Modular multiplicative inverse

Difficulty Level : Hard
Last Updated : 08 Nov, 2020

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++

// 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; 
}

Java

// 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.*/

Python3

# 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. 

C#

// 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.

PHP

<≅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. 
≅> 

Output

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

// 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; 
}

PHP

<≅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 
≅>

Output

Modular multiplicative inverse is 4

Iterative Implementation:

C++

// 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; 
}

Java

// 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.*/

Python3

# 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. 

C#

// 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.

PHP

<≅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 
≅> 

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.

a^m-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++

// 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; 
}

Java

// 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.

Python3

# 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. 

C#

// 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.

PHP

<≅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. 
≅> 

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

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