hcrypto/libtommath/tommath.tex

894 1.  if $a.alloc \ge b$ then return(\textit{MP\_OKAY}) \\
1748 unsigned subtraction algorithm requires the result to be positive.  That is when computing $a - b$ the condition $\vert a \vert \ge \vert b\vert$ must
1759 data type must be able to represent $0 \le x < 2^{32}$ meaning that in this case $\gamma \ge 32$.
1766 \textbf{Input}.   Two mp\_ints $a$ and $b$ ($\vert a \vert \ge \vert b \vert$) \\
1798 passing variables $a$ and $b$ the condition that $\vert a \vert \ge \vert b \vert$ must be met for the algorithm to function correctly.  This
1952 \hspace{3mm}2.1  if $\vert a \vert \ge \vert b \vert$ then do (\textit{mp\_cmp\_mag}) \\
1977 \hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert \ge \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\
2285 4.  If $b \ge lg(\beta)$ then \\
2356 4.  If $b \ge lg(\beta)$ then do \\
2600 digit 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
2790 $tx \ge a.used$ or $ty < 0$ occurs.
3181 3.  If min$(a.used, b.used) \ge TOOM\_MUL\_CUTOFF$ then  \\
3183 4.  else if min$(a.used, b.used) \ge KARATSUBA\_MUL\_CUTOFF$ then \\
3391 $a_5 \cdot a_3$.  Whereas in the multiplication case we would have $5 < a.used$ and $3 \ge 0$ is maintained since we double the sum
3541 1.  If $a.used \ge TOOM\_SQR\_CUTOFF$ then  \\
3543 2.  else if $a.used \ge KARATSUBA\_SQR\_CUTOFF$ then \\
3673 Provided 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
3674 reduction 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
3783 9.  While $a \ge b$ do (\textit{mp\_cmp}) \\
3802 Technically 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
3811 $a \ge b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ only the lower $m+1$ digits are being used to compute the residue, so an implied
3816 performed at most twice, and on average once. However, if $a \ge b^2$ than it will iterate substantially more times than it should.
4087 7.  If $x \ge n$ then \\
4176 10.  If $x \ge n$ then \\
4301 5.  If $x \ge (n - k)$ then \\
4422 8.  If $x \ge n$ then \\
4544 \hspace{11.5mm}($a \ge 0$, $n > 1$, $0 < k < \beta$, $n + k$ is a power of two) \\
4548 2.  While $a \ge n$ do \\
4553 \hspace{3mm}2.5  If $a \ge n$ then do \\
5144 After 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
5315 cases $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ and $\hat k x_s \ge y_t\beta + y_{t-1} - x_s + 1$.  The latter portion of the inequalility
5323 This is trivially true since $x \ge x_s\beta^s$.  Next we replace $\hat kx_s\beta^s$ by the previous inequality for $\hat kx_s$.
5341 Which proves that $y - \hat kx \le x$ and by consequence $\hat k \ge k$ which concludes the proof.  \textbf{QED}
5387 11.  While ($x \ge y$) do \\
5629 \hspace{3mm}7.2  If $\hat w \ge b$ then \\
5834 \hspace{3mm}6.2  If $str_{iy}$ is not in the map or $y \ge r$ then goto step 7. \\
5959 In particular, we would like $a - b$ to decrease in magnitude which implies that $b \ge a$.
5982 The 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
6265 \textbf{Input}.   mp\_int $a$ and $p$, $a \ge 0$, $p \ge 3$, $p \equiv 1 \mbox{ (mod }2\mbox{)}$ \\
6375 \textbf{Input}.   mp\_int $a$ and $b$, $(a, b) = 1$, $p \ge 2$, $0 < a < p$.  \\
6397 8.  If $u \ge v$ then \\
6409 13.  While $C \ge b$ do \\
6546 \textbf{Input}.   mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$.  \\
6583 \textbf{Input}.   mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$.  \\