.  (\textit{mp\_zero}). \\
2105\hspace{3mm}2.2  Return. \\
21063.  $i \leftarrow 0$ \\
21074.  $j \leftarrow b$ \\
21085.  for $n$ from 0 to $a.used - b - 1$ do \\
2109\hspace{3mm}5.1  $a_i \leftarrow a_j$ \\
2110\hspace{3mm}5.2  $i \leftarrow i + 1$ \\
2111\hspace{3mm}5.3  $j \leftarrow j + 1$ \\
21126.  for $n$ from $a.used - b$ to $a.used - 1$ do \\
2113\hspace{3mm}6.1  $a_n \leftarrow 0$ \\
21147.  $a.used \leftarrow a.used - b$ \\
21158.  Return. \\
2116\hline
2117\end{tabular}
2118\end{center}
2119\end{small}
2120\caption{Algorithm mp\_rshd}
2121\end{figure}
2122
2123\textbf{Algorithm mp\_rshd.}
2124This algorithm divides the input in place by the $b$'th power of $x$.  It is analogous to dividing by a $\beta^b$ but much quicker since
2125it does not require single precision division.  This algorithm does not actually return an error code as it cannot fail.  
2126
2127If the input $b$ is less than one the algorithm quickly returns without performing any work.  If the \textbf{used} count is less than or equal
2128to the shift count $b$ then it will simply zero the input and return.
2129
2130After the trivial cases of inputs have been handled the sliding window is setup.  Much like the case of algorithm mp\_lshd a sliding window that
2131is $b$ digits wide is used to copy the digits.  Unlike mp\_lshd the window slides in the opposite direction from the trailing to the leading digit.  
2132Also the digits are copied from the leading to the trailing edge.
2133
2134Once the window copy is complete the upper digits must be zeroed and the \textbf{used} count decremented.
2135
2136EXAM,bn_mp_rshd.c
2137
2138The only noteworthy element of this routine is the lack of a return type since it cannot fail.  Like mp\_lshd() we
2139form a sliding window except we copy in the other direction.  After the window (line @59,for (;@) we then zero
2140the upper digits of the input to make sure the result is correct.
2141
2142\section{Powers of Two}
2143
2144Now that algorithms for moving single bits as well as whole digits exist algorithms for moving the ``in between'' distances are required.  For 
2145example, to quickly multiply by $2^k$ for any $k$ without using a full multiplier algorithm would prove useful.  Instead of performing single
2146shifts $k$ times to achieve a multiplication by $2^{\pm k}$ a mixture of whole digit shifting and partial digit shifting is employed.  
2147
2148\subsection{Multiplication by Power of Two}
2149
2150\newpage\begin{figure}[!here]
2151\begin{small}
2152\begin{center}
2153\begin{tabular}{l}
2154\hline Algorithm \textbf{mp\_mul\_2d}. \\
2155\textbf{Input}.   One mp\_int $a$ and an integer $b$ \\
2156\textbf{Output}.  $c \leftarrow a \cdot 2^b$. \\
2157\hline \\
21581.  $c \leftarrow a$.  (\textit{mp\_copy}) \\
21592.  If $c.alloc < c.used + \lfloor b / lg(\beta) \rfloor + 2$ then grow $c$ accordingly. \\
21603.  If the reallocation failed return(\textit{MP\_MEM}). \\
21614.  If $b \ge lg(\beta)$ then \\
2162\hspace{3mm}4.1  $c \leftarrow c \cdot \beta^{\lfloor b / lg(\beta) \rfloor}$ (\textit{mp\_lshd}). \\
2163\hspace{3mm}4.2  If step 4.1 failed return(\textit{MP\_MEM}). \\
21645.  $d \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\
21656.  If $d \ne 0$ then do \\
2166\hspace{3mm}6.1  $mask \leftarrow 2^d$ \\
2167\hspace{3mm}6.2  $r \leftarrow 0$ \\
2168\hspace{3mm}6.3  for $n$ from $0$ to $c.used - 1$ do \\
2169\hspace{6mm}6.3.1  $rr \leftarrow c_n >> (lg(\beta) - d) \mbox{ (mod }mask\mbox{)}$ \\
2170\hspace{6mm}6.3.2  $c_n \leftarrow (c_n << d) + r \mbox{ (mod }\beta\mbox{)}$ \\
2171\hspace{6mm}6.3.3  $r \leftarrow rr$ \\
2172\hspace{3mm}6.4  If $r > 0$ then do \\
2173\hspace{6mm}6.4.1  $c_{c.used} \leftarrow r$ \\
2174\hspace{6mm}6.4.2  $c.used \leftarrow c.used + 1$ \\
21757.  Return(\textit{MP\_OKAY}). \\
2176\hline
2177\end{tabular}
2178\end{center}
2179\end{small}
2180\caption{Algorithm mp\_mul\_2d}
2181\end{figure}
2182
2183\textbf{Algorithm mp\_mul\_2d.}
2184This algorithm multiplies $a$ by $2^b$ and stores the result in $c$.  The algorithm uses algorithm mp\_lshd and a derivative of algorithm mp\_mul\_2 to
2185quickly compute the product.
2186
2187First the algorithm will multiply $a$ by $x^{\lfloor b / lg(\beta) \rfloor}$ which will ensure that the remainder multiplicand is less than 
2188$\beta$.  For example, if $b = 37$ and $\beta = 2^{28}$ then this step will multiply by $x$ leaving a multiplication by $2^{37 - 28} = 2^{9}$ 
2189left.
2190
2191After the digits have been shifted appropriately at most $lg(\beta) - 1$ shifts are left to perform.  Step 5 calculates the number of remaining shifts 
2192required.  If it is non-zero a modified shift loop is used to calculate the remaining product.  
2193Essentially the loop is a generic version of algorithm mp\_mul\_2 designed to handle any shift count in the range $1 \le x < lg(\beta)$.  The $mask$
2194variable is used to extract the upper $d$ bits to form the carry for the next iteration.  
2195
2196This algorithm is loosely measured as a $O(2n)$ algorithm which means that if the input is $n$-digits that it takes $2n$ ``time'' to 
2197complete.  It is possible to optimize this algorithm down to a $O(n)$ algorithm at a cost of making the algorithm slightly harder to follow.
2198
2199EXAM,bn_mp_mul_2d.c
2200
2201The shifting is performed in--place which means the first step (line @24,a != c@) is to copy the input to the 
2202destination.  We avoid calling mp\_copy() by making sure the mp\_ints are different.  The destination then
2203has to be grown (line @31,grow@) to accomodate the result.
2204
2205If the shift count $b$ is larger than $lg(\beta)$ then a call to mp\_lshd() is used to handle all of the multiples 
2206of $lg(\beta)$.  Leaving only a remaining shift of $lg(\beta) - 1$ or fewer bits left.  Inside the actual shift 
2207loop (lines @45,if@ to @76,}@) we make use of pre--computed values $shift$ and $mask$.   These are used to
2208extract the carry bit(s) to pass into the next iteration of the loop.  The $r$ and $rr$ variables form a 
2209chain between consecutive iterations to propagate the carry.  
2210
2211\subsection{Division by Power of Two}
2212
2213\newpage\begin{figure}[!here]
2214\begin{small}
2215\begin{center}
2216\begin{tabular}{l}
2217\hline Algorithm \textbf{mp\_div\_2d}. \\
2218\textbf{Input}.   One mp\_int $a$ and an integer $b$ \\
2219\textbf{Output}.  $c \leftarrow \lfloor a / 2^b \rfloor, d \leftarrow a \mbox{ (mod }2^b\mbox{)}$. \\
2220\hline \\
22211.  If $b \le 0$ then do \\
2222\hspace{3mm}1.1  $c \leftarrow a$ (\textit{mp\_copy}) \\
2223\hspace{3mm}1.2  $d \leftarrow 0$ (\textit{mp\_zero}) \\
2224\hspace{3mm}1.3  Return(\textit{MP\_OKAY}). \\
22252.  $c \leftarrow a$ \\
22263.  $d \leftarrow a \mbox{ (mod }2^b\mbox{)}$ (\textit{mp\_mod\_2d}) \\
22274.  If $b \ge lg(\beta)$ then do \\
2228\hspace{3mm}4.1  $c \leftarrow \lfloor c/\beta^{\lfloor b/lg(\beta) \rfloor} \rfloor$ (\textit{mp\_rshd}). \\
22295.  $k \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\
22306.  If $k \ne 0$ then do \\
2231\hspace{3mm}6.1  $mask \leftarrow 2^k$ \\
2232\hspace{3mm}6.2  $r \leftarrow 0$ \\
2233\hspace{3mm}6.3  for $n$ from $c.used - 1$ to $0$ do \\
2234\hspace{6mm}6.3.1  $rr \leftarrow c_n \mbox{ (mod }mask\mbox{)}$ \\
2235\hspace{6mm}6.3.2  $c_n \leftarrow (c_n >> k) + (r << (lg(\beta) - k))$ \\
2236\hspace{6mm}6.3.3  $r \leftarrow rr$ \\
22377.  Clamp excess digits of $c$.  (\textit{mp\_clamp}) \\
22388.  Return(\textit{MP\_OKAY}). \\
2239\hline
2240\end{tabular}
2241\end{center}
2242\end{small}
2243\caption{Algorithm mp\_div\_2d}
2244\end{figure}
2245
2246\textbf{Algorithm mp\_div\_2d.}
2247This algorithm will divide an input $a$ by $2^b$ and produce the quotient and remainder.  The algorithm is designed much like algorithm 
2248mp\_mul\_2d by first using whole digit shifts then single precision shifts.  This algorithm will also produce the remainder of the division
2249by using algorithm mp\_mod\_2d.
2250
2251EXAM,bn_mp_div_2d.c
2252
2253The implementation of algorithm mp\_div\_2d is slightly different than the algorithm specifies.  The remainder $d$ may be optionally 
2254ignored by passing \textbf{NULL} as the pointer to the mp\_int variable.    The temporary mp\_int variable $t$ is used to hold the 
2255result of the remainder operation until the end.  This allows $d$ and $a$ to represent the same mp\_int without modifying $a$ before
2256the quotient is obtained.
2257
2258The remainder of the source code is essentially the same as the source code for mp\_mul\_2d.  The only significant difference is
2259the direction of the shifts.
2260
2261\subsection{Remainder of Division by Power of Two}
2262
2263The last algorithm in the series of polynomial basis power of two algorithms is calculating the remainder of division by $2^b$.  This
2264algorithm benefits from the fact that in twos complement arithmetic $a \mbox{ (mod }2^b\mbox{)}$ is the same as $a$ AND $2^b - 1$.  
2265
2266\begin{figure}[!here]
2267\begin{small}
2268\begin{center}
2269\begin{tabular}{l}
2270\hline Algorithm \textbf{mp\_mod\_2d}. \\
2271\textbf{Input}.   One mp\_int $a$ and an integer $b$ \\
2272\textbf{Output}.  $c \leftarrow a \mbox{ (mod }2^b\mbox{)}$. \\
2273\hline \\
22741.  If $b \le 0$ then do \\
2275\hspace{3mm}1.1  $c \leftarrow 0$ (\textit{mp\_zero}) \\
2276\hspace{3mm}1.2  Return(\textit{MP\_OKAY}). \\
22772.  If $b > a.used \cdot lg(\beta)$ then do \\
2278\hspace{3mm}2.1  $c \leftarrow a$ (\textit{mp\_copy}) \\
2279\hspace{3mm}2.2  Return the result of step 2.1. \\
22803.  $c \leftarrow a$ \\
22814.  If step 3 failed return(\textit{MP\_MEM}). \\
22825.  for $n$ from $\lceil b / lg(\beta) \rceil$ to $c.used$ do \\
2283\hspace{3mm}5.1  $c_n \leftarrow 0$ \\
22846.  $k \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\
22857.  $c_{\lfloor b / lg(\beta) \rfloor} \leftarrow c_{\lfloor b / lg(\beta) \rfloor} \mbox{ (mod }2^{k}\mbox{)}$. \\
22868.  Clamp excess digits of $c$.  (\textit{mp\_clamp}) \\
22879.  Return(\textit{MP\_OKAY}). \\
2288\hline
2289\end{tabular}
2290\end{center}
2291\end{small}
2292\caption{Algorithm mp\_mod\_2d}
2293\end{figure}
2294
2295\textbf{Algorithm mp\_mod\_2d.}
2296This algorithm will quickly calculate the value of $a \mbox{ (mod }2^b\mbox{)}$.  First if $b$ is less than or equal to zero the 
2297result is set to zero.  If $b$ is greater than the number of bits in $a$ then it simply copies $a$ to $c$ and returns.  Otherwise, $a$ 
2298is copied to $b$, leading digits are removed and the remaining leading digit is trimed to the exact bit count.
2299
2300EXAM,bn_mp_mod_2d.c
2301
2302We first avoid cases of $b \le 0$ by simply mp\_zero()'ing the destination in such cases.  Next if $2^b$ is larger
2303than the input we just mp\_copy() the input and return right away.  After this point we know we must actually
2304perform some work to produce the remainder.
2305
2306Recalling that reducing modulo $2^k$ and a binary ``and'' with $2^k - 1$ are numerically equivalent we can quickly reduce 
2307the number.  First we zero any digits above the last digit in $2^b$ (line @41,for@).  Next we reduce the 
2308leading digit of both (line @45,&=@) and then mp\_clamp().
2309
2310\section*{Exercises}
2311\begin{tabular}{cl}
2312$\left [ 3 \right ] $ & Devise an algorithm that performs $a \cdot 2^b$ for generic values of $b$ \\
2313                      & in $O(n)$ time. \\
2314                      &\\
2315$\left [ 3 \right ] $ & Devise an efficient algorithm to multiply by small low hamming  \\
2316                      & weight values such as $3$, $5$ and $9$.  Extend it to handle all values \\
2317                      & upto $64$ with a hamming weight less than three. \\
2318                      &\\
2319$\left [ 2 \right ] $ & Modify the preceding algorithm to handle values of the form \\
2320                      & $2^k - 1$ as well. \\
2321                      &\\
2322$\left [ 3 \right ] $ & Using only algorithms mp\_mul\_2, mp\_div\_2 and mp\_add create an \\
2323                      & algorithm to multiply two integers in roughly $O(2n^2)$ time for \\
2324                      & any $n$-bit input.  Note that the time of addition is ignored in the \\
2325                      & calculation.  \\
2326                      & \\
2327$\left [ 5 \right ] $ & Improve the previous algorithm to have a working time of at most \\
2328                      & $O \left (2^{(k-1)}n + \left ({2n^2 \over k} \right ) \right )$ for an appropriate choice of $k$.  Again ignore \\
2329                      & the cost of addition. \\
2330                      & \\
2331$\left [ 2 \right ] $ & Devise a chart to find optimal values of $k$ for the previous problem \\
2332                      & for $n = 64 \ldots 1024$ in steps of $64$. \\
2333                      & \\
2334$\left [ 2 \right ] $ & Using only algorithms mp\_abs and mp\_sub devise another method for \\
2335                      & calculating the result of a signed comparison. \\
2336                      &
2337\end{tabular}
2338
2339\chapter{Multiplication and Squaring}
2340\section{The Multipliers}
2341For most number theoretic problems including certain public key cryptographic algorithms, the ``multipliers'' form the most important subset of 
2342algorithms of any multiple precision integer package.  The set of multiplier algorithms include integer multiplication, squaring and modular reduction 
2343where in each of the algorithms single precision multiplication is the dominant operation performed.  This chapter will discuss integer multiplication 
2344and squaring, leaving modular reductions for the subsequent chapter.  
2345
2346The importance of the multiplier algorithms is for the most part driven by the fact that certain popular public key algorithms are based on modular 
2347exponentiation, that is computing $d \equiv a^b \mbox{ (mod }c\mbox{)}$ for some arbitrary choice of $a$, $b$, $c$ and $d$.  During a modular
2348exponentiation the majority\footnote{Roughly speaking a modular exponentiation will spend about 40\% of the time performing modular reductions, 
234935\% of the time performing squaring and 25\% of the time performing multiplications.} of the processor time is spent performing single precision 
2350multiplications.
2351
2352For centuries general purpose multiplication has required a lengthly $O(n^2)$ process, whereby each digit of one multiplicand has to be multiplied 
2353against every digit of the other multiplicand.  Traditional long-hand multiplication is based on this process;  while the techniques can differ the 
2354overall algorithm used is essentially the same.  Only ``recently'' have faster algorithms been studied.  First Karatsuba multiplication was discovered in 
23551962.  This algorithm can multiply two numbers with considerably fewer single precision multiplications when compared to the long-hand approach.  
2356This technique led to the discovery of polynomial basis algorithms (\textit{good reference?}) and subquently Fourier Transform based solutions.  
2357
2358\section{Multiplication}
2359\subsection{The Baseline Multiplication}
2360\label{sec:basemult}
2361\index{baseline multiplication}
2362Computing the product of two integers in software can be achieved using a trivial adaptation of the standard $O(n^2)$ long-hand multiplication
2363algorithm that school children are taught.  The algorithm is considered an $O(n^2)$ algorithm since for two $n$-digit inputs $n^2$ single precision 
2364multiplications are required.  More specifically for a $m$ and $n$ digit input $m \cdot n$ single precision multiplications are required.  To 
2365simplify most discussions, it will be assumed that the inputs have comparable number of digits.  
2366
2367The ``baseline multiplication'' algorithm is designed to act as the ``catch-all'' algorithm, only to be used when the faster algorithms cannot be 
2368used.  This algorithm does not use any particularly interesting optimizations and should ideally be avoided if possible.    One important 
2369facet of this algorithm, is that it has been modified to only produce a certain amount of output digits as resolution.  The importance of this 
2370modification will become evident during the discussion of Barrett modular reduction.  Recall that for a $n$ and $m$ digit input the product 
2371will be at most $n + m$ digits.  Therefore, this algorithm can be reduced to a full multiplier by having it produce $n + m$ digits of the product.  
2372
2373Recall from ~GAMMA~ the definition of $\gamma$ as the number of bits in the type \textbf{mp\_digit}.  We shall now extend the variable set to 
2374include $\alpha$ which shall represent the number of bits in the type \textbf{mp\_word}.  This implies that $2^{\alpha} > 2 \cdot \beta^2$.  The 
2375constant $\delta = 2^{\alpha - 2lg(\beta)}$ will represent the maximal weight of any column in a product (\textit{see ~COMBA~ for more information}).
2376
2377\newpage\begin{figure}[!here]
2378\begin{small}
2379\begin{center}
2380\begin{tabular}{l}
2381\hline Algorithm \textbf{s\_mp\_mul\_digs}. \\
2382\textbf{Input}.   mp\_int $a$, mp\_int $b$ and an integer $digs$ \\
2383\textbf{Output}.  $c \leftarrow \vert a \vert \cdot \vert b \vert \mbox{ (mod }\beta^{digs}\mbox{)}$. \\
2384\hline \\
23851.  If min$(a.used, b.used) < \delta$ then do \\
2386\hspace{3mm}1.1  Calculate $c = \vert a \vert \cdot \vert b \vert$ by the Comba method (\textit{see algorithm~\ref{fig:COMBAMULT}}).  \\
2387\hspace{3mm}1.2  Return the result of step 1.1 \\
2388\\
2389Allocate and initialize a temporary mp\_int. \\
23902.  Init $t$ to be of size $digs$ \\
23913.  If step 2 failed return(\textit{MP\_MEM}). \\
23924.  $t.used \leftarrow digs$ \\
2393\\
2394Compute the product. \\
23955.  for $ix$ from $0$ to $a.used - 1$ do \\
2396\hspace{3mm}5.1  $u \leftarrow 0$ \\
2397\hspace{3mm}5.2  $pb \leftarrow \mbox{min}(b.used, digs - ix)$ \\
2398\hspace{3mm}5.3  If $pb < 1$ then goto step 6. \\
2399\hspace{3mm}5.4  for $iy$ from $0$ to $pb - 1$ do \\
2400\hspace{6mm}5.4.1  $\hat r \leftarrow t_{iy + ix} + a_{ix} \cdot b_{iy} + u$ \\
2401\hspace{6mm}5.4.2  $t_{iy + ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
2402\hspace{6mm}5.4.3  $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\
2403\hspace{3mm}5.5  if $ix + pb < digs$ then do \\
2404\hspace{6mm}5.5.1  $t_{ix + pb} \leftarrow u$ \\
24056.  Clamp excess digits of $t$. \\
24067.  Swap $c$ with $t$ \\
24078.  Clear $t$ \\
24089.  Return(\textit{MP\_OKAY}). \\
2409\hline
2410\end{tabular}
2411\end{center}
2412\end{small}
2413\caption{Algorithm s\_mp\_mul\_digs}
2414\end{figure}
2415
2416\textbf{Algorithm s\_mp\_mul\_digs.}
2417This algorithm computes the unsigned product of two inputs $a$ and $b$, limited to an output precision of $digs$ digits.  While it may seem
2418a bit awkward to modify the function from its simple $O(n^2)$ description, the usefulness of partial multipliers will arise in a subsequent 
2419algorithm.  The algorithm is loosely based on algorithm 14.12 from \cite[pp. 595]{HAC} and is similar to Algorithm M of Knuth \cite[pp. 268]{TAOCPV2}.  
2420Algorithm s\_mp\_mul\_digs differs from these cited references since it can produce a variable output precision regardless of the precision of the 
2421inputs.
2422
2423The first thing this algorithm checks for is whether a Comba multiplier can be used instead.   If the minimum digit count of either
2424input is less than $\delta$, then the Comba method may be used instead.    After the Comba method is ruled out, the baseline algorithm begins.  A 
2425temporary mp\_int variable $t$ is used to hold the intermediate result of the product.  This allows the algorithm to be used to 
2426compute products when either $a = c$ or $b = c$ without overwriting the inputs.  
2427
2428All of step 5 is the infamous $O(n^2)$ multiplication loop slightly modified to only produce upto $digs$ digits of output.  The $pb$ variable
2429is given the count of digits to read from $b$ inside the nested loop.  If $pb \le 1$ then no more output digits can be produced and the algorithm
2430will exit the loop.  The best way to think of the loops are as a series of $pb \times 1$ multiplications.    That is, in each pass of the 
2431innermost loop $a_{ix}$ is multiplied against $b$ and the result is added (\textit{with an appropriate shift}) to $t$.  
2432
2433For example, consider multiplying $576$ by $241$.  That is equivalent to computing $10^0(1)(576) + 10^1(4)(576) + 10^2(2)(576)$ which is best
2434visualized in the following table.
2435
2436\begin{figure}[here]
2437\begin{center}
2438\begin{tabular}{|c|c|c|c|c|c|l|}
2439\hline   &&          & 5 & 7 & 6 & \\
2440\hline   $\times$&&  & 2 & 4 & 1 & \\
2441\hline &&&&&&\\
2442  &&          & 5 & 7 & 6 & $10^0(1)(576)$ \\
2443  &2 &   3    & 6 & 1 & 6 & $10^1(4)(576) + 10^0(1)(576)$ \\
2444  1 & 3 & 8 & 8 & 1 & 6 &   $10^2(2)(576) + 10^1(4)(576) + 10^0(1)(576)$ \\
2445\hline  
2446\end{tabular}
2447\end{center}
2448\caption{Long-Hand Multiplication Diagram}
2449\end{figure}
2450
2451Each row of the product is added to the result after being shifted to the left (\textit{multiplied by a power of the radix}) by the appropriate 
2452count.  That is in pass $ix$ of the inner loop the product is added starting at the $ix$'th digit of the reult.
2453
2454Step 5.4.1 introduces the hat symbol (\textit{e.g. $\hat r$}) which represents a double precision variable.  The multiplication on that step
2455is assumed to be a double wide output single precision multiplication.  That is, two single precision variables are multiplied to produce a
2456double precision result.  The step is somewhat optimized from a long-hand multiplication algorithm because the carry from the addition in step
24575.4.1 is propagated through the nested loop.  If the carry was not propagated immediately it would overflow the single precision digit 
2458$t_{ix+iy}$ and the result would be lost.  
2459
2460At step 5.5 the nested loop is finished and any carry that was left over should be forwarded.  The carry does not have to be added to the $ix+pb$'th
2461digit since that digit is assumed to be zero at this point.  However, if $ix + pb \ge digs$ the carry is not set as it would make the result
2462exceed the precision requested.
2463
2464EXAM,bn_s_mp_mul_digs.c
2465
2466First we determine (line @30,if@) if the Comba method can be used first since it's faster.  The conditions for 
2467sing the Comba routine are that min$(a.used, b.used) < \delta$ and the number of digits of output is less than 
2468\textbf{MP\_WARRAY}.  This new constant is used to control the stack usage in the Comba routines.  By default it is 
2469set to $\delta$ but can be reduced when memory is at a premium.
2470
2471If we cannot use the Comba method we proceed to setup the baseline routine.  We allocate the the destination mp\_int
2472$t$ (line @36,init@) to the exact size of the output to avoid further re--allocations.  At this point we now 
2473begin the $O(n^2)$ loop.
2474
2475This implementation of multiplication has the caveat that it can be trimmed to only produce a variable number of
2476digits as output.  In each iteration of the outer loop the $pb$ variable is set (line @48,MIN@) to the maximum 
2477number of inner loop iterations.  
2478
2479Inside the inner loop we calculate $\hat r$ as the mp\_word product of the two mp\_digits and the addition of the
2480carry from the previous iteration.  A particularly important observation is that most modern optimizing 
2481C compilers (GCC for instance) can recognize that a $N \times N \rightarrow 2N$ multiplication is all that 
2482is required for the product.  In x86 terms for example, this means using the MUL instruction.
2483
2484Each digit of the product is stored in turn (line @68,tmpt@) and the carry propagated (line @71,>>@) to the 
2485next iteration.
2486
2487\subsection{Faster Multiplication by the ``Comba'' Method}
2488MARK,COMBA
2489
2490One of the huge drawbacks of the ``baseline'' algorithms is that at the $O(n^2)$ level the carry must be 
2491computed and propagated upwards.  This makes the nested loop very sequential and hard to unroll and implement 
2492in parallel.  The ``Comba'' \cite{COMBA} method is named after little known (\textit{in cryptographic venues}) Paul G. 
2493Comba who described a method of implementing fast multipliers that do not require nested carry fixup operations.  As an 
2494interesting aside it seems that Paul Barrett describes a similar technique in his 1986 paper \cite{BARRETT} written 
2495five years before.
2496
2497At the heart of the Comba technique is once again the long-hand algorithm.  Except in this case a slight 
2498twist is placed on how the columns of the result are produced.  In the standard long-hand algorithm rows of products 
2499are produced then added together to form the final result.  In the baseline algorithm the columns are added together 
2500after each iteration to get the result instantaneously.  
2501
2502In the Comba algorithm the columns of the result are produced entirely independently of each other.  That is at 
2503the $O(n^2)$ level a simple multiplication and addition step is performed.  The carries of the columns are propagated 
2504after the nested loop to reduce the amount of work requiored. Succintly the first step of the algorithm is to compute 
2505the product vector $\vec x$ as follows. 
2506
2507\begin{equation}
2508\vec x_n = \sum_{i+j = n} a_ib_j, \forall n \in \lbrace 0, 1, 2, \ldots, i + j \rbrace
2509\end{equation}
2510
2511Where $\vec x_n$ is the $n'th$ column of the output vector.  Consider the following example which computes the vector $\vec x$ for the multiplication
2512of $576$ and $241$.  
2513
2514\newpage\begin{figure}[here]
2515\begin{small}
2516\begin{center}
2517\begin{tabular}{|c|c|c|c|c|c|}
2518  \hline &          & 5 & 7 & 6 & First Input\\
2519  \hline $\times$ & & 2 & 4 & 1 & Second Input\\
2520\hline            &                        & $1 \cdot 5 = 5$   & $1 \cdot 7 = 7$   & $1 \cdot 6 = 6$ & First pass \\
2521                  &  $4 \cdot 5 = 20$      & $4 \cdot 7+5=33$  & $4 \cdot 6+7=31$  & 6               & Second pass \\
2522   $2 \cdot 5 = 10$ &  $2 \cdot 7 + 20 = 34$ & $2 \cdot 6+33=45$ & 31                & 6             & Third pass \\
2523\hline 10 & 34 & 45 & 31 & 6 & Final Result \\   
2524\hline   
2525\end{tabular}
2526\end{center}
2527\end{small}
2528\caption{Comba Multiplication Diagram}
2529\end{figure}
2530
2531At this point the vector $x = \left < 10, 34, 45, 31, 6 \right >$ is the result of the first step of the Comba multipler.  
2532Now the columns must be fixed by propagating the carry upwards.  The resultant vector will have one extra dimension over the input vector which is
2533congruent to adding a leading zero digit.
2534
2535\begin{figure}[!here]
2536\begin{small}
2537\begin{center}
2538\begin{tabular}{l}
2539\hline Algorithm \textbf{Comba Fixup}. \\
2540\textbf{Input}.   Vector $\vec x$ of dimension $k$ \\
2541\textbf{Output}.  Vector $\vec x$ such that the carries have been propagated. \\
2542\hline \\
25431.  for $n$ from $0$ to $k - 1$ do \\
2544\hspace{3mm}1.1 $\vec x_{n+1} \leftarrow \vec x_{n+1} + \lfloor \vec x_{n}/\beta \rfloor$ \\
2545\hspace{3mm}1.2 $\vec x_{n} \leftarrow \vec x_{n} \mbox{ (mod }\beta\mbox{)}$ \\
25462.  Return($\vec x$). \\
2547\hline
2548\end{tabular}
2549\end{center}
2550\end{small}
2551\caption{Algorithm Comba Fixup}
2552\end{figure}
2553
2554With that algorithm and $k = 5$ and $\beta = 10$ the following vector is produced $\vec x= \left < 1, 3, 8, 8, 1, 6 \right >$.  In this case 
2555$241 \cdot 576$ is in fact $138816$ and the procedure succeeded.  If the algorithm is correct and as will be demonstrated shortly more
2556efficient than the baseline algorithm why not simply always use this algorithm?
2557
2558\subsubsection{Column Weight.}
2559At the nested $O(n^2)$ level the Comba method adds the product of two single precision variables to each column of the output 
2560independently.  A serious obstacle is if the carry is lost, due to lack of precision before the algorithm has a chance to fix
2561the carries.  For example, in the multiplication of two three-digit numbers the third column of output will be the sum of
2562three single precision multiplications.  If the precision of the accumulator for the output digits is less then $3 \cdot (\beta - 1)^2$ then
2563an overflow can occur and the carry information will be lost.  For any $m$ and $n$ digit inputs the maximum weight of any column is 
2564min$(m, n)$ which is fairly obvious.
2565
2566The maximum number of terms in any column of a product is known as the ``column weight'' and strictly governs when the algorithm can be used.  Recall
2567from earlier that a double precision type has $\alpha$ bits of resolution and a single precision digit has $lg(\beta)$ bits of precision.  Given these
2568two quantities we must not violate the following
2569
2570\begin{equation}
2571k \cdot \left (\beta - 1 \right )^2 < 2^{\alpha}
2572\end{equation}
2573
2574Which reduces to 
2575
2576\begin{equation}
2577k \cdot \left ( \beta^2 - 2\beta + 1 \right ) < 2^{\alpha}
2578\end{equation}
2579
2580Let $\rho = lg(\beta)$ represent the number of bits in a single precision digit.  By further re-arrangement of the equation the final solution is
2581found.
2582
2583\begin{equation}
2584k  < {{2^{\alpha}} \over {\left (2^{2\rho} - 2^{\rho + 1} + 1 \right )}}
2585\end{equation}
2586
2587The defaults for LibTomMath are $\beta = 2^{28}$ and $\alpha = 2^{64}$ which means that $k$ is bounded by $k < 257$.  In this configuration 
2588the smaller input may not have more than $256$ digits if the Comba method is to be used.  This is quite satisfactory for most applications since 
2589$256$ digits would allow for numbers in the range of $0 \le x < 2^{7168}$ which, is much larger than most public key cryptographic algorithms require.  
2590
2591\newpage\begin{figure}[!here]
2592\begin{small}
2593\begin{center}
2594\begin{tabular}{l}
2595\hline Algorithm \textbf{fast\_s\_mp\_mul\_digs}. \\
2596\textbf{Input}.   mp\_int $a$, mp\_int $b$ and an integer $digs$ \\
2597\textbf{Output}.  $c \leftarrow \vert a \vert \cdot \vert b \vert \mbox{ (mod }\beta^{digs}\mbox{)}$. \\
2598\hline \\
2599Place an array of \textbf{MP\_WARRAY} single precision digits named $W$ on the stack. \\
26001.  If $c.alloc < digs$ then grow $c$ to $digs$ digits. (\textit{mp\_grow}) \\
26012.  If step 1 failed return(\textit{MP\_MEM}).\\
2602\\
26033.  $pa \leftarrow \mbox{MIN}(digs, a.used + b.used)$ \\
2604\\
26054.  $\_ \hat W \leftarrow 0$ \\
26065.  for $ix$ from 0 to $pa - 1$ do \\
2607\hspace{3mm}5.1  $ty \leftarrow \mbox{MIN}(b.used - 1, ix)$ \\
2608\hspace{3mm}5.2  $tx \leftarrow ix - ty$ \\
2609\hspace{3mm}5.3  $iy \leftarrow \mbox{MIN}(a.used - tx, ty + 1)$ \\
2610\hspace{3mm}5.4  for $iz$ from 0 to $iy - 1$ do \\
2611\hspace{6mm}5.4.1  $\_ \hat W \leftarrow \_ \hat W + a_{tx+iy}b_{ty-iy}$ \\
2612\hspace{3mm}5.5  $W_{ix} \leftarrow \_ \hat W (\mbox{mod }\beta)$\\
2613\hspace{3mm}5.6  $\_ \hat W \leftarrow \lfloor \_ \hat W / \beta \rfloor$ \\
2614\\
26156.  $oldused \leftarrow c.used$ \\
26167.  $c.used \leftarrow digs$ \\
26178.  for $ix$ from $0$ to $pa$ do \\
2618\hspace{3mm}8.1  $c_{ix} \leftarrow W_{ix}$ \\
26199.  for $ix$ from $pa + 1$ to $oldused - 1$ do \\
2620\hspace{3mm}9.1 $c_{ix} \leftarrow 0$ \\
2621\\
262210.  Clamp $c$. \\
262311.  Return MP\_OKAY. \\
2624\hline
2625\end{tabular}
2626\end{center}
2627\end{small}
2628\caption{Algorithm fast\_s\_mp\_mul\_digs}
2629\label{fig:COMBAMULT}
2630\end{figure}
2631
2632\textbf{Algorithm fast\_s\_mp\_mul\_digs.}
2633This algorithm performs the unsigned multiplication of $a$ and $b$ using the Comba method limited to $digs$ digits of precision.
2634
2635The outer loop of this algorithm is more complicated than that of the baseline multiplier.  This is because on the inside of the 
2636loop we want to produce one column per pass.  This allows the accumulator $\_ \hat W$ to be placed in CPU registers and
2637reduce the memory bandwidth to two \textbf{mp\_digit} reads per iteration.
