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	Add exercise: [11] Lucas–Lehmer primality test - libzahl - big integer library
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	commit 611db6bdb6a5a08b628f571451a94a1147a0e16f
	parent f2734fde832c3f45d0dc9350c561d4137a422788
	Author: Mattias Andrée <[email protected]>
	Date: Mon, 25 Jul 2016 00:50:52 +0200

	Add exercise: [11] Lucas–Lehmer primality test

	Signed-off-by: Mattias Andrée <[email protected]>

	Diffstat:
	M doc/exercises.tex \| 85 +++++++++++++++++++++++++++++…

	1 file changed, 85 insertions(+), 0 deletions(-)
	---
	diff --git a/doc/exercises.tex b/doc/exercises.tex
	@@ -143,6 +143,43 @@ called a base.)



	+\item {[\textit{11}]} \textbf{Lucas–Lehmer primality test}
	+
	+The Lucas–Lehmer primality test can be used to determine
	+whether a Mersenne numbers $M_n = 2^n - 1$ is a prime (a
	+Mersenne prime). $M_n$ is a prime if and only if
	+$s_{n - 1} \equiv 0 ~(\text{Mod}~M_n)$, where
	+
	+\( \displaystyle{
	+ s_i = \left \{ \begin{array}{ll}
	+ 4 & \text{if} ~ i = 0 \\
	+ s_{i - 1}^2 - 2 & \text{otherwise}.
	+ \end{array} \right .
	+}\)
	+
	+\noindent
	+The Lucas–Lehmer primality test requires that $n \ge 3$,
	+however $M_2 = 2^2 - 1 = 3$ is a prime. Implement a version
	+of the Lucas–Lehmer primality test that takes $n$ as the
	+input. For some more fun, when you are done, you can
	+implement a version that takes $M_n$ as the input.
	+
	+For improved performance, instead of using \texttt{zmod},
	+you can use the recursive function
	+%
	+\( \displaystyle{
	+ k ~\mbox{Mod}~ 2^n - 1 =
	+ \left (
	+ (k ~\mbox{Mod}~ 2^n) + \lfloor k \div 2^n \rfloor
	+ \right ) ~\mbox{Mod}~ 2^n - 1,
	+}\)
	+%
	+where $k ~\mbox{Mod}~ 2^n$ is efficiently calculated
	+using \texttt{zand($k$, $2^n - 1$)}. (This optimisation
	+is not part of the difficulty rating of this problem.)
	+
	+
	+
	\end{enumerate}


	@@ -348,4 +385,52 @@ enum zprimality ptest_fermat(z_t witness, z_t p, int t)



	+\item \textbf{Lucas–Lehmer primality test}
	+
	+\vspace{-1em}
	+\begin{alltt}
	+enum zprimality ptest_llt(z_t n)
	+\{
	+ z_t s, M;
	+ int c;
	+ size_t p;
	+
	+ if ((c = zcmpu(n, 2)) <= 0)
	+ return c ? NONPRIME : PRIME;
	+
	+ if (n->used > 1) \{
	+ \textcolor{c}{/* \textrm{An optimised implementation would not need th…
	+ errno = ENOMEM;
	+ return (enum zprimality)(-1);
	+ \}
	+
	+ zinit(s), zinit(M), zinit(2);
	+
	+ p = (size_t)(n->chars[0]);
	+ zsetu(s, 1), zsetu(M, 0);
	+ zbset(M, M, p, 1), zsub(M, M, s);
	+ zsetu(s, 4);
	+ zsetu(two, 2);
	+
	+ p -= 2;
	+ while (p--) \{
	+ zsqr(s, s);
	+ zsub(s, s, two);
	+ zmod(s, s, M);
	+ \}
	+ c = zzero(s);
	+
	+ zfree(two), zfree(M), zfree(s);
	+ return c ? PRIME : NONPRIME;
	+\}
	+\end{alltt}
	+
	+$M_n$ is composite if $n$ is composite, therefore,
	+if you do not expect prime-only values on $n$, the
	+performance can be improve by using some other
	+primality test (or this same test if $n$ is a
	+Mersenne number) to first check that $n$ is prime.
	+
	+
	+
	\end{enumerate}