【算法编程】基于Miller-Rabin的大素数测试
基本原理：
费尔马小定理:如果p是一个素数,且0
利用费尔马小定理,对于给定的整数n,可以设计素数判定算法,通过计算d=a^(n-1)%n来判断n的素性,当d!=1时,n肯定不是素数,当d=1时,n 很可能是素数.
二次探测定理:如果p是一个素数,且0
利用二次探测定理,可以再利用费尔马小定理计算a^(n-1)%n的过程中增加对整数n的二次探测,一旦发现违背二次探测条件,即得出n不是素数的结论.
如果n是素数,则(n-1)必是偶数,因此可令(n-1)=m*(2^q),其中m是正奇数(若n是偶数,则上面的m*(2^q)一定可以分解成一个正奇数乘以2的k次方的形式 ),q是非负整数,考察下面的测试:
序列:
a^m%n; a^(2m)%n; a^(4m)%n; ……;a^(m*2^q)%n
Miller-Rabin素性测试伪代码描述：
1、找出整数k,q,其中k>0,q是奇数，使（n-1=2kq）。
2、随机选取整数a,1
3、Ifaq mod n=1, printf("该数可能是素数!\n");
4、Forj=0 to k-1 , if a^(2^j*q) mod n = n – 1, printf("该数可能是素数!\n");如果步骤3、4都不成立，则printf("该数肯定不是素数!\n")
5、当该数可能是素数时，随机选取整数a,1
具体代码实现：
1.、BigInt.h文件
#ifndef _BIGNUM_H_
#define _BIGNUM_H_
#define SIZE 128 //一个大整数用个字节保存，最多表示位大整数
#define SIZE_2 2* SIZE
typedef unsigned char UCHAR;
typedef unsigned short USHORT;
UCHAR atox(char ch); //将一个十六进制的字符(4位)转位数字，转换失败返回xff
typedef struct BigNum //大整数结构
{
UCHAR data[SIZE]; //空间为(SIZE * sizeof(UCHAR))，就是SIZE个字节
}BigNum;
BigNum Init(char *str); //初始化大整数，str为十六进制字符串
int GetBitFront(BigNum bignum); //有多少bit (前面的0不算)
int GetBitEnd(BigNum bignum); //有多少0(即只算末尾的0个数)
BigNum MovBitLetf(BigNum bignum, int n);//向左移n位
BigNum MovBitRight(BigNum bignum, int n); //右移n位
int Cmp(BigNum bignum_a, BigNum bignum_b); //大整数比较大小，>返回1，<返回-1，==返回0
BigNum Mod(BigNum bignum_a, BigNum bignum_b); //大整数模运算
BigNum Sub(BigNum bignum_a, BigNum bignum_b); //大整数减法
void print2(BigNum bignum); //以二进制打印
BigNum Mul(BigNum bignum_a, BigNum bignum_b); //大整数乘法
BigNum Div(BigNum bignum_a, BigNum bignum_b); //大整数除法
BigNum Add(BigNum bignum_a, BigNum bignum_b); //大整数加法
void print10(BigNum bignum);//以十进制打印
int b2d(BigNum bignum); //二进制到转十进制
BigNum modMDyn(BigNum a, BigNum power, BigNum mod); //求大整数幂的模
BigNum d2b(int num); //十进制转二进制
int checkprime(BigNum n,BigNum a);
#endif
#include
#include
#include
#include "BigInt.h"
#include "math.h"
#include
void print2(BigNum bignum)//以二进制打印
{
if(GetBitFront(bignum)==0)
printf("0\n");
else
{
for(int i=SIZE-GetBitFront(bignum);i=0;i--)
bignum.data[i]='0';
return bignum;
}
int GetBitFront(BigNum bignum) //有多少bit(前面的0不算)
{
int BitOfBigNum = SIZE;
int i=0;
while ((bignum.data[i] == '0')&& (BitOfBigNum > 0))
{
i++;
BitOfBigNum--;
}
return BitOfBigNum;
}
int GetBitEnd(BigNum bignum) //有多少0(即只算末尾的0个数)
{
int BitOfBigNum = SIZE;
int num=0;
while ((bignum.data[BitOfBigNum -1] == '0') && (BitOfBigNum > 0))