2638
2639The $ty$ variable is set to the minimum count of $ix$ or the number of digits in $b$.  That way if $a$ has more digits than
2640$b$ this will be limited to $b.used - 1$.  The $tx$ variable is set to the to the distance past $b.used$ the variable
2641$ix$ is.  This is used for the immediately subsequent statement where we find $iy$.  
2642
2643The variable $iy$ is the minimum digits we can read from either $a$ or $b$ before running out.  Computing one column at a time
2644means we have to scan one integer upwards and the other downwards.  $a$ starts at $tx$ and $b$ starts at $ty$.  In each
2645pass we are producing the $ix$'th output column and we note that $tx + ty = ix$.  As we move $tx$ upwards we have to 
2646move $ty$ downards so the equality remains valid.  The $iy$ variable is the number of iterations until 
2647$tx \ge a.used$ or $ty < 0$ occurs.
2648
2649After every inner pass we store the lower half of the accumulator into $W_{ix}$ and then propagate the carry of the accumulator
2650into the next round by dividing $\_ \hat W$ by $\beta$.
2651
2652To measure the benefits of the Comba method over the baseline method consider the number of operations that are required.  If the 
2653cost in terms of time of a multiply and addition is $p$ and the cost of a carry propagation is $q$ then a baseline multiplication would require 
2654$O \left ((p + q)n^2 \right )$ time to multiply two $n$-digit numbers.  The Comba method requires only $O(pn^2 + qn)$ time, however in practice, 
2655the speed increase is actually much more.  With $O(n)$ space the algorithm can be reduced to $O(pn + qn)$ time by implementing the $n$ multiply
2656and addition operations in the nested loop in parallel.  
2657
2658EXAM,bn_fast_s_mp_mul_digs.c
2659
2660As per the pseudo--code we first calculate $pa$ (line @47,MIN@) as the number of digits to output.  Next we begin the outer loop
2661to produce the individual columns of the product.  We use the two aliases $tmpx$ and $tmpy$ (lines @61,tmpx@, @62,tmpy@) to point
2662inside the two multiplicands quickly.  
2663
2664The inner loop (lines @70,for@ to @72,}@) of this implementation is where the tradeoff come into play.  Originally this comba 
2665implementation was ``row--major'' which means it adds to each of the columns in each pass.  After the outer loop it would then fix 
2666the carries.  This was very fast except it had an annoying drawback.  You had to read a mp\_word and two mp\_digits and write 
2667one mp\_word per iteration.  On processors such as the Athlon XP and P4 this did not matter much since the cache bandwidth 
2668is very high and it can keep the ALU fed with data.  It did, however, matter on older and embedded cpus where cache is often 
2669slower and also often doesn't exist.  This new algorithm only performs two reads per iteration under the assumption that the 
2670compiler has aliased $\_ \hat W$ to a CPU register.
2671
2672After the inner loop we store the current accumulator in $W$ and shift $\_ \hat W$ (lines @75,W[ix]@, @78,>>@) to forward it as 
2673a carry for the next pass.  After the outer loop we use the final carry (line @82,W[ix]@) as the last digit of the product.  
2674
2675\subsection{Polynomial Basis Multiplication}
2676To break the $O(n^2)$ barrier in multiplication requires a completely different look at integer multiplication.  In the following algorithms
2677the use of polynomial basis representation for two integers $a$ and $b$ as $f(x) = \sum_{i=0}^{n} a_i x^i$ and  
2678$g(x) = \sum_{i=0}^{n} b_i x^i$ respectively, is required.  In this system both $f(x)$ and $g(x)$ have $n + 1$ terms and are of the $n$'th degree.
2679 
2680The product $a \cdot b \equiv f(x)g(x)$ is the polynomial $W(x) = \sum_{i=0}^{2n} w_i x^i$.  The coefficients $w_i$ will
2681directly yield the desired product when $\beta$ is substituted for $x$.  The direct solution to solve for the $2n + 1$ coefficients
2682requires $O(n^2)$ time and would in practice be slower than the Comba technique.
2683
2684However, numerical analysis theory indicates that only $2n + 1$ distinct points in $W(x)$ are required to determine the values of the $2n + 1$ unknown 
2685coefficients.   This means by finding $\zeta_y = W(y)$ for $2n + 1$ small values of $y$ the coefficients of $W(x)$ can be found with 
2686Gaussian elimination.  This technique is also occasionally refered to as the \textit{interpolation technique} (\textit{references please...}) since in 
2687effect an interpolation based on $2n + 1$ points will yield a polynomial equivalent to $W(x)$.  
2688
2689The coefficients of the polynomial $W(x)$ are unknown which makes finding $W(y)$ for any value of $y$ impossible.  However, since 
2690$W(x) = f(x)g(x)$ the equivalent $\zeta_y = f(y) g(y)$ can be used in its place.  The benefit of this technique stems from the 
2691fact that $f(y)$ and $g(y)$ are much smaller than either $a$ or $b$ respectively.  As a result finding the $2n + 1$ relations required 
2692by multiplying $f(y)g(y)$ involves multiplying integers that are much smaller than either of the inputs.
2693
2694When picking points to gather relations there are always three obvious points to choose, $y = 0, 1$ and $ \infty$.  The $\zeta_0$ term
2695is simply the product $W(0) = w_0 = a_0 \cdot b_0$.  The $\zeta_1$ term is the product 
2696$W(1) = \left (\sum_{i = 0}^{n} a_i \right ) \left (\sum_{i = 0}^{n} b_i \right )$.  The third point $\zeta_{\infty}$ is less obvious but rather
2697simple to explain.  The $2n + 1$'th coefficient of $W(x)$ is numerically equivalent to the most significant column in an integer multiplication.  
2698The point at $\infty$ is used symbolically to represent the most significant column, that is $W(\infty) = w_{2n} = a_nb_n$.  Note that the 
2699points at $y = 0$ and $\infty$ yield the coefficients $w_0$ and $w_{2n}$ directly.
2700
2701If more points are required they should be of small values and powers of two such as $2^q$ and the related \textit{mirror points} 
2702$\left (2^q \right )^{2n}  \cdot \zeta_{2^{-q}}$ for small values of $q$.  The term ``mirror point'' stems from the fact that 
2703$\left (2^q \right )^{2n}  \cdot \zeta_{2^{-q}}$ can be calculated in the exact opposite fashion as $\zeta_{2^q}$.  For
2704example, when $n = 2$ and $q = 1$ then following two equations are equivalent to the point $\zeta_{2}$ and its mirror.
2705
2706\begin{eqnarray}
2707\zeta_{2}                  = f(2)g(2) = (4a_2 + 2a_1 + a_0)(4b_2 + 2b_1 + b_0) \nonumber \\
270816 \cdot \zeta_{1 \over 2} = 4f({1\over 2}) \cdot 4g({1 \over 2}) = (a_2 + 2a_1 + 4a_0)(b_2 + 2b_1 + 4b_0)
2709\end{eqnarray}
2710
2711Using such points will allow the values of $f(y)$ and $g(y)$ to be independently calculated using only left shifts.  For example, when $n = 2$ the
2712polynomial $f(2^q)$ is equal to $2^q((2^qa_2) + a_1) + a_0$.  This technique of polynomial representation is known as Horner's method.  
2713
2714As a general rule of the algorithm when the inputs are split into $n$ parts each there are $2n - 1$ multiplications.  Each multiplication is of 
2715multiplicands that have $n$ times fewer digits than the inputs.  The asymptotic running time of this algorithm is 
2716$O \left ( k^{lg_n(2n - 1)} \right )$ for $k$ digit inputs (\textit{assuming they have the same number of digits}).  Figure~\ref{fig:exponent}
2717summarizes the exponents for various values of $n$.
2718
2719\begin{figure}
2720\begin{center}
2721\begin{tabular}{|c|c|c|}
2722\hline \textbf{Split into $n$ Parts} & \textbf{Exponent}  & \textbf{Notes}\\
2723\hline $2$ & $1.584962501$ & This is Karatsuba Multiplication. \\
2724\hline $3$ & $1.464973520$ & This is Toom-Cook Multiplication. \\
2725\hline $4$ & $1.403677461$ &\\
2726\hline $5$ & $1.365212389$ &\\
2727\hline $10$ & $1.278753601$ &\\
2728\hline $100$ & $1.149426538$ &\\
2729\hline $1000$ & $1.100270931$ &\\
2730\hline $10000$ & $1.075252070$ &\\
2731\hline
2732\end{tabular}
2733\end{center}
2734\caption{Asymptotic Running Time of Polynomial Basis Multiplication}
2735\label{fig:exponent}
2736\end{figure}
2737
2738At first it may seem like a good idea to choose $n = 1000$ since the exponent is approximately $1.1$.  However, the overhead
2739of solving for the 2001 terms of $W(x)$ will certainly consume any savings the algorithm could offer for all but exceedingly large
2740numbers.  
2741
2742\subsubsection{Cutoff Point}
2743The polynomial basis multiplication algorithms all require fewer single precision multiplications than a straight Comba approach.  However, 
2744the algorithms incur an overhead (\textit{at the $O(n)$ work level}) since they require a system of equations to be solved.  This makes the
2745polynomial basis approach more costly to use with small inputs.
2746
2747Let $m$ represent the number of digits in the multiplicands (\textit{assume both multiplicands have the same number of digits}).  There exists a 
2748point $y$ such that when $m < y$ the polynomial basis algorithms are more costly than Comba, when $m = y$ they are roughly the same cost and 
2749when $m > y$ the Comba methods are slower than the polynomial basis algorithms.  
2750
2751The exact location of $y$ depends on several key architectural elements of the computer platform in question.
2752
2753\begin{enumerate}
2754\item  The ratio of clock cycles for single precision multiplication versus other simpler operations such as addition, shifting, etc.  For example
2755on the AMD Athlon the ratio is roughly $17 : 1$ while on the Intel P4 it is $29 : 1$.  The higher the ratio in favour of multiplication the lower
2756the cutoff point $y$ will be.  
2757
2758\item  The complexity of the linear system of equations (\textit{for the coefficients of $W(x)$}) is.  Generally speaking as the number of splits
2759grows the complexity grows substantially.  Ideally solving the system will only involve addition, subtraction and shifting of integers.  This
2760directly reflects on the ratio previous mentioned.
2761
2762\item  To a lesser extent memory bandwidth and function call overheads.  Provided the values are in the processor cache this is less of an
2763influence over the cutoff point.
2764
2765\end{enumerate}
2766
2767A clean cutoff point separation occurs when a point $y$ is found such that all of the cutoff point conditions are met.  For example, if the point
2768is too low then there will be values of $m$ such that $m > y$ and the Comba method is still faster.  Finding the cutoff points is fairly simple when
2769a high resolution timer is available.  
2770
2771\subsection{Karatsuba Multiplication}
2772Karatsuba \cite{KARA} multiplication when originally proposed in 1962 was among the first set of algorithms to break the $O(n^2)$ barrier for
2773general purpose multiplication.  Given two polynomial basis representations $f(x) = ax + b$ and $g(x) = cx + d$, Karatsuba proved with 
2774light algebra \cite{KARAP} that the following polynomial is equivalent to multiplication of the two integers the polynomials represent.
2775
2776\begin{equation}
2777f(x) \cdot g(x) = acx^2 + ((a + b)(c + d) - (ac + bd))x + bd
2778\end{equation}
2779
2780Using the observation that $ac$ and $bd$ could be re-used only three half sized multiplications would be required to produce the product.  Applying
2781this algorithm recursively, the work factor becomes $O(n^{lg(3)})$ which is substantially better than the work factor $O(n^2)$ of the Comba technique.  It turns 
2782out what Karatsuba did not know or at least did not publish was that this is simply polynomial basis multiplication with the points 
2783$\zeta_0$, $\zeta_{\infty}$ and $\zeta_{1}$.  Consider the resultant system of equations.
2784
2785\begin{center}
2786\begin{tabular}{rcrcrcrc}
2787$\zeta_{0}$ &      $=$ &  &  &  & & $w_0$ \\
2788$\zeta_{1}$ &      $=$ & $w_2$ & $+$ & $w_1$ & $+$ & $w_0$ \\
2789$\zeta_{\infty}$ & $=$ & $w_2$ &  & &  & \\
2790\end{tabular}
2791\end{center}
2792
2793By adding the first and last equation to the equation in the middle the term $w_1$ can be isolated and all three coefficients solved for.  The simplicity
2794of this system of equations has made Karatsuba fairly popular.  In fact the cutoff point is often fairly low\footnote{With LibTomMath 0.18 it is 70 and 109 digits for the Intel P4 and AMD Athlon respectively.}
2795making it an ideal algorithm to speed up certain public key cryptosystems such as RSA and Diffie-Hellman.  
2796
2797\newpage\begin{figure}[!here]
2798\begin{small}
2799\begin{center}
2800\begin{tabular}{l}
2801\hline Algorithm \textbf{mp\_karatsuba\_mul}. \\
2802\textbf{Input}.   mp\_int $a$ and mp\_int $b$ \\
2803\textbf{Output}.  $c \leftarrow \vert a \vert \cdot \vert b \vert$ \\
2804\hline \\
28051.  Init the following mp\_int variables: $x0$, $x1$, $y0$, $y1$, $t1$, $x0y0$, $x1y1$.\\
28062.  If step 2 failed then return(\textit{MP\_MEM}). \\
2807\\
2808Split the input.  e.g. $a = x1 \cdot \beta^B + x0$ \\
28093.  $B \leftarrow \mbox{min}(a.used, b.used)/2$ \\
28104.  $x0 \leftarrow a \mbox{ (mod }\beta^B\mbox{)}$ (\textit{mp\_mod\_2d}) \\
28115.  $y0 \leftarrow b \mbox{ (mod }\beta^B\mbox{)}$ \\
28126.  $x1 \leftarrow \lfloor a / \beta^B \rfloor$ (\textit{mp\_rshd}) \\
28137.  $y1 \leftarrow \lfloor b / \beta^B \rfloor$ \\
2814\\
2815Calculate the three products. \\
28168.  $x0y0 \leftarrow x0 \cdot y0$ (\textit{mp\_mul}) \\
28179.  $x1y1 \leftarrow x1 \cdot y1$ \\
281810.  $t1 \leftarrow x1 + x0$ (\textit{mp\_add}) \\
281911.  $x0 \leftarrow y1 + y0$ \\
282012.  $t1 \leftarrow t1 \cdot x0$ \\
2821\\
2822Calculate the middle term. \\
282313.  $x0 \leftarrow x0y0 + x1y1$ \\
282414.  $t1 \leftarrow t1 - x0$ (\textit{s\_mp\_sub}) \\
2825\\
2826Calculate the final product. \\
282715.  $t1 \leftarrow t1 \cdot \beta^B$ (\textit{mp\_lshd}) \\
282816.  $x1y1 \leftarrow x1y1 \cdot \beta^{2B}$ \\
282917.  $t1 \leftarrow x0y0 + t1$ \\
283018.  $c \leftarrow t1 + x1y1$ \\
283119.  Clear all of the temporary variables. \\
283220.  Return(\textit{MP\_OKAY}).\\
2833\hline 
2834\end{tabular}
2835\end{center}
2836\end{small}
2837\caption{Algorithm mp\_karatsuba\_mul}
2838\end{figure}
2839
2840\textbf{Algorithm mp\_karatsuba\_mul.}
2841This algorithm computes the unsigned product of two inputs using the Karatsuba multiplication algorithm.  It is loosely based on the description
2842from Knuth \cite[pp. 294-295]{TAOCPV2}.  
2843
2844\index{radix point}
2845In order to split the two inputs into their respective halves, a suitable \textit{radix point} must be chosen.  The radix point chosen must
2846be used for both of the inputs meaning that it must be smaller than the smallest input.  Step 3 chooses the radix point $B$ as half of the 
2847smallest input \textbf{used} count.  After the radix point is chosen the inputs are split into lower and upper halves.  Step 4 and 5 
2848compute the lower halves.  Step 6 and 7 computer the upper halves.  
2849
2850After the halves have been computed the three intermediate half-size products must be computed.  Step 8 and 9 compute the trivial products
2851$x0 \cdot y0$ and $x1 \cdot y1$.  The mp\_int $x0$ is used as a temporary variable after $x1 + x0$ has been computed.  By using $x0$ instead
2852of an additional temporary variable, the algorithm can avoid an addition memory allocation operation.
2853
2854The remaining steps 13 through 18 compute the Karatsuba polynomial through a variety of digit shifting and addition operations.
2855
2856EXAM,bn_mp_karatsuba_mul.c
2857
2858The new coding element in this routine, not  seen in previous routines, is the usage of goto statements.  The conventional
2859wisdom is that goto statements should be avoided.  This is generally true, however when every single function call can fail, it makes sense
2860to handle error recovery with a single piece of code.  Lines @61,if@ to @75,if@ handle initializing all of the temporary variables 
2861required.  Note how each of the if statements goes to a different label in case of failure.  This allows the routine to correctly free only
2862the temporaries that have been successfully allocated so far.
2863
2864The temporary variables are all initialized using the mp\_init\_size routine since they are expected to be large.  This saves the 
2865additional reallocation that would have been necessary.  Also $x0$, $x1$, $y0$ and $y1$ have to be able to hold at least their respective
2866number of digits for the next section of code.
2867
2868The first algebraic portion of the algorithm is to split the two inputs into their halves.  However, instead of using mp\_mod\_2d and mp\_rshd
2869to extract the halves, the respective code has been placed inline within the body of the function.  To initialize the halves, the \textbf{used} and 
2870\textbf{sign} members are copied first.  The first for loop on line @98,for@ copies the lower halves.  Since they are both the same magnitude it 
2871is simpler to calculate both lower halves in a single loop.  The for loop on lines @104,for@ and @109,for@ calculate the upper halves $x1$ and 
2872$y1$ respectively.
2873
2874By inlining the calculation of the halves, the Karatsuba multiplier has a slightly lower overhead and can be used for smaller magnitude inputs.
2875
2876When line @152,err@ is reached, the algorithm has completed succesfully.  The ``error status'' variable $err$ is set to \textbf{MP\_OKAY} so that
2877the same code that handles errors can be used to clear the temporary variables and return.  
2878
2879\subsection{Toom-Cook $3$-Way Multiplication}
2880Toom-Cook $3$-Way \cite{TOOM} multiplication is essentially the polynomial basis algorithm for $n = 2$ except that the points  are 
2881chosen such that $\zeta$ is easy to compute and the resulting system of equations easy to reduce.  Here, the points $\zeta_{0}$, 
2882$16 \cdot \zeta_{1 \over 2}$, $\zeta_1$, $\zeta_2$ and $\zeta_{\infty}$ make up the five required points to solve for the coefficients 
2883of the $W(x)$.
2884
2885With the five relations that Toom-Cook specifies, the following system of equations is formed.
2886
2887\begin{center}
2888\begin{tabular}{rcrcrcrcrcr}
2889$\zeta_0$                    & $=$ & $0w_4$ & $+$ & $0w_3$ & $+$ & $0w_2$ & $+$ & $0w_1$ & $+$ & $1w_0$  \\
2890$16 \cdot \zeta_{1 \over 2}$ & $=$ & $1w_4$ & $+$ & $2w_3$ & $+$ & $4w_2$ & $+$ & $8w_1$ & $+$ & $16w_0$  \\
2891$\zeta_1$                    & $=$ & $1w_4$ & $+$ & $1w_3$ & $+$ & $1w_2$ & $+$ & $1w_1$ & $+$ & $1w_0$  \\
2892$\zeta_2$                    & $=$ & $16w_4$ & $+$ & $8w_3$ & $+$ & $4w_2$ & $+$ & $2w_1$ & $+$ & $1w_0$  \\
2893$\zeta_{\infty}$             & $=$ & $1w_4$ & $+$ & $0w_3$ & $+$ & $0w_2$ & $+$ & $0w_1$ & $+$ & $0w_0$  \\
2894\end{tabular}
2895\end{center}
2896
2897A trivial solution to this matrix requires $12$ subtractions, two multiplications by a small power of two, two divisions by a small power
2898of two, two divisions by three and one multiplication by three.  All of these $19$ sub-operations require less than quadratic time, meaning that
2899the algorithm can be faster than a baseline multiplication.  However, the greater complexity of this algorithm places the cutoff point
2900(\textbf{TOOM\_MUL\_CUTOFF}) where Toom-Cook becomes more efficient much higher than the Karatsuba cutoff point.  
2901
2902\begin{figure}[!here]
2903\begin{small}
2904\begin{center}
2905\begin{tabular}{l}
2906\hline Algorithm \textbf{mp\_toom\_mul}. \\
2907\textbf{Input}.   mp\_int $a$ and mp\_int $b$ \\
2908\textbf{Output}.  $c \leftarrow  a  \cdot  b $ \\
2909\hline \\
2910Split $a$ and $b$ into three pieces.  E.g. $a = a_2 \beta^{2k} + a_1 \beta^{k} + a_0$ \\
29111.  $k \leftarrow \lfloor \mbox{min}(a.used, b.used) / 3 \rfloor$ \\
29122.  $a_0 \leftarrow a \mbox{ (mod }\beta^{k}\mbox{)}$ \\
29133.  $a_1 \leftarrow \lfloor a / \beta^k \rfloor$, $a_1 \leftarrow a_1 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
29144.  $a_2 \leftarrow \lfloor a / \beta^{2k} \rfloor$, $a_2 \leftarrow a_2 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
29155.  $b_0 \leftarrow a \mbox{ (mod }\beta^{k}\mbox{)}$ \\
29166.  $b_1 \leftarrow \lfloor a / \beta^k \rfloor$, $b_1 \leftarrow b_1 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
29177.  $b_2 \leftarrow \lfloor a / \beta^{2k} \rfloor$, $b_2 \leftarrow b_2 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
2918\\
2919Find the five equations for $w_0, w_1, ..., w_4$. \\
29208.  $w_0 \leftarrow a_0 \cdot b_0$ \\
29219.  $w_4 \leftarrow a_2 \cdot b_2$ \\
292210. $tmp_1 \leftarrow 2 \cdot a_0$, $tmp_1 \leftarrow a_1 + tmp_1$, $tmp_1 \leftarrow 2 \cdot tmp_1$, $tmp_1 \leftarrow tmp_1 + a_2$ \\
292311. $tmp_2 \leftarrow 2 \cdot b_0$, $tmp_2 \leftarrow b_1 + tmp_2$, $tmp_2 \leftarrow 2 \cdot tmp_2$, $tmp_2 \leftarrow tmp_2 + b_2$ \\
292412. $w_1 \leftarrow tmp_1 \cdot tmp_2$ \\
292513. $tmp_1 \leftarrow 2 \cdot a_2$, $tmp_1 \leftarrow a_1 + tmp_1$, $tmp_1 \leftarrow 2 \cdot tmp_1$, $tmp_1 \leftarrow tmp_1 + a_0$ \\
292614. $tmp_2 \leftarrow 2 \cdot b_2$, $tmp_2 \leftarrow b_1 + tmp_2$, $tmp_2 \leftarrow 2 \cdot tmp_2$, $tmp_2 \leftarrow tmp_2 + b_0$ \\
292715. $w_3 \leftarrow tmp_1 \cdot tmp_2$ \\
292816. $tmp_1 \leftarrow a_0 + a_1$, $tmp_1 \leftarrow tmp_1 + a_2$, $tmp_2 \leftarrow b_0 + b_1$, $tmp_2 \leftarrow tmp_2 + b_2$ \\
292917. $w_2 \leftarrow tmp_1 \cdot tmp_2$ \\
2930\\
2931Continued on the next page.\\
2932\hline
2933\end{tabular}
2934\end{center}
2935\end{small}
2936\caption{Algorithm mp\_toom\_mul}
2937\end{figure}
2938
2939\newpage\begin{figure}[!here]
2940\begin{small}
2941\begin{center}
2942\begin{tabular}{l}
2943\hline Algorithm \textbf{mp\_toom\_mul} (continued). \\
2944\textbf{Input}.   mp\_int $a$ and mp\_int $b$ \\
2945\textbf{Output}.  $c \leftarrow a \cdot  b $ \\
2946\hline \\
2947Now solve the system of equations. \\
294818. $w_1 \leftarrow w_4 - w_1$, $w_3 \leftarrow w_3 - w_0$ \\
294919. $w_1 \leftarrow \lfloor w_1 / 2 \rfloor$, $w_3 \leftarrow \lfloor w_3 / 2 \rfloor$ \\
295020. $w_2 \leftarrow w_2 - w_0$, $w_2 \leftarrow w_2 - w_4$ \\
295121. $w_1 \leftarrow w_1 - w_2$, $w_3 \leftarrow w_3 - w_2$ \\
295222. $tmp_1 \leftarrow 8 \cdot w_0$, $w_1 \leftarrow w_1 - tmp_1$, $tmp_1 \leftarrow 8 \cdot w_4$, $w_3 \leftarrow w_3 - tmp_1$ \\
295323. $w_2 \leftarrow 3 \cdot w_2$, $w_2 \leftarrow w_2 - w_1$, $w_2 \leftarrow w_2 - w_3$ \\
295424. $w_1 \leftarrow w_1 - w_2$, $w_3 \leftarrow w_3 - w_2$ \\
295525. $w_1 \leftarrow \lfloor w_1 / 3 \rfloor, w_3 \leftarrow \lfloor w_3 / 3 \rfloor$ \\
2956\\
2957Now substitute $\beta^k$ for $x$ by shifting $w_0, w_1, ..., w_4$. \\
295826. for $n$ from $1$ to $4$ do \\
2959\hspace{3mm}26.1  $w_n \leftarrow w_n \cdot \beta^{nk}$ \\
296027. $c \leftarrow w_0 + w_1$, $c \leftarrow c + w_2$, $c \leftarrow c + w_3$, $c \leftarrow c + w_4$ \\
296128. Return(\textit{MP\_OKAY}) \\
2962\hline
2963\end{tabular}
2964\end{center}
2965\end{small}
2966\caption{Algorithm mp\_toom\_mul (continued)}
2967\end{figure}
2968
2969\textbf{Algorithm mp\_toom\_mul.}
2970This algorithm computes the product of two mp\_int variables $a$ and $b$ using the Toom-Cook approach.  Compared to the Karatsuba multiplication, this 
2971algorithm has a lower asymptotic running time of approximately $O(n^{1.464})$ but at an obvious cost in overhead.  In this
2972description, several statements have been compounded to save space.  The intention is that the statements are executed from left to right across
2973any given step.
2974
2975The two inputs $a$ and $b$ are first split into three $k$-digit integers $a_0, a_1, a_2$ and $b_0, b_1, b_2$ respectively.  From these smaller
2976integers the coefficients of the polynomial basis representations $f(x)$ and $g(x)$ are known and can be used to find the relations required.
2977
2978The first two relations $w_0$ and $w_4$ are the points $\zeta_{0}$ and $\zeta_{\infty}$ respectively.  The relation $w_1, w_2$ and $w_3$ correspond
2979to the points $16 \cdot \zeta_{1 \over 2}, \zeta_{2}$ and $\zeta_{1}$ respectively.  These are found using logical shifts to independently find
2980$f(y)$ and $g(y)$ which significantly speeds up the algorithm.
2981
2982After the five relations $w_0, w_1, \ldots, w_4$ have been computed, the system they represent must be solved in order for the unknown coefficients 
2983$w_1, w_2$ and $w_3$ to be isolated.  The steps 18 through 25 perform the system reduction required as previously described.  Each step of
2984the reduction represents the comparable matrix operation that would be performed had this been performed by pencil.  For example, step 18 indicates
2985that row $1$ must be subtracted from row $4$ and simultaneously row $0$ subtracted from row $3$.  
2986
2987Once the coeffients have been isolated, the polynomial $W(x) = \sum_{i=0}^{2n} w_i x^i$ is known.  By substituting $\beta^{k}$ for $x$, the integer 
2988result $a \cdot b$ is produced.
2989
2990EXAM,bn_mp_toom_mul.c
2991
2992The first obvious thing to note is that this algorithm is complicated.  The complexity is worth it if you are multiplying very 
2993large numbers.  For example, a 10,000 digit multiplication takes approximaly 99,282,205 fewer single precision multiplications with
2994Toom--Cook than a Comba or baseline approach (this is a savings of more than 99$\%$).  For most ``crypto'' sized numbers this
2995algorithm is not practical as Karatsuba has a much lower cutoff point.
2996
2997First we split $a$ and $b$ into three roughly equal portions.  This has been accomplished (lines @40,mod@ to @69,rshd@) with 
2998combinations of mp\_rshd() and mp\_mod\_2d() function calls.  At this point $a = a2 \cdot \beta^2 + a1 \cdot \beta + a0$ and similiarly
2999for $b$.  
3000
3001Next we compute the five points $w0, w1, w2, w3$ and $w4$.  Recall that $w0$ and $w4$ can be computed directly from the portions so
3002we get those out of the way first (lines @72,mul@ and @77,mul@).  Next we compute $w1, w2$ and $w3$ using Horners method.
3003
3004After this point we solve for the actual values of $w1, w2$ and $w3$ by reducing the $5 \times 5$ system which is relatively
3005straight forward.  
3006
3007\subsection{Signed Multiplication}
3008Now that algorithms to handle multiplications of every useful dimensions have been developed, a rather simple finishing touch is required.  So far all
3009of the multiplication algorithms have been unsigned multiplications which leaves only a signed multiplication algorithm to be established.  
3010
3011\begin{figure}[!here]
3012\begin{small}
3013\begin{center}
3014\begin{tabular}{l}
3015\hline Algorithm \textbf{mp\_mul}. \\
3016\textbf{Input}.   mp\_int $a$ and mp\_int $b$ \\
3017\textbf{Output}.  $c \leftarrow a \cdot b$ \\
3018\hline \\
30191.  If $a.sign = b.sign$ then \\
3020\hspace{3mm}1.1  $sign = MP\_ZPOS$ \\
30212.  else \\
3022\hspace{3mm}2.1  $sign = MP\_ZNEG$ \\
30233.  If min$(a.used, b.used) \ge TOOM\_MUL\_CUTOFF$ then  \\
3024\hspace{3mm}3.1  $c \leftarrow a \cdot b$ using algorithm mp\_toom\_mul \\
30254.  else if min$(a.used, b.used) \ge KARATSUBA\_MUL\_CUTOFF$ then \\
3026\hspace{3mm}4.1  $c \leftarrow a \cdot b$ using algorithm mp\_karatsuba\_mul \\
30275.  else \\
3028\hspace{3mm}5.1  $digs \leftarrow a.used + b.used + 1$ \\
3029\hspace{3mm}5.2  If $digs < MP\_ARRAY$ and min$(a.used, b.used) \le \delta$ then \\
3030\hspace{6mm}5.2.1  $c \leftarrow a \cdot b \mbox{ (mod }\beta^{digs}\mbox{)}$ using algorithm fast\_s\_mp\_mul\_digs.  \\
3031\hspace{3mm}5.3  else \\
3032\hspace{6mm}5.3.1  $c \leftarrow a \cdot b \mbox{ (mod }\beta^{digs}\mbox{)}$ using algorithm s\_mp\_mul\_digs.  \\
30336.  $c.sign \leftarrow sign$ \\
30347.  Return the result of the unsigned multiplication performed. \\
3035\hline
3036\end{tabular}
3037\end{center}
3038\end{small}
3039\caption{Algorithm mp\_mul}
3040\end{figure}
3041
3042\textbf{Algorithm mp\_mul.}
3043This algorithm performs the signed multiplication of two inputs.  It will make use of any of the three unsigned multiplication algorithms 
3044available when the input is of appropriate size.  The \textbf{sign} of the result is not set until the end of the algorithm since algorithm
3045s\_mp\_mul\_digs will clear it.  
3046
3047EXAM,bn_mp_mul.c
3048
3049The implementation is rather simplistic and is not particularly noteworthy.  Line @22,?@ computes the sign of the result using the ``?'' 
3050operator from the C programming language.  Line @37,<<@ computes $\delta$ using the fact that $1 << k$ is equal to $2^k$.  
3051
3052\section{Squaring}
3053\label{sec:basesquare}
3054
3055Squaring is a special case of multiplication where both multiplicands are equal.  At first it may seem like there is no significant optimization
3056available but in fact there is.  Consider the multiplication of $576$ against $241$.  In total there will be nine single precision multiplications
3057performed which are $1\cdot 6$, $1 \cdot 7$, $1 \cdot 5$, $4 \cdot 6$, $4 \cdot 7$, $4 \cdot 5$, $2 \cdot  6$, $2 \cdot 7$ and $2 \cdot 5$.  Now consider 
3058the multiplication of $123$ against $123$.  The nine products are $3 \cdot 3$, $3 \cdot 2$, $3 \cdot 1$, $2 \cdot 3$, $2 \cdot 2$, $2 \cdot 1$, 
3059$1 \cdot 3$, $1 \cdot 2$ and $1 \cdot 1$.  On closer inspection some of the products are equivalent.  For example, $3 \cdot 2 = 2 \cdot 3$ 
3060and $3 \cdot 1 = 1 \cdot 3$. 
3061
3062For any $n$-digit input, there are ${{\left (n^2 + n \right)}\over 2}$ possible unique single precision multiplications required compared to the $n^2$
3063required for multiplication.  The following diagram gives an example of the operations required.
3064
3065\begin{figure}[here]
3066\begin{center}
3067\begin{tabular}{ccccc|c}
3068&&1&2&3&\\
3069$\times$ &&1&2&3&\\
3070\hline && $3 \cdot 1$ & $3 \cdot 2$ & $3 \cdot 3$ & Row 0\\
3071       & $2 \cdot 1$  & $2 \cdot 2$ & $2 \cdot 3$ && Row 1 \\
3072         $1 \cdot 1$  & $1 \cdot 2$ & $1 \cdot 3$ &&& Row 2 \\
3073\end{tabular}
3074\end{center}
3075\caption{Squaring Optimization Diagram}
3076\end{figure}
3077
3078MARK,SQUARE
3079Starting from zero and numbering the columns from right to left a very simple pattern becomes obvious.  For the purposes of this discussion let $x$
3080represent the number being squared.  The first observation is that in row $k$ the $2k$'th column of the product has a $\left (x_k \right)^2$ term in it.  
3081
3082The second observation is that every column $j$ in row $k$ where $j \ne 2k$ is part of a double product.  Every non-square term of a column will
3083appear twice hence the name ``double product''.  Every odd column is made up entirely of double products.  In fact every column is made up of double 
3084products and at most one square (\textit{see the exercise section}).  
3085
3086The third and final observation is that for row $k$ the first unique non-square term, that is, one that hasn't already appeared in an earlier row, 
3087occurs at column $2k + 1$.  For example, on row $1$ of the previous squaring, column one is part of the double product with column one from row zero. 
3088Column two of row one is a square and column three is the first unique column.
3089
3090\subsection{The Baseline Squaring Algorithm}
3091The baseline squaring algorithm is meant to be a catch-all squaring algorithm.  It will handle any of the input sizes that the faster routines
3092will not handle.  
