Modular multiplicative inverse

Given two integers ‘a’ and ‘m’, find modular multiplicative inverse of ‘a’ under modulo ‘m’.
The modular multiplicative inverse is an integer ‘x’ such that. 

a x ≅ 1 (mod m) 

The value of x should be in {0, 1, 2, … m-1}, i.e., in the range of integer modulo m.
The multiplicative inverse of “a modulo m” exists if and only if a and m are relatively prime (i.e., if gcd(a, m) = 1).

Examples: 

Input:  a = 3, m = 11
Output: 4
Since (4*3) mod 11 = 1, 4 is modulo inverse of 3(under 11).
One might think, 15 also as a valid output as "(15*3) mod 11" 
is also 1, but 15 is not in ring {0, 1, 2, ... 10}, so not 
valid.

Input:  a = 10, m = 17
Output: 12
Since (10*12) mod 17 = 1, 12 is modulo inverse of 10(under 17).

Method 1 (Naive) 
A Naive method is to try all numbers from 1 to m. For every number x, check if (a*x)%m is 1. 

Below is the implementation of this method. 



C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find modular inverse of a under modulo m
#include <iostream>
using namespace std;
  
// A naive method to find modular multiplicative inverse of
// 'a' under modulo 'm'
int modInverse(int a, int m)
{
    a = a % m;
    for (int x = 1; x < m; x++)
        if ((a * x) % m == 1)
            return x;
}
  
// Driver code
int main()
{
    int a = 3, m = 11;
    
    // Function call
    cout << modInverse(a, m);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find modular inverse
// of a under modulo m
import java.io.*;
  
class GFG {
  
    // A naive method to find modulor
    // multiplicative inverse of 'a'
    // under modulo 'm'
    static int modInverse(int a, int m)
    {
        a = a % m;
        for (int x = 1; x < m; x++)
            if ((a * x) % m == 1)
                return x;
        return 1;
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int a = 3, m = 11;
        
        // Function call
        System.out.println(modInverse(a, m));
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to find modular
# inverse of a under modulo m
  
# A naive method to find modulor
# multiplicative inverse of 'a'
# under modulo 'm'
  
  
def modInverse(a, m):
    a = a % m
    for x in range(1, m):
        if ((a * x) % m == 1):
            return x
    return 1
  
  
# Driver Code
a = 3
m = 11
  
# Function call
print(modInverse(a, m))
  
# This code is contributed by Nikita Tiwari.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find modular inverse
// of a under modulo m
using System;
  
class GFG {
  
    // A naive method to find modulor
    // multiplicative inverse of 'a'
    // under modulo 'm'
    static int modInverse(int a, int m)
    {
        a = a % m;
        for (int x = 1; x < m; x++)
            if ((a * x) % m == 1)
                return x;
        return 1;
    }
  
    // Driver Code
    public static void Main()
    {
        int a = 3, m = 11;
        
        // Function call
        Console.WriteLine(modInverse(a, m));
    }
}
  
// This code is contributed by anuj_67.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<≅php
// PHP program to find modular 
// inverse of a under modulo m
  
// A naive method to find modulor
// multiplicative inverse of
// 'a' under modulo 'm'
function modInverse( $a, $m)
{
    $a = $a % $m;
    for ($x = 1; $x < $m; $x++)
    if (($a * $x) % $m == 1)
        return $x;
}
  
    // Driver Code
    $a = 3; 
    $m = 11;
  
    // Function call
    echo modInverse($a, $m);
  
// This code is contributed by anuj_67.
≅>

chevron_right


Output

4

 Time Complexity: O(m).

Method 2 (Works when m and a are coprime) 
The idea is to use Extended Euclidean algorithms that takes two integers ‘a’ and ‘b’, finds their gcd and also find ‘x’ and ‘y’ such that 

ax + by = gcd(a, b)

To find multiplicative inverse of ‘a’ under ‘m’, we put b = m in above formula. Since we know that a and m are relatively prime, we can put value of gcd as 1.

ax + my = 1 

If we take modulo m on both sides, we get

ax + my ≅ 1 (mod m)

We can remove the second term on left side as ‘my (mod m)’ would always be 0 for an integer y. 

ax  ≅ 1 (mod m) 

So the ‘x’ that we can find using Extended Euclid Algorithm is the multiplicative inverse of ‘a’

Below is the implementation of the above algorithm.  



