Find Square Root under Modulo p | Set 2 (Shanks Tonelli algorithm)
Given a number ‘n’ and a prime ‘p’, find square root of n under modulo p if it exists.
Examples:
Input: n = 2, p = 113 Output: 62 62^2 = 3844 and 3844 % 113 = 2 Input: n = 2, p = 7 Output: 3 or 4 3 and 4 both are square roots of 2 under modulo 7 because (3*3) % 7 = 2 and (4*4) % 7 = 2 Input: n = 2, p = 5 Output: Square root doesn't exist
We have discussed Euler’s criterion to check if square root exists or not. We have also discussed a solution that works only when p is in form of 4*i + 3
In this post, Shank Tonelli’s algorithm is discussed that works for all types of inputs.
Algorithm steps to find modular square root using shank Tonelli’s algorithm :
1) Calculate n ^ ((p – 1) / 2) (mod p), it must be 1 or p-1, if it is p-1, then modular square root is not possible.
2) Then after write p-1 as (s * 2^e) for some integer s and e, where s must be an odd number and both s and e should be positive.
3) Then find a number q such that q ^ ((p – 1) / 2) (mod p) = -1
4) Initialize variable x, b, g and r by following values
x = n ^ ((s + 1) / 2 (first guess of square root) b = n ^ s g = q ^ s r = e (exponent e will decrease after each updation)
5) Now loop until m > 0 and update value of x, which will be our final answer.
Find least integer m such that b^(2^m) = 1(mod p) and 0 <= m <= r – 1
If m = 0, then we found correct answer and return x as result
Else update x, b, g, r as below
x = x * g ^ (2 ^ (r – m - 1))
b = b * g ^(2 ^ (r - m))
g = g ^ (2 ^ (r - m))
r = m
so if m becomes 0 or b becomes 1, we terminate and print the result. This loop guarantees to terminate because value of m is decreased each time after updation.
Following is the implementation of above algorithm.
C++
// C++ program to implement Shanks Tonelli algorithm for // finding Modular Square Roots #include <bits/stdc++.h> using namespace std; // utility function to find pow(base, exponent) % modulus int pow(int base, int exponent, int modulus) { int result = 1; base = base % modulus; while (exponent > 0) { if (exponent % 2 == 1) result = (result * base)% modulus; exponent = exponent >> 1; base = (base * base) % modulus; } return result; } // utility function to find gcd int gcd(int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } // Returns k such that b^k = 1 (mod p) int order(int p, int b) { if (gcd(p, b) != 1) { printf("p and b are not co-prime.\n"); return -1; } // Initializing k with first odd prime number int k = 3; while (1) { if (pow(b, k, p) == 1) return k; k++; } } // function return p - 1 (= x argument) as x * 2^e, // where x will be odd sending e as reference because // updation is needed in actual e int convertx2e(int x, int& e) { e = 0; while (x % 2 == 0) { x /= 2; e++; } return x; } // Main function for finding the modular square root int STonelli(int n, int p) { // a and p should be coprime for finding the modular // square root if (gcd(n, p) != 1) { printf("a and p are not coprime\n"); return -1; } // If below expression return (p - 1) then modular // square root is not possible if (pow(n, (p - 1) / 2, p) == (p - 1)) { printf("no sqrt possible\n"); return -1; } // expressing p - 1, in terms of s * 2^e, where s // is odd number int s, e; s = convertx2e(p - 1, e); // finding smallest q such that q ^ ((p - 1) / 2) // (mod p) = p - 1 int q; for (q = 2; ; q++) { // q - 1 is in place of (-1 % p) if (pow(q, (p - 1) / 2, p) == (p - 1)) break; } // Initializing variable x, b and g int x = pow(n, (s + 1) / 2, p); int b = pow(n, s, p); int g = pow(q, s, p); int r = e; // keep looping until b become 1 or m becomes 0 while (1) { int m; for (m = 0; m < r; m++) { if (order(p, b) == -1) return -1; // finding m such that b^ (2^m) = 1 if (order(p, b) == pow(2, m)) break; } if (m == 0) return x; // updating value of x, g and b according to // algorithm x = (x * pow(g, pow(2, r - m - 1), p)) % p; g = pow(g, pow(2, r - m), p); b = (b * g) % p; if (b == 1) return x; r = m; } } // driver program to test above function int main() { int n = 2; // p should be prime int p = 113; int x = STonelli(n, p); if (x == -1) printf("Modular square root is not exist\n"); else printf("Modular square root of %d and %d is %d\n", n, p, x); } |