3093
3094\begin{figure}[!here]
3095\begin{small}
3096\begin{center}
3097\begin{tabular}{l}
3098\hline Algorithm \textbf{s\_mp\_sqr}. \\
3099\textbf{Input}.   mp\_int $a$ \\
3100\textbf{Output}.  $b \leftarrow a^2$ \\
3101\hline \\
31021.  Init a temporary mp\_int of at least $2 \cdot a.used +1$ digits.  (\textit{mp\_init\_size}) \\
31032.  If step 1 failed return(\textit{MP\_MEM}) \\
31043.  $t.used \leftarrow 2 \cdot a.used + 1$ \\
31054.  For $ix$ from 0 to $a.used - 1$ do \\
3106\hspace{3mm}Calculate the square. \\
3107\hspace{3mm}4.1  $\hat r \leftarrow t_{2ix} + \left (a_{ix} \right )^2$ \\
3108\hspace{3mm}4.2  $t_{2ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
3109\hspace{3mm}Calculate the double products after the square. \\
3110\hspace{3mm}4.3  $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\
3111\hspace{3mm}4.4  For $iy$ from $ix + 1$ to $a.used - 1$ do \\
3112\hspace{6mm}4.4.1  $\hat r \leftarrow 2 \cdot a_{ix}a_{iy} + t_{ix + iy} + u$ \\
3113\hspace{6mm}4.4.2  $t_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
3114\hspace{6mm}4.4.3  $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\
3115\hspace{3mm}Set the last carry. \\
3116\hspace{3mm}4.5  While $u > 0$ do \\
3117\hspace{6mm}4.5.1  $iy \leftarrow iy + 1$ \\
3118\hspace{6mm}4.5.2  $\hat r \leftarrow t_{ix + iy} + u$ \\
3119\hspace{6mm}4.5.3  $t_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
3120\hspace{6mm}4.5.4  $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\
31215.  Clamp excess digits of $t$.  (\textit{mp\_clamp}) \\
31226.  Exchange $b$ and $t$. \\
31237.  Clear $t$ (\textit{mp\_clear}) \\
31248.  Return(\textit{MP\_OKAY}) \\
3125\hline
3126\end{tabular}
3127\end{center}
3128\end{small}
3129\caption{Algorithm s\_mp\_sqr}
3130\end{figure}
3131
3132\textbf{Algorithm s\_mp\_sqr.}
3133This algorithm computes the square of an input using the three observations on squaring.  It is based fairly faithfully on  algorithm 14.16 of HAC
3134\cite[pp.596-597]{HAC}.  Similar to algorithm s\_mp\_mul\_digs, a temporary mp\_int is allocated to hold the result of the squaring.  This allows the 
3135destination mp\_int to be the same as the source mp\_int.
3136
3137The outer loop of this algorithm begins on step 4. It is best to think of the outer loop as walking down the rows of the partial results, while
3138the inner loop computes the columns of the partial result.  Step 4.1 and 4.2 compute the square term for each row, and step 4.3 and 4.4 propagate
3139the carry and compute the double products.  
3140
3141The requirement that a mp\_word be able to represent the range $0 \le x < 2 \beta^2$ arises from this
3142very algorithm.  The product $a_{ix}a_{iy}$ will lie in the range $0 \le x \le \beta^2 - 2\beta + 1$ which is obviously less than $\beta^2$ meaning that
3143when it is multiplied by two, it can be properly represented by a mp\_word.
3144
3145Similar to algorithm s\_mp\_mul\_digs, after every pass of the inner loop, the destination is correctly set to the sum of all of the partial 
3146results calculated so far.  This involves expensive carry propagation which will be eliminated in the next algorithm.  
3147
3148EXAM,bn_s_mp_sqr.c
3149
3150Inside the outer loop (line @32,for@) the square term is calculated on line @35,r =@.  The carry (line @42,>>@) has been
3151extracted from the mp\_word accumulator using a right shift.  Aliases for $a_{ix}$ and $t_{ix+iy}$ are initialized 
3152(lines @45,tmpx@ and @48,tmpt@) to simplify the inner loop.  The doubling is performed using two
3153additions (line @57,r + r@) since it is usually faster than shifting, if not at least as fast.  
3154
3155The important observation is that the inner loop does not begin at $iy = 0$ like for multiplication.  As such the inner loops
3156get progressively shorter as the algorithm proceeds.  This is what leads to the savings compared to using a multiplication to
3157square a number. 
3158
3159\subsection{Faster Squaring by the ``Comba'' Method}
3160A major drawback to the baseline method is the requirement for single precision shifting inside the $O(n^2)$ nested loop.  Squaring has an additional
3161drawback that it must double the product inside the inner loop as well.  As for multiplication, the Comba technique can be used to eliminate these
3162performance hazards.
3163
3164The first obvious solution is to make an array of mp\_words which will hold all of the columns.  This will indeed eliminate all of the carry
3165propagation operations from the inner loop.  However, the inner product must still be doubled $O(n^2)$ times.  The solution stems from the simple fact
3166that $2a + 2b + 2c = 2(a + b + c)$.  That is the sum of all of the double products is equal to double the sum of all the products.  For example,
3167$ab + ba + ac + ca = 2ab + 2ac = 2(ab + ac)$.  
3168
3169However, we cannot simply double all of the columns, since the squares appear only once per row.  The most practical solution is to have two 
3170mp\_word arrays.  One array will hold the squares and the other array will hold the double products.  With both arrays the doubling and 
3171carry propagation can be moved to a $O(n)$ work level outside the $O(n^2)$ level.  In this case, we have an even simpler solution in mind.
3172
3173\newpage\begin{figure}[!here]
3174\begin{small}
3175\begin{center}
3176\begin{tabular}{l}
3177\hline Algorithm \textbf{fast\_s\_mp\_sqr}. \\
3178\textbf{Input}.   mp\_int $a$ \\
3179\textbf{Output}.  $b \leftarrow a^2$ \\
3180\hline \\
3181Place an array of \textbf{MP\_WARRAY} mp\_digits named $W$ on the stack. \\
31821.  If $b.alloc < 2a.used + 1$ then grow $b$ to $2a.used + 1$ digits.  (\textit{mp\_grow}). \\
31832.  If step 1 failed return(\textit{MP\_MEM}). \\
3184\\
31853.  $pa \leftarrow 2 \cdot a.used$ \\
31864.  $\hat W1 \leftarrow 0$ \\
31875.  for $ix$ from $0$ to $pa - 1$ do \\
3188\hspace{3mm}5.1  $\_ \hat W \leftarrow 0$ \\
3189\hspace{3mm}5.2  $ty \leftarrow \mbox{MIN}(a.used - 1, ix)$ \\
3190\hspace{3mm}5.3  $tx \leftarrow ix - ty$ \\
3191\hspace{3mm}5.4  $iy \leftarrow \mbox{MIN}(a.used - tx, ty + 1)$ \\
3192\hspace{3mm}5.5  $iy \leftarrow \mbox{MIN}(iy, \lfloor \left (ty - tx + 1 \right )/2 \rfloor)$ \\
3193\hspace{3mm}5.6  for $iz$ from $0$ to $iz - 1$ do \\
3194\hspace{6mm}5.6.1  $\_ \hat W \leftarrow \_ \hat W + a_{tx + iz}a_{ty - iz}$ \\
3195\hspace{3mm}5.7  $\_ \hat W \leftarrow 2 \cdot \_ \hat W  + \hat W1$ \\
3196\hspace{3mm}5.8  if $ix$ is even then \\
3197\hspace{6mm}5.8.1  $\_ \hat W \leftarrow \_ \hat W + \left ( a_{\lfloor ix/2 \rfloor}\right )^2$ \\
3198\hspace{3mm}5.9  $W_{ix} \leftarrow \_ \hat W (\mbox{mod }\beta)$ \\
3199\hspace{3mm}5.10  $\hat W1 \leftarrow \lfloor \_ \hat W / \beta \rfloor$ \\
3200\\
32016.  $oldused \leftarrow b.used$ \\
32027.  $b.used \leftarrow 2 \cdot a.used$ \\
32038.  for $ix$ from $0$ to $pa - 1$ do \\
3204\hspace{3mm}8.1  $b_{ix} \leftarrow W_{ix}$ \\
32059.  for $ix$ from $pa$ to $oldused - 1$ do \\
3206\hspace{3mm}9.1  $b_{ix} \leftarrow 0$ \\
320710.  Clamp excess digits from $b$.  (\textit{mp\_clamp}) \\
320811.  Return(\textit{MP\_OKAY}). \\ 
3209\hline
3210\end{tabular}
3211\end{center}
3212\end{small}
3213\caption{Algorithm fast\_s\_mp\_sqr}
3214\end{figure}
3215
3216\textbf{Algorithm fast\_s\_mp\_sqr.}
3217This algorithm computes the square of an input using the Comba technique.  It is designed to be a replacement for algorithm 
3218s\_mp\_sqr when the number of input digits is less than \textbf{MP\_WARRAY} and less than $\delta \over 2$.  
3219This algorithm is very similar to the Comba multiplier except with a few key differences we shall make note of.
3220
3221First, we have an accumulator and carry variables $\_ \hat W$ and $\hat W1$ respectively.  This is because the inner loop
3222products are to be doubled.  If we had added the previous carry in we would be doubling too much.  Next we perform an
3223addition MIN condition on $iy$ (step 5.5) to prevent overlapping digits.  For example, $a_3 \cdot a_5$ is equal
3224$a_5 \cdot a_3$.  Whereas in the multiplication case we would have $5 < a.used$ and $3 \ge 0$ is maintained since we double the sum
3225of the products just outside the inner loop we have to avoid doing this.  This is also a good thing since we perform
3226fewer multiplications and the routine ends up being faster.
3227
3228Finally the last difference is the addition of the ``square'' term outside the inner loop (step 5.8).  We add in the square
3229only to even outputs and it is the square of the term at the $\lfloor ix / 2 \rfloor$ position.
3230
3231EXAM,bn_fast_s_mp_sqr.c
3232
3233This implementation is essentially a copy of Comba multiplication with the appropriate changes added to make it faster for 
3234the special case of squaring.  
3235
3236\subsection{Polynomial Basis Squaring}
3237The same algorithm that performs optimal polynomial basis multiplication can be used to perform polynomial basis squaring.  The minor exception
3238is that $\zeta_y = f(y)g(y)$ is actually equivalent to $\zeta_y = f(y)^2$ since $f(y) = g(y)$.  Instead of performing $2n + 1$
3239multiplications to find the $\zeta$ relations, squaring operations are performed instead.  
3240
3241\subsection{Karatsuba Squaring}
3242Let $f(x) = ax + b$ represent the polynomial basis representation of a number to square.  
3243Let $h(x) = \left ( f(x) \right )^2$ represent the square of the polynomial.  The Karatsuba equation can be modified to square a 
3244number with the following equation.
3245
3246\begin{equation}
3247h(x) = a^2x^2 + \left ((a + b)^2 - (a^2 + b^2) \right )x + b^2
3248\end{equation}
3249
3250Upon closer inspection this equation only requires the calculation of three half-sized squares: $a^2$, $b^2$ and $(a + b)^2$.  As in 
3251Karatsuba multiplication, this algorithm can be applied recursively on the input and will achieve an asymptotic running time of 
3252$O \left ( n^{lg(3)} \right )$.
3253
3254If the asymptotic times of Karatsuba squaring and multiplication are the same, why not simply use the multiplication algorithm 
3255instead?  The answer to this arises from the cutoff point for squaring.  As in multiplication there exists a cutoff point, at which the 
3256time required for a Comba based squaring and a Karatsuba based squaring meet.  Due to the overhead inherent in the Karatsuba method, the cutoff 
3257point is fairly high.  For example, on an AMD Athlon XP processor with $\beta = 2^{28}$, the cutoff point is around 127 digits.  
3258
3259Consider squaring a 200 digit number with this technique.  It will be split into two 100 digit halves which are subsequently squared.  
3260The 100 digit halves will not be squared using Karatsuba, but instead using the faster Comba based squaring algorithm.  If Karatsuba multiplication
3261were used instead, the 100 digit numbers would be squared with a slower Comba based multiplication.  
3262
3263\newpage\begin{figure}[!here]
3264\begin{small}
3265\begin{center}
3266\begin{tabular}{l}
3267\hline Algorithm \textbf{mp\_karatsuba\_sqr}. \\
3268\textbf{Input}.   mp\_int $a$ \\
3269\textbf{Output}.  $b \leftarrow a^2$ \\
3270\hline \\
32711.  Initialize the following temporary mp\_ints:  $x0$, $x1$, $t1$, $t2$, $x0x0$ and $x1x1$. \\
32722.  If any of the initializations on step 1 failed return(\textit{MP\_MEM}). \\
3273\\
3274Split the input.  e.g. $a = x1\beta^B + x0$ \\
32753.  $B \leftarrow \lfloor a.used / 2 \rfloor$ \\
32764.  $x0 \leftarrow a \mbox{ (mod }\beta^B\mbox{)}$ (\textit{mp\_mod\_2d}) \\
32775.  $x1 \leftarrow \lfloor a / \beta^B \rfloor$ (\textit{mp\_lshd}) \\
3278\\
3279Calculate the three squares. \\
32806.  $x0x0 \leftarrow x0^2$ (\textit{mp\_sqr}) \\
32817.  $x1x1 \leftarrow x1^2$ \\
32828.  $t1 \leftarrow x1 + x0$ (\textit{s\_mp\_add}) \\
32839.  $t1 \leftarrow t1^2$ \\
3284\\
3285Compute the middle term. \\
328610.  $t2 \leftarrow x0x0 + x1x1$ (\textit{s\_mp\_add}) \\
328711.  $t1 \leftarrow t1 - t2$ \\
3288\\
3289Compute final product. \\
329012.  $t1 \leftarrow t1\beta^B$ (\textit{mp\_lshd}) \\
329113.  $x1x1 \leftarrow x1x1\beta^{2B}$ \\
329214.  $t1 \leftarrow t1 + x0x0$ \\
329315.  $b \leftarrow t1 + x1x1$ \\
329416.  Return(\textit{MP\_OKAY}). \\
3295\hline
3296\end{tabular}
3297\end{center}
3298\end{small}
3299\caption{Algorithm mp\_karatsuba\_sqr}
3300\end{figure}
3301
3302\textbf{Algorithm mp\_karatsuba\_sqr.}
3303This algorithm computes the square of an input $a$ using the Karatsuba technique.  This algorithm is very similar to the Karatsuba based
3304multiplication algorithm with the exception that the three half-size multiplications have been replaced with three half-size squarings.
3305
3306The radix point for squaring is simply placed exactly in the middle of the digits when the input has an odd number of digits, otherwise it is
3307placed just below the middle.  Step 3, 4 and 5 compute the two halves required using $B$
3308as the radix point.  The first two squares in steps 6 and 7 are rather straightforward while the last square is of a more compact form.
3309
3310By expanding $\left (x1 + x0 \right )^2$, the $x1^2$ and $x0^2$ terms in the middle disappear, that is $(x0 - x1)^2 - (x1^2 + x0^2)  = 2 \cdot x0 \cdot x1$.
3311Now if $5n$ single precision additions and a squaring of $n$-digits is faster than multiplying two $n$-digit numbers and doubling then
3312this method is faster.  Assuming no further recursions occur, the difference can be estimated with the following inequality.
3313
3314Let $p$ represent the cost of a single precision addition and $q$ the cost of a single precision multiplication both in terms of time\footnote{Or
3315machine clock cycles.}. 
3316
3317\begin{equation}
33185pn +{{q(n^2 + n)} \over 2} \le pn + qn^2
3319\end{equation}
3320
3321For example, on an AMD Athlon XP processor $p = {1 \over 3}$ and $q = 6$.  This implies that the following inequality should hold.
3322\begin{center}
3323\begin{tabular}{rcl}
3324${5n \over 3} + 3n^2 + 3n$     & $<$ & ${n \over 3} + 6n^2$ \\
3325${5 \over 3} + 3n + 3$     & $<$ & ${1 \over 3} + 6n$ \\
3326${13 \over 9}$     & $<$ & $n$ \\
3327\end{tabular}
3328\end{center}
3329
3330This results in a cutoff point around $n = 2$.  As a consequence it is actually faster to compute the middle term the ``long way'' on processors
3331where multiplication is substantially slower\footnote{On the Athlon there is a 1:17 ratio between clock cycles for addition and multiplication.  On
3332the Intel P4 processor this ratio is 1:29 making this method even more beneficial.  The only common exception is the ARMv4 processor which has a
3333ratio of 1:7.  } than simpler operations such as addition.  
3334
3335EXAM,bn_mp_karatsuba_sqr.c
3336
3337This implementation is largely based on the implementation of algorithm mp\_karatsuba\_mul.  It uses the same inline style to copy and 
3338shift the input into the two halves.  The loop from line @54,{@ to line @70,}@ has been modified since only one input exists.  The \textbf{used}
3339count of both $x0$ and $x1$ is fixed up and $x0$ is clamped before the calculations begin.  At this point $x1$ and $x0$ are valid equivalents
3340to the respective halves as if mp\_rshd and mp\_mod\_2d had been used.  
3341
3342By inlining the copy and shift operations the cutoff point for Karatsuba multiplication can be lowered.  On the Athlon the cutoff point
3343is exactly at the point where Comba squaring can no longer be used (\textit{128 digits}).  On slower processors such as the Intel P4
3344it is actually below the Comba limit (\textit{at 110 digits}).
3345
3346This routine uses the same error trap coding style as mp\_karatsuba\_sqr.  As the temporary variables are initialized errors are 
3347redirected to the error trap higher up.  If the algorithm completes without error the error code is set to \textbf{MP\_OKAY} and 
3348mp\_clears are executed normally.
3349
3350\subsection{Toom-Cook Squaring}
3351The Toom-Cook squaring algorithm mp\_toom\_sqr is heavily based on the algorithm mp\_toom\_mul with the exception that squarings are used
3352instead of multiplication to find the five relations.  The reader is encouraged to read the description of the latter algorithm and try to 
3353derive their own Toom-Cook squaring algorithm.  
3354
3355\subsection{High Level Squaring}
3356\newpage\begin{figure}[!here]
3357\begin{small}
3358\begin{center}
3359\begin{tabular}{l}
3360\hline Algorithm \textbf{mp\_sqr}. \\
3361\textbf{Input}.   mp\_int $a$ \\
3362\textbf{Output}.  $b \leftarrow a^2$ \\
3363\hline \\
33641.  If $a.used \ge TOOM\_SQR\_CUTOFF$ then  \\
3365\hspace{3mm}1.1  $b \leftarrow a^2$ using algorithm mp\_toom\_sqr \\
33662.  else if $a.used \ge KARATSUBA\_SQR\_CUTOFF$ then \\
3367\hspace{3mm}2.1  $b \leftarrow a^2$ using algorithm mp\_karatsuba\_sqr \\
33683.  else \\
3369\hspace{3mm}3.1  $digs \leftarrow a.used + b.used + 1$ \\
3370\hspace{3mm}3.2  If $digs < MP\_ARRAY$ and $a.used \le \delta$ then \\
3371\hspace{6mm}3.2.1  $b \leftarrow a^2$ using algorithm fast\_s\_mp\_sqr.  \\
3372\hspace{3mm}3.3  else \\
3373\hspace{6mm}3.3.1  $b \leftarrow a^2$ using algorithm s\_mp\_sqr.  \\
33744.  $b.sign \leftarrow MP\_ZPOS$ \\
33755.  Return the result of the unsigned squaring performed. \\
3376\hline
3377\end{tabular}
3378\end{center}
3379\end{small}
3380\caption{Algorithm mp\_sqr}
3381\end{figure}
3382
3383\textbf{Algorithm mp\_sqr.}
3384This algorithm computes the square of the input using one of four different algorithms.  If the input is very large and has at least
3385\textbf{TOOM\_SQR\_CUTOFF} or \textbf{KARATSUBA\_SQR\_CUTOFF} digits then either the Toom-Cook or the Karatsuba Squaring algorithm is used.  If
3386neither of the polynomial basis algorithms should be used then either the Comba or baseline algorithm is used.  
3387
3388EXAM,bn_mp_sqr.c
3389
3390\section*{Exercises}
3391\begin{tabular}{cl}
3392$\left [ 3 \right ] $ & Devise an efficient algorithm for selection of the radix point to handle inputs \\
3393                      & that have different number of digits in Karatsuba multiplication. \\
3394                      & \\
3395$\left [ 2 \right ] $ & In ~SQUARE~ the fact that every column of a squaring is made up \\
3396                      & of double products and at most one square is stated.  Prove this statement. \\
3397                      & \\                      
3398$\left [ 3 \right ] $ & Prove the equation for Karatsuba squaring. \\
3399                      & \\
3400$\left [ 1 \right ] $ & Prove that Karatsuba squaring requires $O \left (n^{lg(3)} \right )$ time. \\
3401                      & \\ 
3402$\left [ 2 \right ] $ & Determine the minimal ratio between addition and multiplication clock cycles \\
3403                      & required for equation $6.7$ to be true.  \\
3404                      & \\
3405$\left [ 3 \right ] $ & Implement a threaded version of Comba multiplication (and squaring) where you \\
3406                      & compute subsets of the columns in each thread.  Determine a cutoff point where \\
3407                      & it is effective and add the logic to mp\_mul() and mp\_sqr(). \\
3408                      &\\
3409$\left [ 4 \right ] $ & Same as the previous but also modify the Karatsuba and Toom-Cook.  You must \\
3410                      & increase the throughput of mp\_exptmod() for random odd moduli in the range \\
3411                      & $512 \ldots 4096$ bits significantly ($> 2x$) to complete this challenge. \\
3412                      & \\
3413\end{tabular}
3414
3415\chapter{Modular Reduction}
3416MARK,REDUCTION
3417\section{Basics of Modular Reduction}
3418\index{modular residue}
3419Modular reduction is an operation that arises quite often within public key cryptography algorithms and various number theoretic algorithms, 
3420such as factoring.  Modular reduction algorithms are the third class of algorithms of the ``multipliers'' set.  A number $a$ is said to be \textit{reduced}
3421modulo another number $b$ by finding the remainder of the division $a/b$.  Full integer division with remainder is a topic to be covered 
3422in~\ref{sec:division}.
3423
3424Modular reduction is equivalent to solving for $r$ in the following equation.  $a = bq + r$ where $q = \lfloor a/b \rfloor$.  The result 
3425$r$ is said to be ``congruent to $a$ modulo $b$'' which is also written as $r \equiv a \mbox{ (mod }b\mbox{)}$.  In other vernacular $r$ is known as the 
3426``modular residue'' which leads to ``quadratic residue''\footnote{That's fancy talk for $b \equiv a^2 \mbox{ (mod }p\mbox{)}$.} and
3427other forms of residues.  
3428
3429Modular reductions are normally used to create either finite groups, rings or fields.  The most common usage for performance driven modular reductions 
3430is in modular exponentiation algorithms.  That is to compute $d = a^b \mbox{ (mod }c\mbox{)}$ as fast as possible.  This operation is used in the 
3431RSA and Diffie-Hellman public key algorithms, for example.  Modular multiplication and squaring also appears as a fundamental operation in 
3432elliptic curve cryptographic algorithms.  As will be discussed in the subsequent chapter there exist fast algorithms for computing modular 
3433exponentiations without having to perform (\textit{in this example}) $b - 1$ multiplications.  These algorithms will produce partial results in the 
3434range $0 \le x < c^2$ which can be taken advantage of to create several efficient algorithms.   They have also been used to create redundancy check 
3435algorithms known as CRCs, error correction codes such as Reed-Solomon and solve a variety of number theoeretic problems.  
3436
3437\section{The Barrett Reduction}
3438The Barrett reduction algorithm \cite{BARRETT} was inspired by fast division algorithms which multiply by the reciprocal to emulate
3439division.  Barretts observation was that the residue $c$ of $a$ modulo $b$ is equal to 
3440
3441\begin{equation}
3442c = a - b \cdot \lfloor a/b \rfloor
3443\end{equation}
3444
3445Since algorithms such as modular exponentiation would be using the same modulus extensively, typical DSP\footnote{It is worth noting that Barrett's paper 
3446targeted the DSP56K processor.}  intuition would indicate the next step would be to replace $a/b$ by a multiplication by the reciprocal.  However, 
3447DSP intuition on its own will not work as these numbers are considerably larger than the precision of common DSP floating point data types.  
3448It would take another common optimization to optimize the algorithm.
3449
3450\subsection{Fixed Point Arithmetic}
3451The trick used to optimize the above equation is based on a technique of emulating floating point data types with fixed precision integers.  Fixed
3452point arithmetic would become very popular as it greatly optimize the ``3d-shooter'' genre of games in the mid 1990s when floating point units were 
3453fairly slow if not unavailable.   The idea behind fixed point arithmetic is to take a normal $k$-bit integer data type and break it into $p$-bit 
3454integer and a $q$-bit fraction part (\textit{where $p+q = k$}).  
3455
3456In this system a $k$-bit integer $n$ would actually represent $n/2^q$.  For example, with $q = 4$ the integer $n = 37$ would actually represent the
3457value $2.3125$.  To multiply two fixed point numbers the integers are multiplied using traditional arithmetic and subsequently normalized by 
3458moving the implied decimal point back to where it should be.  For example, with $q = 4$ to multiply the integers $9$ and $5$ they must be converted 
3459to fixed point first by multiplying by $2^q$.  Let $a = 9(2^q)$ represent the fixed point representation of $9$ and $b = 5(2^q)$ represent the 
3460fixed point representation of $5$.  The product $ab$ is equal to $45(2^{2q})$ which when normalized by dividing by $2^q$ produces $45(2^q)$.  
3461
3462This technique became popular since a normal integer multiplication and logical shift right are the only required operations to perform a multiplication
3463of two fixed point numbers.  Using fixed point arithmetic, division can be easily approximated by multiplying by the reciprocal.  If $2^q$ is 
3464equivalent to one than $2^q/b$ is equivalent to the fixed point approximation of $1/b$ using real arithmetic.  Using this fact dividing an integer 
3465$a$ by another integer $b$ can be achieved with the following expression.
3466
3467\begin{equation}
3468\lfloor a / b \rfloor \mbox{ }\approx\mbox{ } \lfloor (a \cdot \lfloor 2^q / b \rfloor)/2^q \rfloor
3469\end{equation}
3470
3471The precision of the division is proportional to the value of $q$.  If the divisor $b$ is used frequently as is the case with 
3472modular exponentiation pre-computing $2^q/b$ will allow a division to be performed with a multiplication and a right shift.  Both operations
3473are considerably faster than division on most processors.  
3474
3475Consider dividing $19$ by $5$.  The correct result is $\lfloor 19/5 \rfloor = 3$.  With $q = 3$ the reciprocal is $\lfloor 2^q/5 \rfloor = 1$ which
3476leads to a product of $19$ which when divided by $2^q$ produces $2$.  However, with $q = 4$ the reciprocal is $\lfloor 2^q/5 \rfloor = 3$ and
3477the result of the emulated division is $\lfloor 3 \cdot 19 / 2^q \rfloor = 3$ which is correct.  The value of $2^q$ must be close to or ideally
3478larger than the dividend.  In effect if $a$ is the dividend then $q$ should allow $0 \le \lfloor a/2^q \rfloor \le 1$ in order for this approach
3479to work correctly.  Plugging this form of divison into the original equation the following modular residue equation arises.
3480
3481\begin{equation}
3482c = a - b \cdot \lfloor (a \cdot \lfloor 2^q / b \rfloor)/2^q \rfloor
3483\end{equation}
3484
3485Using the notation from \cite{BARRETT} the value of $\lfloor 2^q / b \rfloor$ will be represented by the $\mu$ symbol.  Using the $\mu$
3486variable also helps re-inforce the idea that it is meant to be computed once and re-used.
3487
3488\begin{equation}
3489c = a - b \cdot \lfloor (a \cdot \mu)/2^q \rfloor
3490\end{equation}
3491
3492Provided that $2^q \ge a$ this algorithm will produce a quotient that is either exactly correct or off by a value of one.  In the context of Barrett
3493reduction the value of $a$ is bound by $0 \le a \le (b - 1)^2$ meaning that $2^q \ge b^2$ is sufficient to ensure the reciprocal will have enough
3494precision.  
3495
3496Let $n$ represent the number of digits in $b$.  This algorithm requires approximately $2n^2$ single precision multiplications to produce the quotient and 
3497another $n^2$ single precision multiplications to find the residue.  In total $3n^2$ single precision multiplications are required to 
3498reduce the number.  
3499
3500For example, if $b = 1179677$ and $q = 41$ ($2^q > b^2$), then the reciprocal $\mu$ is equal to $\lfloor 2^q / b \rfloor = 1864089$.  Consider reducing
3501$a = 180388626447$ modulo $b$ using the above reduction equation.  The quotient using the new formula is $\lfloor (a \cdot \mu) / 2^q \rfloor = 152913$.
3502By subtracting $152913b$ from $a$ the correct residue $a \equiv 677346 \mbox{ (mod }b\mbox{)}$ is found.
3503
3504\subsection{Choosing a Radix Point}
3505Using the fixed point representation a modular reduction can be performed with $3n^2$ single precision multiplications.  If that were the best
3506that could be achieved a full division\footnote{A division requires approximately $O(2cn^2)$ single precision multiplications for a small value of $c$.  
3507See~\ref{sec:division} for further details.} might as well be used in its place.  The key to optimizing the reduction is to reduce the precision of
3508the initial multiplication that finds the quotient.  
3509
3510Let $a$ represent the number of which the residue is sought.  Let $b$ represent the modulus used to find the residue.  Let $m$ represent
3511the number of digits in $b$.  For the purposes of this discussion we will assume that the number of digits in $a$ is $2m$, which is generally true if 
3512two $m$-digit numbers have been multiplied.  Dividing $a$ by $b$ is the same as dividing a $2m$ digit integer by a $m$ digit integer.  Digits below the 
3513$m - 1$'th digit of $a$ will contribute at most a value of $1$ to the quotient because $\beta^k < b$ for any $0 \le k \le m - 1$.  Another way to
3514express this is by re-writing $a$ as two parts.  If $a' \equiv a \mbox{ (mod }b^m\mbox{)}$ and $a'' = a - a'$ then 
3515${a \over b} \equiv {{a' + a''} \over b}$ which is equivalent to ${a' \over b} + {a'' \over b}$.  Since $a'$ is bound to be less than $b$ the quotient
3516is bound by $0 \le {a' \over b} < 1$.
3517
3518Since the digits of $a'$ do not contribute much to the quotient the observation is that they might as well be zero.  However, if the digits 
3519``might as well be zero'' they might as well not be there in the first place.  Let $q_0 = \lfloor a/\beta^{m-1} \rfloor$ represent the input
3520with the irrelevant digits trimmed.  Now the modular reduction is trimmed to the almost equivalent equation
3521
3522\begin{equation}
3523c = a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor
3524\end{equation}
3525
3526Note that the original divisor $2^q$ has been replaced with $\beta^{m+1}$ where in this case $q$ is a multiple of $lg(\beta)$. Also note that the 
3527exponent on the divisor when added to the amount $q_0$ was shifted by equals $2m$.  If the optimization had not been performed the divisor 
3528would have the exponent $2m$ so in the end the exponents do ``add up''. Using the above equation the quotient 
3529$\lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ can be off from the true quotient by at most two.  The original fixed point quotient can be off
3530by as much as one (\textit{provided the radix point is chosen suitably}) and now that the lower irrelevent digits have been trimmed the quotient
3531can be off by an additional value of one for a total of at most two.  This implies that 
3532$0 \le a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor < 3b$.  By first subtracting $b$ times the quotient and then conditionally subtracting 
3533$b$ once or twice the residue is found.
3534
3535The quotient is now found using $(m + 1)(m) = m^2 + m$ single precision multiplications and the residue with an additional $m^2$ single
3536precision multiplications, ignoring the subtractions required.  In total $2m^2 + m$ single precision multiplications are required to find the residue.  
3537This is considerably faster than the original attempt.
3538
3539For example, let $\beta = 10$ represent the radix of the digits.  Let $b = 9999$ represent the modulus which implies $m = 4$. Let $a = 99929878$ 
3540represent the value of which the residue is desired.  In this case $q = 8$ since $10^7 < 9999^2$ meaning that $\mu = \lfloor \beta^{q}/b \rfloor = 10001$.  
3541With the new observation the multiplicand for the quotient is equal to $q_0 = \lfloor a / \beta^{m - 1} \rfloor = 99929$.  The quotient is then 
3542$\lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor = 9993$.  Subtracting $9993b$ from $a$ and the correct residue $a \equiv 9871 \mbox{ (mod }b\mbox{)}$ 
3543is found.  
3544
3545\subsection{Trimming the Quotient}
3546So far the reduction algorithm has been optimized from $3m^2$ single precision multiplications down to $2m^2 + m$ single precision multiplications.  As 
3547it stands now the algorithm is already fairly fast compared to a full integer division algorithm.  However, there is still room for
3548optimization.  
3549
3550After the first multiplication inside the quotient ($q_0 \cdot \mu$) the value is shifted right by $m + 1$ places effectively nullifying the lower
3551half of the product.  It would be nice to be able to remove those digits from the product to effectively cut down the number of single precision 
3552multiplications.  If the number of digits in the modulus $m$ is far less than $\beta$ a full product is not required for the algorithm to work properly.  
3553In fact the lower $m - 2$ digits will not affect the upper half of the product at all and do not need to be computed.  
3554
3555The value of $\mu$ is a $m$-digit number and $q_0$ is a $m + 1$ digit number.  Using a full multiplier $(m + 1)(m) = m^2 + m$ single precision
3556multiplications would be required.  Using a multiplier that will only produce digits at and above the $m - 1$'th digit reduces the number
3557of single precision multiplications to ${m^2 + m} \over 2$ single precision multiplications.  
3558
3559\subsection{Trimming the Residue}
3560After the quotient has been calculated it is used to reduce the input.  As previously noted the algorithm is not exact and it can be off by a small
3561multiple of the modulus, that is $0 \le a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor < 3b$.  If $b$ is $m$ digits than the 
3562result of reduction equation is a value of at most $m + 1$ digits (\textit{provided $3 < \beta$}) implying that the upper $m - 1$ digits are
3563implicitly zero.  
3564
3565The next optimization arises from this very fact.  Instead of computing $b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ using a full
3566$O(m^2)$ multiplication algorithm only the lower $m+1$ digits of the product have to be computed.  Similarly the value of $a$ can
3567be reduced modulo $\beta^{m+1}$ before the multiple of $b$ is subtracted which simplifes the subtraction as well.  A multiplication that produces 
3568only the lower $m+1$ digits requires ${m^2 + 3m - 2} \over 2$ single precision multiplications.  
3569
3570With both optimizations in place the algorithm is the algorithm Barrett proposed.  It requires $m^2 + 2m - 1$ single precision multiplications which
3571is considerably faster than the straightforward $3m^2$ method.  