C

filter_none

edit
close

play_arrow

link
brightness_4
code

// C program to find multiplicative modulo inverse using
// Extended Euclid algorithm.
#include <stdio.h>
  
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y);
  
// Function to find modulo inverse of a
void modInverse(int a, int m)
{
    int x, y;
    int g = gcdExtended(a, m, &x, &y);
    if (g != 1)
        printf("Inverse doesn't exist");
    else 
    {
        // m is added to handle negative x
        int res = (x % m + m) % m;
        printf("Modular multiplicative inverse is %d", res);
    }
}
  
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y)
{
    // Base Case
    if (a == 0) 
    {
        *x = 0, *y = 1;
        return b;
    }
  
    int x1, y1; // To store results of recursive call
    int gcd = gcdExtended(b % a, a, &x1, &y1);
  
    // Update x and y using results of recursive
    // call
    *x = y1 - (b / a) * x1;
    *y = x1;
  
    return gcd;
}
  
// Driver Code
int main()
{
    int a = 3, m = 11;
    
    // Function call
    modInverse(a, m);
    return 0;
}

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<≅php
// PHP program to find multiplicative modulo 
// inverse using Extended Euclid algorithm.
// Function to find modulo inverse of a
function modInverse($a, $m)
{
    $x = 0;
    $y = 0;
    $g = gcdExtended($a, $m, $x, $y);
    if ($g != 1)
        echo "Inverse doesn't exist";
    else
    {
        // m is added to handle negative x
        $res = ($x % $m + $m) % $m;
        echo "Modular multiplicative "
                   "inverse is " . $res;
    }
}
  
// function for extended Euclidean Algorithm
function gcdExtended($a, $b, &$x, &$y)
{
    // Base Case
    if ($a == 0)
    {
        $x = 0;
        $y = 1;
        return $b;
    }
  
    $x1;
    $y1; // To store results of recursive call
    $gcd = gcdExtended($b%$a, $a, $x1, $y1);
  
    // Update x and y using results of 
    // recursive call
    $x = $y1 - (int)($b/$a) * $x1;
    $y = $x1;
  
    return $gcd;
}
  
// Driver Code
$a = 3;
$m = 11;
  
// Function call
modInverse($a, $m);
  
// This code is contributed by chandan_jnu
≅>

chevron_right


Output

Modular multiplicative inverse is 4

 Iterative Implementation: 

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// Iterative C++ program to find modular
// inverse using extended Euclid algorithm
#include <stdio.h>
  
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
int modInverse(int a, int m)
{
    int m0 = m;
    int y = 0, x = 1;
  
    if (m == 1)
        return 0;
  
    while (a > 1) {
        // q is quotient
        int q = a / m;
        int t = m;
  
        // m is remainder now, process same as
        // Euclid's algo
        m = a % m, a = t;
        t = y;
  
        // Update y and x
        y = x - q * y;
        x = t;
    }
  
    // Make x positive
    if (x < 0)
        x += m0;
  
    return x;
}
  
// Driver Code
int main()
{
    int a = 3, m = 11;
  
    // Function call
    printf("Modular multiplicative inverse is %d\n",
           modInverse(a, m));
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Iterative Java program to find modular
// inverse using extended Euclid algorithm
  
class GFG {
  
    // Returns modulo inverse of a with
    // respect to m using extended Euclid
    // Algorithm Assumption: a and m are
    // coprimes, i.e., gcd(a, m) = 1
    static int modInverse(int a, int m)
    {
        int m0 = m;
        int y = 0, x = 1;
  
        if (m == 1)
            return 0;
  
        while (a > 1) {
            // q is quotient
            int q = a / m;
  
            int t = m;
  
            // m is remainder now, process
            // same as Euclid's algo
            m = a % m;
            a = t;
            t = y;
  
            // Update x and y
            y = x - q * y;
            x = t;
        }
  
        // Make x positive
        if (x < 0)
            x += m0;
  
        return x;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int a = 3, m = 11;
          
        // Function call
        System.out.println("Modular multiplicative "
                           + "inverse is "
                           + modInverse(a, m));
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Iterative Python 3 program to find
# modular inverse using extended
# Euclid algorithm
  
# Returns modulo inverse of a with
# respect to m using extended Euclid
# Algorithm Assumption: a and m are
# coprimes, i.e., gcd(a, m) = 1
  
  
def modInverse(a, m):
    m0 = m
    y = 0
    x = 1
  
    if (m == 1):
        return 0
  
    while (a > 1):
  