{
num++;
BitOfBigNum--;
}
return num;
}
BigNum MovBitLetf(BigNum bignum, int n)//向左移n位
{
int bignum_len =GetBitFront(bignum);
for (int i =SIZE- bignum_len; i=SIZE-bignum_len; i--)
{
if (i<0)
{
continue;
}
bignum.data[i] =bignum.data[i-n];
}
for (i =0; i 返回1，<返回-1，==返回0
{
int bignum_a_len =GetBitFront(bignum_a);
int bignum_b_len =GetBitFront(bignum_b);
if(bignum_a_len>bignum_b_len)return 1;
if(bignum_a_len bignum_b.data[i])
{
return 1;
}
if (bignum_a.data[i]< bignum_b.data[i])
{
return -1;
}
}
}
return 0;
}
BigNum Sub(BigNum bignum_a, BigNum bignum_b) //大整数减法
{
BigNum bignum_c;
int temp=0;
int temp1=0;
int carry = 0;
int i;
int j=0;
for (i = SIZE-1; i >=0; i--)
{
temp = bignum_a.data[i] -bignum_b.data[i] -carry;
temp1=temp;
if(temp==-1)
temp1=1;
if(temp==-2)
temp1=0;
bignum_c.data[i] =temp1+48;
if(temp<0)
carry=1;
else
carry=0;
j++;
}
return bignum_c;
}
BigNum Mod(BigNum bignum_a, BigNum bignum_b) //大整数模运算
{
BigNum bignum_c =Init("0");
BigNum B;
B = bignum_b;
int bignum_a_len;
int bignum_b_len;
int bignum_c_len;
if (Cmp(bignum_b, bignum_c) == 0)
{
printf("错误！除数为\n");
return bignum_c;
}
bignum_a_len =GetBitFront(bignum_a);
bignum_b_len =GetBitFront(bignum_b);
bignum_c_len = bignum_a_len -bignum_b_len;
while (bignum_c_len >= 0)
{
B = MovBitLetf(bignum_b,bignum_c_len);
int m=0;
m=Cmp(bignum_a, B);
while (Cmp(bignum_a, B) !=-1)//大于等于
{
bignum_a =Sub(bignum_a, B);
}
bignum_c_len--;
}
return bignum_a;
}
BigNum Mul(BigNum bignum_a, BigNum bignum_b) //大整数乘法
{
BigNum bignum_c =Init("0");
BigNum bignum=Init("0");
int wei=0;
wei=GetBitFront(bignum_a)+GetBitFront(bignum_b)-1;
int carry[SIZE_2];
int carry1[SIZE_2];
int mod[SIZE_2];
for(int k=0;k<=SIZE_2;k++)
{
carry[k]=0;
carry1[k]=0;
mod[k]=0;
}
int i=0;
int j=0;
for(i=SIZE-1;i>=0;i--)
{
for(j=SIZE-1;j>=0;j--)
carry[i+j+1]=(bignum_a.data[i]-48)*(bignum_b.data[j]-48)+carry[i+j+1];
}
for(k=SIZE_2-1;k>=0;k--)
{
if(k==SIZE_2-1)
carry1[k]=carry[k];
else
carry1[k]=carry1[k+1]/2+carry[k];
}
wei=GetBitFront(bignum_a)+GetBitFront(bignum_b)-1;
bignum=d2b(carry1[SIZE_2-wei]);
for(i=SIZE-1,j=SIZE_2-wei;i>=0&&j>=0;i--,j--)
carry1[j]=bignum.data[i]-48;
for(k=0;k=k;i--,j--)
{
bignum_c.data[i]=carry1[j]%2+48;
}
return bignum_c;
}
BigNum Div(BigNum bignum_a, BigNum bignum_b) //大整数除法
{
BigNum bignum_c =Init("0");