3572
3573\subsection{The Barrett Algorithm}
3574\newpage\begin{figure}[!here]
3575\begin{small}
3576\begin{center}
3577\begin{tabular}{l}
3578\hline Algorithm \textbf{mp\_reduce}. \\
3579\textbf{Input}.   mp\_int $a$, mp\_int $b$ and $\mu = \lfloor \beta^{2m}/b \rfloor, m = \lceil lg_{\beta}(b) \rceil, (0 \le a < b^2, b > 1)$ \\
3580\textbf{Output}.  $a \mbox{ (mod }b\mbox{)}$ \\
3581\hline \\
3582Let $m$ represent the number of digits in $b$.  \\
35831.  Make a copy of $a$ and store it in $q$.  (\textit{mp\_init\_copy}) \\
35842.  $q \leftarrow \lfloor q / \beta^{m - 1} \rfloor$ (\textit{mp\_rshd}) \\
3585\\
3586Produce the quotient. \\
35873.  $q \leftarrow q \cdot \mu$  (\textit{note: only produce digits at or above $m-1$}) \\
35884.  $q \leftarrow \lfloor q / \beta^{m + 1} \rfloor$ \\
3589\\
3590Subtract the multiple of modulus from the input. \\
35915.  $a \leftarrow a \mbox{ (mod }\beta^{m+1}\mbox{)}$ (\textit{mp\_mod\_2d}) \\
35926.  $q \leftarrow q \cdot b \mbox{ (mod }\beta^{m+1}\mbox{)}$ (\textit{s\_mp\_mul\_digs}) \\
35937.  $a \leftarrow a - q$ (\textit{mp\_sub}) \\
3594\\
3595Add $\beta^{m+1}$ if a carry occured. \\
35968.  If $a < 0$ then (\textit{mp\_cmp\_d}) \\
3597\hspace{3mm}8.1  $q \leftarrow 1$ (\textit{mp\_set}) \\
3598\hspace{3mm}8.2  $q \leftarrow q \cdot \beta^{m+1}$ (\textit{mp\_lshd}) \\
3599\hspace{3mm}8.3  $a \leftarrow a + q$ \\
3600\\
3601Now subtract the modulus if the residue is too large (e.g. quotient too small). \\
36029.  While $a \ge b$ do (\textit{mp\_cmp}) \\
3603\hspace{3mm}9.1  $c \leftarrow a - b$ \\
360410.  Clear $q$. \\
360511.  Return(\textit{MP\_OKAY}) \\
3606\hline
3607\end{tabular}
3608\end{center}
3609\end{small}
3610\caption{Algorithm mp\_reduce}
3611\end{figure}
3612
3613\textbf{Algorithm mp\_reduce.}
3614This algorithm will reduce the input $a$ modulo $b$ in place using the Barrett algorithm.  It is loosely based on algorithm 14.42 of HAC
3615\cite[pp.  602]{HAC} which is based on the paper from Paul Barrett \cite{BARRETT}.  The algorithm has several restrictions and assumptions which must 
3616be adhered to for the algorithm to work.
3617
3618First the modulus $b$ is assumed to be positive and greater than one.  If the modulus were less than or equal to one than subtracting
3619a multiple of it would either accomplish nothing or actually enlarge the input.  The input $a$ must be in the range $0 \le a < b^2$ in order
3620for the quotient to have enough precision.  If $a$ is the product of two numbers that were already reduced modulo $b$, this will not be a problem.
3621Technically the algorithm will still work if $a \ge b^2$ but it will take much longer to finish.  The value of $\mu$ is passed as an argument to this 
3622algorithm and is assumed to be calculated and stored before the algorithm is used.  
3623
3624Recall that the multiplication for the quotient on step 3 must only produce digits at or above the $m-1$'th position.  An algorithm called 
3625$s\_mp\_mul\_high\_digs$ which has not been presented is used to accomplish this task.  The algorithm is based on $s\_mp\_mul\_digs$ except that
3626instead of stopping at a given level of precision it starts at a given level of precision.  This optimal algorithm can only be used if the number
3627of digits in $b$ is very much smaller than $\beta$.  
3628
3629While it is known that 
3630$a \ge b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ only the lower $m+1$ digits are being used to compute the residue, so an implied 
3631``borrow'' from the higher digits might leave a negative result.  After the multiple of the modulus has been subtracted from $a$ the residue must be 
3632fixed up in case it is negative.  The invariant $\beta^{m+1}$ must be added to the residue to make it positive again.  
3633
3634The while loop at step 9 will subtract $b$ until the residue is less than $b$.  If the algorithm is performed correctly this step is 
3635performed at most twice, and on average once. However, if $a \ge b^2$ than it will iterate substantially more times than it should.
3636
3637EXAM,bn_mp_reduce.c
3638
3639The first multiplication that determines the quotient can be performed by only producing the digits from $m - 1$ and up.  This essentially halves
3640the number of single precision multiplications required.  However, the optimization is only safe if $\beta$ is much larger than the number of digits
3641in the modulus.  In the source code this is evaluated on lines @36,if@ to @44,}@ where algorithm s\_mp\_mul\_high\_digs is used when it is
3642safe to do so.  
3643
3644\subsection{The Barrett Setup Algorithm}
3645In order to use algorithm mp\_reduce the value of $\mu$ must be calculated in advance.  Ideally this value should be computed once and stored for
3646future use so that the Barrett algorithm can be used without delay.  
3647
3648\newpage\begin{figure}[!here]
3649\begin{small}
3650\begin{center}
3651\begin{tabular}{l}
3652\hline Algorithm \textbf{mp\_reduce\_setup}. \\
3653\textbf{Input}.   mp\_int $a$ ($a > 1$)  \\
3654\textbf{Output}.  $\mu \leftarrow \lfloor \beta^{2m}/a \rfloor$ \\
3655\hline \\
36561.  $\mu \leftarrow 2^{2 \cdot lg(\beta) \cdot  m}$ (\textit{mp\_2expt}) \\
36572.  $\mu \leftarrow \lfloor \mu / b \rfloor$ (\textit{mp\_div}) \\
36583.  Return(\textit{MP\_OKAY}) \\
3659\hline
3660\end{tabular}
3661\end{center}
3662\end{small}
3663\caption{Algorithm mp\_reduce\_setup}
3664\end{figure}
3665
3666\textbf{Algorithm mp\_reduce\_setup.}
3667This algorithm computes the reciprocal $\mu$ required for Barrett reduction.  First $\beta^{2m}$ is calculated as $2^{2 \cdot lg(\beta) \cdot  m}$ which
3668is equivalent and much faster.  The final value is computed by taking the integer quotient of $\lfloor \mu / b \rfloor$.
3669
3670EXAM,bn_mp_reduce_setup.c
3671
3672This simple routine calculates the reciprocal $\mu$ required by Barrett reduction.  Note the extended usage of algorithm mp\_div where the variable
3673which would received the remainder is passed as NULL.  As will be discussed in~\ref{sec:division} the division routine allows both the quotient and the 
3674remainder to be passed as NULL meaning to ignore the value.  
3675
3676\section{The Montgomery Reduction}
3677Montgomery reduction\footnote{Thanks to Niels Ferguson for his insightful explanation of the algorithm.} \cite{MONT} is by far the most interesting 
3678form of reduction in common use.  It computes a modular residue which is not actually equal to the residue of the input yet instead equal to a 
3679residue times a constant.  However, as perplexing as this may sound the algorithm is relatively simple and very efficient.  
3680
3681Throughout this entire section the variable $n$ will represent the modulus used to form the residue.  As will be discussed shortly the value of
3682$n$ must be odd.  The variable $x$ will represent the quantity of which the residue is sought.  Similar to the Barrett algorithm the input
3683is restricted to $0 \le x < n^2$.  To begin the description some simple number theory facts must be established.
3684
3685\textbf{Fact 1.}  Adding $n$ to $x$ does not change the residue since in effect it adds one to the quotient $\lfloor x / n \rfloor$.  Another way
3686to explain this is that $n$ is (\textit{or multiples of $n$ are}) congruent to zero modulo $n$.  Adding zero will not change the value of the residue.  
3687
3688\textbf{Fact 2.}  If $x$ is even then performing a division by two in $\Z$ is congruent to $x \cdot 2^{-1} \mbox{ (mod }n\mbox{)}$.  Actually
3689this is an application of the fact that if $x$ is evenly divisible by any $k \in \Z$ then division in $\Z$ will be congruent to 
3690multiplication by $k^{-1}$ modulo $n$.  
3691
3692From these two simple facts the following simple algorithm can be derived.
3693
3694\newpage\begin{figure}[!here]
3695\begin{small}
3696\begin{center}
3697\begin{tabular}{l}
3698\hline Algorithm \textbf{Montgomery Reduction}. \\
3699\textbf{Input}.   Integer $x$, $n$ and $k$ \\
3700\textbf{Output}.  $2^{-k}x \mbox{ (mod }n\mbox{)}$ \\
3701\hline \\
37021.  for $t$ from $1$ to $k$ do \\
3703\hspace{3mm}1.1  If $x$ is odd then \\
3704\hspace{6mm}1.1.1  $x \leftarrow x + n$ \\
3705\hspace{3mm}1.2  $x \leftarrow x/2$ \\
37062.  Return $x$. \\
3707\hline
3708\end{tabular}
3709\end{center}
3710\end{small}
3711\caption{Algorithm Montgomery Reduction}
3712\end{figure}
3713
3714The algorithm reduces the input one bit at a time using the two congruencies stated previously.  Inside the loop $n$, which is odd, is
3715added to $x$ if $x$ is odd.  This forces $x$ to be even which allows the division by two in $\Z$ to be congruent to a modular division by two.  Since
3716$x$ is assumed to be initially much larger than $n$ the addition of $n$ will contribute an insignificant magnitude to $x$.  Let $r$ represent the 
3717final result of the Montgomery algorithm.  If $k > lg(n)$ and $0 \le x < n^2$ then the final result is limited to 
3718$0 \le r < \lfloor x/2^k \rfloor + n$.  As a result at most a single subtraction is required to get the residue desired.
3719
3720\begin{figure}[here]
3721\begin{small}
3722\begin{center}
3723\begin{tabular}{|c|l|}
3724\hline \textbf{Step number ($t$)} & \textbf{Result ($x$)} \\
3725\hline $1$ & $x + n = 5812$, $x/2 = 2906$ \\
3726\hline $2$ & $x/2 = 1453$ \\
3727\hline $3$ & $x + n = 1710$, $x/2 = 855$ \\
3728\hline $4$ & $x + n = 1112$, $x/2 = 556$ \\
3729\hline $5$ & $x/2 = 278$ \\
3730\hline $6$ & $x/2 = 139$ \\
3731\hline $7$ & $x + n = 396$, $x/2 = 198$ \\
3732\hline $8$ & $x/2 = 99$ \\
3733\hline $9$ & $x + n = 356$, $x/2 = 178$ \\
3734\hline
3735\end{tabular}
3736\end{center}
3737\end{small}
3738\caption{Example of Montgomery Reduction (I)}
3739\label{fig:MONT1}
3740\end{figure}
3741
3742Consider the example in figure~\ref{fig:MONT1} which reduces $x = 5555$ modulo $n = 257$ when $k = 9$ (note $\beta^k = 512$ which is larger than $n$).  The result of 
3743the algorithm $r = 178$ is congruent to the value of $2^{-9} \cdot 5555 \mbox{ (mod }257\mbox{)}$.  When $r$ is multiplied by $2^9$ modulo $257$ the correct residue 
3744$r \equiv 158$ is produced.  
3745
3746Let $k = \lfloor lg(n) \rfloor + 1$ represent the number of bits in $n$.  The current algorithm requires $2k^2$ single precision shifts
3747and $k^2$ single precision additions.  At this rate the algorithm is most certainly slower than Barrett reduction and not terribly useful.  
3748Fortunately there exists an alternative representation of the algorithm.
3749
3750\begin{figure}[!here]
3751\begin{small}
3752\begin{center}
3753\begin{tabular}{l}
3754\hline Algorithm \textbf{Montgomery Reduction} (modified I). \\
3755\textbf{Input}.   Integer $x$, $n$ and $k$ ($2^k > n$) \\
3756\textbf{Output}.  $2^{-k}x \mbox{ (mod }n\mbox{)}$ \\
3757\hline \\
37581.  for $t$ from $1$ to $k$ do \\
3759\hspace{3mm}1.1  If the $t$'th bit of $x$ is one then \\
3760\hspace{6mm}1.1.1  $x \leftarrow x + 2^tn$ \\
37612.  Return $x/2^k$. \\
3762\hline
3763\end{tabular}
3764\end{center}
3765\end{small}
3766\caption{Algorithm Montgomery Reduction (modified I)}
3767\end{figure}
3768
3769This algorithm is equivalent since $2^tn$ is a multiple of $n$ and the lower $k$ bits of $x$ are zero by step 2.  The number of single
3770precision shifts has now been reduced from $2k^2$ to $k^2 + k$ which is only a small improvement.
3771
3772\begin{figure}[here]
3773\begin{small}
3774\begin{center}
3775\begin{tabular}{|c|l|r|}
3776\hline \textbf{Step number ($t$)} & \textbf{Result ($x$)} & \textbf{Result ($x$) in Binary} \\
3777\hline -- & $5555$ & $1010110110011$ \\
3778\hline $1$ & $x + 2^{0}n = 5812$ &  $1011010110100$ \\
3779\hline $2$ & $5812$ & $1011010110100$ \\
3780\hline $3$ & $x + 2^{2}n = 6840$ & $1101010111000$ \\
3781\hline $4$ & $x + 2^{3}n = 8896$ & $10001011000000$ \\
3782\hline $5$ & $8896$ & $10001011000000$ \\
3783\hline $6$ & $8896$ & $10001011000000$ \\
3784\hline $7$ & $x + 2^{6}n = 25344$ & $110001100000000$ \\
3785\hline $8$ & $25344$ & $110001100000000$ \\
3786\hline $9$ & $x + 2^{7}n = 91136$ & $10110010000000000$ \\
3787\hline -- & $x/2^k = 178$ & \\
3788\hline
3789\end{tabular}
3790\end{center}
3791\end{small}
3792\caption{Example of Montgomery Reduction (II)}
3793\label{fig:MONT2}
3794\end{figure}
3795
3796Figure~\ref{fig:MONT2} demonstrates the modified algorithm reducing $x = 5555$ modulo $n = 257$ with $k = 9$. 
3797With this algorithm a single shift right at the end is the only right shift required to reduce the input instead of $k$ right shifts inside the 
3798loop.  Note that for the iterations $t = 2, 5, 6$ and $8$ where the result $x$ is not changed.  In those iterations the $t$'th bit of $x$ is 
3799zero and the appropriate multiple of $n$ does not need to be added to force the $t$'th bit of the result to zero.  
3800
3801\subsection{Digit Based Montgomery Reduction}
3802Instead of computing the reduction on a bit-by-bit basis it is actually much faster to compute it on digit-by-digit basis.  Consider the
3803previous algorithm re-written to compute the Montgomery reduction in this new fashion.
3804
3805\begin{figure}[!here]
3806\begin{small}
3807\begin{center}
3808\begin{tabular}{l}
3809\hline Algorithm \textbf{Montgomery Reduction} (modified II). \\
3810\textbf{Input}.   Integer $x$, $n$ and $k$ ($\beta^k > n$) \\
3811\textbf{Output}.  $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\
3812\hline \\
38131.  for $t$ from $0$ to $k - 1$ do \\
3814\hspace{3mm}1.1  $x \leftarrow x + \mu n \beta^t$ \\
38152.  Return $x/\beta^k$. \\
3816\hline
3817\end{tabular}
3818\end{center}
3819\end{small}
3820\caption{Algorithm Montgomery Reduction (modified II)}
3821\end{figure}
3822
3823The value $\mu n \beta^t$ is a multiple of the modulus $n$ meaning that it will not change the residue.  If the first digit of 
3824the value $\mu n \beta^t$ equals the negative (modulo $\beta$) of the $t$'th digit of $x$ then the addition will result in a zero digit.  This
3825problem breaks down to solving the following congruency.  
3826
3827\begin{center}
3828\begin{tabular}{rcl}
3829$x_t + \mu n_0$ & $\equiv$ & $0 \mbox{ (mod }\beta\mbox{)}$ \\
3830$\mu n_0$ & $\equiv$ & $-x_t \mbox{ (mod }\beta\mbox{)}$ \\
3831$\mu$ & $\equiv$ & $-x_t/n_0 \mbox{ (mod }\beta\mbox{)}$ \\
3832\end{tabular}
3833\end{center}
3834
3835In each iteration of the loop on step 1 a new value of $\mu$ must be calculated.  The value of $-1/n_0 \mbox{ (mod }\beta\mbox{)}$ is used 
3836extensively in this algorithm and should be precomputed.  Let $\rho$ represent the negative of the modular inverse of $n_0$ modulo $\beta$.  
3837
3838For example, let $\beta = 10$ represent the radix.  Let $n = 17$ represent the modulus which implies $k = 2$ and $\rho \equiv 7$.  Let $x = 33$ 
3839represent the value to reduce.
3840
3841\newpage\begin{figure}
3842\begin{center}
3843\begin{tabular}{|c|c|c|}
3844\hline \textbf{Step ($t$)} & \textbf{Value of $x$} & \textbf{Value of $\mu$} \\
3845\hline --                 & $33$ & --\\
3846\hline $0$                 & $33 + \mu n = 50$ & $1$ \\
3847\hline $1$                 & $50 + \mu n \beta = 900$ & $5$ \\
3848\hline
3849\end{tabular}
3850\end{center}
3851\caption{Example of Montgomery Reduction}
3852\end{figure}
3853
3854The final result $900$ is then divided by $\beta^k$ to produce the final result $9$.  The first observation is that $9 \nequiv x \mbox{ (mod }n\mbox{)}$ 
3855which implies the result is not the modular residue of $x$ modulo $n$.  However, recall that the residue is actually multiplied by $\beta^{-k}$ in
3856the algorithm.  To get the true residue the value must be multiplied by $\beta^k$.  In this case $\beta^k \equiv 15 \mbox{ (mod }n\mbox{)}$ and
3857the correct residue is $9 \cdot 15 \equiv 16 \mbox{ (mod }n\mbox{)}$.  
3858
3859\subsection{Baseline Montgomery Reduction}
3860The baseline Montgomery reduction algorithm will produce the residue for any size input.  It is designed to be a catch-all algororithm for 
3861Montgomery reductions.  
3862
3863\newpage\begin{figure}[!here]
3864\begin{small}
3865\begin{center}
3866\begin{tabular}{l}
3867\hline Algorithm \textbf{mp\_montgomery\_reduce}. \\
3868\textbf{Input}.   mp\_int $x$, mp\_int $n$ and a digit $\rho \equiv -1/n_0 \mbox{ (mod }n\mbox{)}$. \\
3869\hspace{11.5mm}($0 \le x < n^2, n > 1, (n, \beta) = 1, \beta^k > n$) \\
3870\textbf{Output}.  $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\
3871\hline \\
38721.  $digs \leftarrow 2n.used + 1$ \\
38732.  If $digs < MP\_ARRAY$ and $m.used < \delta$ then \\
3874\hspace{3mm}2.1  Use algorithm fast\_mp\_montgomery\_reduce instead. \\
3875\\
3876Setup $x$ for the reduction. \\
38773.  If $x.alloc < digs$ then grow $x$ to $digs$ digits. \\
38784.  $x.used \leftarrow digs$ \\
3879\\
3880Eliminate the lower $k$ digits. \\
38815.  For $ix$ from $0$ to $k - 1$ do \\
3882\hspace{3mm}5.1  $\mu \leftarrow x_{ix} \cdot \rho \mbox{ (mod }\beta\mbox{)}$ \\
3883\hspace{3mm}5.2  $u \leftarrow 0$ \\
3884\hspace{3mm}5.3  For $iy$ from $0$ to $k - 1$ do \\
3885\hspace{6mm}5.3.1  $\hat r \leftarrow \mu n_{iy} + x_{ix + iy} + u$ \\
3886\hspace{6mm}5.3.2  $x_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
3887\hspace{6mm}5.3.3  $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\
3888\hspace{3mm}5.4  While $u > 0$ do \\
3889\hspace{6mm}5.4.1  $iy \leftarrow iy + 1$ \\
3890\hspace{6mm}5.4.2  $x_{ix + iy} \leftarrow x_{ix + iy} + u$ \\
3891\hspace{6mm}5.4.3  $u \leftarrow \lfloor x_{ix+iy} / \beta \rfloor$ \\
3892\hspace{6mm}5.4.4  $x_{ix + iy} \leftarrow x_{ix+iy} \mbox{ (mod }\beta\mbox{)}$ \\
3893\\
3894Divide by $\beta^k$ and fix up as required. \\
38956.  $x \leftarrow \lfloor x / \beta^k \rfloor$ \\
38967.  If $x \ge n$ then \\
3897\hspace{3mm}7.1  $x \leftarrow x - n$ \\
38988.  Return(\textit{MP\_OKAY}). \\
3899\hline
3900\end{tabular}
3901\end{center}
3902\end{small}
3903\caption{Algorithm mp\_montgomery\_reduce}
3904\end{figure}
3905
3906\textbf{Algorithm mp\_montgomery\_reduce.}
3907This algorithm reduces the input $x$ modulo $n$ in place using the Montgomery reduction algorithm.  The algorithm is loosely based
3908on algorithm 14.32 of \cite[pp.601]{HAC} except it merges the multiplication of $\mu n \beta^t$ with the addition in the inner loop.  The
3909restrictions on this algorithm are fairly easy to adapt to.  First $0 \le x < n^2$ bounds the input to numbers in the same range as 
3910for the Barrett algorithm.  Additionally if $n > 1$ and $n$ is odd there will exist a modular inverse $\rho$.  $\rho$ must be calculated in
3911advance of this algorithm.  Finally the variable $k$ is fixed and a pseudonym for $n.used$.  
3912
3913Step 2 decides whether a faster Montgomery algorithm can be used.  It is based on the Comba technique meaning that there are limits on
3914the size of the input.  This algorithm is discussed in ~COMBARED~.
3915
3916Step 5 is the main reduction loop of the algorithm.  The value of $\mu$ is calculated once per iteration in the outer loop.  The inner loop
3917calculates $x + \mu n \beta^{ix}$ by multiplying $\mu n$ and adding the result to $x$ shifted by $ix$ digits.  Both the addition and
3918multiplication are performed in the same loop to save time and memory.  Step 5.4 will handle any additional carries that escape the inner loop.
3919
3920Using a quick inspection this algorithm requires $n$ single precision multiplications for the outer loop and $n^2$ single precision multiplications 
3921in the inner loop.  In total $n^2 + n$ single precision multiplications which compares favourably to Barrett at $n^2 + 2n - 1$ single precision
3922multiplications.  
3923
3924EXAM,bn_mp_montgomery_reduce.c
3925
3926This is the baseline implementation of the Montgomery reduction algorithm.  Lines @30,digs@ to @35,}@ determine if the Comba based
3927routine can be used instead.  Line @47,mu@ computes the value of $\mu$ for that particular iteration of the outer loop.  
3928
3929The multiplication $\mu n \beta^{ix}$ is performed in one step in the inner loop.  The alias $tmpx$ refers to the $ix$'th digit of $x$ and
3930the alias $tmpn$ refers to the modulus $n$.  
3931
3932\subsection{Faster ``Comba'' Montgomery Reduction}
3933MARK,COMBARED
3934
3935The Montgomery reduction requires fewer single precision multiplications than a Barrett reduction, however it is much slower due to the serial
3936nature of the inner loop.  The Barrett reduction algorithm requires two slightly modified multipliers which can be implemented with the Comba
3937technique.  The Montgomery reduction algorithm cannot directly use the Comba technique to any significant advantage since the inner loop calculates
3938a $k \times 1$ product $k$ times. 
3939
3940The biggest obstacle is that at the $ix$'th iteration of the outer loop the value of $x_{ix}$ is required to calculate $\mu$.  This means the 
3941carries from $0$ to $ix - 1$ must have been propagated upwards to form a valid $ix$'th digit.  The solution as it turns out is very simple.  
3942Perform a Comba like multiplier and inside the outer loop just after the inner loop fix up the $ix + 1$'th digit by forwarding the carry.  
3943
3944With this change in place the Montgomery reduction algorithm can be performed with a Comba style multiplication loop which substantially increases
3945the speed of the algorithm.  
3946
3947\newpage\begin{figure}[!here]
3948\begin{small}
3949\begin{center}
3950\begin{tabular}{l}
3951\hline Algorithm \textbf{fast\_mp\_montgomery\_reduce}. \\
3952\textbf{Input}.   mp\_int $x$, mp\_int $n$ and a digit $\rho \equiv -1/n_0 \mbox{ (mod }n\mbox{)}$. \\
3953\hspace{11.5mm}($0 \le x < n^2, n > 1, (n, \beta) = 1, \beta^k > n$) \\
3954\textbf{Output}.  $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\
3955\hline \\
3956Place an array of \textbf{MP\_WARRAY} mp\_word variables called $\hat W$ on the stack. \\
39571.  if $x.alloc < n.used + 1$ then grow $x$ to $n.used + 1$ digits. \\
3958Copy the digits of $x$ into the array $\hat W$ \\
39592.  For $ix$ from $0$ to $x.used - 1$ do \\
3960\hspace{3mm}2.1  $\hat W_{ix} \leftarrow x_{ix}$ \\
39613.  For $ix$ from $x.used$ to $2n.used - 1$ do \\
3962\hspace{3mm}3.1  $\hat W_{ix} \leftarrow 0$ \\
3963Elimiate the lower $k$ digits. \\
39644.  for $ix$ from $0$ to $n.used - 1$ do \\
3965\hspace{3mm}4.1  $\mu \leftarrow \hat W_{ix} \cdot \rho \mbox{ (mod }\beta\mbox{)}$ \\
3966\hspace{3mm}4.2  For $iy$ from $0$ to $n.used - 1$ do \\
3967\hspace{6mm}4.2.1  $\hat W_{iy + ix} \leftarrow \hat W_{iy + ix} + \mu \cdot n_{iy}$ \\
3968\hspace{3mm}4.3  $\hat W_{ix + 1} \leftarrow \hat W_{ix + 1} + \lfloor \hat W_{ix} / \beta \rfloor$ \\
3969Propagate carries upwards. \\
39705.  for $ix$ from $n.used$ to $2n.used + 1$ do \\
3971\hspace{3mm}5.1  $\hat W_{ix + 1} \leftarrow \hat W_{ix + 1} + \lfloor \hat W_{ix} / \beta \rfloor$ \\
3972Shift right and reduce modulo $\beta$ simultaneously. \\
39736.  for $ix$ from $0$ to $n.used + 1$ do \\
3974\hspace{3mm}6.1  $x_{ix} \leftarrow \hat W_{ix + n.used} \mbox{ (mod }\beta\mbox{)}$ \\
3975Zero excess digits and fixup $x$. \\
39767.  if $x.used > n.used + 1$ then do \\
3977\hspace{3mm}7.1  for $ix$ from $n.used + 1$ to $x.used - 1$ do \\
3978\hspace{6mm}7.1.1  $x_{ix} \leftarrow 0$ \\
39798.  $x.used \leftarrow n.used + 1$ \\
39809.  Clamp excessive digits of $x$. \\
398110.  If $x \ge n$ then \\
3982\hspace{3mm}10.1  $x \leftarrow x - n$ \\
398311.  Return(\textit{MP\_OKAY}). \\
3984\hline
3985\end{tabular}
3986\end{center}
3987\end{small}
3988\caption{Algorithm fast\_mp\_montgomery\_reduce}
3989\end{figure}
3990
3991\textbf{Algorithm fast\_mp\_montgomery\_reduce.}
3992This algorithm will compute the Montgomery reduction of $x$ modulo $n$ using the Comba technique.  It is on most computer platforms significantly
3993faster than algorithm mp\_montgomery\_reduce and algorithm mp\_reduce (\textit{Barrett reduction}).  The algorithm has the same restrictions
3994on the input as the baseline reduction algorithm.  An additional two restrictions are imposed on this algorithm.  The number of digits $k$ in the 
3995the modulus $n$ must not violate $MP\_WARRAY > 2k +1$ and $n < \delta$.   When $\beta = 2^{28}$ this algorithm can be used to reduce modulo
3996a modulus of at most $3,556$ bits in length.  
3997
3998As in the other Comba reduction algorithms there is a $\hat W$ array which stores the columns of the product.  It is initially filled with the
3999contents of $x$ with the excess digits zeroed.  The reduction loop is very similar the to the baseline loop at heart.  The multiplication on step
40004.1 can be single precision only since $ab \mbox{ (mod }\beta\mbox{)} \equiv (a \mbox{ mod }\beta)(b \mbox{ mod }\beta)$.  Some multipliers such
4001as those on the ARM processors take a variable length time to complete depending on the number of bytes of result it must produce.  By performing
4002a single precision multiplication instead half the amount of time is spent.
4003
4004Also note that digit $\hat W_{ix}$ must have the carry from the $ix - 1$'th digit propagated upwards in order for this to work.  That is what step
40054.3 will do.  In effect over the $n.used$ iterations of the outer loop the $n.used$'th lower columns all have the their carries propagated forwards.  Note
4006how the upper bits of those same words are not reduced modulo $\beta$.  This is because those values will be discarded shortly and there is no
4007point.
4008
4009Step 5 will propagate the remainder of the carries upwards.  On step 6 the columns are reduced modulo $\beta$ and shifted simultaneously as they are
4010stored in the destination $x$.  
4011
4012EXAM,bn_fast_mp_montgomery_reduce.c
4013
4014The $\hat W$ array is first filled with digits of $x$ on line @49,for@ then the rest of the digits are zeroed on line @54,for@.  Both loops share
4015the same alias variables to make the code easier to read.  
4016
4017The value of $\mu$ is calculated in an interesting fashion.  First the value $\hat W_{ix}$ is reduced modulo $\beta$ and cast to a mp\_digit.  This
4018forces the compiler to use a single precision multiplication and prevents any concerns about loss of precision.   Line @101,>>@ fixes the carry 
4019for the next iteration of the loop by propagating the carry from $\hat W_{ix}$ to $\hat W_{ix+1}$.
4020
4021The for loop on line @113,for@ propagates the rest of the carries upwards through the columns.  The for loop on line @126,for@ reduces the columns
4022modulo $\beta$ and shifts them $k$ places at the same time.  The alias $\_ \hat W$ actually refers to the array $\hat W$ starting at the $n.used$'th
4023digit, that is $\_ \hat W_{t} = \hat W_{n.used + t}$.  
4024
4025\subsection{Montgomery Setup}
4026To calculate the variable $\rho$ a relatively simple algorithm will be required.  
4027
4028\begin{figure}[!here]
4029\begin{small}
4030\begin{center}
4031\begin{tabular}{l}
4032\hline Algorithm \textbf{mp\_montgomery\_setup}. \\
4033\textbf{Input}.   mp\_int $n$ ($n > 1$ and $(n, 2) = 1$) \\
4034\textbf{Output}.  $\rho \equiv -1/n_0 \mbox{ (mod }\beta\mbox{)}$ \\
4035\hline \\
40361.  $b \leftarrow n_0$ \\
40372.  If $b$ is even return(\textit{MP\_VAL}) \\
40383.  $x \leftarrow (((b + 2) \mbox{ AND } 4) << 1) + b$ \\
40394.  for $k$ from 0 to $\lceil lg(lg(\beta)) \rceil - 2$ do \\
4040\hspace{3mm}4.1  $x \leftarrow x \cdot (2 - bx)$ \\
40415.  $\rho \leftarrow \beta - x \mbox{ (mod }\beta\mbox{)}$ \\
40426.  Return(\textit{MP\_OKAY}). \\
4043\hline
4044\end{tabular}
4045\end{center}
4046\end{small}
4047\caption{Algorithm mp\_montgomery\_setup} 
4048\end{figure}
4049
4050\textbf{Algorithm mp\_montgomery\_setup.}
4051This algorithm will calculate the value of $\rho$ required within the Montgomery reduction algorithms.  It uses a very interesting trick 
4052to calculate $1/n_0$ when $\beta$ is a power of two.  
4053
4054EXAM,bn_mp_montgomery_setup.c
4055
4056This source code computes the value of $\rho$ required to perform Montgomery reduction.  It has been modified to avoid performing excess
4057multiplications when $\beta$ is not the default 28-bits.  
4058
4059\section{The Diminished Radix Algorithm}
4060The Diminished Radix method of modular reduction \cite{DRMET} is a fairly clever technique which can be more efficient than either the Barrett
4061or Montgomery methods for certain forms of moduli.  The technique is based on the following simple congruence.
4062
4063\begin{equation}
4064(x \mbox{ mod } n) + k \lfloor x / n \rfloor \equiv x \mbox{ (mod }(n - k)\mbox{)}
4065\end{equation}
4066
4067This observation was used in the MMB \cite{MMB} block cipher to create a diffusion primitive.  It used the fact that if $n = 2^{31}$ and $k=1$ that 
4068then a x86 multiplier could produce the 62-bit product and use  the ``shrd'' instruction to perform a double-precision right shift.  The proof
4069of the above equation is very simple.  First write $x$ in the product form.
4070
4071\begin{equation}
4072x = qn + r
4073\end{equation}
4074
4075Now reduce both sides modulo $(n - k)$.
4076
4077\begin{equation}
4078x \equiv qk + r  \mbox{ (mod }(n-k)\mbox{)}
4079\end{equation}
4080
4081The variable $n$ reduces modulo $n - k$ to $k$.  By putting $q = \lfloor x/n \rfloor$ and $r = x \mbox{ mod } n$ 
4082into the equation the original congruence is reproduced, thus concluding the proof.  The following algorithm is based on this observation.
4083
4084\begin{figure}[!here]
4085\begin{small}
4086\begin{center}
4087\begin{tabular}{l}
4088\hline Algorithm \textbf{Diminished Radix Reduction}. \\
4089\textbf{Input}.   Integer $x$, $n$, $k$ \\
4090\textbf{Output}.  $x \mbox{ mod } (n - k)$ \\
4091\hline \\
40921.  $q \leftarrow \lfloor x / n \rfloor$ \\
40932.  $q \leftarrow k \cdot q$ \\
40943.  $x \leftarrow x \mbox{ (mod }n\mbox{)}$ \\
40954.  $x \leftarrow x + q$ \\
40965.  If $x \ge (n - k)$ then \\
4097\hspace{3mm}5.1  $x \leftarrow x - (n - k)$ \\
4098\hspace{3mm}5.2  Goto step 1. \\
40996.  Return $x$ \\
4100\hline
4101\end{tabular}
4102\end{center}
4103\end{small}
4104\caption{Algorithm Diminished Radix Reduction}
4105\label{fig:DR}
4106\end{figure}
4107
4108This algorithm will reduce $x$ modulo $n - k$ and return the residue.  If $0 \le x < (n - k)^2$ then the algorithm will loop almost always
4109once or twice and occasionally three times.  For simplicity sake the value of $x$ is bounded by the following simple polynomial.
4110
4111\begin{equation} 
41120 \le x < n^2 + k^2 - 2nk
4113\end{equation}
4114
4115The true bound is  $0 \le x < (n - k - 1)^2$ but this has quite a few more terms.  The value of $q$ after step 1 is bounded by the following.
4116
4117\begin{equation}
4118q < n - 2k - k^2/n
4119\end{equation}
4120
4121Since $k^2$ is going to be considerably smaller than $n$ that term will always be zero.  The value of $x$ after step 3 is bounded trivially as
4122$0 \le x < n$.  By step four the sum $x + q$ is bounded by 
4123
4124\begin{equation}
41250 \le q + x < (k + 1)n - 2k^2 - 1
4126\end{equation}
4127
4128With a second pass $q$ will be loosely bounded by $0 \le q < k^2$ after step 2 while $x$ will still be loosely bounded by $0 \le x < n$ after step 3.  After the second pass it is highly unlike that the
4129sum in step 4 will exceed $n - k$.  In practice fewer than three passes of the algorithm are required to reduce virtually every input in the 
4130range $0 \le x < (n - k - 1)^2$.  