        # q is quotient
        q = a // m
  
        t = m
  
        # m is remainder now, process
        # same as Euclid's algo
        m = a % m
        a = t
        t = y
  
        # Update x and y
        y = x - q * y
        x = t
  
    # Make x positive
    if (x < 0):
        x = x + m0
  
    return x
  
  
# Driver code
a = 3
m = 11
  
# Function call
print("Modular multiplicative inverse is",
      modInverse(a, m))
  
# This code is contributed by Nikita tiwari.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// Iterative C# program to find modular
// inverse using extended Euclid algorithm
using System;
class GFG {
  
    // Returns modulo inverse of a with
    // respect to m using extended Euclid
    // Algorithm Assumption: a and m are
    // coprimes, i.e., gcd(a, m) = 1
    static int modInverse(int a, int m)
    {
        int m0 = m;
        int y = 0, x = 1;
  
        if (m == 1)
            return 0;
  
        while (a > 1) {
            // q is quotient
            int q = a / m;
  
            int t = m;
  
            // m is remainder now, process
            // same as Euclid's algo
            m = a % m;
            a = t;
            t = y;
  
            // Update x and y
            y = x - q * y;
            x = t;
        }
  
        // Make x positive
        if (x < 0)
            x += m0;
  
        return x;
    }
  
    // Driver Code
    public static void Main()
    {
        int a = 3, m = 11;
        
        // Function call
        Console.WriteLine("Modular multiplicative "
                          + "inverse is "
                          + modInverse(a, m));
    }
}
  
// This code is contributed by anuj_67.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<≅php
// Iterative PHP program to find modular
// inverse using extended Euclid algorithm
  
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
function modInverse($a, $m)
{
    $m0 = $m;
    $y = 0;
    $x = 1;
  
    if ($m == 1)
    return 0;
  
    while ($a > 1)
    {
          
        // q is quotient
        $q = (int) ($a / $m);
        $t = $m;
  
        // m is remainder now,
        // process same as
        // Euclid's algo
        $m = $a % $m;
        $a = $t;
        $t = $y;
  
        // Update y and x
        $y = $x - $q * $y;
        $x = $t;
    }
  
    // Make x positive
    if ($x < 0)
    $x += $m0;
  
    return $x;
}
  
    // Driver Code
    $a = 3;
    $m = 11;
  
    // Function call
    echo "Modular multiplicative inverse is\n",
                            modInverse($a, $m);
          
// This code is contributed by Anuj_67
≅>

chevron_right


Output

Modular multiplicative inverse is 4

Time Complexity: O(Log m)
  
Method 3 (Works when m is prime) 
If we know m is prime, then we can also use Fermats’s little theorem to find the inverse. 

am-1 ≅ 1 (mod m)  

If we multiply both sides with a-1, we get 

a-1 ≅ a m-2 (mod m)

Below is the implementation of the above idea. 

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find modular inverse of a under modulo m
// This program works only if m is prime.
#include <iostream>
using namespace std;
  
// To find GCD of a and b
int gcd(int a, int b);
  
// To compute x raised to power y under modulo m
int power(int x, unsigned int y, unsigned int m);
  
// Function to find modular inverse of a under modulo m
// Assumption: m is prime
void modInverse(int a, int m)
{
    int g = gcd(a, m);
    if (g != 1)
        cout << "Inverse doesn't exist";
    else 
    {
        // If a and m are relatively prime, then modulo
        // inverse is a^(m-2) mode m
        cout << "Modular multiplicative inverse is "
             << power(a, m - 2, m);
    }
}
  
// To compute x^y under modulo m
int power(int x, unsigned int y, unsigned int m)
{
    if (y == 0)
        return 1;
    int p = power(x, y / 2, m) % m;
    p = (p * p) % m;
  
    return (y % 2 == 0) ? p : (x * p) % m;
}
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Driver code
int main()
{
    int a = 3, m = 11;
  
    // Function call
    modInverse(a, m);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find modular
// inverse of a under modulo m
// This program works only if
// m is prime.
import java.io.*;
  
class GFG {
  
    // Function to find modular inverse of a
    // under modulo m Assumption: m is prime
    static void modInverse(int a, int m)
    {
        int g = gcd(a, m);
        if (g != 1)
            System.out.println("Inverse doesn't exist");
        else 
        {
            // If a and m are relatively prime, then modulo
            // inverse is a^(m-2) mode m
            System.out.println(
                "Modular multiplicative inverse is "
                + power(a, m - 2, m));
        }
    }
  
