Pollard_rho 因数分解

Int64以内Rabin-Miller强伪素数测试和Pollard 因数分解的算法实现

选取随机数\(a\) 随机数\(b\),检查\(gcd(a - b, n)\)是否大于1,若大于1则\(a - b\)\(n\)的一个因数

实现1:floyd判环

利用多项式\(f(x)\)迭代出\({x_0, x_1, \dots, x_k}\)

设定\(x = y = x_0\)的初始值,选用多项式进行迭代,每次:\(x = f(x)\), \(y = f(f(y))\),即:\(x = x_k, y = x_{2k}\)\(x == y\)时出现循环

\(x = y = 2\)\(f(n) = n^2 + a\)

typedef long long ll;

ll mul_mod(ll a, ll b, ll m){
    ll ans = 0, exp = a;
    while(a >= m) a -= m;
    while(b){
        if(b & 1){
            ans += exp;
            while(ans >= m) ans -= m;
        }
        exp += exp;
        while(exp >= m) exp -= m;
        b >>= 1;
    }
    return ans;
}

ll pollard_rho(ll n, int a){
    ll x = 2, y = 2, d = 1;
    while(d == 1){
        x = mul_mod(x, x, n) + a;
        y = mul_mod(y, y, n) + a;
        y = mul_mod(y, y, n) + a;
        d = __gcd((x >= y ? x - y : y - x), n);
    }
    if(d == n) return pollard_rho(n, a + 1);
    return d;
}

实现2: brent判环(更高效)

不同于floyd每次计算\(x_k, x_{2k}\)进行判断,brent每次只计算\(x_k\),当k是2的方幂时,\(y = x_k\),每次计算\(d = gcd(x_k - y, n)\)

typedef long long ll;

ll mul_mod(ll a, ll b, ll m){
    ll ans = 0, exp = a;
    while(a >= m) a -= m;
    while(b){
        if(b & 1){
            ans += exp;
            while(ans >= m) ans -= m;
        }
        exp += exp;
        while(exp >= m) exp -= m;
        b >>= 1;
    }
    return ans;
}

ll pollard_rho(ll n, int a){
    ll x = 2, y = 2, d = 1, k = 0, i = 1;
    while(d == 1){
        ++k;
        x = mul_mod(x, x, n) + a;
        d = __gcd(x >= y ? x - y : y - x, n);
        if(k == i){
            y = x;
            i <<= 1;
        }
    }
    if(d == n) return pollard_rho(n, a + 1);
    return d;
}
posted @ 2017-01-25 13:26  book丶book丶  阅读(2047)  评论(0编辑  收藏  举报