4131
4132\begin{figure}
4133\begin{small}
4134\begin{center}
4135\begin{tabular}{|l|}
4136\hline
4137$x = 123456789, n = 256, k = 3$ \\
4138\hline $q \leftarrow \lfloor x/n \rfloor = 482253$ \\
4139$q \leftarrow q*k = 1446759$ \\
4140$x \leftarrow x \mbox{ mod } n = 21$ \\
4141$x \leftarrow x + q = 1446780$ \\
4142$x \leftarrow x - (n - k) = 1446527$ \\
4143\hline 
4144$q \leftarrow \lfloor x/n \rfloor = 5650$ \\
4145$q \leftarrow q*k = 16950$ \\
4146$x \leftarrow x \mbox{ mod } n = 127$ \\
4147$x \leftarrow x + q = 17077$ \\
4148$x \leftarrow x - (n - k) = 16824$ \\
4149\hline 
4150$q \leftarrow \lfloor x/n \rfloor = 65$ \\
4151$q \leftarrow q*k = 195$ \\
4152$x \leftarrow x \mbox{ mod } n = 184$ \\
4153$x \leftarrow x + q = 379$ \\
4154$x \leftarrow x - (n - k) = 126$ \\
4155\hline
4156\end{tabular}
4157\end{center}
4158\end{small}
4159\caption{Example Diminished Radix Reduction}
4160\label{fig:EXDR}
4161\end{figure}
4162
4163Figure~\ref{fig:EXDR} demonstrates the reduction of $x = 123456789$ modulo $n - k = 253$ when $n = 256$ and $k = 3$.  Note that even while $x$
4164is considerably larger than $(n - k - 1)^2 = 63504$ the algorithm still converges on the modular residue exceedingly fast.  In this case only
4165three passes were required to find the residue $x \equiv 126$.
4166
4167
4168\subsection{Choice of Moduli}
4169On the surface this algorithm looks like a very expensive algorithm.  It requires a couple of subtractions followed by multiplication and other
4170modular reductions.  The usefulness of this algorithm becomes exceedingly clear when an appropriate modulus is chosen.
4171
4172Division in general is a very expensive operation to perform.  The one exception is when the division is by a power of the radix of representation used.  
4173Division by ten for example is simple for pencil and paper mathematics since it amounts to shifting the decimal place to the right.  Similarly division 
4174by two (\textit{or powers of two}) is very simple for binary computers to perform.  It would therefore seem logical to choose $n$ of the form $2^p$ 
4175which would imply that $\lfloor x / n \rfloor$ is a simple shift of $x$ right $p$ bits.  
4176
4177However, there is one operation related to division of power of twos that is even faster than this.  If $n = \beta^p$ then the division may be 
4178performed by moving whole digits to the right $p$ places.  In practice division by $\beta^p$ is much faster than division by $2^p$ for any $p$.  
4179Also with the choice of $n = \beta^p$ reducing $x$ modulo $n$ merely requires zeroing the digits above the $p-1$'th digit of $x$.  
4180
4181Throughout the next section the term ``restricted modulus'' will refer to a modulus of the form $\beta^p - k$ whereas the term ``unrestricted
4182modulus'' will refer to a modulus of the form $2^p - k$.  The word ``restricted'' in this case refers to the fact that it is based on the 
4183$2^p$ logic except $p$ must be a multiple of $lg(\beta)$.  
4184
4185\subsection{Choice of $k$}
4186Now that division and reduction (\textit{step 1 and 3 of figure~\ref{fig:DR}}) have been optimized to simple digit operations the multiplication by $k$
4187in step 2 is the most expensive operation.  Fortunately the choice of $k$ is not terribly limited.  For all intents and purposes it might
4188as well be a single digit.  The smaller the value of $k$ is the faster the algorithm will be.  
4189
4190\subsection{Restricted Diminished Radix Reduction}
4191The restricted Diminished Radix algorithm can quickly reduce an input modulo a modulus of the form $n = \beta^p - k$.  This algorithm can reduce 
4192an input $x$ within the range $0 \le x < n^2$ using only a couple passes of the algorithm demonstrated in figure~\ref{fig:DR}.  The implementation
4193of this algorithm has been optimized to avoid additional overhead associated with a division by $\beta^p$, the multiplication by $k$ or the addition 
4194of $x$ and $q$.  The resulting algorithm is very efficient and can lead to substantial improvements over Barrett and Montgomery reduction when modular 
4195exponentiations are performed.
4196
4197\newpage\begin{figure}[!here]
4198\begin{small}
4199\begin{center}
4200\begin{tabular}{l}
4201\hline Algorithm \textbf{mp\_dr\_reduce}. \\
4202\textbf{Input}.   mp\_int $x$, $n$ and a mp\_digit $k = \beta - n_0$ \\
4203\hspace{11.5mm}($0 \le x < n^2$, $n > 1$, $0 < k < \beta$) \\
4204\textbf{Output}.  $x \mbox{ mod } n$ \\
4205\hline \\
42061.  $m \leftarrow n.used$ \\
42072.  If $x.alloc < 2m$ then grow $x$ to $2m$ digits. \\
42083.  $\mu \leftarrow 0$ \\
42094.  for $i$ from $0$ to $m - 1$ do \\
4210\hspace{3mm}4.1  $\hat r \leftarrow k \cdot x_{m+i} + x_{i} + \mu$ \\
4211\hspace{3mm}4.2  $x_{i} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
4212\hspace{3mm}4.3  $\mu \leftarrow \lfloor \hat r / \beta \rfloor$ \\
42135.  $x_{m} \leftarrow \mu$ \\
42146.  for $i$ from $m + 1$ to $x.used - 1$ do \\
4215\hspace{3mm}6.1  $x_{i} \leftarrow 0$ \\
42167.  Clamp excess digits of $x$. \\
42178.  If $x \ge n$ then \\
4218\hspace{3mm}8.1  $x \leftarrow x - n$ \\
4219\hspace{3mm}8.2  Goto step 3. \\
42209.  Return(\textit{MP\_OKAY}). \\
4221\hline
4222\end{tabular}
4223\end{center}
4224\end{small}
4225\caption{Algorithm mp\_dr\_reduce}
4226\end{figure}
4227
4228\textbf{Algorithm mp\_dr\_reduce.}
4229This algorithm will perform the Dimished Radix reduction of $x$ modulo $n$.  It has similar restrictions to that of the Barrett reduction
4230with the addition that $n$ must be of the form $n = \beta^m - k$ where $0 < k <\beta$.  
4231
4232This algorithm essentially implements the pseudo-code in figure~\ref{fig:DR} except with a slight optimization.  The division by $\beta^m$, multiplication by $k$
4233and addition of $x \mbox{ mod }\beta^m$ are all performed simultaneously inside the loop on step 4.  The division by $\beta^m$ is emulated by accessing
4234the term at the $m+i$'th position which is subsequently multiplied by $k$ and added to the term at the $i$'th position.  After the loop the $m$'th
4235digit is set to the carry and the upper digits are zeroed.  Steps 5 and 6 emulate the reduction modulo $\beta^m$ that should have happend to 
4236$x$ before the addition of the multiple of the upper half.  
4237
4238At step 8 if $x$ is still larger than $n$ another pass of the algorithm is required.  First $n$ is subtracted from $x$ and then the algorithm resumes
4239at step 3.  
4240
4241EXAM,bn_mp_dr_reduce.c
4242
4243The first step is to grow $x$ as required to $2m$ digits since the reduction is performed in place on $x$.  The label on line @49,top:@ is where
4244the algorithm will resume if further reduction passes are required.  In theory it could be placed at the top of the function however, the size of
4245the modulus and question of whether $x$ is large enough are invariant after the first pass meaning that it would be a waste of time.  
4246
4247The aliases $tmpx1$ and $tmpx2$ refer to the digits of $x$ where the latter is offset by $m$ digits.  By reading digits from $x$ offset by $m$ digits
4248a division by $\beta^m$ can be simulated virtually for free.  The loop on line @61,for@ performs the bulk of the work (\textit{corresponds to step 4 of algorithm 7.11})
4249in this algorithm.
4250
4251By line @68,mu@ the pointer $tmpx1$ points to the $m$'th digit of $x$ which is where the final carry will be placed.  Similarly by line @71,for@ the 
4252same pointer will point to the $m+1$'th digit where the zeroes will be placed.  
4253
4254Since the algorithm is only valid if both $x$ and $n$ are greater than zero an unsigned comparison suffices to determine if another pass is required.  
4255With the same logic at line @82,sub@ the value of $x$ is known to be greater than or equal to $n$ meaning that an unsigned subtraction can be used
4256as well.  Since the destination of the subtraction is the larger of the inputs the call to algorithm s\_mp\_sub cannot fail and the return code
4257does not need to be checked.
4258
4259\subsubsection{Setup}
4260To setup the restricted Diminished Radix algorithm the value $k = \beta - n_0$ is required.  This algorithm is not really complicated but provided for
4261completeness.
4262
4263\begin{figure}[!here]
4264\begin{small}
4265\begin{center}
4266\begin{tabular}{l}
4267\hline Algorithm \textbf{mp\_dr\_setup}. \\
4268\textbf{Input}.   mp\_int $n$ \\
4269\textbf{Output}.  $k = \beta - n_0$ \\
4270\hline \\
42711.  $k \leftarrow \beta - n_0$ \\
4272\hline
4273\end{tabular}
4274\end{center}
4275\end{small}
4276\caption{Algorithm mp\_dr\_setup}
4277\end{figure}
4278
4279EXAM,bn_mp_dr_setup.c
4280
4281\subsubsection{Modulus Detection}
4282Another algorithm which will be useful is the ability to detect a restricted Diminished Radix modulus.  An integer is said to be
4283of restricted Diminished Radix form if all of the digits are equal to $\beta - 1$ except the trailing digit which may be any value.
4284
4285\begin{figure}[!here]
4286\begin{small}
4287\begin{center}
4288\begin{tabular}{l}
4289\hline Algorithm \textbf{mp\_dr\_is\_modulus}. \\
4290\textbf{Input}.   mp\_int $n$ \\
4291\textbf{Output}.  $1$ if $n$ is in D.R form, $0$ otherwise \\
4292\hline
42931.  If $n.used < 2$ then return($0$). \\
42942.  for $ix$ from $1$ to $n.used - 1$ do \\
4295\hspace{3mm}2.1  If $n_{ix} \ne \beta - 1$ return($0$). \\
42963.  Return($1$). \\
4297\hline
4298\end{tabular}
4299\end{center}
4300\end{small}
4301\caption{Algorithm mp\_dr\_is\_modulus}
4302\end{figure}
4303
4304\textbf{Algorithm mp\_dr\_is\_modulus.}
4305This algorithm determines if a value is in Diminished Radix form.  Step 1 rejects obvious cases where fewer than two digits are
4306in the mp\_int.  Step 2 tests all but the first digit to see if they are equal to $\beta - 1$.  If the algorithm manages to get to
4307step 3 then $n$ must be of Diminished Radix form.
4308
4309EXAM,bn_mp_dr_is_modulus.c
4310
4311\subsection{Unrestricted Diminished Radix Reduction}
4312The unrestricted Diminished Radix algorithm allows modular reductions to be performed when the modulus is of the form $2^p - k$.  This algorithm
4313is a straightforward adaptation of algorithm~\ref{fig:DR}.
4314
4315In general the restricted Diminished Radix reduction algorithm is much faster since it has considerably lower overhead.  However, this new
4316algorithm is much faster than either Montgomery or Barrett reduction when the moduli are of the appropriate form.
4317
4318\begin{figure}[!here]
4319\begin{small}
4320\begin{center}
4321\begin{tabular}{l}
4322\hline Algorithm \textbf{mp\_reduce\_2k}. \\
4323\textbf{Input}.   mp\_int $a$ and $n$.  mp\_digit $k$  \\
4324\hspace{11.5mm}($a \ge 0$, $n > 1$, $0 < k < \beta$, $n + k$ is a power of two) \\
4325\textbf{Output}.  $a \mbox{ (mod }n\mbox{)}$ \\
4326\hline
43271.  $p \leftarrow \lceil lg(n) \rceil$  (\textit{mp\_count\_bits}) \\
43282.  While $a \ge n$ do \\
4329\hspace{3mm}2.1  $q \leftarrow \lfloor a / 2^p \rfloor$ (\textit{mp\_div\_2d}) \\
4330\hspace{3mm}2.2  $a \leftarrow a \mbox{ (mod }2^p\mbox{)}$ (\textit{mp\_mod\_2d}) \\
4331\hspace{3mm}2.3  $q \leftarrow q \cdot k$ (\textit{mp\_mul\_d}) \\
4332\hspace{3mm}2.4  $a \leftarrow a - q$ (\textit{s\_mp\_sub}) \\
4333\hspace{3mm}2.5  If $a \ge n$ then do \\
4334\hspace{6mm}2.5.1  $a \leftarrow a - n$ \\
43353.  Return(\textit{MP\_OKAY}). \\
4336\hline
4337\end{tabular}
4338\end{center}
4339\end{small}
4340\caption{Algorithm mp\_reduce\_2k}
4341\end{figure}
4342
4343\textbf{Algorithm mp\_reduce\_2k.}
4344This algorithm quickly reduces an input $a$ modulo an unrestricted Diminished Radix modulus $n$.  Division by $2^p$ is emulated with a right
4345shift which makes the algorithm fairly inexpensive to use.  
4346
4347EXAM,bn_mp_reduce_2k.c
4348
4349The algorithm mp\_count\_bits calculates the number of bits in an mp\_int which is used to find the initial value of $p$.  The call to mp\_div\_2d
4350on line @31,mp_div_2d@ calculates both the quotient $q$ and the remainder $a$ required.  By doing both in a single function call the code size
4351is kept fairly small.  The multiplication by $k$ is only performed if $k > 1$. This allows reductions modulo $2^p - 1$ to be performed without
4352any multiplications.  
4353
4354The unsigned s\_mp\_add, mp\_cmp\_mag and s\_mp\_sub are used in place of their full sign counterparts since the inputs are only valid if they are 
4355positive.  By using the unsigned versions the overhead is kept to a minimum.  
4356
4357\subsubsection{Unrestricted Setup}
4358To setup this reduction algorithm the value of $k = 2^p - n$ is required.  
4359
4360\begin{figure}[!here]
4361\begin{small}
4362\begin{center}
4363\begin{tabular}{l}
4364\hline Algorithm \textbf{mp\_reduce\_2k\_setup}. \\
4365\textbf{Input}.   mp\_int $n$   \\
4366\textbf{Output}.  $k = 2^p - n$ \\
4367\hline
43681.  $p \leftarrow \lceil lg(n) \rceil$  (\textit{mp\_count\_bits}) \\
43692.  $x \leftarrow 2^p$ (\textit{mp\_2expt}) \\
43703.  $x \leftarrow x - n$ (\textit{mp\_sub}) \\
43714.  $k \leftarrow x_0$ \\
43725.  Return(\textit{MP\_OKAY}). \\
4373\hline
4374\end{tabular}
4375\end{center}
4376\end{small}
4377\caption{Algorithm mp\_reduce\_2k\_setup}
4378\end{figure}
4379
4380\textbf{Algorithm mp\_reduce\_2k\_setup.}
4381This algorithm computes the value of $k$ required for the algorithm mp\_reduce\_2k.  By making a temporary variable $x$ equal to $2^p$ a subtraction
4382is sufficient to solve for $k$.  Alternatively if $n$ has more than one digit the value of $k$ is simply $\beta - n_0$.  
4383
4384EXAM,bn_mp_reduce_2k_setup.c
4385
4386\subsubsection{Unrestricted Detection}
4387An integer $n$ is a valid unrestricted Diminished Radix modulus if either of the following are true.
4388
4389\begin{enumerate}
4390\item  The number has only one digit.
4391\item  The number has more than one digit and every bit from the $\beta$'th to the most significant is one.
4392\end{enumerate}
4393
4394If either condition is true than there is a power of two $2^p$ such that $0 < 2^p - n < \beta$.   If the input is only
4395one digit than it will always be of the correct form.  Otherwise all of the bits above the first digit must be one.  This arises from the fact
4396that there will be value of $k$ that when added to the modulus causes a carry in the first digit which propagates all the way to the most
4397significant bit.  The resulting sum will be a power of two.
4398
4399\begin{figure}[!here]
4400\begin{small}
4401\begin{center}
4402\begin{tabular}{l}
4403\hline Algorithm \textbf{mp\_reduce\_is\_2k}. \\
4404\textbf{Input}.   mp\_int $n$   \\
4405\textbf{Output}.  $1$ if of proper form, $0$ otherwise \\
4406\hline
44071.  If $n.used = 0$ then return($0$). \\
44082.  If $n.used = 1$ then return($1$). \\
44093.  $p \leftarrow \lceil lg(n) \rceil$  (\textit{mp\_count\_bits}) \\
44104.  for $x$ from $lg(\beta)$ to $p$ do \\
4411\hspace{3mm}4.1  If the ($x \mbox{ mod }lg(\beta)$)'th bit of the $\lfloor x / lg(\beta) \rfloor$ of $n$ is zero then return($0$). \\
44125.  Return($1$). \\
4413\hline
4414\end{tabular}
4415\end{center}
4416\end{small}
4417\caption{Algorithm mp\_reduce\_is\_2k}
4418\end{figure}
4419
4420\textbf{Algorithm mp\_reduce\_is\_2k.}
4421This algorithm quickly determines if a modulus is of the form required for algorithm mp\_reduce\_2k to function properly.  
4422
4423EXAM,bn_mp_reduce_is_2k.c
4424
4425
4426
4427\section{Algorithm Comparison}
4428So far three very different algorithms for modular reduction have been discussed.  Each of the algorithms have their own strengths and weaknesses
4429that makes having such a selection very useful.  The following table sumarizes the three algorithms along with comparisons of work factors.  Since
4430all three algorithms have the restriction that $0 \le x < n^2$ and $n > 1$ those limitations are not included in the table.  
4431
4432\begin{center}
4433\begin{small}
4434\begin{tabular}{|c|c|c|c|c|c|}
4435\hline \textbf{Method} & \textbf{Work Required} & \textbf{Limitations} & \textbf{$m = 8$} & \textbf{$m = 32$} & \textbf{$m = 64$} \\
4436\hline Barrett    & $m^2 + 2m - 1$ & None              & $79$ & $1087$ & $4223$ \\
4437\hline Montgomery & $m^2 + m$      & $n$ must be odd   & $72$ & $1056$ & $4160$ \\
4438\hline D.R.       & $2m$           & $n = \beta^m - k$ & $16$ & $64$   & $128$  \\
4439\hline
4440\end{tabular}
4441\end{small}
4442\end{center}
4443
4444In theory Montgomery and Barrett reductions would require roughly the same amount of time to complete.  However, in practice since Montgomery
4445reduction can be written as a single function with the Comba technique it is much faster.  Barrett reduction suffers from the overhead of
4446calling the half precision multipliers, addition and division by $\beta$ algorithms.
4447
4448For almost every cryptographic algorithm Montgomery reduction is the algorithm of choice.  The one set of algorithms where Diminished Radix reduction truly
4449shines are based on the discrete logarithm problem such as Diffie-Hellman \cite{DH} and ElGamal \cite{ELGAMAL}.  In these algorithms
4450primes of the form $\beta^m - k$ can be found and shared amongst users.  These primes will allow the Diminished Radix algorithm to be used in
4451modular exponentiation to greatly speed up the operation.
4452
4453
4454
4455\section*{Exercises}
4456\begin{tabular}{cl}
4457$\left [ 3 \right ]$ & Prove that the ``trick'' in algorithm mp\_montgomery\_setup actually \\
4458                     & calculates the correct value of $\rho$. \\
4459                     & \\
4460$\left [ 2 \right ]$ & Devise an algorithm to reduce modulo $n + k$ for small $k$ quickly.  \\
4461                     & \\
4462$\left [ 4 \right ]$ & Prove that the pseudo-code algorithm ``Diminished Radix Reduction'' \\
4463                     & (\textit{figure~\ref{fig:DR}}) terminates.  Also prove the probability that it will \\
4464                     & terminate within $1 \le k \le 10$ iterations. \\
4465                     & \\
4466\end{tabular}                     
4467
4468
4469\chapter{Exponentiation}
4470Exponentiation is the operation of raising one variable to the power of another, for example, $a^b$.  A variant of exponentiation, computed
4471in a finite field or ring, is called modular exponentiation.  This latter style of operation is typically used in public key 
4472cryptosystems such as RSA and Diffie-Hellman.  The ability to quickly compute modular exponentiations is of great benefit to any
4473such cryptosystem and many methods have been sought to speed it up.
4474
4475\section{Exponentiation Basics}
4476A trivial algorithm would simply multiply $a$ against itself $b - 1$ times to compute the exponentiation desired.  However, as $b$ grows in size
4477the number of multiplications becomes prohibitive.  Imagine what would happen if $b$ $\approx$ $2^{1024}$ as is the case when computing an RSA signature
4478with a $1024$-bit key.  Such a calculation could never be completed as it would take simply far too long.
4479
4480Fortunately there is a very simple algorithm based on the laws of exponents.  Recall that $lg_a(a^b) = b$ and that $lg_a(a^ba^c) = b + c$ which
4481are two trivial relationships between the base and the exponent.  Let $b_i$ represent the $i$'th bit of $b$ starting from the least 
4482significant bit.  If $b$ is a $k$-bit integer than the following equation is true.
4483
4484\begin{equation}
4485a^b = \prod_{i=0}^{k-1} a^{2^i \cdot b_i}
4486\end{equation}
4487
4488By taking the base $a$ logarithm of both sides of the equation the following equation is the result.
4489
4490\begin{equation}
4491b = \sum_{i=0}^{k-1}2^i \cdot b_i
4492\end{equation}
4493
4494The term $a^{2^i}$ can be found from the $i - 1$'th term by squaring the term since $\left ( a^{2^i} \right )^2$ is equal to
4495$a^{2^{i+1}}$.  This observation forms the basis of essentially all fast exponentiation algorithms.  It requires $k$ squarings and on average
4496$k \over 2$ multiplications to compute the result.  This is indeed quite an improvement over simply multiplying by $a$ a total of $b-1$ times.
4497
4498While this current method is a considerable speed up there are further improvements to be made.  For example, the $a^{2^i}$ term does not need to 
4499be computed in an auxilary variable.  Consider the following equivalent algorithm.
4500
4501\begin{figure}[!here]
4502\begin{small}
4503\begin{center}
4504\begin{tabular}{l}
4505\hline Algorithm \textbf{Left to Right Exponentiation}. \\
4506\textbf{Input}.   Integer $a$, $b$ and $k$ \\
4507\textbf{Output}.  $c = a^b$ \\
4508\hline \\
45091.  $c \leftarrow 1$ \\
45102.  for $i$ from $k - 1$ to $0$ do \\
4511\hspace{3mm}2.1  $c \leftarrow c^2$ \\
4512\hspace{3mm}2.2  $c \leftarrow c \cdot a^{b_i}$ \\
45133.  Return $c$. \\
4514\hline
4515\end{tabular}
4516\end{center}
4517\end{small}
4518\caption{Left to Right Exponentiation}
4519\label{fig:LTOR}
4520\end{figure}
4521
4522This algorithm starts from the most significant bit and works towards the least significant bit.  When the $i$'th bit of $b$ is set $a$ is
4523multiplied against the current product.  In each iteration the product is squared which doubles the exponent of the individual terms of the
4524product.  
4525
4526For example, let $b = 101100_2 \equiv 44_{10}$.  The following chart demonstrates the actions of the algorithm.
4527
4528\newpage\begin{figure}
4529\begin{center}
4530\begin{tabular}{|c|c|}
4531\hline \textbf{Value of $i$} & \textbf{Value of $c$} \\
4532\hline - & $1$ \\
4533\hline $5$ & $a$ \\
4534\hline $4$ & $a^2$ \\
4535\hline $3$ & $a^4 \cdot a$ \\
4536\hline $2$ & $a^8 \cdot a^2 \cdot a$ \\
4537\hline $1$ & $a^{16} \cdot a^4 \cdot a^2$ \\
4538\hline $0$ & $a^{32} \cdot a^8 \cdot a^4$ \\
4539\hline
4540\end{tabular}
4541\end{center}
4542\caption{Example of Left to Right Exponentiation}
4543\end{figure}
4544
4545When the product $a^{32} \cdot a^8 \cdot a^4$ is simplified it is equal $a^{44}$ which is the desired exponentiation.  This particular algorithm is 
4546called ``Left to Right'' because it reads the exponent in that order.  All of the exponentiation algorithms that will be presented are of this nature.  
4547
4548\subsection{Single Digit Exponentiation}
4549The first algorithm in the series of exponentiation algorithms will be an unbounded algorithm where the exponent is a single digit.  It is intended 
4550to be used when a small power of an input is required (\textit{e.g. $a^5$}).  It is faster than simply multiplying $b - 1$ times for all values of 
4551$b$ that are greater than three.  
4552
4553\newpage\begin{figure}[!here]
4554\begin{small}
4555\begin{center}
4556\begin{tabular}{l}
4557\hline Algorithm \textbf{mp\_expt\_d}. \\
4558\textbf{Input}.   mp\_int $a$ and mp\_digit $b$ \\
4559\textbf{Output}.  $c = a^b$ \\
4560\hline \\
45611.  $g \leftarrow a$ (\textit{mp\_init\_copy}) \\
45622.  $c \leftarrow 1$ (\textit{mp\_set}) \\
45633.  for $x$ from 1 to $lg(\beta)$ do \\
4564\hspace{3mm}3.1  $c \leftarrow c^2$ (\textit{mp\_sqr}) \\
4565\hspace{3mm}3.2  If $b$ AND $2^{lg(\beta) - 1} \ne 0$ then \\
4566\hspace{6mm}3.2.1  $c \leftarrow c \cdot g$ (\textit{mp\_mul}) \\
4567\hspace{3mm}3.3  $b \leftarrow b << 1$ \\
45684.  Clear $g$. \\
45695.  Return(\textit{MP\_OKAY}). \\
4570\hline
4571\end{tabular}
4572\end{center}
4573\end{small}
4574\caption{Algorithm mp\_expt\_d}
4575\end{figure}
4576
4577\textbf{Algorithm mp\_expt\_d.}
4578This algorithm computes the value of $a$ raised to the power of a single digit $b$.  It uses the left to right exponentiation algorithm to
4579quickly compute the exponentiation.  It is loosely based on algorithm 14.79 of HAC \cite[pp. 615]{HAC} with the difference that the 
4580exponent is a fixed width.  
4581
4582A copy of $a$ is made first to allow destination variable $c$ be the same as the source variable $a$.  The result is set to the initial value of 
4583$1$ in the subsequent step.
4584
4585Inside the loop the exponent is read from the most significant bit first down to the least significant bit.  First $c$ is invariably squared
4586on step 3.1.  In the following step if the most significant bit of $b$ is one the copy of $a$ is multiplied against $c$.  The value
4587of $b$ is shifted left one bit to make the next bit down from the most signficant bit the new most significant bit.  In effect each
4588iteration of the loop moves the bits of the exponent $b$ upwards to the most significant location.
4589
4590EXAM,bn_mp_expt_d.c
4591
4592Line @29,mp_set@ sets the initial value of the result to $1$.  Next the loop on line @31,for@ steps through each bit of the exponent starting from
4593the most significant down towards the least significant. The invariant squaring operation placed on line @333,mp_sqr@ is performed first.  After 
4594the squaring the result $c$ is multiplied by the base $g$ if and only if the most significant bit of the exponent is set.  The shift on line
4595@47,<<@ moves all of the bits of the exponent upwards towards the most significant location.  
4596
4597\section{$k$-ary Exponentiation}
4598When calculating an exponentiation the most time consuming bottleneck is the multiplications which are in general a small factor
4599slower than squaring.  Recall from the previous algorithm that $b_{i}$ refers to the $i$'th bit of the exponent $b$.  Suppose instead it referred to
4600the $i$'th $k$-bit digit of the exponent of $b$.  For $k = 1$ the definitions are synonymous and for $k > 1$ algorithm~\ref{fig:KARY}
4601computes the same exponentiation.  A group of $k$ bits from the exponent is called a \textit{window}.  That is it is a small window on only a
4602portion of the entire exponent.  Consider the following modification to the basic left to right exponentiation algorithm.
4603
4604\begin{figure}[!here]
4605\begin{small}
4606\begin{center}
4607\begin{tabular}{l}
4608\hline Algorithm \textbf{$k$-ary Exponentiation}. \\
4609\textbf{Input}.   Integer $a$, $b$, $k$ and $t$ \\
4610\textbf{Output}.  $c = a^b$ \\
4611\hline \\
46121.  $c \leftarrow 1$ \\
46132.  for $i$ from $t - 1$ to $0$ do \\
4614\hspace{3mm}2.1  $c \leftarrow c^{2^k} $ \\
4615\hspace{3mm}2.2  Extract the $i$'th $k$-bit word from $b$ and store it in $g$. \\
4616\hspace{3mm}2.3  $c \leftarrow c \cdot a^g$ \\
46173.  Return $c$. \\
4618\hline
4619\end{tabular}
4620\end{center}
4621\end{small}
4622\caption{$k$-ary Exponentiation}
4623\label{fig:KARY}
4624\end{figure}
4625
4626The squaring on step 2.1 can be calculated by squaring the value $c$ successively $k$ times.  If the values of $a^g$ for $0 < g < 2^k$ have been
4627precomputed this algorithm requires only $t$ multiplications and $tk$ squarings.  The table can be generated with $2^{k - 1} - 1$ squarings and
4628$2^{k - 1} + 1$ multiplications.  This algorithm assumes that the number of bits in the exponent is evenly divisible by $k$.  
4629However, when it is not the remaining $0 < x \le k - 1$ bits can be handled with algorithm~\ref{fig:LTOR}.
4630
4631Suppose $k = 4$ and $t = 100$.  This modified algorithm will require $109$ multiplications and $408$ squarings to compute the exponentiation.  The
4632original algorithm would on average have required $200$ multiplications and $400$ squrings to compute the same value.  The total number of squarings
4633has increased slightly but the number of multiplications has nearly halved.
4634
4635\subsection{Optimal Values of $k$}
4636An optimal value of $k$ will minimize $2^{k} + \lceil n / k \rceil + n - 1$ for a fixed number of bits in the exponent $n$.  The simplest
4637approach is to brute force search amongst the values $k = 2, 3, \ldots, 8$ for the lowest result.  Table~\ref{fig:OPTK} lists optimal values of $k$
4638for various exponent sizes and compares the number of multiplication and squarings required against algorithm~\ref{fig:LTOR}.  
4639
4640\begin{figure}[here]
4641\begin{center}
4642\begin{small}
4643\begin{tabular}{|c|c|c|c|c|c|}
4644\hline \textbf{Exponent (bits)} & \textbf{Optimal $k$} & \textbf{Work at $k$} & \textbf{Work with ~\ref{fig:LTOR}} \\
4645\hline $16$ & $2$ & $27$ & $24$ \\
4646\hline $32$ & $3$ & $49$ & $48$ \\
4647\hline $64$ & $3$ & $92$ & $96$ \\
4648\hline $128$ & $4$ & $175$ & $192$ \\
4649\hline $256$ & $4$ & $335$ & $384$ \\
4650\hline $512$ & $5$ & $645$ & $768$ \\
4651\hline $1024$ & $6$ & $1257$ & $1536$ \\
4652\hline $2048$ & $6$ & $2452$ & $3072$ \\
4653\hline $4096$ & $7$ & $4808$ & $6144$ \\
4654\hline
4655\end{tabular}
4656\end{small}
4657\end{center}
4658\caption{Optimal Values of $k$ for $k$-ary Exponentiation}
4659\label{fig:OPTK}
4660\end{figure}
4661
4662\subsection{Sliding-Window Exponentiation}
4663A simple modification to the previous algorithm is only generate the upper half of the table in the range $2^{k-1} \le g < 2^k$.  Essentially
4664this is a table for all values of $g$ where the most significant bit of $g$ is a one.  However, in order for this to be allowed in the 
4665algorithm values of $g$ in the range $0 \le g < 2^{k-1}$ must be avoided.  
4666
4667Table~\ref{fig:OPTK2} lists optimal values of $k$ for various exponent sizes and compares the work required against algorithm~\ref{fig:KARY}.  
4668
4669\begin{figure}[here]
4670\begin{center}
4671\begin{small}
4672\begin{tabular}{|c|c|c|c|c|c|}
4673\hline \textbf{Exponent (bits)} & \textbf{Optimal $k$} & \textbf{Work at $k$} & \textbf{Work with ~\ref{fig:KARY}} \\
4674\hline $16$ & $3$ & $24$ & $27$ \\
4675\hline $32$ & $3$ & $45$ & $49$ \\
4676\hline $64$ & $4$ & $87$ & $92$ \\
4677\hline $128$ & $4$ & $167$ & $175$ \\
4678\hline $256$ & $5$ & $322$ & $335$ \\
4679\hline $512$ & $6$ & $628$ & $645$ \\
4680\hline $1024$ & $6$ & $1225$ & $1257$ \\
4681\hline $2048$ & $7$ & $2403$ & $2452$ \\
4682\hline $4096$ & $8$ & $4735$ & $4808$ \\
4683\hline
4684\end{tabular}
4685\end{small}
4686\end{center}
4687\caption{Optimal Values of $k$ for Sliding Window Exponentiation}
4688\label{fig:OPTK2}
4689\end{figure}
4690
4691\newpage\begin{figure}[!here]
4692\begin{small}
4693\begin{center}
4694\begin{tabular}{l}
4695\hline Algorithm \textbf{Sliding Window $k$-ary Exponentiation}. \\
4696\textbf{Input}.   Integer $a$, $b$, $k$ and $t$ \\
4697\textbf{Output}.  $c = a^b$ \\
4698\hline \\
46991.  $c \leftarrow 1$ \\
47002.  for $i$ from $t - 1$ to $0$ do \\
4701\hspace{3mm}2.1  If the $i$'th bit of $b$ is a zero then \\
4702\hspace{6mm}2.1.1   $c \leftarrow c^2$ \\
4703\hspace{3mm}2.2  else do \\
4704\hspace{6mm}2.2.1  $c \leftarrow c^{2^k}$ \\
4705\hspace{6mm}2.2.2  Extract the $k$ bits from $(b_{i}b_{i-1}\ldots b_{i-(k-1)})$ and store it in $g$. \\
4706\hspace{6mm}2.2.3  $c \leftarrow c \cdot a^g$ \\
4707\hspace{6mm}2.2.4  $i \leftarrow i - k$ \\
47083.  Return $c$. \\
4709\hline
4710\end{tabular}
4711\end{center}
4712\end{small}
4713\caption{Sliding Window $k$-ary Exponentiation}
4714\end{figure}
4715
4716Similar to the previous algorithm this algorithm must have a special handler when fewer than $k$ bits are left in the exponent.  While this
4717algorithm requires the same number of squarings it can potentially have fewer multiplications.  The pre-computed table $a^g$ is also half
4718the size as the previous table.  
