Lucas-Lehmer test: Difference between revisions
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Mersenne primes:
M2 M3 M5 M7 M13 M17 M19 M31
=={{header|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
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Revision as of 04:39, 16 February 2008
Lucas-Lehmer test
You are encouraged to solve this task according to the task description, using any language you may know.
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 precision.
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=20001 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 = ENTIER(SHORTEN long log(SHORTEN LONG LONG REAL(long long max int))/log(2)/2); printf(($" Finding Mersenne primes in M[2.."g(0)"]: "l$,upb)); FOR p FROM 2 TO upb DO IF is prime(p) THEN IF is mersenne prime(p) THEN printf (($" M"g(0)$,p)) FI FI OD )
Output:
Finding Mersenne primes in M[2..33253]: 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