Tonelli–Shanks Algorithm 二次剩余系解法 (Ural 1132. Square Root)

wikipedia上的解释和证明：http://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm

The Tonelli–Shanks algorithm (referred to by Shanks as the RESSOL algorithm) is used within modular arithmetic to solve a congruence of the form

$x^2 \equiv n \pmod p$

where n is a quadratic residue (mod p), and p is an odd prime.

Tonelli–Shanks cannot be used for composite moduli; finding square roots modulo composite numbers is a computational problem equivalent to integer factorization.^[1]

An equivalent, but slightly more redundant version of this algorithm was developed by Alberto Tonelli in 1891. The version discussed here was developed independently by Daniel Shanksin 1973, who explained:

"My tardiness in learning of these historical references was because I had lent Volume 1 of Dickson's History to a friend and it was never returned."^[2]

(Note: All $\equiv$ are taken to mean $\pmod p$ , unless indicated otherwise).[edit]The algorithm

Inputs: p, an odd prime. n, an integer which is a quadratic residue (mod p), meaning that the Legendre symbol $\bigl(\tfrac{n}{p}\bigr)=1$ .

Outputs: R, an integer satisfying $R^2 \equiv n$ .

Factor out powers of 2 from p − 1, defining Q and S as: $p-1 = Q2^S$ with Q odd. Note that if $S = 1$ , i.e. $p \equiv 3 \pmod 4$ , then solutions are given directly by $R \equiv \pm n^{\frac{p+1}{4}}$ .
Select a z such that the Legendre symbol $\bigl(\tfrac{z}{p}\bigr)=-1$ (that is, z should be a quadratic non-residue modulo p), and set $c \equiv z^Q$ .
Let $R \equiv n^{\frac{Q+1}{2}}, t\equiv n^Q, M = S.$
Loop:
1. If $t \equiv 1$ , return R.
2. Otherwise, find the lowest i, $0 < i < M$ , such that $t^{2^i} \equiv 1$ ; e.g. via repeated squaring.
3. Let $b \equiv c^{2^{M-i-1}}$ , and set $R \equiv Rb, \; t \equiv tb^2, c \equiv b^2$ and $M =\; i$ .

Once you have solved the congruence with R the second solution is p − R.

Example

Solving the congruence $x^2 \equiv 10 \pmod {13}$ . It is clear that $13$ is odd, and since $10^{\frac{13-1}{2}} = 10^6 \equiv 1 \pmod {13}$ , 10 is a quadratic residue (by Euler's criterion).

Step 1: Observe $p-1 = 12 = 3 \cdot 2^2$ so $Q=3$ , $S=2$ .

Step 2: Take $z=2$ as the quadratic nonresidue (2 is a quadratic nonresidue since $2^{\frac{13-1}{2}} = -1 \pmod {13}$ (again, Euler's criterion)). Set $c = 2^3 \equiv 8 \pmod {13}.$

Step 3: $R=10^2 \equiv -4, \; t\equiv 10^3 \equiv -1 \pmod {13}, M = 2.$

Step 4: Now we start the loop: so ; i.e.
- Let $b \equiv 8^{2^{2-1-1}} \equiv 8 \pmod {13}$ , so $b^2 \equiv 8^2 \equiv -1 \pmod {13}$ .
- Set $R=-4\cdot8 \equiv 7 \pmod {13}$ . Set $t \equiv -1 \cdot -1 \equiv 1 \pmod {13}$ , and $M =\;1.$
- We restart the loop, and since $t \equiv 1 \pmod{13}$ we are done, returning $R\equiv7 \pmod {13}.$

Indeed, observe that $7^2 = 49 \equiv 10 \pmod {13}$ and naturally also $(-7)^2 \equiv 6^2 \equiv 10 \pmod {13}$ . So the algorithm yields two solutions to our congruence.

Proof

First write $p-1=Q2^S$ . Now write $r \equiv n^{\frac{Q+1}{2}}\pmod p$ and $t \equiv n^Q \pmod p$ , observing that $r^2 \equiv nt \pmod p$ . This latter congruence will be true after every iteration of the algorithm's main loop. If at any point, $t \equiv 1 \pmod p$ then $r^2 \equiv n \pmod p$ and the algorithm terminates with $R \equiv \pm r \pmod p$ .

