Lucas-Lehmer test
You are encouraged to solve this task according to the task description, using any language you may know.
Lucas-Lehmer Test: for p a prime, the Mersenne number 2**p-1 is prime if and only if 2**p-1 divides S(p-1) where S(n+1)=S(n)**2-2, and S(1)=4.
The following programs calculate all Mersenne primes up to the implementation's maximium precision, or the 45th Mersenne prime. (Which ever comes first).
Ada
with Ada.Text_Io; use Ada.Text_Io;
with Ada.Integer_Text_Io; use Ada.Integer_Text_Io;
procedure Lucas_Lehmer_Test is
type Ull is mod 2**64;
function Mersenne(Item : Integer) return Boolean is
S : Ull := 4;
MP : Ull := 2**Item - 1;
begin
if Item = 2 then
return True;
else
for I in 3..Item loop
S := (S * S - 2) mod MP;
end loop;
return S = 0;
end if;
end Mersenne;
Upper_Bound : constant Integer := 64;
begin
Put_Line(" Mersenne primes:");
for P in 2..Upper_Bound loop
if Mersenne(P) then
Put(" M");
Put(Item => P, Width => 1);
end if;
end loop;
end Lucas_Lehmer_Test;
Output:
Mersenne primes: M2 M3 M5 M7 M13 M17 M19 M31
ALGOL 68
main:(
PRAGMAT stack=1M precision=20000 PRAGMAT
PROC is prime = ( INT p )BOOL:
IF p = 2 THEN TRUE
ELIF p <= 1 OR p MOD 2 = 0 THEN FALSE
ELSE
BOOL prime := TRUE;
FOR i FROM 3 BY 2 TO ENTIER sqrt(p)
WHILE prime := p MOD i /= 0 DO SKIP OD;
prime
FI;
PROC is mersenne prime = ( INT p )BOOL:
IF p = 2 THEN TRUE
ELSE
LONG LONG INT m p := LONG LONG 2 ** p - 1, s := 4;
FROM 3 TO p DO
s := (s ** 2 - 2) MOD m p
OD;
s = 0
FI;
INT upb prime = ( long long bits width - 1 ) OVER 2; # no unsigned #
INT upb count = 45; # find 45 mprimes if INT has enough bits #
printf(($" Finding Mersenne primes in M[2.."g(0)"]: "l$,upb prime));
INT count:=0;
FOR p FROM 2 TO upb prime WHILE
IF is prime(p) THEN
IF is mersenne prime(p) THEN
printf (($" M"g(0)$,p));
count +:= 1
FI
FI;
count <= upb count
DO SKIP OD
)
Output:
Finding Mersenne primes in M[2..33252]: M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213 M19937 M21701 M23209
See also: http://www.xs4all.nl/~jmvdveer/mersenne.a68.html
C
Compiler options: gcc -std=c99 -lm Lucas-Lehmer_test.c -o Lucas-Lehmer_test
#include <math.h>
#include <stdio.h>
#include <limits.h>
#pragma precision=log10l(ULLONG_MAX)/2
typedef enum { FALSE=0, TRUE=1 } BOOL;
BOOL is_prime( int p ){
if( p == 2 ) return TRUE;
else if( p <= 1 || p % 2 == 0 ) return FALSE;
else {
BOOL prime = TRUE;
const int to = sqrt(p);
int i;
for(i = 3; i <= to; i+=2)
if (!(prime = p % i))break;
return prime;
}
}
BOOL is_mersenne_prime( int p ){
if( p == 2 ) return TRUE;
else {
const long long unsigned m_p = ( 1LLU << p ) - 1;
long long unsigned s = 4;
int i;
for (i = 3; i <= p; i++){
s = (s * s - 2) % m_p;
}
return s == 0;
}
}
int main(int argc, char **argv){
const int upb = log2l(ULLONG_MAX)/2;
int p;
printf(" Mersenne primes:\n");
for( p = 2; p <= upb; p += 1 ){
if( is_prime(p) && is_mersenne_prime(p) ){
printf (" M%u",p);
}
}
printf("\n");
}
Output:
Mersenne primes: M2 M3 M5 M7 M13 M17 M19 M31
C++
#include <iostream>
#include <gmpxx.h>
mpz_class pow2(mpz_class exp);
bool is_mersenne_prime(mpz_class p);
int main()
{
mpz_class maxcount(45);
mpz_class found(0);
mpz_class check(0);
for( mpz_nextprime(check.get_mpz_t(), check.get_mpz_t());
found < maxcount;
mpz_nextprime(check.get_mpz_t(), check.get_mpz_t()))
{
//std::cout << "P" << check << " " << std::flush;
if( is_mersenne_prime(check) )
{
++found;
std::cout << "M" << check << " " << std::flush;
}
}
}
bool is_mersenne_prime(mpz_class p)
{
if( 2 == p )
