Miller_robin Determination of large prime numbers

According to acdreamer's blog http://blog.csdn.net/acdreamers/article/details/7913786

On 51nod, I saw a prime number in the range of 10 ^ 30. It is definitely not good to create a table. We should use Miller-Robin to determine the addition of the biginteger in Java.

First, based on the Fermat small theorem, when gcd (A, P) = 1, a ^ (PM) % P = 1. Therefore, after a random number of rand (2, P-1) is generated, if the input is a prime number, then there must be a POW (p-1) % P = 1. Here we use the Fast Power modulo. After a probe, if the result is not 1, P must not be a prime number; otherwise, 75% of the probability is a prime number, so there are 10 probes

1-(0.75) ^ 10 = the probability of 0.94369 is a prime number (as if so ). Of course, you can find 20 more times, with a success rate of 0.996828.

 

There is also an optimization here, that is, according to the second probe theorem described in the above blog x ^ 2 = 1 (mod P), when P is a prime number, X must be 1 or P-1, then there is an optimization, first filtering out the two power of the P-1, assuming that the P-1 = m * 2 ^ sqr, first calculating x = POW (a, m) % P, then, check whether y = x * x % P = 1, but X is not 1, P-1 indicates that P is not a prime number.

The Code is as follows:

Import Java. io. *; import Java. util. *; import Java. math. biginteger; public class minller {public static final int iter = 10; public static biginteger fast_fac (biginteger A, biginteger N, biginteger mod) {biginteger two = biginteger. valueof (2), ans; A =. moD (MOD); ans = biginteger. one; while (! (N. equals (biginteger. zero) {If (n. moD (two )). equals (biginteger. one) {ans = (ans. multiply ()). moD (MOD); n = n. subtract (biginteger. one);} n = n. divide (two); A = (. multiply ()). moD (MOD);} return ans;} public static Boolean millerrobin (biginteger p) {biginteger two = biginteger. valueof (2), a, X, N, y = biginteger. zero, P_1; random RND = new random (); P_1 = P. subtract (biginteger. one); If (P. equals (biginteger. one) r Eturn false; If (P. equals (two) return true; If (P. moD (two )). equals (biginteger. zero) return false; int Max, sqr = 0; // filter out the power of 2 n = P. subtract (biginteger. one); While (n. moD (two )). equals (biginteger. zero) {++ sqr; n = n. divide (two);} If (N. compareto (biginteger. valueof (10000) = 1) max = 10000; else max = n. intvalue (); For (INT I = 0; I <ITER; ++ I) {A = biginteger. valueof (RND. nextint (MAX-2) + 2); X = fast_fac (A, N, P); If (X. equals (biginteger. one) continue; For (Int J = 0; j <sqr; ++ J) {Y = (X. multiply (x )). moD (p); If (Y. equals (biginteger. one )&&! X. Equals (biginteger. One )&&! X. equals (P_1) return false; X = y;} If (Y. equals (biginteger. one) = false) return false;} return true;} public static void main (string [] argv) {using CIN = new using (system. in); While (CIN. hasnextbiginteger () {biginteger n = cin. nextbiginteger (); If (millerrobin (N) system. out. println ("yes"); else system. out. println ("no ");}}}

Miller_robin Determination of large prime numbers