    // To compute x^y under modulo m
    static int power(int x, int y, int m)
    {
        if (y == 0)
            return 1;
  
        int p = power(x, y / 2, m) % m;
        p = (p * p) % m;
  
        if (y % 2 == 0)
            return p;
        else
            return (x * p) % m;
    }
  
    // Function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int a = 3, m = 11;
         
        // Function call
        modInverse(a, m);
    }
}
  
// This code is contributed by Nikita Tiwari.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find modular
# inverse of a under modulo m
  
# This program works only if m is prime.
  
# Function to find modular
# inverse of a under modulo m
# Assumption: m is prime
  
  
def modInverse(a, m):
  
    g = gcd(a, m)
  
    if (g != 1):
        print("Inverse doesn't exist")
  
    else:
  
        # If a and m are relatively prime,
        # then modulo inverse is a^(m-2) mode m
        print("Modular multiplicative inverse is ",
              power(a, m - 2, m))
  
# To compute x^y under modulo m
  
  
def power(x, y, m):
  
    if (y == 0):
        return 1
  
    p = power(x, y // 2, m) % m
    p = (p * p) % m
  
    if(y % 2 == 0):
        return p
    else:
        return ((x * p) % m)
  
# Function to return gcd of a and b
  
  
def gcd(a, b):
    if (a == 0):
        return b
  
    return gcd(b % a, a)
  
  
# Driver Code
a = 3
m = 11
  
# Function call
modInverse(a, m)
  
  
# This code is contributed by Nikita Tiwari.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find modular
// inverse of a under modulo m
// This program works only if
// m is prime.
using System;
class GFG {
  
    // Function to find modular
    // inverse of a under modulo
    // m Assumption: m is prime
    static void modInverse(int a, int m)
    {
        int g = gcd(a, m);
        if (g != 1)
            Console.Write("Inverse doesn't exist");
        else {
            // If a and m are relatively
            // prime, then modulo inverse
            // is a^(m-2) mode m
            Console.Write(
                "Modular multiplicative inverse is "
                + power(a, m - 2, m));
        }
    }
  
    // To compute x^y under
    // modulo m
    static int power(int x, int y, int m)
    {
        if (y == 0)
            return 1;
  
        int p = power(x, y / 2, m) % m;
        p = (p * p) % m;
  
        if (y % 2 == 0)
            return p;
        else
            return (x * p) % m;
    }
  
    // Function to return
    // gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
  
    // Driver Code
    public static void Main()
    {
        int a = 3, m = 11;
        
        // Function call
        modInverse(a, m);
    }
}
  
// This code is contributed by nitin mittal.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<≅php
// PHP program to find modular 
// inverse of a under modulo m
// This program works only if m 
// is prime.
  
// Function to find modular inverse
// of a under modulo m
// Assumption: m is prime
function modInverse( $a, $m)
{
    $g = gcd($a, $m);
    if ($g != 1)
        echo "Inverse doesn't exist";
    else
    {
          
        // If a and m are relatively 
        // prime, then modulo inverse
        // is a^(m-2) mode m
        echo "Modular multiplicative inverse is "
                        , power($a, $m - 2, $m);
    }
}
  
// To compute x^y under modulo m
function power( $x, $y, $m)
{
    if ($y == 0)
        return 1;
    $p = power($x, $y / 2, $m) % $m;
    $p = ($p * $p) % $m;
  
    return ($y % 2 == 0)? $p : ($x * $p) % $m;
}
  
// Function to return gcd of a and b
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// Driver Code
$a = 3;
$m = 11;
  
// Function call
modInverse($a, $m);
      
// This code is contributed by anuj_67.
≅>

chevron_right


Output

Modular multiplicative inverse is 4

Time Complexity: O(Log m)

We have discussed three methods to find multiplicative inverse modulo m. 
1) Naive Method, O(m) 
2) Extended Euler’s GCD algorithm, O(Log m) [Works when a and m are coprime] 
3) Fermat’s Little theorem, O(Log m) [Works when ‘m’ is prime]

Applications: 
Computation of the modular multiplicative inverse is an essential step in RSA public-key encryption method.

References: 
https://en.wikipedia.org/wiki/Modular_multiplicative_inverse 
http://e-maxx.ru/algo/reverse_element
This article is contributed by Ankur. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up