4719
4720Consider the exponent $b = 111101011001000_2 \equiv 31432_{10}$ with $k = 3$ using both algorithms.  The first algorithm will divide the exponent up as 
4721the following five $3$-bit words $b \equiv \left ( 111, 101, 011, 001, 000 \right )_{2}$.  The second algorithm will break the 
4722exponent as $b \equiv \left ( 111, 101, 0, 110, 0, 100, 0 \right )_{2}$.  The single digit $0$ in the second representation are where
4723a single squaring took place instead of a squaring and multiplication.  In total the first method requires $10$ multiplications and $18$ 
4724squarings.  The second method requires $8$ multiplications and $18$ squarings.  
4725
4726In general the sliding window method is never slower than the generic $k$-ary method and often it is slightly faster.  
4727
4728\section{Modular Exponentiation}
4729
4730Modular exponentiation is essentially computing the power of a base within a finite field or ring.  For example, computing 
4731$d \equiv a^b \mbox{ (mod }c\mbox{)}$ is a modular exponentiation.  Instead of first computing $a^b$ and then reducing it 
4732modulo $c$ the intermediate result is reduced modulo $c$ after every squaring or multiplication operation.  
4733
4734This guarantees that any intermediate result is bounded by $0 \le d \le c^2 - 2c + 1$ and can be reduced modulo $c$ quickly using
4735one of the algorithms presented in ~REDUCTION~.  
4736
4737Before the actual modular exponentiation algorithm can be written a wrapper algorithm must be written first.  This algorithm
4738will allow the exponent $b$ to be negative which is computed as $c \equiv \left (1 / a \right )^{\vert b \vert} \mbox{(mod }d\mbox{)}$. The
4739value of $(1/a) \mbox{ mod }c$ is computed using the modular inverse (\textit{see \ref{sec;modinv}}).  If no inverse exists the algorithm
4740terminates with an error.  
4741
4742\begin{figure}[!here]
4743\begin{small}
4744\begin{center}
4745\begin{tabular}{l}
4746\hline Algorithm \textbf{mp\_exptmod}. \\
4747\textbf{Input}.   mp\_int $a$, $b$ and $c$ \\
4748\textbf{Output}.  $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\
4749\hline \\
47501.  If $c.sign = MP\_NEG$ return(\textit{MP\_VAL}). \\
47512.  If $b.sign = MP\_NEG$ then \\
4752\hspace{3mm}2.1  $g' \leftarrow g^{-1} \mbox{ (mod }c\mbox{)}$ \\
4753\hspace{3mm}2.2  $x' \leftarrow \vert x \vert$ \\
4754\hspace{3mm}2.3  Compute $d \equiv g'^{x'} \mbox{ (mod }c\mbox{)}$ via recursion. \\
47553.  if $p$ is odd \textbf{OR} $p$ is a D.R. modulus then \\
4756\hspace{3mm}3.1  Compute $y \equiv g^{x} \mbox{ (mod }p\mbox{)}$ via algorithm mp\_exptmod\_fast. \\
47574.  else \\
4758\hspace{3mm}4.1  Compute $y \equiv g^{x} \mbox{ (mod }p\mbox{)}$ via algorithm s\_mp\_exptmod. \\
4759\hline
4760\end{tabular}
4761\end{center}
4762\end{small}
4763\caption{Algorithm mp\_exptmod}
4764\end{figure}
4765
4766\textbf{Algorithm mp\_exptmod.}
4767The first algorithm which actually performs modular exponentiation is algorithm s\_mp\_exptmod.  It is a sliding window $k$-ary algorithm 
4768which uses Barrett reduction to reduce the product modulo $p$.  The second algorithm mp\_exptmod\_fast performs the same operation 
4769except it uses either Montgomery or Diminished Radix reduction.  The two latter reduction algorithms are clumped in the same exponentiation
4770algorithm since their arguments are essentially the same (\textit{two mp\_ints and one mp\_digit}).  
4771
4772EXAM,bn_mp_exptmod.c
4773
4774In order to keep the algorithms in a known state the first step on line @29,if@ is to reject any negative modulus as input.  If the exponent is
4775negative the algorithm tries to perform a modular exponentiation with the modular inverse of the base $G$.  The temporary variable $tmpG$ is assigned
4776the modular inverse of $G$ and $tmpX$ is assigned the absolute value of $X$.  The algorithm will recuse with these new values with a positive
4777exponent.
4778
4779If the exponent is positive the algorithm resumes the exponentiation.  Line @63,dr_@ determines if the modulus is of the restricted Diminished Radix 
4780form.  If it is not line @65,reduce@ attempts to determine if it is of a unrestricted Diminished Radix form.  The integer $dr$ will take on one
4781of three values.
4782
4783\begin{enumerate}
4784\item $dr = 0$ means that the modulus is not of either restricted or unrestricted Diminished Radix form.
4785\item $dr = 1$ means that the modulus is of restricted Diminished Radix form.
4786\item $dr = 2$ means that the modulus is of unrestricted Diminished Radix form.
4787\end{enumerate}
4788
4789Line @69,if@ determines if the fast modular exponentiation algorithm can be used.  It is allowed if $dr \ne 0$ or if the modulus is odd.  Otherwise,
4790the slower s\_mp\_exptmod algorithm is used which uses Barrett reduction.  
4791
4792\subsection{Barrett Modular Exponentiation}
4793
4794\newpage\begin{figure}[!here]
4795\begin{small}
4796\begin{center}
4797\begin{tabular}{l}
4798\hline Algorithm \textbf{s\_mp\_exptmod}. \\
4799\textbf{Input}.   mp\_int $a$, $b$ and $c$ \\
4800\textbf{Output}.  $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\
4801\hline \\
48021.  $k \leftarrow lg(x)$ \\
48032.  $winsize \leftarrow  \left \lbrace \begin{array}{ll}
4804                              2 &  \mbox{if }k \le 7 \\
4805                              3 &  \mbox{if }7 < k \le 36 \\
4806                              4 &  \mbox{if }36 < k \le 140 \\
4807                              5 &  \mbox{if }140 < k \le 450 \\
4808                              6 &  \mbox{if }450 < k \le 1303 \\
4809                              7 &  \mbox{if }1303 < k \le 3529 \\
4810                              8 &  \mbox{if }3529 < k \\
4811                              \end{array} \right .$ \\
48123.  Initialize $2^{winsize}$ mp\_ints in an array named $M$ and one mp\_int named $\mu$ \\
48134.  Calculate the $\mu$ required for Barrett Reduction (\textit{mp\_reduce\_setup}). \\
48145.  $M_1 \leftarrow g \mbox{ (mod }p\mbox{)}$ \\
4815\\
4816Setup the table of small powers of $g$.  First find $g^{2^{winsize}}$ and then all multiples of it. \\
48176.  $k \leftarrow 2^{winsize - 1}$ \\
48187.  $M_{k} \leftarrow M_1$ \\
48198.  for $ix$ from 0 to $winsize - 2$ do \\
4820\hspace{3mm}8.1  $M_k \leftarrow \left ( M_k \right )^2$ (\textit{mp\_sqr})  \\
4821\hspace{3mm}8.2  $M_k \leftarrow M_k \mbox{ (mod }p\mbox{)}$ (\textit{mp\_reduce}) \\
48229.  for $ix$ from $2^{winsize - 1} + 1$ to $2^{winsize} - 1$ do \\
4823\hspace{3mm}9.1  $M_{ix} \leftarrow M_{ix - 1} \cdot M_{1}$ (\textit{mp\_mul}) \\
4824\hspace{3mm}9.2  $M_{ix} \leftarrow M_{ix} \mbox{ (mod }p\mbox{)}$ (\textit{mp\_reduce}) \\
482510.  $res \leftarrow 1$ \\
4826\\
4827Start Sliding Window. \\
482811.  $mode \leftarrow 0, bitcnt \leftarrow 1, buf \leftarrow 0, digidx \leftarrow x.used - 1, bitcpy \leftarrow 0, bitbuf \leftarrow 0$ \\
482912.  Loop \\
4830\hspace{3mm}12.1  $bitcnt \leftarrow bitcnt - 1$ \\
4831\hspace{3mm}12.2  If $bitcnt = 0$ then do \\
4832\hspace{6mm}12.2.1  If $digidx = -1$ goto step 13. \\
4833\hspace{6mm}12.2.2  $buf \leftarrow x_{digidx}$ \\
4834\hspace{6mm}12.2.3  $digidx \leftarrow digidx - 1$ \\
4835\hspace{6mm}12.2.4  $bitcnt \leftarrow lg(\beta)$ \\
4836Continued on next page. \\
4837\hline
4838\end{tabular}
4839\end{center}
4840\end{small}
4841\caption{Algorithm s\_mp\_exptmod}
4842\end{figure}
4843
4844\newpage\begin{figure}[!here]
4845\begin{small}
4846\begin{center}
4847\begin{tabular}{l}
4848\hline Algorithm \textbf{s\_mp\_exptmod} (\textit{continued}). \\
4849\textbf{Input}.   mp\_int $a$, $b$ and $c$ \\
4850\textbf{Output}.  $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\
4851\hline \\
4852\hspace{3mm}12.3  $y \leftarrow (buf >> (lg(\beta) - 1))$ AND $1$ \\
4853\hspace{3mm}12.4  $buf \leftarrow buf << 1$ \\
4854\hspace{3mm}12.5  if $mode = 0$ and $y = 0$ then goto step 12. \\
4855\hspace{3mm}12.6  if $mode = 1$ and $y = 0$ then do \\
4856\hspace{6mm}12.6.1  $res \leftarrow res^2$ \\
4857\hspace{6mm}12.6.2  $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\
4858\hspace{6mm}12.6.3  Goto step 12. \\
4859\hspace{3mm}12.7  $bitcpy \leftarrow bitcpy + 1$ \\
4860\hspace{3mm}12.8  $bitbuf \leftarrow bitbuf + (y << (winsize - bitcpy))$ \\
4861\hspace{3mm}12.9  $mode \leftarrow 2$ \\
4862\hspace{3mm}12.10  If $bitcpy = winsize$ then do \\
4863\hspace{6mm}Window is full so perform the squarings and single multiplication. \\
4864\hspace{6mm}12.10.1  for $ix$ from $0$ to $winsize -1$ do \\
4865\hspace{9mm}12.10.1.1  $res \leftarrow res^2$ \\
4866\hspace{9mm}12.10.1.2  $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\
4867\hspace{6mm}12.10.2  $res \leftarrow res \cdot M_{bitbuf}$ \\
4868\hspace{6mm}12.10.3  $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\
4869\hspace{6mm}Reset the window. \\
4870\hspace{6mm}12.10.4  $bitcpy \leftarrow 0, bitbuf \leftarrow 0, mode \leftarrow 1$ \\
4871\\
4872No more windows left.  Check for residual bits of exponent. \\
487313.  If $mode = 2$ and $bitcpy > 0$ then do \\
4874\hspace{3mm}13.1  for $ix$ form $0$ to $bitcpy - 1$ do \\
4875\hspace{6mm}13.1.1  $res \leftarrow res^2$ \\
4876\hspace{6mm}13.1.2  $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\
4877\hspace{6mm}13.1.3  $bitbuf \leftarrow bitbuf << 1$ \\
4878\hspace{6mm}13.1.4  If $bitbuf$ AND $2^{winsize} \ne 0$ then do \\
4879\hspace{9mm}13.1.4.1  $res \leftarrow res \cdot M_{1}$ \\
4880\hspace{9mm}13.1.4.2  $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\
488114.  $y \leftarrow res$ \\
488215.  Clear $res$, $mu$ and the $M$ array. \\
488316.  Return(\textit{MP\_OKAY}). \\
4884\hline
4885\end{tabular}
4886\end{center}
4887\end{small}
4888\caption{Algorithm s\_mp\_exptmod (continued)}
4889\end{figure}
4890
4891\textbf{Algorithm s\_mp\_exptmod.}
4892This algorithm computes the $x$'th power of $g$ modulo $p$ and stores the result in $y$.  It takes advantage of the Barrett reduction
4893algorithm to keep the product small throughout the algorithm.
4894
4895The first two steps determine the optimal window size based on the number of bits in the exponent.  The larger the exponent the 
4896larger the window size becomes.  After a window size $winsize$ has been chosen an array of $2^{winsize}$ mp\_int variables is allocated.  This
4897table will hold the values of $g^x \mbox{ (mod }p\mbox{)}$ for $2^{winsize - 1} \le x < 2^{winsize}$.  
4898
4899After the table is allocated the first power of $g$ is found.  Since $g \ge p$ is allowed it must be first reduced modulo $p$ to make
4900the rest of the algorithm more efficient.  The first element of the table at $2^{winsize - 1}$ is found by squaring $M_1$ successively $winsize - 2$
4901times.  The rest of the table elements are found by multiplying the previous element by $M_1$ modulo $p$.
4902
4903Now that the table is available the sliding window may begin.  The following list describes the functions of all the variables in the window.
4904\begin{enumerate}
4905\item The variable $mode$ dictates how the bits of the exponent are interpreted.  
4906\begin{enumerate}
4907   \item When $mode = 0$ the bits are ignored since no non-zero bit of the exponent has been seen yet.  For example, if the exponent were simply 
4908         $1$ then there would be $lg(\beta) - 1$ zero bits before the first non-zero bit.  In this case bits are ignored until a non-zero bit is found.  
4909   \item When $mode = 1$ a non-zero bit has been seen before and a new $winsize$-bit window has not been formed yet.  In this mode leading $0$ bits 
4910         are read and a single squaring is performed.  If a non-zero bit is read a new window is created.  
4911   \item When $mode = 2$ the algorithm is in the middle of forming a window and new bits are appended to the window from the most significant bit
4912         downwards.
4913\end{enumerate}
4914\item The variable $bitcnt$ indicates how many bits are left in the current digit of the exponent left to be read.  When it reaches zero a new digit
4915      is fetched from the exponent.
4916\item The variable $buf$ holds the currently read digit of the exponent. 
4917\item The variable $digidx$ is an index into the exponents digits.  It starts at the leading digit $x.used - 1$ and moves towards the trailing digit.
4918\item The variable $bitcpy$ indicates how many bits are in the currently formed window.  When it reaches $winsize$ the window is flushed and
4919      the appropriate operations performed.
4920\item The variable $bitbuf$ holds the current bits of the window being formed.  
4921\end{enumerate}
4922
4923All of step 12 is the window processing loop.  It will iterate while there are digits available form the exponent to read.  The first step
4924inside this loop is to extract a new digit if no more bits are available in the current digit.  If there are no bits left a new digit is
4925read and if there are no digits left than the loop terminates.  
4926
4927After a digit is made available step 12.3 will extract the most significant bit of the current digit and move all other bits in the digit
4928upwards.  In effect the digit is read from most significant bit to least significant bit and since the digits are read from leading to 
4929trailing edges the entire exponent is read from most significant bit to least significant bit.
4930
4931At step 12.5 if the $mode$ and currently extracted bit $y$ are both zero the bit is ignored and the next bit is read.  This prevents the 
4932algorithm from having to perform trivial squaring and reduction operations before the first non-zero bit is read.  Step 12.6 and 12.7-10 handle
4933the two cases of $mode = 1$ and $mode = 2$ respectively.  
4934
4935FIGU,expt_state,Sliding Window State Diagram
4936
4937By step 13 there are no more digits left in the exponent.  However, there may be partial bits in the window left.  If $mode = 2$ then 
4938a Left-to-Right algorithm is used to process the remaining few bits.  
4939
4940EXAM,bn_s_mp_exptmod.c
4941
4942Lines @31,if@ through @45,}@ determine the optimal window size based on the length of the exponent in bits.  The window divisions are sorted
4943from smallest to greatest so that in each \textbf{if} statement only one condition must be tested.  For example, by the \textbf{if} statement 
4944on line @37,if@ the value of $x$ is already known to be greater than $140$.  
4945
4946The conditional piece of code beginning on line @42,ifdef@ allows the window size to be restricted to five bits.  This logic is used to ensure
4947the table of precomputed powers of $G$ remains relatively small.  
4948
4949The for loop on line @60,for@ initializes the $M$ array while lines @71,mp_init@ and @75,mp_reduce@ through @85,}@ initialize the reduction
4950function that will be used for this modulus.
4951
4952-- More later.
4953
4954\section{Quick Power of Two}
4955Calculating $b = 2^a$ can be performed much quicker than with any of the previous algorithms.  Recall that a logical shift left $m << k$ is
4956equivalent to $m \cdot 2^k$.  By this logic when $m = 1$ a quick power of two can be achieved.
4957
4958\begin{figure}[!here]
4959\begin{small}
4960\begin{center}
4961\begin{tabular}{l}
4962\hline Algorithm \textbf{mp\_2expt}. \\
4963\textbf{Input}.   integer $b$ \\
4964\textbf{Output}.  $a \leftarrow 2^b$ \\
4965\hline \\
49661.  $a \leftarrow 0$ \\
49672.  If $a.alloc < \lfloor b / lg(\beta) \rfloor + 1$ then grow $a$ appropriately. \\
49683.  $a.used \leftarrow \lfloor b / lg(\beta) \rfloor + 1$ \\
49694.  $a_{\lfloor b / lg(\beta) \rfloor} \leftarrow 1 << (b \mbox{ mod } lg(\beta))$ \\
49705.  Return(\textit{MP\_OKAY}). \\
4971\hline
4972\end{tabular}
4973\end{center}
4974\end{small}
4975\caption{Algorithm mp\_2expt}
4976\end{figure}
4977
4978\textbf{Algorithm mp\_2expt.}
4979
4980EXAM,bn_mp_2expt.c
4981
4982\chapter{Higher Level Algorithms}
4983
4984This chapter discusses the various higher level algorithms that are required to complete a well rounded multiple precision integer package.  These
4985routines are less performance oriented than the algorithms of chapters five, six and seven but are no less important.  
4986
4987The first section describes a method of integer division with remainder that is universally well known.  It provides the signed division logic
4988for the package.  The subsequent section discusses a set of algorithms which allow a single digit to be the 2nd operand for a variety of operations.  
4989These algorithms serve mostly to simplify other algorithms where small constants are required.  The last two sections discuss how to manipulate 
4990various representations of integers.  For example, converting from an mp\_int to a string of character.
4991
4992\section{Integer Division with Remainder}
4993\label{sec:division}
4994
4995Integer division aside from modular exponentiation is the most intensive algorithm to compute.  Like addition, subtraction and multiplication
4996the basis of this algorithm is the long-hand division algorithm taught to school children.  Throughout this discussion several common variables
4997will be used.  Let $x$ represent the divisor and $y$ represent the dividend.  Let $q$ represent the integer quotient $\lfloor y / x \rfloor$ and 
4998let $r$ represent the remainder $r = y - x \lfloor y / x \rfloor$.  The following simple algorithm will be used to start the discussion.
4999
5000\newpage\begin{figure}[!here]
5001\begin{small}
5002\begin{center}
5003\begin{tabular}{l}
5004\hline Algorithm \textbf{Radix-$\beta$ Integer Division}. \\
5005\textbf{Input}.   integer $x$ and $y$ \\
5006\textbf{Output}.  $q = \lfloor y/x\rfloor, r = y - xq$ \\
5007\hline \\
50081.  $q \leftarrow 0$ \\
50092.  $n \leftarrow \vert \vert y \vert \vert - \vert \vert x \vert \vert$ \\
50103.  for $t$ from $n$ down to $0$ do \\
5011\hspace{3mm}3.1  Maximize $k$ such that $kx\beta^t$ is less than or equal to $y$ and $(k + 1)x\beta^t$ is greater. \\
5012\hspace{3mm}3.2  $q \leftarrow q + k\beta^t$ \\
5013\hspace{3mm}3.3  $y \leftarrow y - kx\beta^t$ \\
50144.  $r \leftarrow y$ \\
50155.  Return($q, r$) \\
5016\hline
5017\end{tabular}
5018\end{center}
5019\end{small}
5020\caption{Algorithm Radix-$\beta$ Integer Division}
5021\label{fig:raddiv}
5022\end{figure}
5023
5024As children we are taught this very simple algorithm for the case of $\beta = 10$.  Almost instinctively several optimizations are taught for which
5025their reason of existing are never explained.  For this example let $y = 5471$ represent the dividend and $x = 23$ represent the divisor.
5026
5027To find the first digit of the quotient the value of $k$ must be maximized such that $kx\beta^t$ is less than or equal to $y$ and 
5028simultaneously $(k + 1)x\beta^t$ is greater than $y$.  Implicitly $k$ is the maximum value the $t$'th digit of the quotient may have.  The habitual method
5029used to find the maximum is to ``eyeball'' the two numbers, typically only the leading digits and quickly estimate a quotient.  By only using leading
5030digits a much simpler division may be used to form an educated guess at what the value must be.  In this case $k = \lfloor 54/23\rfloor = 2$ quickly 
5031arises as a possible  solution.  Indeed $2x\beta^2 = 4600$ is less than $y = 5471$ and simultaneously $(k + 1)x\beta^2 = 6900$ is larger than $y$.  
5032As a  result $k\beta^2$ is added to the quotient which now equals $q = 200$ and $4600$ is subtracted from $y$ to give a remainder of $y = 841$.
5033
5034Again this process is repeated to produce the quotient digit $k = 3$ which makes the quotient $q = 200 + 3\beta = 230$ and the remainder 
5035$y = 841 - 3x\beta = 181$.  Finally the last iteration of the loop produces $k = 7$ which leads to the quotient $q = 230 + 7 = 237$ and the
5036remainder $y = 181 - 7x = 20$.  The final quotient and remainder found are $q = 237$ and $r = y = 20$ which are indeed correct since 
5037$237 \cdot 23 + 20 = 5471$ is true.  
5038
5039\subsection{Quotient Estimation}
5040\label{sec:divest}
5041As alluded to earlier the quotient digit $k$ can be estimated from only the leading digits of both the divisor and dividend.  When $p$ leading
5042digits are used from both the divisor and dividend to form an estimation the accuracy of the estimation rises as $p$ grows.  Technically
5043speaking the estimation is based on assuming the lower $\vert \vert y \vert \vert - p$ and $\vert \vert x \vert \vert - p$ lower digits of the
5044dividend and divisor are zero.  
5045
5046The value of the estimation may off by a few values in either direction and in general is fairly correct.  A simplification \cite[pp. 271]{TAOCPV2}
5047of the estimation technique is to use $t + 1$ digits of the dividend and $t$ digits of the divisor, in particularly when $t = 1$.  The estimate 
5048using this technique is never too small.  For the following proof let $t = \vert \vert y \vert \vert - 1$ and $s = \vert \vert x \vert \vert - 1$ 
5049represent the most significant digits of the dividend and divisor respectively.
5050
5051\textbf{Proof.}\textit{  The quotient $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ is greater than or equal to 
5052$k = \lfloor y / (x \cdot \beta^{\vert \vert y \vert \vert - \vert \vert x \vert \vert - 1}) \rfloor$. }
5053The first obvious case is when $\hat k = \beta - 1$ in which case the proof is concluded since the real quotient cannot be larger.  For all other 
5054cases $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ and $\hat k x_s \ge y_t\beta + y_{t-1} - x_s + 1$.  The latter portion of the inequalility
5055$-x_s + 1$ arises from the fact that a truncated integer division will give the same quotient for at most $x_s - 1$ values.  Next a series of 
5056inequalities will prove the hypothesis.
5057
5058\begin{equation}
5059y - \hat k x \le y - \hat k x_s\beta^s
5060\end{equation}
5061
5062This is trivially true since $x \ge x_s\beta^s$.  Next we replace $\hat kx_s\beta^s$ by the previous inequality for $\hat kx_s$.  
5063
5064\begin{equation}
5065y - \hat k x \le y_t\beta^t + \ldots + y_0 - (y_t\beta^t + y_{t-1}\beta^{t-1} - x_s\beta^t + \beta^s)
5066\end{equation}
5067
5068By simplifying the previous inequality the following inequality is formed.
5069
5070\begin{equation}
5071y - \hat k x \le y_{t-2}\beta^{t-2} + \ldots + y_0 + x_s\beta^s - \beta^s
5072\end{equation}
5073
5074Subsequently,
5075
5076\begin{equation}
5077y_{t-2}\beta^{t-2} + \ldots +  y_0  + x_s\beta^s - \beta^s < x_s\beta^s \le x
5078\end{equation}
5079
5080Which proves that $y - \hat kx \le x$ and by consequence $\hat k \ge k$ which concludes the proof.  \textbf{QED}
5081
5082
5083\subsection{Normalized Integers}
5084For the purposes of division a normalized input is when the divisors leading digit $x_n$ is greater than or equal to $\beta / 2$.  By multiplying both
5085$x$ and $y$ by $j = \lfloor (\beta / 2) / x_n \rfloor$ the quotient remains unchanged and the remainder is simply $j$ times the original
5086remainder.  The purpose of normalization is to ensure the leading digit of the divisor is sufficiently large such that the estimated quotient will
5087lie in the domain of a single digit.  Consider the maximum dividend $(\beta - 1) \cdot \beta + (\beta - 1)$ and the minimum divisor $\beta / 2$.  
5088
5089\begin{equation} 
5090{{\beta^2 - 1} \over { \beta / 2}} \le 2\beta - {2 \over \beta} 
5091\end{equation}
5092
5093At most the quotient approaches $2\beta$, however, in practice this will not occur since that would imply the previous quotient digit was too small.  
5094
5095\subsection{Radix-$\beta$ Division with Remainder}
5096\newpage\begin{figure}[!here]
5097\begin{small}
5098\begin{center}
5099\begin{tabular}{l}
5100\hline Algorithm \textbf{mp\_div}. \\
5101\textbf{Input}.   mp\_int $a, b$ \\
5102\textbf{Output}.  $c = \lfloor a/b \rfloor$, $d = a - bc$ \\
5103\hline \\
51041.  If $b = 0$ return(\textit{MP\_VAL}). \\
51052.  If $\vert a \vert < \vert b \vert$ then do \\
5106\hspace{3mm}2.1  $d \leftarrow a$ \\
5107\hspace{3mm}2.2  $c \leftarrow 0$ \\
5108\hspace{3mm}2.3  Return(\textit{MP\_OKAY}). \\
5109\\
5110Setup the quotient to receive the digits. \\
51113.  Grow $q$ to $a.used + 2$ digits. \\
51124.  $q \leftarrow 0$ \\
51135.  $x \leftarrow \vert a \vert , y \leftarrow \vert b \vert$ \\
51146.  $sign \leftarrow  \left \lbrace \begin{array}{ll}
5115                              MP\_ZPOS &  \mbox{if }a.sign = b.sign \\
5116                              MP\_NEG  &  \mbox{otherwise} \\
5117                              \end{array} \right .$ \\
5118\\
5119Normalize the inputs such that the leading digit of $y$ is greater than or equal to $\beta / 2$. \\
51207.  $norm \leftarrow (lg(\beta) - 1) - (\lceil lg(y) \rceil \mbox{ (mod }lg(\beta)\mbox{)})$ \\
51218.  $x \leftarrow x \cdot 2^{norm}, y \leftarrow y \cdot 2^{norm}$ \\
5122\\
5123Find the leading digit of the quotient. \\
51249.  $n \leftarrow x.used - 1, t \leftarrow y.used - 1$ \\
512510.  $y \leftarrow y \cdot \beta^{n - t}$ \\
512611.  While ($x \ge y$) do \\
5127\hspace{3mm}11.1  $q_{n - t} \leftarrow q_{n - t} + 1$ \\
5128\hspace{3mm}11.2  $x \leftarrow x - y$ \\
512912.  $y \leftarrow \lfloor y / \beta^{n-t} \rfloor$ \\
5130\\
5131Continued on the next page. \\
5132\hline
5133\end{tabular}
5134\end{center}
5135\end{small}
5136\caption{Algorithm mp\_div}
5137\end{figure}
5138
5139\newpage\begin{figure}[!here]
5140\begin{small}
5141\begin{center}
5142\begin{tabular}{l}
5143\hline Algorithm \textbf{mp\_div} (continued). \\
5144\textbf{Input}.   mp\_int $a, b$ \\
5145\textbf{Output}.  $c = \lfloor a/b \rfloor$, $d = a - bc$ \\
5146\hline \\
5147Now find the remainder fo the digits. \\
514813.  for $i$ from $n$ down to $(t + 1)$ do \\
5149\hspace{3mm}13.1  If $i > x.used$ then jump to the next iteration of this loop. \\
5150\hspace{3mm}13.2  If $x_{i} = y_{t}$ then \\
5151\hspace{6mm}13.2.1  $q_{i - t - 1} \leftarrow \beta - 1$ \\
5152\hspace{3mm}13.3  else \\
5153\hspace{6mm}13.3.1  $\hat r \leftarrow x_{i} \cdot \beta + x_{i - 1}$ \\
5154\hspace{6mm}13.3.2  $\hat r \leftarrow \lfloor \hat r / y_{t} \rfloor$ \\
5155\hspace{6mm}13.3.3  $q_{i - t - 1} \leftarrow \hat r$ \\
5156\hspace{3mm}13.4  $q_{i - t - 1} \leftarrow q_{i - t - 1} + 1$ \\
5157\\
5158Fixup quotient estimation. \\
5159\hspace{3mm}13.5  Loop \\
5160\hspace{6mm}13.5.1  $q_{i - t - 1} \leftarrow q_{i - t - 1} - 1$ \\
5161\hspace{6mm}13.5.2  t$1 \leftarrow 0$ \\
5162\hspace{6mm}13.5.3  t$1_0 \leftarrow y_{t - 1}, $ t$1_1 \leftarrow y_t,$ t$1.used \leftarrow 2$ \\
5163\hspace{6mm}13.5.4  $t1 \leftarrow t1 \cdot q_{i - t - 1}$ \\
5164\hspace{6mm}13.5.5  t$2_0 \leftarrow x_{i - 2}, $ t$2_1 \leftarrow x_{i - 1}, $ t$2_2 \leftarrow x_i, $ t$2.used \leftarrow 3$ \\
5165\hspace{6mm}13.5.6  If $\vert t1 \vert > \vert t2 \vert$ then goto step 13.5. \\
5166\hspace{3mm}13.6  t$1 \leftarrow y \cdot q_{i - t - 1}$ \\
5167\hspace{3mm}13.7  t$1 \leftarrow $ t$1 \cdot \beta^{i - t - 1}$ \\
5168\hspace{3mm}13.8  $x \leftarrow x - $ t$1$ \\
5169\hspace{3mm}13.9  If $x.sign = MP\_NEG$ then \\
5170\hspace{6mm}13.10  t$1 \leftarrow y$ \\
5171\hspace{6mm}13.11  t$1 \leftarrow $ t$1 \cdot \beta^{i - t - 1}$ \\
5172\hspace{6mm}13.12  $x \leftarrow x + $ t$1$ \\
5173\hspace{6mm}13.13  $q_{i - t - 1} \leftarrow q_{i - t - 1} - 1$ \\
5174\\
5175Finalize the result. \\
517614.  Clamp excess digits of $q$ \\
517715.  $c \leftarrow q, c.sign \leftarrow sign$ \\
517816.  $x.sign \leftarrow a.sign$ \\
517917.  $d \leftarrow \lfloor x / 2^{norm} \rfloor$ \\
518018.  Return(\textit{MP\_OKAY}). \\
5181\hline
5182\end{tabular}
5183\end{center}
5184\end{small}
5185\caption{Algorithm mp\_div (continued)}
5186\end{figure}
5187\textbf{Algorithm mp\_div.}
5188This algorithm will calculate quotient and remainder from an integer division given a dividend and divisor.  The algorithm is a signed
5189division and will produce a fully qualified quotient and remainder.
5190
5191First the divisor $b$ must be non-zero which is enforced in step one.  If the divisor is larger than the dividend than the quotient is implicitly 
5192zero and the remainder is the dividend.  
5193
5194After the first two trivial cases of inputs are handled the variable $q$ is setup to receive the digits of the quotient.  Two unsigned copies of the
5195divisor $y$ and dividend $x$ are made as well.  The core of the division algorithm is an unsigned division and will only work if the values are
5196positive.  Now the two values $x$ and $y$ must be normalized such that the leading digit of $y$ is greater than or equal to $\beta / 2$.  
5197This is performed by shifting both to the left by enough bits to get the desired normalization.  
5198
5199At this point the division algorithm can begin producing digits of the quotient.  Recall that maximum value of the estimation used is 
5200$2\beta - {2 \over \beta}$ which means that a digit of the quotient must be first produced by another means.  In this case $y$ is shifted
5201to the left (\textit{step ten}) so that it has the same number of digits as $x$.  The loop on step eleven will subtract multiples of the 
5202shifted copy of $y$ until $x$ is smaller.  Since the leading digit of $y$ is greater than or equal to $\beta/2$ this loop will iterate at most two
5203times to produce the desired leading digit of the quotient.  
5204
5205Now the remainder of the digits can be produced.  The equation $\hat q = \lfloor {{x_i \beta + x_{i-1}}\over y_t} \rfloor$ is used to fairly
5206accurately approximate the true quotient digit.  The estimation can in theory produce an estimation as high as $2\beta - {2 \over \beta}$ but by
5207induction the upper quotient digit is correct (\textit{as established on step eleven}) and the estimate must be less than $\beta$.  
5208
5209Recall from section~\ref{sec:divest} that the estimation is never too low but may be too high.  The next step of the estimation process is
5210to refine the estimation.  The loop on step 13.5 uses $x_i\beta^2 + x_{i-1}\beta + x_{i-2}$ and $q_{i - t - 1}(y_t\beta + y_{t-1})$ as a higher
5211order approximation to adjust the quotient digit.
5212
5213After both phases of estimation the quotient digit may still be off by a value of one\footnote{This is similar to the error introduced
5214by optimizing Barrett reduction.}.  Steps 13.6 and 13.7 subtract the multiple of the divisor from the dividend (\textit{Similar to step 3.3 of
5215algorithm~\ref{fig:raddiv}} and then subsequently add a multiple of the divisor if the quotient was too large.  
5216
5217Now that the quotient has been determine finializing the result is a matter of clamping the quotient, fixing the sizes and de-normalizing the 
5218remainder.  An important aspect of this algorithm seemingly overlooked in other descriptions such as that of Algorithm 14.20 HAC \cite[pp. 598]{HAC}
5219is that when the estimations are being made (\textit{inside the loop on step 13.5}) that the digits $y_{t-1}$, $x_{i-2}$ and $x_{i-1}$ may lie 
5220outside their respective boundaries.  For example, if $t = 0$ or $i \le 1$ then the digits would be undefined.  In those cases the digits should
5221respectively be replaced with a zero.  
5222
5223EXAM,bn_mp_div.c
5224
5225The implementation of this algorithm differs slightly from the pseudo code presented previously.  In this algorithm either of the quotient $c$ or
5226remainder $d$ may be passed as a \textbf{NULL} pointer which indicates their value is not desired.  For example, the C code to call the division
5227algorithm with only the quotient is 
5228
5229\begin{verbatim}
5230mp_div(&a, &b, &c, NULL);  /* c = [a/b] */
5231\end{verbatim}
5232
5233Lines @108,if@ and @113,if@ handle the two trivial cases of inputs which are division by zero and dividend smaller than the divisor 
5234respectively.  After the two trivial cases all of the temporary variables are initialized.  Line @147,neg@ determines the sign of 
5235the quotient and line @148,sign@ ensures that both $x$ and $y$ are positive.  