return true;
else
{
mpz_class div = pow2(p) - mpz_class(1);
mpz_class s(4);
mpz_class s(4);
for( mpz_class i(3);
i <= p;
++i )
{
s = (s * s - mpz_class(2)) % div ;
}
return ( s == mpz_class(0) );
}
}
mpz_class pow2(mpz_class exp)
{
// Unfortunately, GMP doesn't have a left-shift method.
// It also doesn't have a pow() equivalent that takes arbitrary-precision exponents.
// So we have to do it the hard (and presumably slow) way.
mpz_class ret(2);
mpz_class ret(2);
for(mpz_class i(1); i < exp; ++i)
ret *= mpz_class(2);
ret *= mpz_class(2);
//std::cout << "pow2( " << exp << " ) = " << ret << std::endl;
return ret;
}
Output: (Incomplete; It takes a long time.)
M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213 M19937 M21701 M23209
Haskell
module Main
where
main = printMersennes $ take 45 $ filter lucasLehmer $ sieve [2..]
s mp 1 = 4 `mod` mp
s mp n = ((s mp $ n-1)^2-2) `mod` mp
lucasLehmer 2 = True
lucasLehmer p = s (2^p-1) (p-1) == 0
printMersennes [] = return ()
printMersennes (x:xs) = do putStrLn $ "M"++(show x)
printMersennes xs
It is pointed out on the Sieve of Eratosthenes page that the following "sieve" is inefficient. Nonetheless it takes very little time compared to the Lucas-Lehmer test itself.
sieve (p:xs) = p : sieve [x | x <- xs, x `mod` p > 0]
It takes about 30 minutes to get up to:
M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213
Java
We use arbitrary-precision integers in order to be able to test any arbitrary prime.
import java.math.BigInteger;
public class Mersenne
{
public static boolean isPrime(int p) {
if (p == 2)
return true;
else if (p <= 1 || p % 2 == 0)
return false;
else {
int to = (int)Math.sqrt(p);
for (int i = 3; i <= to; i += 2)
if (p % i == 0)
return false;
return true;
}
}
public static boolean isMersennePrime(int p) {
if (p == 2)
return true;
else {
BigInteger m_p = BigInteger.ONE.shiftLeft(p).subtract(BigInteger.ONE);
BigInteger s = BigInteger.valueOf(4);
for (int i = 3; i <= p; i++)
s = s.multiply(s).subtract(BigInteger.valueOf(2)).mod(m_p);
return s.equals(BigInteger.ZERO);
}
}
// an arbitrary upper bound can be given as an argument
public static void main(String[] args) {
int upb;
if (args.length == 0)
upb = 500;
else
upb = Integer.parseInt(args[0]);
System.out.println(" Finding Mersenne primes in M[2.." + upb + "]: ");
for (int p = 3; p <= upb; p += 2)
if (isPrime(p) && isMersennePrime(p))
System.out.print(" M" + p);
System.out.println();
}
}
Output (after about eight hours):
Finding Mersenne primes in M[2..2147483647]: M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213
Oz
Oz's multiple precision number system use GMP core. <ocaml>%% compile : ozc -x <file.oz> functor import
Application System
define
fun {Arg Idx Default}
Cmd = {Application.getArgs plain}
Len = {Length Cmd}
in
if Len < Idx then
Default
else
{StringToInt {Nth Cmd Idx}}
end
end
fun {LLtest N}
Mp = {Pow 2 N} - 1
fun {S K} X T
in
if K == 1 then 4
else
T = {S K-1}
X = T * T - 2
X mod Mp
end
end
in
if N == 2 then
true
else
{S N-1} == 0
end
end
proc {FindLL X}
fun {Sieve Ls}
case Ls of nil then nil
[] X|Xs then
fun {DIV M} M mod X \= 0 end
in
X|{Sieve {Filter Xs DIV}}
end
end
in
if {IsList X} then
case X of nil then skip
[] M|Ms then
{System.printInfo "M"#M#" "}
{FindLL Ms}
end
else
{FindLL {Filter {Sieve 2|{List.number 3 X 2}} LLtest}}
end
end
Num = {Arg 1 607}
{FindLL Num}
{Application.exit 0}
end</ocaml>
Pop11
Checking large numbers takes a lot of time so we limit p to be smaller than 1000.
