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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int gcd( int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
int order( int p, int b)
{
if (gcd(p, b) != 1)
{
printf ( "p and b are not co-prime.\n" );
return -1;
}
int k = 3;
while (1)
{
if ( pow (b, k, p) == 1)
return k;
k++;
}
}
int convertx2e( int x, int & e)
{
e = 0;
while (x % 2 == 0)
{
x /= 2;
e++;
}
return x;
}
int STonelli( int n, int p)
{
if (gcd(n, p) != 1)
{
printf ( "a and p are not coprime\n" );
return -1;
}
if ( pow (n, (p - 1) / 2, p) == (p - 1))
{
printf ( "no sqrt possible\n" );
return -1;
}
int s, e;
s = convertx2e(p - 1, e);
int q;
for (q = 2; ; q++)
{
if ( pow (q, (p - 1) / 2, p) == (p - 1))
break ;
}
int x = pow (n, (s + 1) / 2, p);
int b = pow (n, s, p);
int g = pow (q, s, p);
int r = e;
while (1)
{
int m;
for (m = 0; m < r; m++)
{
if (order(p, b) == -1)
return -1;
if (order(p, b) == pow (2, m))
break ;
}
if (m == 0)
return x;
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;
}
}
int main()
{
int n = 2;
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);
}
|
Java
import java.util.*;
class GFG
{
static int z = 0 ;
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;
}
static int gcd( int a, int b)
{
if (b == 0 )
return a;
else
return gcd(b, a % b);
}
static int order( int p, int b)
{
if (gcd(p, b) != 1 )
{
System.out.println( "p and b are" +
"not co-prime." );
return - 1 ;
}
int k = 3 ;
while ( true )
{
if (pow1(b, k, p) == 1 )
return k;
k++;
}
}
static int convertx2e( int x)
{
z = 0 ;
while (x % 2 == 0 )
{
x /= 2 ;
z++;
}
return x;
}
static int STonelli( int n, int p)
{
if (gcd(n, p) != 1 )
{
System.out.println( "a and p are not coprime" );
return - 1 ;
}
if (pow1(n, (p - 1 ) / 2 , p) == (p - 1 ))
{
System.out.println( "no sqrt possible" );
return - 1 ;
}
int s, e;
s = convertx2e(p - 1 );
e = z;
int q;
for (q = 2 ; ; q++)
{
if (pow1(q, (p - 1 ) / 2 , p) == (p - 1 ))
break ;
}
int x = pow1(n, (s + 1 ) / 2 , p);
int b = pow1(n, s, p);
int g = pow1(q, s, p);
int r = e;
while ( true )
{
int m;
for (m = 0 ; m < r; m++)
{
if (order(p, b) == - 1 )
return - 1 ;
if (order(p, b) == Math.pow( 2 , m))
break ;
}
if (m == 0 )
return x;
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;
}
}
public static void main (String[] args)
{
int n = 2 ;
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" );
}
}
|
Python3
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;
def gcd(a, b):
if (b = = 0 ):
return a;
else :
return gcd(b, a % b);
def order(p, b):
if (gcd(p, b) ! = 1 ):
print ( "p and b are not co-prime.\n" );
return - 1 ;
k = 3 ;
while ( True ):
if (pow1(b, k, p) = = 1 ):
return k;
k + = 1 ;
def convertx2e(x):
z = 0 ;
while (x % 2 = = 0 ):
x = x / 2 ;
z + = 1 ;
return [x, z];
def STonelli(n, p):
if (gcd(n, p) ! = 1 ):
print ( "a and p are not coprime\n" );
return - 1 ;
if (pow1(n, (p - 1 ) / 2 , p) = = (p - 1 )):
print ( "no sqrt possible\n" );
return - 1 ;
ar = convertx2e(p - 1 );
s = ar[ 0 ];
e = ar[ 1 ];
q = 2 ;
while ( True ):
if (pow1(q, (p - 1 ) / 2 , p) = = (p - 1 )):
break ;
q + = 1 ;
x = pow1(n, (s + 1 ) / 2 , p);
b = pow1(n, s, p);
g = pow1(q, s, p);
r = e;
while ( True ):
m = 0 ;
while (m < r):
if (order(p, b) = = - 1 ):
return - 1 ;
if (order(p, b) = = pow ( 2 , m)):
break ;
m + = 1 ;
if (m = = 0 ):
return x;
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;
n = 2 ;
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);
|
C#
using System;
class GFG
{
static int z=0;
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;
}
static int gcd( int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
static int order( int p, int b)
{
if (gcd(p, b) != 1)