5236
5237The number of bits in the leading digit is calculated on line @151,norm@.  Implictly an mp\_int with $r$ digits will require $lg(\beta)(r-1) + k$ bits
5238of precision which when reduced modulo $lg(\beta)$ produces the value of $k$.  In this case $k$ is the number of bits in the leading digit which is
5239exactly what is required.  For the algorithm to operate $k$ must equal $lg(\beta) - 1$ and when it does not the inputs must be normalized by shifting
5240them to the left by $lg(\beta) - 1 - k$ bits.
5241
5242Throughout the variables $n$ and $t$ will represent the highest digit of $x$ and $y$ respectively.  These are first used to produce the 
5243leading digit of the quotient.  The loop beginning on line @184,for@ will produce the remainder of the quotient digits.
5244
5245The conditional ``continue'' on line @186,continue@ is used to prevent the algorithm from reading past the leading edge of $x$ which can occur when the
5246algorithm eliminates multiple non-zero digits in a single iteration.  This ensures that $x_i$ is always non-zero since by definition the digits
5247above the $i$'th position $x$ must be zero in order for the quotient to be precise\footnote{Precise as far as integer division is concerned.}.  
5248
5249Lines @214,t1@, @216,t1@ and @222,t2@ through @225,t2@ manually construct the high accuracy estimations by setting the digits of the two mp\_int 
5250variables directly.  
5251
5252\section{Single Digit Helpers}
5253
5254This section briefly describes a series of single digit helper algorithms which come in handy when working with small constants.  All of 
5255the helper functions assume the single digit input is positive and will treat them as such.
5256
5257\subsection{Single Digit Addition and Subtraction}
5258
5259Both addition and subtraction are performed by ``cheating'' and using mp\_set followed by the higher level addition or subtraction 
5260algorithms.   As a result these algorithms are subtantially simpler with a slight cost in performance.
5261
5262\newpage\begin{figure}[!here]
5263\begin{small}
5264\begin{center}
5265\begin{tabular}{l}
5266\hline Algorithm \textbf{mp\_add\_d}. \\
5267\textbf{Input}.   mp\_int $a$ and a mp\_digit $b$ \\
5268\textbf{Output}.  $c = a + b$ \\
5269\hline \\
52701.  $t \leftarrow b$ (\textit{mp\_set}) \\
52712.  $c \leftarrow a + t$ \\
52723.  Return(\textit{MP\_OKAY}) \\
5273\hline
5274\end{tabular}
5275\end{center}
5276\end{small}
5277\caption{Algorithm mp\_add\_d}
5278\end{figure}
5279
5280\textbf{Algorithm mp\_add\_d.}
5281This algorithm initiates a temporary mp\_int with the value of the single digit and uses algorithm mp\_add to add the two values together.
5282
5283EXAM,bn_mp_add_d.c
5284
5285Clever use of the letter 't'.
5286
5287\subsubsection{Subtraction}
5288The single digit subtraction algorithm mp\_sub\_d is essentially the same except it uses mp\_sub to subtract the digit from the mp\_int.
5289
5290\subsection{Single Digit Multiplication}
5291Single digit multiplication arises enough in division and radix conversion that it ought to be implement as a special case of the baseline
5292multiplication algorithm.  Essentially this algorithm is a modified version of algorithm s\_mp\_mul\_digs where one of the multiplicands
5293only has one digit.
5294
5295\begin{figure}[!here]
5296\begin{small}
5297\begin{center}
5298\begin{tabular}{l}
5299\hline Algorithm \textbf{mp\_mul\_d}. \\
5300\textbf{Input}.   mp\_int $a$ and a mp\_digit $b$ \\
5301\textbf{Output}.  $c = ab$ \\
5302\hline \\
53031.  $pa \leftarrow a.used$ \\
53042.  Grow $c$ to at least $pa + 1$ digits. \\
53053.  $oldused \leftarrow c.used$ \\
53064.  $c.used \leftarrow pa + 1$ \\
53075.  $c.sign \leftarrow a.sign$ \\
53086.  $\mu \leftarrow 0$ \\
53097.  for $ix$ from $0$ to $pa - 1$ do \\
5310\hspace{3mm}7.1  $\hat r \leftarrow \mu + a_{ix}b$ \\
5311\hspace{3mm}7.2  $c_{ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\
5312\hspace{3mm}7.3  $\mu \leftarrow \lfloor \hat r / \beta \rfloor$ \\
53138.  $c_{pa} \leftarrow \mu$ \\
53149.  for $ix$ from $pa + 1$ to $oldused$ do \\
5315\hspace{3mm}9.1  $c_{ix} \leftarrow 0$ \\
531610.  Clamp excess digits of $c$. \\
531711.  Return(\textit{MP\_OKAY}). \\
5318\hline
5319\end{tabular}
5320\end{center}
5321\end{small}
5322\caption{Algorithm mp\_mul\_d}
5323\end{figure}
5324\textbf{Algorithm mp\_mul\_d.}
5325This algorithm quickly multiplies an mp\_int by a small single digit value.  It is specially tailored to the job and has a minimal of overhead.  
5326Unlike the full multiplication algorithms this algorithm does not require any significnat temporary storage or memory allocations.  
5327
5328EXAM,bn_mp_mul_d.c
5329
5330In this implementation the destination $c$ may point to the same mp\_int as the source $a$ since the result is written after the digit is 
5331read from the source.  This function uses pointer aliases $tmpa$ and $tmpc$ for the digits of $a$ and $c$ respectively.  
5332
5333\subsection{Single Digit Division}
5334Like the single digit multiplication algorithm, single digit division is also a fairly common algorithm used in radix conversion.  Since the
5335divisor is only a single digit a specialized variant of the division algorithm can be used to compute the quotient.  
5336
5337\newpage\begin{figure}[!here]
5338\begin{small}
5339\begin{center}
5340\begin{tabular}{l}
5341\hline Algorithm \textbf{mp\_div\_d}. \\
5342\textbf{Input}.   mp\_int $a$ and a mp\_digit $b$ \\
5343\textbf{Output}.  $c = \lfloor a / b \rfloor, d = a - cb$ \\
5344\hline \\
53451.  If $b = 0$ then return(\textit{MP\_VAL}).\\
53462.  If $b = 3$ then use algorithm mp\_div\_3 instead. \\
53473.  Init $q$ to $a.used$ digits.  \\
53484.  $q.used \leftarrow a.used$ \\
53495.  $q.sign \leftarrow a.sign$ \\
53506.  $\hat w \leftarrow 0$ \\
53517.  for $ix$ from $a.used - 1$ down to $0$ do \\
5352\hspace{3mm}7.1  $\hat w \leftarrow \hat w \beta + a_{ix}$ \\
5353\hspace{3mm}7.2  If $\hat w \ge b$ then \\
5354\hspace{6mm}7.2.1  $t \leftarrow \lfloor \hat w / b \rfloor$ \\
5355\hspace{6mm}7.2.2  $\hat w \leftarrow \hat w \mbox{ (mod }b\mbox{)}$ \\
5356\hspace{3mm}7.3  else\\
5357\hspace{6mm}7.3.1  $t \leftarrow 0$ \\
5358\hspace{3mm}7.4  $q_{ix} \leftarrow t$ \\
53598.  $d \leftarrow \hat w$ \\
53609.  Clamp excess digits of $q$. \\
536110.  $c \leftarrow q$ \\
536211.  Return(\textit{MP\_OKAY}). \\
5363\hline
5364\end{tabular}
5365\end{center}
5366\end{small}
5367\caption{Algorithm mp\_div\_d}
5368\end{figure}
5369\textbf{Algorithm mp\_div\_d.}
5370This algorithm divides the mp\_int $a$ by the single mp\_digit $b$ using an optimized approach.  Essentially in every iteration of the
5371algorithm another digit of the dividend is reduced and another digit of quotient produced.  Provided $b < \beta$ the value of $\hat w$
5372after step 7.1 will be limited such that $0 \le \lfloor \hat w / b \rfloor < \beta$.  
5373
5374If the divisor $b$ is equal to three a variant of this algorithm is used which is called mp\_div\_3.  It replaces the division by three with
5375a multiplication by $\lfloor \beta / 3 \rfloor$ and the appropriate shift and residual fixup.  In essence it is much like the Barrett reduction
5376from chapter seven.  
5377
5378EXAM,bn_mp_div_d.c
5379
5380Like the implementation of algorithm mp\_div this algorithm allows either of the quotient or remainder to be passed as a \textbf{NULL} pointer to
5381indicate the respective value is not required.  This allows a trivial single digit modular reduction algorithm, mp\_mod\_d to be created.
5382
5383The division and remainder on lines @44,/@ and @45,%@ can be replaced often by a single division on most processors.  For example, the 32-bit x86 based 
5384processors can divide a 64-bit quantity by a 32-bit quantity and produce the quotient and remainder simultaneously.  Unfortunately the GCC 
5385compiler does not recognize that optimization and will actually produce two function calls to find the quotient and remainder respectively.  
5386
5387\subsection{Single Digit Root Extraction}
5388
5389Finding the $n$'th root of an integer is fairly easy as far as numerical analysis is concerned.  Algorithms such as the Newton-Raphson approximation 
5390(\ref{eqn:newton}) series will converge very quickly to a root for any continuous function $f(x)$.  
5391
5392\begin{equation}
5393x_{i+1} = x_i - {f(x_i) \over f'(x_i)}
5394\label{eqn:newton}
5395\end{equation}
5396
5397In this case the $n$'th root is desired and $f(x) = x^n - a$ where $a$ is the integer of which the root is desired.  The derivative of $f(x)$ is 
5398simply $f'(x) = nx^{n - 1}$.  Of particular importance is that this algorithm will be used over the integers not over the a more continuous domain
5399such as the real numbers.  As a result the root found can be above the true root by few and must be manually adjusted.  Ideally at the end of the 
5400algorithm the $n$'th root $b$ of an integer $a$ is desired such that $b^n \le a$.  
5401
5402\newpage\begin{figure}[!here]
5403\begin{small}
5404\begin{center}
5405\begin{tabular}{l}
5406\hline Algorithm \textbf{mp\_n\_root}. \\
5407\textbf{Input}.   mp\_int $a$ and a mp\_digit $b$ \\
5408\textbf{Output}.  $c^b \le a$ \\
5409\hline \\
54101.  If $b$ is even and $a.sign = MP\_NEG$ return(\textit{MP\_VAL}). \\
54112.  $sign \leftarrow a.sign$ \\
54123.  $a.sign \leftarrow MP\_ZPOS$ \\
54134.  t$2 \leftarrow 2$ \\
54145.  Loop \\
5415\hspace{3mm}5.1  t$1 \leftarrow $ t$2$ \\
5416\hspace{3mm}5.2  t$3 \leftarrow $ t$1^{b - 1}$ \\
5417\hspace{3mm}5.3  t$2 \leftarrow $ t$3 $ $\cdot$ t$1$ \\
5418\hspace{3mm}5.4  t$2 \leftarrow $ t$2 - a$ \\
5419\hspace{3mm}5.5  t$3 \leftarrow $ t$3 \cdot b$ \\
5420\hspace{3mm}5.6  t$3 \leftarrow \lfloor $t$2 / $t$3 \rfloor$ \\
5421\hspace{3mm}5.7  t$2 \leftarrow $ t$1 - $ t$3$ \\
5422\hspace{3mm}5.8  If t$1 \ne $ t$2$ then goto step 5.  \\
54236.  Loop \\
5424\hspace{3mm}6.1  t$2 \leftarrow $ t$1^b$ \\
5425\hspace{3mm}6.2  If t$2 > a$ then \\
5426\hspace{6mm}6.2.1  t$1 \leftarrow $ t$1 - 1$ \\
5427\hspace{6mm}6.2.2  Goto step 6. \\
54287.  $a.sign \leftarrow sign$ \\
54298.  $c \leftarrow $ t$1$ \\
54309.  $c.sign \leftarrow sign$  \\
543110.  Return(\textit{MP\_OKAY}).  \\
5432\hline
5433\end{tabular}
5434\end{center}
5435\end{small}
5436\caption{Algorithm mp\_n\_root}
5437\end{figure}
5438\textbf{Algorithm mp\_n\_root.}
5439This algorithm finds the integer $n$'th root of an input using the Newton-Raphson approach.  It is partially optimized based on the observation
5440that the numerator of ${f(x) \over f'(x)}$ can be derived from a partial denominator.  That is at first the denominator is calculated by finding
5441$x^{b - 1}$.  This value can then be multiplied by $x$ and have $a$ subtracted from it to find the numerator.  This saves a total of $b - 1$ 
5442multiplications by t$1$ inside the loop.  
5443
5444The initial value of the approximation is t$2 = 2$ which allows the algorithm to start with very small values and quickly converge on the
5445root.  Ideally this algorithm is meant to find the $n$'th root of an input where $n$ is bounded by $2 \le n \le 5$.  
5446
5447EXAM,bn_mp_n_root.c
5448
5449\section{Random Number Generation}
5450
5451Random numbers come up in a variety of activities from public key cryptography to simple simulations and various randomized algorithms.  Pollard-Rho 
5452factoring for example, can make use of random values as starting points to find factors of a composite integer.  In this case the algorithm presented
5453is solely for simulations and not intended for cryptographic use.  
5454
5455\newpage\begin{figure}[!here]
5456\begin{small}
5457\begin{center}
5458\begin{tabular}{l}
5459\hline Algorithm \textbf{mp\_rand}. \\
5460\textbf{Input}.   An integer $b$ \\
5461\textbf{Output}.  A pseudo-random number of $b$ digits \\
5462\hline \\
54631.  $a \leftarrow 0$ \\
54642.  If $b \le 0$ return(\textit{MP\_OKAY}) \\
54653.  Pick a non-zero random digit $d$. \\
54664.  $a \leftarrow a + d$ \\
54675.  for $ix$ from 1 to $d - 1$ do \\
5468\hspace{3mm}5.1  $a \leftarrow a \cdot \beta$ \\
5469\hspace{3mm}5.2  Pick a random digit $d$. \\
5470\hspace{3mm}5.3  $a \leftarrow a + d$ \\
54716.  Return(\textit{MP\_OKAY}). \\
5472\hline
5473\end{tabular}
5474\end{center}
5475\end{small}
5476\caption{Algorithm mp\_rand}
5477\end{figure}
5478\textbf{Algorithm mp\_rand.}
5479This algorithm produces a pseudo-random integer of $b$ digits.  By ensuring that the first digit is non-zero the algorithm also guarantees that the
5480final result has at least $b$ digits.  It relies heavily on a third-part random number generator which should ideally generate uniformly all of
5481the integers from $0$ to $\beta - 1$.  
5482
5483EXAM,bn_mp_rand.c
5484
5485\section{Formatted Representations}
5486The ability to emit a radix-$n$ textual representation of an integer is useful for interacting with human parties.  For example, the ability to
5487be given a string of characters such as ``114585'' and turn it into the radix-$\beta$ equivalent would make it easier to enter numbers
5488into a program.
5489
5490\subsection{Reading Radix-n Input}
5491For the purposes of this text we will assume that a simple lower ASCII map (\ref{fig:ASC}) is used for the values of from $0$ to $63$ to 
5492printable characters.  For example, when the character ``N'' is read it represents the integer $23$.  The first $16$ characters of the
5493map are for the common representations up to hexadecimal.  After that they match the ``base64'' encoding scheme which are suitable chosen
5494such that they are printable.  While outputting as base64 may not be too helpful for human operators it does allow communication via non binary
5495mediums.
5496
5497\newpage\begin{figure}[here]
5498\begin{center}
5499\begin{tabular}{cc|cc|cc|cc}
5500\hline \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} &  \textbf{Value} & \textbf{Char} \\
5501\hline 
55020 & 0 & 1 & 1 & 2 & 2 & 3 & 3 \\
55034 & 4 & 5 & 5 & 6 & 6 & 7 & 7 \\
55048 & 8 & 9 & 9 & 10 & A & 11 & B \\
550512 & C & 13 & D & 14 & E & 15 & F \\
550616 & G & 17 & H & 18 & I & 19 & J \\
550720 & K & 21 & L & 22 & M & 23 & N \\
550824 & O & 25 & P & 26 & Q & 27 & R \\
550928 & S & 29 & T & 30 & U & 31 & V \\
551032 & W & 33 & X & 34 & Y & 35 & Z \\
551136 & a & 37 & b & 38 & c & 39 & d \\
551240 & e & 41 & f & 42 & g & 43 & h \\
551344 & i & 45 & j & 46 & k & 47 & l \\
551448 & m & 49 & n & 50 & o & 51 & p \\
551552 & q & 53 & r & 54 & s & 55 & t \\
551656 & u & 57 & v & 58 & w & 59 & x \\
551760 & y & 61 & z & 62 & $+$ & 63 & $/$ \\
5518\hline
5519\end{tabular}
5520\end{center}
5521\caption{Lower ASCII Map}
5522\label{fig:ASC}
5523\end{figure}
5524
5525\newpage\begin{figure}[!here]
5526\begin{small}
5527\begin{center}
5528\begin{tabular}{l}
5529\hline Algorithm \textbf{mp\_read\_radix}. \\
5530\textbf{Input}.   A string $str$ of length $sn$ and radix $r$. \\
5531\textbf{Output}.  The radix-$\beta$ equivalent mp\_int. \\
5532\hline \\
55331.  If $r < 2$ or $r > 64$ return(\textit{MP\_VAL}). \\
55342.  $ix \leftarrow 0$ \\
55353.  If $str_0 =$ ``-'' then do \\
5536\hspace{3mm}3.1  $ix \leftarrow ix + 1$ \\
5537\hspace{3mm}3.2  $sign \leftarrow MP\_NEG$ \\
55384.  else \\
5539\hspace{3mm}4.1  $sign \leftarrow MP\_ZPOS$ \\
55405.  $a \leftarrow 0$ \\
55416.  for $iy$ from $ix$ to $sn - 1$ do \\
5542\hspace{3mm}6.1  Let $y$ denote the position in the map of $str_{iy}$. \\
5543\hspace{3mm}6.2  If $str_{iy}$ is not in the map or $y \ge r$ then goto step 7. \\
5544\hspace{3mm}6.3  $a \leftarrow a \cdot r$ \\
5545\hspace{3mm}6.4  $a \leftarrow a + y$ \\
55467.  If $a \ne 0$ then $a.sign \leftarrow sign$ \\
55478.  Return(\textit{MP\_OKAY}). \\
5548\hline
5549\end{tabular}
5550\end{center}
5551\end{small}
5552\caption{Algorithm mp\_read\_radix}
5553\end{figure}
5554\textbf{Algorithm mp\_read\_radix.}
5555This algorithm will read an ASCII string and produce the radix-$\beta$ mp\_int representation of the same integer.  A minus symbol ``-'' may precede the 
5556string  to indicate the value is negative, otherwise it is assumed to be positive.  The algorithm will read up to $sn$ characters from the input
5557and will stop when it reads a character it cannot map the algorithm stops reading characters from the string.  This allows numbers to be embedded
5558as part of larger input without any significant problem.
5559
5560EXAM,bn_mp_read_radix.c
5561
5562\subsection{Generating Radix-$n$ Output}
5563Generating radix-$n$ output is fairly trivial with a division and remainder algorithm.  
5564
5565\newpage\begin{figure}[!here]
5566\begin{small}
5567\begin{center}
5568\begin{tabular}{l}
5569\hline Algorithm \textbf{mp\_toradix}. \\
5570\textbf{Input}.   A mp\_int $a$ and an integer $r$\\
5571\textbf{Output}.  The radix-$r$ representation of $a$ \\
5572\hline \\
55731.  If $r < 2$ or $r > 64$ return(\textit{MP\_VAL}). \\
55742.  If $a = 0$ then $str = $ ``$0$'' and return(\textit{MP\_OKAY}).  \\
55753.  $t \leftarrow a$ \\
55764.  $str \leftarrow$ ``'' \\
55775.  if $t.sign = MP\_NEG$ then \\
5578\hspace{3mm}5.1  $str \leftarrow str + $ ``-'' \\
5579\hspace{3mm}5.2  $t.sign = MP\_ZPOS$ \\
55806.  While ($t \ne 0$) do \\
5581\hspace{3mm}6.1  $d \leftarrow t \mbox{ (mod }r\mbox{)}$ \\
5582\hspace{3mm}6.2  $t \leftarrow \lfloor t / r \rfloor$ \\
5583\hspace{3mm}6.3  Look up $d$ in the map and store the equivalent character in $y$. \\
5584\hspace{3mm}6.4  $str \leftarrow str + y$ \\
55857.  If $str_0 = $``$-$'' then \\
5586\hspace{3mm}7.1  Reverse the digits $str_1, str_2, \ldots str_n$. \\
55878.  Otherwise \\
5588\hspace{3mm}8.1  Reverse the digits $str_0, str_1, \ldots str_n$. \\
55899.  Return(\textit{MP\_OKAY}).\\
5590\hline
5591\end{tabular}
5592\end{center}
5593\end{small}
5594\caption{Algorithm mp\_toradix}
5595\end{figure}
5596\textbf{Algorithm mp\_toradix.}
5597This algorithm computes the radix-$r$ representation of an mp\_int $a$.  The ``digits'' of the representation are extracted by reducing 
5598successive powers of $\lfloor a / r^k \rfloor$ the input modulo $r$ until $r^k > a$.  Note that instead of actually dividing by $r^k$ in
5599each iteration the quotient $\lfloor a / r \rfloor$ is saved for the next iteration.  As a result a series of trivial $n \times 1$ divisions
5600are required instead of a series of $n \times k$ divisions.  One design flaw of this approach is that the digits are produced in the reverse order 
5601(see~\ref{fig:mpradix}).  To remedy this flaw the digits must be swapped or simply ``reversed''.
5602
5603\begin{figure}
5604\begin{center}
5605\begin{tabular}{|c|c|c|}
5606\hline \textbf{Value of $a$} & \textbf{Value of $d$} & \textbf{Value of $str$} \\
5607\hline $1234$ & -- & -- \\
5608\hline $123$  & $4$ & ``4'' \\
5609\hline $12$   & $3$ & ``43'' \\
5610\hline $1$    & $2$ & ``432'' \\
5611\hline $0$    & $1$ & ``4321'' \\
5612\hline
5613\end{tabular}
5614\end{center}
5615\caption{Example of Algorithm mp\_toradix.}
5616\label{fig:mpradix}
5617\end{figure}
5618
5619EXAM,bn_mp_toradix.c
5620
5621\chapter{Number Theoretic Algorithms}
5622This chapter discusses several fundamental number theoretic algorithms such as the greatest common divisor, least common multiple and Jacobi 
5623symbol computation.  These algorithms arise as essential components in several key cryptographic algorithms such as the RSA public key algorithm and
5624various Sieve based factoring algorithms.
5625
5626\section{Greatest Common Divisor}
5627The greatest common divisor of two integers $a$ and $b$, often denoted as $(a, b)$ is the largest integer $k$ that is a proper divisor of
5628both $a$ and $b$.  That is, $k$ is the largest integer such that $0 \equiv a \mbox{ (mod }k\mbox{)}$ and $0 \equiv b \mbox{ (mod }k\mbox{)}$ occur
5629simultaneously.
5630
5631The most common approach (cite) is to reduce one input modulo another.  That is if $a$ and $b$ are divisible by some integer $k$ and if $qa + r = b$ then
5632$r$ is also divisible by $k$.  The reduction pattern follows $\left < a , b \right > \rightarrow \left < b, a \mbox{ mod } b \right >$.  
5633
5634\newpage\begin{figure}[!here]
5635\begin{small}
5636\begin{center}
5637\begin{tabular}{l}
5638\hline Algorithm \textbf{Greatest Common Divisor (I)}. \\
5639\textbf{Input}.   Two positive integers $a$ and $b$ greater than zero. \\
5640\textbf{Output}.  The greatest common divisor $(a, b)$.  \\
5641\hline \\
56421.  While ($b > 0$) do \\
5643\hspace{3mm}1.1  $r \leftarrow a \mbox{ (mod }b\mbox{)}$ \\
5644\hspace{3mm}1.2  $a \leftarrow b$ \\
5645\hspace{3mm}1.3  $b \leftarrow r$ \\
56462.  Return($a$). \\
5647\hline
5648\end{tabular}
5649\end{center}
5650\end{small}
5651\caption{Algorithm Greatest Common Divisor (I)}
5652\label{fig:gcd1}
5653\end{figure}
5654
5655This algorithm will quickly converge on the greatest common divisor since the residue $r$ tends diminish rapidly.  However, divisions are
5656relatively expensive operations to perform and should ideally be avoided.  There is another approach based on a similar relationship of 
5657greatest common divisors.  The faster approach is based on the observation that if $k$ divides both $a$ and $b$ it will also divide $a - b$.  
5658In particular, we would like $a - b$ to decrease in magnitude which implies that $b \ge a$.  
5659
5660\begin{figure}[!here]
5661\begin{small}
5662\begin{center}
5663\begin{tabular}{l}
5664\hline Algorithm \textbf{Greatest Common Divisor (II)}. \\
5665\textbf{Input}.   Two positive integers $a$ and $b$ greater than zero. \\
5666\textbf{Output}.  The greatest common divisor $(a, b)$.  \\
5667\hline \\
56681.  While ($b > 0$) do \\
5669\hspace{3mm}1.1  Swap $a$ and $b$ such that $a$ is the smallest of the two. \\
5670\hspace{3mm}1.2  $b \leftarrow b - a$ \\
56712.  Return($a$). \\
5672\hline
5673\end{tabular}
5674\end{center}
5675\end{small}
5676\caption{Algorithm Greatest Common Divisor (II)}
5677\label{fig:gcd2}
5678\end{figure}
5679
5680\textbf{Proof} \textit{Algorithm~\ref{fig:gcd2} will return the greatest common divisor of $a$ and $b$.}
5681The algorithm in figure~\ref{fig:gcd2} will eventually terminate since $b \ge a$ the subtraction in step 1.2 will be a value less than $b$.  In other
5682words in every iteration that tuple $\left < a, b \right >$ decrease in magnitude until eventually $a = b$.  Since both $a$ and $b$ are always 
5683divisible by the greatest common divisor (\textit{until the last iteration}) and in the last iteration of the algorithm $b = 0$, therefore, in the 
5684second to last iteration of the algorithm $b = a$ and clearly $(a, a) = a$ which concludes the proof.  \textbf{QED}.
5685
5686As a matter of practicality algorithm \ref{fig:gcd1} decreases far too slowly to be useful.  Specially if $b$ is much larger than $a$ such that 
5687$b - a$ is still very much larger than $a$.  A simple addition to the algorithm is to divide $b - a$ by a power of some integer $p$ which does
5688not divide the greatest common divisor but will divide $b - a$.  In this case ${b - a} \over p$ is also an integer and still divisible by
5689the greatest common divisor.
5690
5691However, instead of factoring $b - a$ to find a suitable value of $p$ the powers of $p$ can be removed from $a$ and $b$ that are in common first.  
5692Then inside the loop whenever $b - a$ is divisible by some power of $p$ it can be safely removed.  
5693
5694\begin{figure}[!here]
5695\begin{small}
5696\begin{center}
5697\begin{tabular}{l}
5698\hline Algorithm \textbf{Greatest Common Divisor (III)}. \\
5699\textbf{Input}.   Two positive integers $a$ and $b$ greater than zero. \\
5700\textbf{Output}.  The greatest common divisor $(a, b)$.  \\
5701\hline \\
57021.  $k \leftarrow 0$ \\
57032.  While $a$ and $b$ are both divisible by $p$ do \\
5704\hspace{3mm}2.1  $a \leftarrow \lfloor a / p \rfloor$ \\
5705\hspace{3mm}2.2  $b \leftarrow \lfloor b / p \rfloor$ \\
5706\hspace{3mm}2.3  $k \leftarrow k + 1$ \\
57073.  While $a$ is divisible by $p$ do \\
5708\hspace{3mm}3.1  $a \leftarrow \lfloor a / p \rfloor$ \\
57094.  While $b$ is divisible by $p$ do \\
5710\hspace{3mm}4.1  $b \leftarrow \lfloor b / p \rfloor$ \\
57115.  While ($b > 0$) do \\
5712\hspace{3mm}5.1  Swap $a$ and $b$ such that $a$ is the smallest of the two. \\
5713\hspace{3mm}5.2  $b \leftarrow b - a$ \\
5714\hspace{3mm}5.3  While $b$ is divisible by $p$ do \\
5715\hspace{6mm}5.3.1  $b \leftarrow \lfloor b / p \rfloor$ \\
57166.  Return($a \cdot p^k$). \\
5717\hline
5718\end{tabular}
5719\end{center}
5720\end{small}
5721\caption{Algorithm Greatest Common Divisor (III)}
5722\label{fig:gcd3}
5723\end{figure}
5724
5725This algorithm is based on the first except it removes powers of $p$ first and inside the main loop to ensure the tuple $\left < a, b \right >$ 
5726decreases more rapidly.  The first loop on step two removes powers of $p$ that are in common.  A count, $k$, is kept which will present a common
5727divisor of $p^k$.  After step two the remaining common divisor of $a$ and $b$ cannot be divisible by $p$.  This means that $p$ can be safely 
5728divided out of the difference $b - a$ so long as the division leaves no remainder.  
5729
5730In particular the value of $p$ should be chosen such that the division on step 5.3.1 occur often.  It also helps that division by $p$ be easy
5731to compute.  The ideal choice of $p$ is two since division by two amounts to a right logical shift.  Another important observation is that by
5732step five both $a$ and $b$ are odd.  Therefore, the diffrence $b - a$ must be even which means that each iteration removes one bit from the 
5733largest of the pair.
5734
5735\subsection{Complete Greatest Common Divisor}
5736The algorithms presented so far cannot handle inputs which are zero or negative.  The following algorithm can handle all input cases properly
5737and will produce the greatest common divisor.
5738
5739\newpage\begin{figure}[!here]
5740\begin{small}
5741\begin{center}
5742\begin{tabular}{l}
5743\hline Algorithm \textbf{mp\_gcd}. \\
5744\textbf{Input}.   mp\_int $a$ and $b$ \\
5745\textbf{Output}.  The greatest common divisor $c = (a, b)$.  \\
5746\hline \\
57471.  If $a = 0$ then \\
5748\hspace{3mm}1.1  $c \leftarrow \vert b \vert $ \\
5749\hspace{3mm}1.2  Return(\textit{MP\_OKAY}). \\
57502.  If $b = 0$ then \\
5751\hspace{3mm}2.1  $c \leftarrow \vert a \vert $ \\
5752\hspace{3mm}2.2  Return(\textit{MP\_OKAY}). \\
57533.  $u \leftarrow \vert a \vert, v \leftarrow \vert b \vert$ \\
57544.  $k \leftarrow 0$ \\
57555.  While $u.used > 0$ and $v.used > 0$ and $u_0 \equiv v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
5756\hspace{3mm}5.1  $k \leftarrow k + 1$ \\
5757\hspace{3mm}5.2  $u \leftarrow \lfloor u / 2 \rfloor$ \\
5758\hspace{3mm}5.3  $v \leftarrow \lfloor v / 2 \rfloor$ \\
57596.  While $u.used > 0$ and $u_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
5760\hspace{3mm}6.1  $u \leftarrow \lfloor u / 2 \rfloor$ \\
57617.  While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
5762\hspace{3mm}7.1  $v \leftarrow \lfloor v / 2 \rfloor$ \\
57638.  While $v.used > 0$ \\
5764\hspace{3mm}8.1  If $\vert u \vert > \vert v \vert$ then \\
5765\hspace{6mm}8.1.1  Swap $u$ and $v$. \\
5766\hspace{3mm}8.2  $v \leftarrow \vert v \vert - \vert u \vert$ \\
5767\hspace{3mm}8.3  While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
5768\hspace{6mm}8.3.1  $v \leftarrow \lfloor v / 2 \rfloor$ \\
57699.  $c \leftarrow u \cdot 2^k$ \\
577010.  Return(\textit{MP\_OKAY}). \\
5771\hline
5772\end{tabular}
5773\end{center}
5774\end{small}
5775\caption{Algorithm mp\_gcd}
5776\end{figure}
5777\textbf{Algorithm mp\_gcd.}
5778This algorithm will produce the greatest common divisor of two mp\_ints $a$ and $b$.  The algorithm was originally based on Algorithm B of
5779Knuth \cite[pp. 338]{TAOCPV2} but has been modified to be simpler to explain.  In theory it achieves the same asymptotic working time as
5780Algorithm B and in practice this appears to be true.  
5781
5782The first two steps handle the cases where either one of or both inputs are zero.  If either input is zero the greatest common divisor is the 
5783largest input or zero if they are both zero.  If the inputs are not trivial than $u$ and $v$ are assigned the absolute values of 
5784$a$ and $b$ respectively and the algorithm will proceed to reduce the pair.
5785
5786Step five will divide out any common factors of two and keep track of the count in the variable $k$.  After this step, two is no longer a
5787factor of the remaining greatest common divisor between $u$ and $v$ and can be safely evenly divided out of either whenever they are even.  Step 
5788six and seven ensure that the $u$ and $v$ respectively have no more factors of two.  At most only one of the while--loops will iterate since 
5789they cannot both be even.
5790
5791By step eight both of $u$ and $v$ are odd which is required for the inner logic.  First the pair are swapped such that $v$ is equal to
5792or greater than $u$.  This ensures that the subtraction on step 8.2 will always produce a positive and even result.  Step 8.3 removes any
5793factors of two from the difference $u$ to ensure that in the next iteration of the loop both are once again odd.
5794
5795After $v = 0$ occurs the variable $u$ has the greatest common divisor of the pair $\left < u, v \right >$ just after step six.  The result
5796must be adjusted by multiplying by the common factors of two ($2^k$) removed earlier.  
5797
5798EXAM,bn_mp_gcd.c
5799
5800This function makes use of the macros mp\_iszero and mp\_iseven.  The former evaluates to $1$ if the input mp\_int is equivalent to the 
5801integer zero otherwise it evaluates to $0$.  The latter evaluates to $1$ if the input mp\_int represents a non-zero even integer otherwise
5802it evaluates to $0$.  Note that just because mp\_iseven may evaluate to $0$ does not mean the input is odd, it could also be zero.  The three 
5803trivial cases of inputs are handled on lines @23,zero@ through @29,}@.  After those lines the inputs are assumed to be non-zero.
5804
5805Lines @32,if@ and @36,if@ make local copies $u$ and $v$ of the inputs $a$ and $b$ respectively.  At this point the common factors of two 
5806must be divided out of the two inputs.  The block starting at line @43,common@ removes common factors of two by first counting the number of trailing
5807zero bits in both.  The local integer $k$ is used to keep track of how many factors of $2$ are pulled out of both values.  It is assumed that 
5808the number of factors will not exceed the maximum value of a C ``int'' data type\footnote{Strictly speaking no array in C may have more than 
5809entries than are accessible by an ``int'' so this is not a limitation.}.  
5810
5811At this point there are no more common factors of two in the two values.  The divisions by a power of two on lines @60,div_2d@ and @67,div_2d@ remove 
5812any independent factors of two such that both $u$ and $v$ are guaranteed to be an odd integer before hitting the main body of the algorithm.  The while loop
5813on line @72, while@ performs the reduction of the pair until $v$ is equal to zero.  The unsigned comparison and subtraction algorithms are used in
5814place of the full signed routines since both values are guaranteed to be positive and the result of the subtraction is guaranteed to be non-negative.