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Java
// Java program to implement Shanks // Tonelli algorithm for finding // Modular Square Roots class GFG { static int z = 0; // utility function to find // pow(base, exponent) % modulus static int pow1(int base1, int exponent, int modulus) { int result = 1; base1 = base1 % modulus; while (exponent > 0) { if (exponent % 2 == 1) result = (result * base1) % modulus; exponent = exponent >> 1; base1 = (base1 * base1) % modulus; } return result; } // utility function to find gcd static int gcd(int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } // Returns k such that b^k = 1 (mod p) static int order(int p, int b) { if (gcd(p, b) != 1) { System.out.println("p and b are" + "not co-prime."); return -1; } // Initializing k with first // odd prime number int k = 3; while (true) { if (pow1(b, k, p) == 1) return k; k++; } } // function return p - 1 (= x argument) // as x * 2^e, where x will be odd // sending e as reference because // updation is needed in actual e static int convertx2e(int x) { z = 0; while (x % 2 == 0) { x /= 2; z++; } return x; } // Main function for finding // the modular square root static int STonelli(int n, int p) { // a and p should be coprime for // finding the modular square root if (gcd(n, p) != 1) { System.out.println("a and p are not coprime"); return -1; } // If below expression return (p - 1) then modular // square root is not possible if (pow1(n, (p - 1) / 2, p) == (p - 1)) { System.out.println("no sqrt possible"); return -1; } // expressing p - 1, in terms of // s * 2^e, where s is odd number int s, e; s = convertx2e(p - 1); e = z; // finding smallest q such that q ^ ((p - 1) / 2) // (mod p) = p - 1 int q; for (q = 2; ; q++) { // q - 1 is in place of (-1 % p) if (pow1(q, (p - 1) / 2, p) == (p - 1)) break; } // Initializing variable x, b and g int x = pow1(n, (s + 1) / 2, p); int b = pow1(n, s, p); int g = pow1(q, s, p); int r = e; // keep looping until b // become 1 or m becomes 0 while (true) { int m; for (m = 0; m < r; m++) { if (order(p, b) == -1) return -1; // finding m such that b^ (2^m) = 1 if (order(p, b) == Math.pow(2, m)) break; } if (m == 0) return x; // updating value of x, g and b // according to algorithm x = (x * pow1(g, (int)Math.pow(2, r - m - 1), p)) % p; g = pow1(g, (int)Math.pow(2, r - m), p); b = (b * g) % p; if (b == 1) return x; r = m; } } // Driver code public static void main (String[] args) { int n = 2; // p should be prime int p = 113; int x = STonelli(n, p); if (x == -1) System.out.println("Modular square" + "root is not exist\n"); else System.out.println("Modular square root of " + n + " and " + p + " is " + x + "\n"); } } // This code is contributed by mits |
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Python3
# Python3 program to implement Shanks Tonelli
# algorithm for finding Modular Square Roots
# utility function to find pow(base,
# exponent) % modulus
def pow1(base, exponent, modulus):
result = 1;
base = base % modulus;
while (exponent > 0):
if (exponent % 2 == 1):
result = (result * base) % modulus;
exponent = int(exponent) >> 1;
base = (base * base) % modulus;
return result;
# utility function to find gcd
def gcd(a, b):
if (b == 0):
return a;
else:
return gcd(b, a % b);