{
Console.WriteLine( "p and b are" +
"not co-prime." );
return -1;
}
int k = 3;
while ( true )
{
if (pow1(b, k, p) == 1)
return k;
k++;
}
}
static int convertx2e( int x)
{
z = 0;
while (x % 2 == 0)
{
x /= 2;
z++;
}
return x;
}
static int STonelli( int n, int p)
{
if (gcd(n, p) != 1)
{
Console.WriteLine( "a and p are not coprime" );
return -1;
}
if (pow1(n, (p - 1) / 2, p) == (p - 1))
{
Console.WriteLine( "no sqrt possible" );
return -1;
}
int s, e;
s = convertx2e(p - 1);
e=z;
int q;
for (q = 2; ; q++)
{
if (pow1(q, (p - 1) / 2, p) == (p - 1))
break ;
}
int x = pow1(n, (s + 1) / 2, p);
int b = pow1(n, s, p);
int g = pow1(q, s, p);
int r = e;
while ( true )
{
int m;
for (m = 0; m < r; m++)
{
if (order(p, b) == -1)
return -1;
if (order(p, b) == Math.Pow(2, m))
break ;
}
if (m == 0)
return x;
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;
}
}
static void Main()
{
int n = 2;
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);
}
}
|
PHP
<?php
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 ;
}
function gcd( $a , $b )
{
if ( $b == 0)
return $a ;
else
return gcd( $b , $a % $b );
}
function order( $p , $b )
{
if (gcd( $p , $b ) != 1)
{
print ( "p and b are not co-prime.\n" );
return -1;
}
$k = 3;
while (1)
{
if (pow1( $b , $k , $p ) == 1)
return $k ;
$k ++;
}
}
function convertx2e( $x , & $e )
{
$e = 0;
while ( $x % 2 == 0)
{
$x = (int)( $x / 2);
$e ++;
}
return $x ;
}
function STonelli( $n , $p )
{
if (gcd( $n , $p ) != 1)
{
print ( "a and p are not coprime\n" );
return -1;
}
if (pow1( $n , ( $p - 1) / 2, $p ) == ( $p - 1))
{
printf( "no sqrt possible\n" );
return -1;
}
$e = 0;
$s = convertx2e( $p - 1, $e );
$q = 2;
for (; ; $q ++)
{
if (pow1( $q , ( $p - 1) / 2, $p ) == ( $p - 1))
break ;
}
$x = pow1( $n , ( $s + 1) / 2, $p );
$b = pow1( $n , $s , $p );
$g = pow1( $q , $s , $p );
$r = $e ;
while (1)
{
$m = 0;
for (; $m < $r ; $m ++)
{
if (order( $p , $b ) == -1)
return -1;
if (order( $p , $b ) == pow(2, $m ))
break ;
}
if ( $m == 0)
return $x ;
$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 ;
}
}
$n = 2;
$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" );
?>
|
Javascript
<script>
let z = 0;
function pow1(base1,
exponent, modulus)
{
let 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;
}
function gcd(a, b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
function order(p, b)
{
if (gcd(p, b) != 1)
{
document.write( "p and b are" +
"not co-prime." + "<br/>" );
return -1;
}
let k = 3;
while ( true )
{
if (pow1(b, k, p) == 1)
return k;
k++;
}
}
function convertx2e(x)
{
z = 0;
while (x % 2 == 0)
{
x /= 2;
z++;
}
return x;
}
function STonelli(n, p)
{
if (gcd(n, p) != 1)
{
System.out.prletln( "a and p are not coprime" );
return -1;
}
if (pow1(n, (p - 1) / 2, p) == (p - 1))
{
document.write( "no sqrt possible" + "<br/>" );
return -1;
}
let s, e;
s = convertx2e(p - 1);
e = z;
let q;
for (q = 2; ; q++)
{
if (pow1(q, (p - 1) / 2, p) == (p - 1))
break ;
}
let x = pow1(n, (s + 1) / 2, p);
let b = pow1(n, s, p);
let g = pow1(q, s, p);
let r = e;
while ( true )
{
let m;
for (m = 0; m < r; m++)
{
if (order(p, b) == -1)
return -1;
if (order(p, b) == Math.pow(2, m))
break ;
}
if (m == 0)
return x;
x = (x * pow1(g, Math.pow(2,
r - m - 1), p)) % p;
g = pow1(g, Math.pow(2, r - m), p);
b = (b * g) % p;
if (b == 1)
return x;
r = m;
}
}
let n = 2;
let p = 113;
let x = STonelli(n, p);
if (x == -1)
document.write( "Modular square" +
"root is not exist\n" );
else
document.write( "Modular square root of " +
n + " and " + p + " is " +
x + "\n" );
</script>
|
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
Last Updated :
20 Dec, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...