5815
5816\section{Least Common Multiple}
5817The least common multiple of a pair of integers is their product divided by their greatest common divisor.  For two integers $a$ and $b$ the
5818least common multiple is normally denoted as $[ a, b ]$ and numerically equivalent to ${ab} \over {(a, b)}$.  For example, if $a = 2 \cdot 2 \cdot 3 = 12$
5819and $b = 2 \cdot 3 \cdot 3 \cdot 7 = 126$ the least common multiple is ${126 \over {(12, 126)}} = {126 \over 6} = 21$.
5820
5821The least common multiple arises often in coding theory as well as number theory.  If two functions have periods of $a$ and $b$ respectively they will
5822collide, that is be in synchronous states, after only $[ a, b ]$ iterations.  This is why, for example, random number generators based on 
5823Linear Feedback Shift Registers (LFSR) tend to use registers with periods which are co-prime (\textit{e.g. the greatest common divisor is one.}).  
5824Similarly in number theory if a composite $n$ has two prime factors $p$ and $q$ then maximal order of any unit of $\Z/n\Z$ will be $[ p - 1, q - 1] $.
5825
5826\begin{figure}[!here]
5827\begin{small}
5828\begin{center}
5829\begin{tabular}{l}
5830\hline Algorithm \textbf{mp\_lcm}. \\
5831\textbf{Input}.   mp\_int $a$ and $b$ \\
5832\textbf{Output}.  The least common multiple $c = [a, b]$.  \\
5833\hline \\
58341.  $c \leftarrow (a, b)$ \\
58352.  $t \leftarrow a \cdot b$ \\
58363.  $c \leftarrow \lfloor t / c \rfloor$ \\
58374.  Return(\textit{MP\_OKAY}). \\
5838\hline
5839\end{tabular}
5840\end{center}
5841\end{small}
5842\caption{Algorithm mp\_lcm}
5843\end{figure}
5844\textbf{Algorithm mp\_lcm.}
5845This algorithm computes the least common multiple of two mp\_int inputs $a$ and $b$.  It computes the least common multiple directly by
5846dividing the product of the two inputs by their greatest common divisor.
5847
5848EXAM,bn_mp_lcm.c
5849
5850\section{Jacobi Symbol Computation}
5851To explain the Jacobi Symbol we shall first discuss the Legendre function\footnote{Arrg.  What is the name of this?} off which the Jacobi symbol is 
5852defined.  The Legendre function computes whether or not an integer $a$ is a quadratic residue modulo an odd prime $p$.  Numerically it is
5853equivalent to equation \ref{eqn:legendre}.
5854
5855\textit{-- Tom, don't be an ass, cite your source here...!}
5856
5857\begin{equation}
5858a^{(p-1)/2} \equiv \begin{array}{rl}
5859                              -1 &  \mbox{if }a\mbox{ is a quadratic non-residue.} \\
5860                              0  &  \mbox{if }a\mbox{ divides }p\mbox{.} \\
5861                              1  &  \mbox{if }a\mbox{ is a quadratic residue}. 
5862                              \end{array} \mbox{ (mod }p\mbox{)}
5863\label{eqn:legendre}                              
5864\end{equation}
5865
5866\textbf{Proof.} \textit{Equation \ref{eqn:legendre} correctly identifies the residue status of an integer $a$ modulo a prime $p$.}
5867An integer $a$ is a quadratic residue if the following equation has a solution.
5868
5869\begin{equation}
5870x^2 \equiv a \mbox{ (mod }p\mbox{)}
5871\label{eqn:root}
5872\end{equation}
5873
5874Consider the following equation.
5875
5876\begin{equation}
58770 \equiv x^{p-1} - 1 \equiv \left \lbrace \left (x^2 \right )^{(p-1)/2} - a^{(p-1)/2} \right \rbrace + \left ( a^{(p-1)/2} - 1 \right ) \mbox{ (mod }p\mbox{)}
5878\label{eqn:rooti}
5879\end{equation}
5880
5881Whether equation \ref{eqn:root} has a solution or not equation \ref{eqn:rooti} is always true.  If $a^{(p-1)/2} - 1 \equiv 0 \mbox{ (mod }p\mbox{)}$
5882then the quantity in the braces must be zero.  By reduction,
5883
5884\begin{eqnarray}
5885\left (x^2 \right )^{(p-1)/2} - a^{(p-1)/2} \equiv 0  \nonumber \\
5886\left (x^2 \right )^{(p-1)/2} \equiv a^{(p-1)/2} \nonumber \\
5887x^2 \equiv a \mbox{ (mod }p\mbox{)} 
5888\end{eqnarray}
5889
5890As a result there must be a solution to the quadratic equation and in turn $a$ must be a quadratic residue.  If $a$ does not divide $p$ and $a$
5891is not a quadratic residue then the only other value $a^{(p-1)/2}$ may be congruent to is $-1$ since
5892\begin{equation}
58930 \equiv a^{p - 1} - 1 \equiv (a^{(p-1)/2} + 1)(a^{(p-1)/2} - 1) \mbox{ (mod }p\mbox{)}
5894\end{equation}
5895One of the terms on the right hand side must be zero.  \textbf{QED}
5896
5897\subsection{Jacobi Symbol}
5898The Jacobi symbol is a generalization of the Legendre function for any odd non prime moduli $p$ greater than 2.  If $p = \prod_{i=0}^n p_i$ then
5899the Jacobi symbol $\left ( { a \over p } \right )$ is equal to the following equation.
5900
5901\begin{equation}
5902\left ( { a \over p } \right ) = \left ( { a \over p_0} \right ) \left ( { a \over p_1} \right ) \ldots \left ( { a \over p_n} \right )
5903\end{equation}
5904
5905By inspection if $p$ is prime the Jacobi symbol is equivalent to the Legendre function.  The following facts\footnote{See HAC \cite[pp. 72-74]{HAC} for
5906further details.} will be used to derive an efficient Jacobi symbol algorithm.  Where $p$ is an odd integer greater than two and $a, b \in \Z$ the
5907following are true.  
5908
5909\begin{enumerate}
5910\item $\left ( { a \over p} \right )$ equals $-1$, $0$ or $1$. 
5911\item $\left ( { ab \over p} \right ) = \left ( { a \over p} \right )\left ( { b \over p} \right )$.
5912\item If $a \equiv b$ then $\left ( { a \over p} \right ) = \left ( { b \over p} \right )$.
5913\item $\left ( { 2 \over p} \right )$ equals $1$ if $p \equiv 1$ or $7 \mbox{ (mod }8\mbox{)}$.  Otherwise, it equals $-1$.
5914\item $\left ( { a \over p} \right ) \equiv \left ( { p \over a} \right ) \cdot (-1)^{(p-1)(a-1)/4}$.  More specifically 
5915$\left ( { a \over p} \right ) = \left ( { p \over a} \right )$ if $p \equiv a \equiv 1 \mbox{ (mod }4\mbox{)}$.  
5916\end{enumerate}
5917
5918Using these facts if $a = 2^k \cdot a'$ then
5919
5920\begin{eqnarray}
5921\left ( { a \over p } \right ) = \left ( {{2^k} \over p } \right ) \left ( {a' \over p} \right ) \nonumber \\
5922                               = \left ( {2 \over p } \right )^k \left ( {a' \over p} \right ) 
5923\label{eqn:jacobi}
5924\end{eqnarray}
5925
5926By fact five, 
5927
5928\begin{equation}
5929\left ( { a \over p } \right ) = \left ( { p \over a } \right ) \cdot (-1)^{(p-1)(a-1)/4} 
5930\end{equation}
5931
5932Subsequently by fact three since $p \equiv (p \mbox{ mod }a) \mbox{ (mod }a\mbox{)}$ then 
5933
5934\begin{equation}
5935\left ( { a \over p } \right ) = \left ( { {p \mbox{ mod } a} \over a } \right ) \cdot (-1)^{(p-1)(a-1)/4} 
5936\end{equation}
5937
5938By putting both observations into equation \ref{eqn:jacobi} the following simplified equation is formed.
5939
5940\begin{equation}
5941\left ( { a \over p } \right ) = \left ( {2 \over p } \right )^k \left ( {{p\mbox{ mod }a'} \over a'} \right )  \cdot (-1)^{(p-1)(a'-1)/4} 
5942\end{equation}
5943
5944The value of $\left ( {{p \mbox{ mod }a'} \over a'} \right )$ can be found by using the same equation recursively.  The value of 
5945$\left ( {2 \over p } \right )^k$ equals $1$ if $k$ is even otherwise it equals $\left ( {2 \over p } \right )$.  Using this approach the 
5946factors of $p$ do not have to be known.  Furthermore, if $(a, p) = 1$ then the algorithm will terminate when the recursion requests the 
5947Jacobi symbol computation of $\left ( {1 \over a'} \right )$ which is simply $1$.  
5948
5949\newpage\begin{figure}[!here]
5950\begin{small}
5951\begin{center}
5952\begin{tabular}{l}
5953\hline Algorithm \textbf{mp\_jacobi}. \\
5954\textbf{Input}.   mp\_int $a$ and $p$, $a \ge 0$, $p \ge 3$, $p \equiv 1 \mbox{ (mod }2\mbox{)}$ \\
5955\textbf{Output}.  The Jacobi symbol $c = \left ( {a \over p } \right )$. \\
5956\hline \\
59571.  If $a = 0$ then \\
5958\hspace{3mm}1.1  $c \leftarrow 0$ \\
5959\hspace{3mm}1.2  Return(\textit{MP\_OKAY}). \\
59602.  If $a = 1$ then \\
5961\hspace{3mm}2.1  $c \leftarrow 1$ \\
5962\hspace{3mm}2.2  Return(\textit{MP\_OKAY}). \\
59633.  $a' \leftarrow a$ \\
59644.  $k \leftarrow 0$ \\
59655.  While $a'.used > 0$ and $a'_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
5966\hspace{3mm}5.1  $k \leftarrow k + 1$ \\
5967\hspace{3mm}5.2  $a' \leftarrow \lfloor a' / 2 \rfloor$ \\
59686.  If $k \equiv 0 \mbox{ (mod }2\mbox{)}$ then \\
5969\hspace{3mm}6.1  $s \leftarrow 1$ \\
59707.  else \\
5971\hspace{3mm}7.1  $r \leftarrow p_0 \mbox{ (mod }8\mbox{)}$ \\
5972\hspace{3mm}7.2  If $r = 1$ or $r = 7$ then \\
5973\hspace{6mm}7.2.1  $s \leftarrow 1$ \\
5974\hspace{3mm}7.3  else \\
5975\hspace{6mm}7.3.1  $s \leftarrow -1$ \\
59768.  If $p_0 \equiv a'_0 \equiv 3 \mbox{ (mod }4\mbox{)}$ then \\
5977\hspace{3mm}8.1  $s \leftarrow -s$ \\
59789.  If $a' \ne 1$ then \\
5979\hspace{3mm}9.1  $p' \leftarrow p \mbox{ (mod }a'\mbox{)}$ \\
5980\hspace{3mm}9.2  $s \leftarrow s \cdot \mbox{mp\_jacobi}(p', a')$ \\
598110.  $c \leftarrow s$ \\
598211.  Return(\textit{MP\_OKAY}). \\
5983\hline
5984\end{tabular}
5985\end{center}
5986\end{small}
5987\caption{Algorithm mp\_jacobi}
5988\end{figure}
5989\textbf{Algorithm mp\_jacobi.}
5990This algorithm computes the Jacobi symbol for an arbitrary positive integer $a$ with respect to an odd integer $p$ greater than three.  The algorithm
5991is based on algorithm 2.149 of HAC \cite[pp. 73]{HAC}.  
5992
5993Step numbers one and two handle the trivial cases of $a = 0$ and $a = 1$ respectively.  Step five determines the number of two factors in the
5994input $a$.  If $k$ is even than the term $\left ( { 2 \over p } \right )^k$ must always evaluate to one.  If $k$ is odd than the term evaluates to one 
5995if $p_0$ is congruent to one or seven modulo eight, otherwise it evaluates to $-1$. After the the $\left ( { 2 \over p } \right )^k$ term is handled 
5996the $(-1)^{(p-1)(a'-1)/4}$ is computed and multiplied against the current product $s$.  The latter term evaluates to one if both $p$ and $a'$ 
5997are congruent to one modulo four, otherwise it evaluates to negative one.
5998
5999By step nine if $a'$ does not equal one a recursion is required.  Step 9.1 computes $p' \equiv p \mbox{ (mod }a'\mbox{)}$ and will recurse to compute
6000$\left ( {p' \over a'} \right )$ which is multiplied against the current Jacobi product.
6001
6002EXAM,bn_mp_jacobi.c
6003
6004As a matter of practicality the variable $a'$ as per the pseudo-code is reprensented by the variable $a1$ since the $'$ symbol is not valid for a C 
6005variable name character. 
6006
6007The two simple cases of $a = 0$ and $a = 1$ are handled at the very beginning to simplify the algorithm.  If the input is non-trivial the algorithm
6008has to proceed compute the Jacobi.  The variable $s$ is used to hold the current Jacobi product.  Note that $s$ is merely a C ``int'' data type since
6009the values it may obtain are merely $-1$, $0$ and $1$.  
6010
6011After a local copy of $a$ is made all of the factors of two are divided out and the total stored in $k$.  Technically only the least significant
6012bit of $k$ is required, however, it makes the algorithm simpler to follow to perform an addition. In practice an exclusive-or and addition have the same 
6013processor requirements and neither is faster than the other.
6014
6015Line @59, if@ through @70, }@ determines the value of $\left ( { 2 \over p } \right )^k$.  If the least significant bit of $k$ is zero than
6016$k$ is even and the value is one.  Otherwise, the value of $s$ depends on which residue class $p$ belongs to modulo eight.  The value of
6017$(-1)^{(p-1)(a'-1)/4}$ is compute and multiplied against $s$ on lines @73, if@ through @75, }@.  
6018
6019Finally, if $a1$ does not equal one the algorithm must recurse and compute $\left ( {p' \over a'} \right )$.  
6020
6021\textit{-- Comment about default $s$ and such...}
6022
6023\section{Modular Inverse}
6024\label{sec:modinv}
6025The modular inverse of a number actually refers to the modular multiplicative inverse.  Essentially for any integer $a$ such that $(a, p) = 1$ there
6026exist another integer $b$ such that $ab \equiv 1 \mbox{ (mod }p\mbox{)}$.  The integer $b$ is called the multiplicative inverse of $a$ which is
6027denoted as $b = a^{-1}$.  Technically speaking modular inversion is a well defined operation for any finite ring or field not just for rings and 
6028fields of integers.  However, the former will be the matter of discussion.
6029
6030The simplest approach is to compute the algebraic inverse of the input.  That is to compute $b \equiv a^{\Phi(p) - 1}$.  If $\Phi(p)$ is the 
6031order of the multiplicative subgroup modulo $p$ then $b$ must be the multiplicative inverse of $a$.  The proof of which is trivial.
6032
6033\begin{equation}
6034ab \equiv a \left (a^{\Phi(p) - 1} \right ) \equiv a^{\Phi(p)} \equiv a^0 \equiv 1 \mbox{ (mod }p\mbox{)}
6035\end{equation}
6036
6037However, as simple as this approach may be it has two serious flaws.  It requires that the value of $\Phi(p)$ be known which if $p$ is composite 
6038requires all of the prime factors.  This approach also is very slow as the size of $p$ grows.  
6039
6040A simpler approach is based on the observation that solving for the multiplicative inverse is equivalent to solving the linear 
6041Diophantine\footnote{See LeVeque \cite[pp. 40-43]{LeVeque} for more information.} equation.
6042
6043\begin{equation}
6044ab + pq = 1
6045\end{equation}
6046
6047Where $a$, $b$, $p$ and $q$ are all integers.  If such a pair of integers $ \left < b, q \right >$ exist than $b$ is the multiplicative inverse of 
6048$a$ modulo $p$.  The extended Euclidean algorithm (Knuth \cite[pp. 342]{TAOCPV2}) can be used to solve such equations provided $(a, p) = 1$.  
6049However, instead of using that algorithm directly a variant known as the binary Extended Euclidean algorithm will be used in its place.  The
6050binary approach is very similar to the binary greatest common divisor algorithm except it will produce a full solution to the Diophantine 
6051equation.  
6052
6053\subsection{General Case}
6054\newpage\begin{figure}[!here]
6055\begin{small}
6056\begin{center}
6057\begin{tabular}{l}
6058\hline Algorithm \textbf{mp\_invmod}. \\
6059\textbf{Input}.   mp\_int $a$ and $b$, $(a, b) = 1$, $p \ge 2$, $0 < a < p$.  \\
6060\textbf{Output}.  The modular inverse $c \equiv a^{-1} \mbox{ (mod }b\mbox{)}$. \\
6061\hline \\
60621.  If $b \le 0$ then return(\textit{MP\_VAL}). \\
60632.  If $b_0 \equiv 1 \mbox{ (mod }2\mbox{)}$ then use algorithm fast\_mp\_invmod. \\
60643.  $x \leftarrow \vert a \vert, y \leftarrow b$ \\
60654.  If $x_0 \equiv y_0  \equiv 0 \mbox{ (mod }2\mbox{)}$ then return(\textit{MP\_VAL}). \\
60665.  $B \leftarrow 0, C \leftarrow 0, A \leftarrow 1, D \leftarrow 1$ \\
60676.  While $u.used > 0$ and $u_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
6068\hspace{3mm}6.1  $u \leftarrow \lfloor u / 2 \rfloor$ \\
6069\hspace{3mm}6.2  If ($A.used > 0$ and $A_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) or ($B.used > 0$ and $B_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) then \\
6070\hspace{6mm}6.2.1  $A \leftarrow A + y$ \\
6071\hspace{6mm}6.2.2  $B \leftarrow B - x$ \\
6072\hspace{3mm}6.3  $A \leftarrow \lfloor A / 2 \rfloor$ \\
6073\hspace{3mm}6.4  $B \leftarrow \lfloor B / 2 \rfloor$ \\
60747.  While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
6075\hspace{3mm}7.1  $v \leftarrow \lfloor v / 2 \rfloor$ \\
6076\hspace{3mm}7.2  If ($C.used > 0$ and $C_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) or ($D.used > 0$ and $D_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) then \\
6077\hspace{6mm}7.2.1  $C \leftarrow C + y$ \\
6078\hspace{6mm}7.2.2  $D \leftarrow D - x$ \\
6079\hspace{3mm}7.3  $C \leftarrow \lfloor C / 2 \rfloor$ \\
6080\hspace{3mm}7.4  $D \leftarrow \lfloor D / 2 \rfloor$ \\
60818.  If $u \ge v$ then \\
6082\hspace{3mm}8.1  $u \leftarrow u - v$ \\
6083\hspace{3mm}8.2  $A \leftarrow A - C$ \\
6084\hspace{3mm}8.3  $B \leftarrow B - D$ \\
60859.  else \\
6086\hspace{3mm}9.1  $v \leftarrow v - u$ \\
6087\hspace{3mm}9.2  $C \leftarrow C - A$ \\
6088\hspace{3mm}9.3  $D \leftarrow D - B$ \\
608910.  If $u \ne 0$ goto step 6. \\
609011.  If $v \ne 1$ return(\textit{MP\_VAL}). \\
609112.  While $C \le 0$ do \\
6092\hspace{3mm}12.1  $C \leftarrow C + b$ \\
609313.  While $C \ge b$ do \\
6094\hspace{3mm}13.1  $C \leftarrow C - b$ \\
609514.  $c \leftarrow C$ \\
609615.  Return(\textit{MP\_OKAY}). \\
6097\hline
6098\end{tabular}
6099\end{center}
6100\end{small}
6101\end{figure}
6102\textbf{Algorithm mp\_invmod.}
6103This algorithm computes the modular multiplicative inverse of an integer $a$ modulo an integer $b$.  This algorithm is a variation of the 
6104extended binary Euclidean algorithm from HAC \cite[pp. 608]{HAC}.  It has been modified to only compute the modular inverse and not a complete
6105Diophantine solution.  
6106
6107If $b \le 0$ than the modulus is invalid and MP\_VAL is returned.  Similarly if both $a$ and $b$ are even then there cannot be a multiplicative
6108inverse for $a$ and the error is reported.  
6109
6110The astute reader will observe that steps seven through nine are very similar to the binary greatest common divisor algorithm mp\_gcd.  In this case
6111the other variables to the Diophantine equation are solved.  The algorithm terminates when $u = 0$ in which case the solution is
6112
6113\begin{equation}
6114Ca + Db = v
6115\end{equation}
6116
6117If $v$, the greatest common divisor of $a$ and $b$ is not equal to one then the algorithm will report an error as no inverse exists.  Otherwise, $C$
6118is the modular inverse of $a$.  The actual value of $C$ is congruent to, but not necessarily equal to, the ideal modular inverse which should lie 
6119within $1 \le a^{-1} < b$.  Step numbers twelve and thirteen adjust the inverse until it is in range.  If the original input $a$ is within $0 < a < p$ 
6120then only a couple of additions or subtractions will be required to adjust the inverse.
6121
6122EXAM,bn_mp_invmod.c
6123
6124\subsubsection{Odd Moduli}
6125
6126When the modulus $b$ is odd the variables $A$ and $C$ are fixed and are not required to compute the inverse.  In particular by attempting to solve
6127the Diophantine $Cb + Da = 1$ only $B$ and $D$ are required to find the inverse of $a$.  
6128
6129The algorithm fast\_mp\_invmod is a direct adaptation of algorithm mp\_invmod with all all steps involving either $A$ or $C$ removed.  This 
6130optimization will halve the time required to compute the modular inverse.
6131
6132\section{Primality Tests}
6133
6134A non-zero integer $a$ is said to be prime if it is not divisible by any other integer excluding one and itself.  For example, $a = 7$ is prime 
6135since the integers $2 \ldots 6$ do not evenly divide $a$.  By contrast, $a = 6$ is not prime since $a = 6 = 2 \cdot 3$. 
6136
6137Prime numbers arise in cryptography considerably as they allow finite fields to be formed.  The ability to determine whether an integer is prime or
6138not quickly has been a viable subject in cryptography and number theory for considerable time.  The algorithms that will be presented are all
6139probablistic algorithms in that when they report an integer is composite it must be composite.  However, when the algorithms report an integer is
6140prime the algorithm may be incorrect.  
6141
6142As will be discussed it is possible to limit the probability of error so well that for practical purposes the probablity of error might as 
6143well be zero.  For the purposes of these discussions let $n$ represent the candidate integer of which the primality is in question.
6144
6145\subsection{Trial Division}
6146
6147Trial division means to attempt to evenly divide a candidate integer by small prime integers.  If the candidate can be evenly divided it obviously
6148cannot be prime.  By dividing by all primes $1 < p \le \sqrt{n}$ this test can actually prove whether an integer is prime.  However, such a test
6149would require a prohibitive amount of time as $n$ grows.
6150
6151Instead of dividing by every prime, a smaller, more mangeable set of primes may be used instead.  By performing trial division with only a subset
6152of the primes less than $\sqrt{n} + 1$ the algorithm cannot prove if a candidate is prime.  However, often it can prove a candidate is not prime.
6153
6154The benefit of this test is that trial division by small values is fairly efficient.  Specially compared to the other algorithms that will be
6155discussed shortly.  The probability that this approach correctly identifies a composite candidate when tested with all primes upto $q$ is given by
6156$1 - {1.12 \over ln(q)}$.  The graph (\ref{pic:primality}, will be added later) demonstrates the probability of success for the range 
6157$3 \le q \le 100$.  
6158
6159At approximately $q = 30$ the gain of performing further tests diminishes fairly quickly.  At $q = 90$ further testing is generally not going to 
6160be of any practical use.  In the case of LibTomMath the default limit $q = 256$ was chosen since it is not too high and will eliminate 
6161approximately $80\%$ of all candidate integers.  The constant \textbf{PRIME\_SIZE} is equal to the number of primes in the test base.  The 
6162array \_\_prime\_tab is an array of the first \textbf{PRIME\_SIZE} prime numbers.  
6163
6164\begin{figure}[!here]
6165\begin{small}
6166\begin{center}
6167\begin{tabular}{l}
6168\hline Algorithm \textbf{mp\_prime\_is\_divisible}. \\
6169\textbf{Input}.   mp\_int $a$ \\
6170\textbf{Output}.  $c = 1$ if $n$ is divisible by a small prime, otherwise $c = 0$.  \\
6171\hline \\
61721.  for $ix$ from $0$ to $PRIME\_SIZE$ do \\
6173\hspace{3mm}1.1  $d \leftarrow n \mbox{ (mod }\_\_prime\_tab_{ix}\mbox{)}$ \\
6174\hspace{3mm}1.2  If $d = 0$ then \\
6175\hspace{6mm}1.2.1  $c \leftarrow 1$ \\
6176\hspace{6mm}1.2.2  Return(\textit{MP\_OKAY}). \\
61772.  $c \leftarrow 0$ \\
61783.  Return(\textit{MP\_OKAY}). \\
6179\hline
6180\end{tabular}
6181\end{center}
6182\end{small}
6183\caption{Algorithm mp\_prime\_is\_divisible}
6184\end{figure}
6185\textbf{Algorithm mp\_prime\_is\_divisible.}
6186This algorithm attempts to determine if a candidate integer $n$ is composite by performing trial divisions.  
6187
6188EXAM,bn_mp_prime_is_divisible.c
6189
6190The algorithm defaults to a return of $0$ in case an error occurs.  The values in the prime table are all specified to be in the range of a 
6191mp\_digit.  The table \_\_prime\_tab is defined in the following file.
6192
6193EXAM,bn_prime_tab.c
6194
6195Note that there are two possible tables.  When an mp\_digit is 7-bits long only the primes upto $127$ may be included, otherwise the primes
6196upto $1619$ are used.  Note that the value of \textbf{PRIME\_SIZE} is a constant dependent on the size of a mp\_digit. 
6197
6198\subsection{The Fermat Test}
6199The Fermat test is probably one the oldest tests to have a non-trivial probability of success.  It is based on the fact that if $n$ is in 
6200fact prime then $a^{n} \equiv a \mbox{ (mod }n\mbox{)}$ for all $0 < a < n$.  The reason being that if $n$ is prime than the order of
6201the multiplicative sub group is $n - 1$.  Any base $a$ must have an order which divides $n - 1$ and as such $a^n$ is equivalent to 
6202$a^1 = a$.  
6203
6204If $n$ is composite then any given base $a$ does not have to have a period which divides $n - 1$.  In which case 
6205it is possible that $a^n \nequiv a \mbox{ (mod }n\mbox{)}$.  However, this test is not absolute as it is possible that the order
6206of a base will divide $n - 1$ which would then be reported as prime.  Such a base yields what is known as a Fermat pseudo-prime.  Several 
6207integers known as Carmichael numbers will be a pseudo-prime to all valid bases.  Fortunately such numbers are extremely rare as $n$ grows
6208in size.
6209
6210\begin{figure}[!here]
6211\begin{small}
6212\begin{center}
6213\begin{tabular}{l}
6214\hline Algorithm \textbf{mp\_prime\_fermat}. \\
6215\textbf{Input}.   mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$.  \\
6216\textbf{Output}.  $c = 1$ if $b^a \equiv b \mbox{ (mod }a\mbox{)}$, otherwise $c = 0$.  \\
6217\hline \\
62181.  $t \leftarrow b^a \mbox{ (mod }a\mbox{)}$ \\
62192.  If $t = b$ then \\
6220\hspace{3mm}2.1  $c = 1$ \\
62213.  else \\
6222\hspace{3mm}3.1  $c = 0$ \\
62234.  Return(\textit{MP\_OKAY}). \\
6224\hline
6225\end{tabular}
6226\end{center}
6227\end{small}
6228\caption{Algorithm mp\_prime\_fermat}
6229\end{figure}
6230\textbf{Algorithm mp\_prime\_fermat.}
6231This algorithm determines whether an mp\_int $a$ is a Fermat prime to the base $b$ or not.  It uses a single modular exponentiation to
6232determine the result.  
6233
6234EXAM,bn_mp_prime_fermat.c
6235
6236\subsection{The Miller-Rabin Test}
6237The Miller-Rabin (citation) test is another primality test which has tighter error bounds than the Fermat test specifically with sequentially chosen 
6238candidate  integers.  The algorithm is based on the observation that if $n - 1 = 2^kr$ and if $b^r \nequiv \pm 1$ then after upto $k - 1$ squarings the 
6239value must be equal to $-1$.  The squarings are stopped as soon as $-1$ is observed.  If the value of $1$ is observed first it means that
6240some value not congruent to $\pm 1$ when squared equals one which cannot occur if $n$ is prime.
6241
6242\begin{figure}[!here]
6243\begin{small}
6244\begin{center}
6245\begin{tabular}{l}
6246\hline Algorithm \textbf{mp\_prime\_miller\_rabin}. \\
6247\textbf{Input}.   mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$.  \\
6248\textbf{Output}.  $c = 1$ if $a$ is a Miller-Rabin prime to the base $a$, otherwise $c = 0$.  \\
6249\hline
62501.  $a' \leftarrow a - 1$ \\
62512.  $r  \leftarrow n1$    \\
62523.  $c \leftarrow 0, s  \leftarrow 0$ \\
62534.  While $r.used > 0$ and $r_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\
6254\hspace{3mm}4.1  $s \leftarrow s + 1$ \\
6255\hspace{3mm}4.2  $r \leftarrow \lfloor r / 2 \rfloor$ \\
62565.  $y \leftarrow b^r \mbox{ (mod }a\mbox{)}$ \\
62576.  If $y \nequiv \pm 1$ then \\
6258\hspace{3mm}6.1  $j \leftarrow 1$ \\
6259\hspace{3mm}6.2  While $j \le (s - 1)$ and $y \nequiv a'$ \\
6260\hspace{6mm}6.2.1  $y \leftarrow y^2 \mbox{ (mod }a\mbox{)}$ \\
6261\hspace{6mm}6.2.2  If $y = 1$ then goto step 8. \\
6262\hspace{6mm}6.2.3  $j \leftarrow j + 1$ \\
6263\hspace{3mm}6.3  If $y \nequiv a'$ goto step 8. \\
62647.  $c \leftarrow 1$\\
62658.  Return(\textit{MP\_OKAY}). \\
6266\hline
6267\end{tabular}
6268\end{center}
6269\end{small}
6270\caption{Algorithm mp\_prime\_miller\_rabin}
6271\end{figure}
6272\textbf{Algorithm mp\_prime\_miller\_rabin.}
6273This algorithm performs one trial round of the Miller-Rabin algorithm to the base $b$.  It will set $c = 1$ if the algorithm cannot determine
6274if $b$ is composite or $c = 0$ if $b$ is provably composite.  The values of $s$ and $r$ are computed such that $a' = a - 1 = 2^sr$.  
6275
6276If the value $y \equiv b^r$ is congruent to $\pm 1$ then the algorithm cannot prove if $a$ is composite or not.  Otherwise, the algorithm will
6277square $y$ upto $s - 1$ times stopping only when $y \equiv -1$.  If $y^2 \equiv 1$ and $y \nequiv \pm 1$ then the algorithm can report that $a$
6278is provably composite.  If the algorithm performs $s - 1$ squarings and $y \nequiv -1$ then $a$ is provably composite.  If $a$ is not provably 
6279composite then it is \textit{probably} prime.
6280
6281EXAM,bn_mp_prime_miller_rabin.c
6282
6283
6284
6285
6286\backmatter
6287\appendix
6288\begin{thebibliography}{ABCDEF}
6289\bibitem[1]{TAOCPV2}
6290Donald Knuth, \textit{The Art of Computer Programming}, Third Edition, Volume Two, Seminumerical Algorithms, Addison-Wesley, 1998
6291
6292\bibitem[2]{HAC}
6293A. Menezes, P. van Oorschot, S. Vanstone, \textit{Handbook of Applied Cryptography}, CRC Press, 1996
6294
6295\bibitem[3]{ROSE}
6296Michael Rosing, \textit{Implementing Elliptic Curve Cryptography}, Manning Publications, 1999
6297
6298\bibitem[4]{COMBA}
6299Paul G. Comba, \textit{Exponentiation Cryptosystems on the IBM PC}. IBM Systems Journal 29(4): 526-538 (1990)
6300
6301\bibitem[5]{KARA}
6302A. Karatsuba, Doklay Akad. Nauk SSSR 145 (1962), pp.293-294
6303
6304\bibitem[6]{KARAP}
6305Andre Weimerskirch and Christof Paar, \textit{Generalizations of the Karatsuba Algorithm for Polynomial Multiplication}, Submitted to Design, Codes and Cryptography, March 2002
6306
6307\bibitem[7]{BARRETT}
6308Paul Barrett, \textit{Implementing the Rivest Shamir and Adleman Public Key Encryption Algorithm on a Standard Digital Signal Processor}, Advances in Cryptology, Crypto '86, Springer-Verlag.
6309
6310\bibitem[8]{MONT}
6311P.L.Montgomery. \textit{Modular multiplication without trial division}. Mathematics of Computation, 44(170):519-521, April 1985.
6312
6313\bibitem[9]{DRMET}
6314Chae Hoon Lim and Pil Joong Lee, \textit{Generating Efficient Primes for Discrete Log Cryptosystems}, POSTECH Information Research Laboratories
6315
6316\bibitem[10]{MMB}
6317J. Daemen and R. Govaerts and J. Vandewalle, \textit{Block ciphers based on Modular Arithmetic}, State and {P}rogress in the {R}esearch of {C}ryptography, 1993, pp. 80-89
6318
6319\bibitem[11]{RSAREF}
6320R.L. Rivest, A. Shamir, L. Adleman, \textit{A Method for Obtaining Digital Signatures and Public-Key Cryptosystems}
6321
6322\bibitem[12]{DHREF}
6323Whitfield Diffie, Martin E. Hellman, \textit{New Directions in Cryptography}, IEEE Transactions on Information Theory, 1976
6324
6325\bibitem[13]{IEEE}
6326IEEE Standard for Binary Floating-Point Arithmetic (ANSI/IEEE Std 754-1985)
6327
6328\bibitem[14]{GMP}
6329GNU Multiple Precision (GMP), \url{http://www.swox.com/gmp/}
6330
6331\bibitem[15]{MPI}
6332Multiple Precision Integer Library (MPI), Michael Fromberger, \url{http://thayer.dartmouth.edu/~sting/mpi/}
6333
6334\bibitem[16]{OPENSSL}
6335OpenSSL Cryptographic Toolkit, \url{http://openssl.org}
6336
6337\bibitem[17]{LIP}
6338Large Integer Package, \url{http://home.hetnet.nl/~ecstr/LIP.zip}
6339
6340\bibitem[18]{ISOC}
6341JTC1/SC22/WG14, ISO/IEC 9899:1999, ``A draft rationale for the C99 standard.''
6342
6343\bibitem[19]{JAVA}
6344The Sun Java Website, \url{http://java.sun.com/}
6345
6346\end{thebibliography}
6347
6348\input{tommath.ind}
6349
6350\end{document}
6351