If $t \not\equiv 1 \pmod p$ , then consider $z$ , a quadratic non-residue of $p$ . Let $c \equiv z^Q \pmod p$ . Then $c^{2^S} \equiv (z^Q)^{2^S} \equiv z^{2^SQ}\equiv z^{p-1} \equiv 1 \pmod p$ and $c^{2^{S-1}} \equiv z^\frac{p-1}{2}\equiv -1 \pmod p$ , which shows that the order of $c$ is $2^S$ .

Similarly we have $t^{2^S} \equiv 1 \pmod p$ , so the order of $t$ divides $2^S$ . Suppose the order of $t$ is $2^{S'}$ . Since $n$ is a square modulo $p$ , $t \equiv n^Q \pmod p$ is also a square, and hence $S'\leq S-1$ .

Now we set $b \equiv c^{2^{S-S'-1}} \pmod p$ and with this $r' \equiv br \pmod p$ , $c' \equiv b^2 \pmod p$ and $t' \equiv c't \pmod p$ . As before, $r'^2 \equiv nt' \pmod p$ holds; however with this construction both $t$ and $c'$ have order $2^{S'}$ . This implies that $t'$ has order $2^{S''}$ with $S'' < S'$ .

If $S'' \equiv 0 \pmod p$ then $t' \equiv 1 \pmod p$ , and the algorithm stops, returning $R \equiv \pm r' \pmod p$ . Else, we restart the loop with analogous definitions of $b'$ , $r''$ , $c''$ and $t''$ until we arrive at an $S^{(j)'}$ that equals 0. Since the sequence of S is strictly decreasing the algorithm terminates.

算法的大体过程已经说的很清楚，然后模拟那个过程就可以了，不过对于Ural上的这个题，要对n = 2特判一下。详见代码：

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;


using namespace std;

LL MOD;

LL mod_exp(LL a, LL b) {
    LL res = 1;
    while(b > 0) {
        if(b&1)    res = (res*a)%MOD;
        a = (a*a)%MOD;
        b >>= 1;
    }
    return res;
}

LL solve(LL n, LL p) {
    int Q = p - 1, S = 0;
    while(Q%2 == 0) {
        Q >>= 1;
        S++;
    }
    if(S == 1) {return mod_exp(n, (p + 1)/4);}
    int z;
    while(1) {
        z = 1 + rand()%(p - 1);
        if(mod_exp(z, (p - 1)/2) != 1)   break;
    }
    LL c = mod_exp(z, Q);
    LL R = mod_exp(n, (Q + 1)/2);
    LL t = mod_exp(n, Q);
    LL M = S, b, i;
    while(1) {
        if(t%p == 1)  break;
        for(i = 1; i < M; ++i) {
            if(mod_exp(t, 1<<i) == 1)    break;
        }
        b = mod_exp(c, 1<<(M-i-1));
        R = (R*b)%p;
        t = (t*b*b)%p;
        c = (b*b)%p;
        M = i;
    }
    return (R%p + p)%p;
}

int main() {
    //Read();

    int T, a, n, i;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &a, &n);
        if(n == 2) {    //***
            if(a%n == 1)    printf("1\n");
            else    puts("No root");
            continue;
        }
        MOD = n;
        if(mod_exp(a, (n-1)/2) != 1)    {puts("No root"); continue; }

        i = solve(a, n);
        if(i == n - i)  printf("%d\n", i);
        else    printf("%d %d\n", min(i, n - i), max(i, n - i));
    }
    return 0;
}

POJ 1808 勒让德符号(Legendre symbol)判定

View Code

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;


using namespace std;

LL MOD;

LL mod_exp(LL a, LL b) {
    LL res = 1;
    while(b > 0) {
        if(b&1)    res = (res*a)%MOD;
        a = (a*a)%MOD;
        b >>= 1;
    }
    return res;
}

int main() {
    //Read();

    LL a, p;
    int T, cas = 0;
    scanf("%d", &T);
    while(T--) {
        scanf("%lld%lld", &a, &p);
        MOD = p;
        printf("Scenario #%d:\n", ++cas);
        if(mod_exp((a%p+p)%p, (p - 1)/2) == 1)  puts("1");    //注意a有可能是负数
        else    puts("-1");
        cout << endl;
    }
}