define Lucas_Lehmer_Test(p);
lvars mp = 2**p - 1, sn = 4, i;
for i from 2 to p - 1 do
(sn*sn - 2) rem mp -> sn;
endfor;
sn = 0;
enddefine;
lvars p = 3;
printf('M2', '%p\n');
while p < 1000 do
if Lucas_Lehmer_Test(p) then
printf('M', '%p');
printf(p, '%p\n');
endif;
p + 2 -> p;
endwhile;
The output (obtained in few seconds) is:
M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607
Python
from sys import stdout
from math import sqrt, log
fi = od = None
def is_prime ( p ):
if p == 2: return True
elif p <= 1 or p % 2 == 0: return False
else:
prime = True;
for i in range(3, int(sqrt(p))+1, 2 ): # do
if p % i == 0: return False
else:
return True
fi
fi;
def is_mersenne_prime ( p ):
if p == 2: return True
else:
m_p = ( 1 << p ) - 1; s = 4;
for i in range(3, p+1): # do
s = (s ** 2 - 2) % m_p
od;
return s == 0
fi;
precision = 20000; # maximum requested number of decimal places of 2 ** MP-1 #
long_bits_width = precision / log(2) * log(10);
upb_prime = int( long_bits_width - 1 ) / 2; # no unsigned #
upb_count = 45; # find 45 mprimes if int was given enough bits #
print ((" Finding Mersenne primes in M[2..%d]:"%upb_prime));
count=0;
for p in range(2, upb_prime+1): # do
if is_prime(p) and is_mersenne_prime(p):
print("M%d"%p),
stdout.flush();
count += 1
fi;
if count >= upb_count: break
od
print
Output:
Finding Mersenne primes in M[2..33218]: M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127 M521 M607 M1279 M2203 M2281 M3217 M4253 M4423 M9689 M9941 M11213 M19937 M21701 M23209
Scheme
;;;The heart of the algorithm
(define (S n)
(let ((m (- (expt 2 n) 1)))
(let loop ((c (- n 2)) (a 4))
(if (zero? c)
a
(loop (- c 1) (remainder (- (* a a) 2) m))))))
(define (mersenne-prime? n)
(if (= n 2)
#t
(zero? (S n))))
;;;Trivial unoptimized implementation for the base primes
(define (next-prime x)
(if (prime? (+ x 1))
(+ x 1)
(next-prime (+ x 1))))
(define (prime? x)
(let loop ((c 2))
(cond ((>= c x) #t)
((zero? (remainder x c)) #f)
(else (loop (+ c 1))))))
;;Main loop
(let loop ((i 45) (p 2))
(if (not (zero? i))
(if (mersenne-prime? p)
(begin
(display "M") (display p) (display " ")
(loop (- i 1) (next-prime p)))
(loop i (next-prime p)))))
M2 M3 M5 M7 M13...