# Returns k such that b^k = 1 (mod p)
def order(p, b):
if (gcd(p, b) != 1):
print(“p and b are not co-prime.\n”);
return -1;
# Initializing k with first
# odd prime number
k = 3;
while (True):
if (pow1(b, k, p) == 1):
return k;
k += 1;
# function return p – 1 (= x argument) as
# x * 2^e, where x will be odd sending e
# as reference because updation is needed
# in actual e
def convertx2e(x):
z = 0;
while (x % 2 == 0):
x = x / 2;
z += 1;
return [x, z];
# Main function for finding the
# modular square root
def STonelli(n, p):
# a and p should be coprime for
# finding the modular square root
if (gcd(n, p) != 1):
print(“a and p are not coprime\n”);
return -1;
# If below expression return (p – 1) then
# modular square root is not possible
if (pow1(n, (p – 1) / 2, p) == (p – 1)):
print(“no sqrt possible\n”);
return -1;
# expressing p – 1, in terms of s * 2^e,
# where s is odd number
ar = convertx2e(p – 1);
s = ar[0];
e = ar[1];
# finding smallest q such that
# q ^ ((p – 1) / 2) (mod p) = p – 1
q = 2;
while (True):
# q – 1 is in place of (-1 % p)
if (pow1(q, (p – 1) / 2, p) == (p – 1)):
break;
q += 1;
# Initializing variable x, b and g
x = pow1(n, (s + 1) / 2, p);
b = pow1(n, s, p);
g = pow1(q, s, p);
r = e;
# keep looping until b become
# 1 or m becomes 0
while (True):
m = 0;
while (m < r):
if (order(p, b) == -1):
return -1;
# finding m such that b^ (2^m) = 1
if (order(p, b) == pow(2, m)):
break;
m += 1;
if (m == 0):
return x;
# updating value of x, g and b
# according to algorithm
x = (x * pow1(g, pow(2, r - m - 1), p)) % p;
g = pow1(g, pow(2, r - m), p);
b = (b * g) % p;
if (b == 1):
return x;
r = m;
# Driver Code
n = 2;
# p should be prime
p = 113;
x = STonelli(n, p);
if (x == -1):
print("Modular square root is not exist\n");
else:
print("Modular square root of", n,
"and", p, "is", x);
# This code is contributed by mits
[tabby title="C#"]
// C# program to implement Shanks // Tonelli algorithm for finding // Modular Square Roots using System; class GFG { static int z=0; // utility function to find // pow(base, exponent) % modulus static int pow1(int base1, int exponent, int modulus) { int result = 1; base1 = base1 % modulus; while (exponent > 0) { if (exponent % 2 == 1) result = (result * base1) % modulus; exponent = exponent >> 1; base1 = (base1 * base1) % modulus; } return result; } // utility function to find gcd static int gcd(int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } // Returns k such that b^k = 1 (mod p) static int order(int p, int b) { if (gcd(p, b) != 1) { Console.WriteLine("p and b are" + "not co-prime."); return -1; } // Initializing k with // first odd prime number int k = 3; while (true) { if (pow1(b, k, p) == 1) return k; k++; } } // function return p - 1 (= x argument) // as x * 2^e, where x will be odd sending // e as reference because updation is // needed in actual e static int convertx2e(int x) { z = 0; while (x % 2 == 0) { x /= 2; z++; } return x; } // Main function for finding // the modular square root static int STonelli(int n, int p) { // a and p should be coprime for // finding the modular square root if (gcd(n, p) != 1) { Console.WriteLine("a and p are not coprime"); return -1; } // If below expression return (p - 1) then // modular square root is not possible if (pow1(n, (p - 1) / 2, p) == (p - 1)) { Console.WriteLine("no sqrt possible"); return -1; } // expressing p - 1, in terms of s * 2^e, // where s is odd number int s, e; s = convertx2e(p - 1); e=z; // finding smallest q such that q ^ ((p - 1) / 2) // (mod p) = p - 1 int q; for (q = 2; ; q++) { // q - 1 is in place of (-1 % p) if (pow1(q, (p - 1) / 2, p) == (p - 1)) break; } // Initializing variable x, b and g int x = pow1(n, (s + 1) / 2, p); int b = pow1(n, s, p); int g = pow1(q, s, p); int r = e; // keep looping until b become // 1 or m becomes 0 while (true) { int m; for (m = 0; m < r; m++) { if (order(p, b) == -1) return -1; // finding m such that b^ (2^m) = 1 if (order(p, b) == Math.Pow(2, m)) break; } if (m == 0) return x; // updating value of x, g and b // according to algorithm x = (x * pow1(g, (int)Math.Pow(2, r - m - 1), p)) % p; g = pow1(g, (int)Math.Pow(2, r - m), p); b = (b * g) % p; if (b == 1) return x; r = m; } } // Driver code static void Main() { int n = 2; // p should be prime int p = 113; int x = STonelli(n, p); if (x == -1) Console.WriteLine("Modular square root" + "is not exist\n"); else Console.WriteLine("Modular square root of" + "{0} and {1} is {2}\n", n, p, x); } } // This code is contributed by mits |
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PHP
<?php // PHP program to implement Shanks Tonelli // algorithm for finding Modular Square Roots // utility function to find pow(base, // exponent) % modulus function pow1($base, $exponent, $modulus) { $result = 1; $base = $base % $modulus; while ($exponent > 0) { if ($exponent % 2 == 1) $result = ($result * $base) % $modulus; $exponent = $exponent >> 1; $base = ($base * $base) % $modulus; } return $result; } // utility function to find gcd function gcd($a, $b) { if ($b == 0) return $a; else return gcd($b, $a % $b); } // Returns k such that b^k = 1 (mod p) function order($p, $b) { if (gcd($p, $b) != 1) { print("p and b are not co-prime.\n"); return -1; } // Initializing k with first // odd prime number $k = 3; while (1) { if (pow1($b, $k, $p) == 1) return $k; $k++; } } // function return p - 1 (= x argument) as // x * 2^e, where x will be odd sending e // as reference because updation is needed // in actual e function convertx2e($x, &$e) { $e = 0; while ($x % 2 == 0) { $x = (int)($x / 2); $e++; } return $x; } // Main function for finding the // modular square root function STonelli($n, $p) { // a and p should be coprime for // finding the modular square root if (gcd($n, $p) != 1) { print("a and p are not coprime\n"); return -1; } // If below expression return (p - 1) then // modular square root is not possible if (pow1($n, ($p - 1) / 2, $p) == ($p - 1)) { printf("no sqrt possible\n"); return -1; } // expressing p - 1, in terms of s * 2^e, // where s is odd number $e = 0; $s = convertx2e($p - 1, $e); // finding smallest q such that // q ^ ((p - 1) / 2) (mod p) = p - 1 $q = 2; for (; ; $q++) { // q - 1 is in place of (-1 % p) if (pow1($q, ($p - 1) / 2, $p) == ($p - 1)) break; } // Initializing variable x, b and g $x = pow1($n, ($s + 1) / 2, $p); $b = pow1($n, $s, $p); $g = pow1($q, $s, $p); $r = $e; // keep looping until b become // 1 or m becomes 0 while (1) { $m = 0; for (; $m < $r; $m++) { if (order($p, $b) == -1) return -1; // finding m such that b^ (2^m) = 1 if (order($p, $b) == pow(2, $m)) break; } if ($m == 0) return $x; // updating value of x, g and b // according to algorithm $x = ($x * pow1($g, pow(2, $r - $m - 1), $p)) % $p; $g = pow1($g, pow(2, $r - $m), $p); $b = ($b * $g) % $p; if ($b == 1) return $x; $r = $m; } } // Driver Code $n = 2; // p should be prime $p = 113; $x = STonelli($n, $p); if ($x == -1) print("Modular square root is not exist\n"); else print("Modular square root of " . "$n and $p is $x\n"); // This code is contributed by mits ?> |
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Output :
Modular square root of 2 and 113 is 62
For more detail about above algorithm please visit :
http://cs.indstate.edu/~sgali1/Shanks_Tonelli.pdf
For detail of example (2, 113) see :
http://www.math.vt.edu/people/brown/class_homepages/shanks_tonelli.pdf
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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