BigNum B;
int bignum_a_len;
int bignum_b_len;
int bignum_c_len;
if (Cmp(bignum_b, bignum_c) == 0)
{
printf("错误！除数为\n");
return bignum_c;
}
bignum_a_len =GetBitFront(bignum_a);
bignum_b_len = GetBitFront(bignum_b);
bignum_c_len = bignum_a_len -bignum_b_len;
while (bignum_c_len >= 0)
{
B = MovBitLetf(bignum_b,bignum_c_len);
while (Cmp(bignum_a, B) !=-1)
{
bignum_a =Sub(bignum_a, B);
bignum_c.data[SIZE-1-bignum_c_len]++;
}
bignum_c_len--;
}
return bignum_c;
}
BigNum Add(BigNum bignum_a, BigNum bignum_b) //大整数加法
{
BigNum bignum_c;
int temp;
int carry = 0;
int i;
for (i = SIZE-1; i>=0; i--)
{
temp = bignum_a.data[i]-48+ bignum_b.data[i]-48 + carry;
if(temp==2)
{
temp=0;
carry=1;
}
else if(temp==3)
{
temp=1;
carry=1;
}
else carry=0;
bignum_c.data[i] = temp+48;
}
return bignum_c;
}
int b2d(BigNum bignum) //二进制转十进制
{
int n=0;
int j=0;
int result=0;
n=GetBitFront(bignum);
for(int i=SIZE-1;i>=0;i--)
{
result=result+(bignum.data[i]-48)*pow(2,j);
j++;
}
return result;
}
void print10(BigNum bignum) //打印十进制大整数
{
int temp[SIZE];
int i = 0;
int j;
BigNum c;
while (Cmp(bignum,Init("0")) == 1)
{
c=Mod(bignum,Init("1010"));
temp[i] = b2d(c);
bignum = Div(bignum,Init("1010"));
i++;
}
for (j = i - 1; j >= 0; j--)
{
printf("%d",temp[j]);
}
printf("\n");
}
BigNum modMDyn(BigNum a, BigNum power, BigNum mod) //求大整数幂的模
{
BigNum temp;
BigNum result;
BigNum t1;
temp=Mod(a,mod);
result=Init("1");
for(inti=SIZE-1;i>=SIZE-GetBitFront(power);i--)
{
if(power.data[i]=='1')
{
t1=Mul(result,temp);
result=Mod(Mul(result,temp),mod);
}
temp=Mod(Mul(temp,temp),mod);
}
return result;
}
BigNum d2b(int num) //十进制转二进制
{
BigNum bignum;
bignum=Init("0");
int a=0;
int b=0;
int i=1;
while(num>0)
{
a=num%2;
num=num/2;
bignum.data[SIZE-i]=a+48;
i++;
}
return bignum;
}
int checkprime(BigNum n,BigNum a)
{
BigNum k;
BigNum q;
// BigNum a;
BigNum n1;//n1=n-1
BigNum num1;//num1为常数1
BigNum num2;//num2为常数2
BigNum k2;//2^k
BigNum k22;
int k1=0; //末尾0的个数
num1=Init("1");
num2=Init("10");
k=Init("0");
q=Init("0");
n1=Init("0");
k22=Init("10");
// a=Init("1010");//选择的数
n1=Sub(n,num1);
k1=GetBitEnd(n1);
k=d2b(k1);
q=MovBitRight(n1,k1);
k2=Div(n1,q);
if(Cmp(modMDyn(a,q,n),num1)==0)
{
// print2(n);
// printf("该数可能是素数!\n");
return 1;
}
for(int i=0;i 运行结果如下：
原文：http://blog.csdn.net/tengweitw/article/details/23952839
作者：nineheadedbird