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;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Lucas_Lehmer_Test is
function Mersenne(Item : Integer) return Boolean is
S : Long_Long_Integer := 4;
MP : Long_Long_Integer := 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 := Integer(Float'Rounding(Log(10.0) / Log(2.0) * 10_000_000.0));
M_Count : Natural := 0;
begin
Put_Line(" Mersenne primes:");
for P in 2..Upper_Bound loop
if Mersenne(P) then
Put(" M");
Put(Item => P, Width => 1);
M_Count := M_Count + 1;
exit when M_Count = 45;
end if;
end loop;
end Lucas_Lehmer_Test;
Output: (Output was truncated by an arithmetic overflow exception.)
Mersenne primes: M2 M3 M5 M7 M13 M17 M19 M31
ALGOL 68
Interpretor: algol68g-mk11
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 * s - 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
#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);
s = s % 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");
}
Compiler version: gcc version 4.1.2 20070925 (Red Hat 4.1.2-27)
Compiler options: gcc -std=c99 -lm Lucas-Lehmer_test.c -o Lucas-Lehmer_test
Output:
Mersenne primes: M2 M3 M5 M7 M13 M17 M19 M31
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 = 2; p <= upb; p++)
if (isPrime(p) && isMersennePrime(p))
System.out.print(" M" + p);
System.out.println();
}
}
Output:
Finding Mersenne primes in M[2..500]: M2 M3 M5 M7 M13 M17 M19 M31 M61 M89 M107 M127
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 ):
# print "p:",p
if p % i == 0 : return False
else:
return True
fi
fi;
def is_mersenne_prime ( p ):
if p == 2 : return True
else:
m_p = 2 ** p - 1; s = 4;
for i in range(3, p+1):
s = (s * s - 2) % m_p
od;
return s == 0
fi;
precision = 20000;
long_long_bits_width = precision / log(2) * log(10);
upb_prime = int( long_long_bits_width - 1 ) / 2; # no unsigned #
upb_count = 45; # find 45 mprimes if int has enough bits #
print ((" Finding Mersenne primes in M[2..%d]: "%upb_prime));
count=0;
for p in range(2, upb_prime+1):
if is_prime(p) :
if is_mersenne_prime(p) :
stdout.write(" M%d"%p);
stdout.flush();
count += 1
fi
fi;
if count >= upb_count : break; fi
od
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