Modular multiplicative inverse from 1 to n

Give a positive integer n, find modular multiplicative inverse of all integer from 1 to n with respect to a big prime number, say, ‘prime’.

The modular multiplicative inverse of a is an integer ‘x’ such that.

 a x ≡ 1 (mod prime)

Examples :

Input : n = 10, prime = 17
Output : 1 9 6 13 7 3 5 15 2 12
Explanation :
For 1, modular inverse is 1 as (1 * 1)%17 is 1
For 2, modular inverse is 9 as (2 * 9)%17 is 1
For 3, modular inverse is 6 as (3 * 6)%17 is 1
....... 

Input : n = 5, prime = 7
Output : 1 4 5 2 3

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to one by one find modular inverse for every number.

C++

// C++ program to find modular inverse of  
// all numbers from 1 to n using naive 
// method 
#include<iostream> 
using namespace std; 
   
// A naive method to find modular 
// multiplicative inverse of 'a' 
// under modulo 'prime' 
int modInverse(int a, int prime) 
{ 
    a = a % prime; 
    for (int x=1; x<prime; x++) 
       if ((a*x) % prime == 1) 
          return x; 
       
    return -1; 
} 
  
void printModIverses(int n, int prime) 
{ 
    for (int i=1; i<=n; i++) 
      cout << modInverse(i, prime) << " "; 
} 
   
// Driver Program 
int main() 
{ 
    int n = 10, prime = 17; 
    printModIverses(n, prime); 
    return 0; 
} 

Java

// Java program to find modular inverse of  
// all numbers from 1 to n using naive 
// method 
import java.io.*; 
  
class GFG { 
      
    // A naive method to find modular 
    // multiplicative inverse of 'a' 
    // under modulo 'prime' 
    static int modInverse(int a, int prime) 
    { 
        a = a % prime; 
        for (int x = 1; x  <prime; x++) 
        if ((a * x) % prime == 1) 
            return x; 
          
        return -1; 
    } 
      
    static void printModIverses(int n, int prime) 
    { 
        for (int i = 1; i <= n; i++) 
        System.out.print(modInverse(i, prime) + " "); 
    } 
      
    // Driver Program 
    public static void main(String args[]) 
    { 
        int n = 10, prime = 17; 
        printModIverses(n, prime); 
    } 
} 
  
  
// This code is contributed by Nikita Tiwari. 

Python3

# Python 3 program to find 
# modular inverse of  
# all numbers from 1 
# to n using naive 
# method 
  
  
# A naive method to find modular 
# multiplicative inverse of 'a' 
# under modulo 'prime' 
  
def modInverse(a, prime) : 
    a = a % prime 
    for x in range(1,prime) : 
        if ((a*x) % prime == 1) : 
            return x 
        
    return -1
      
   
def printModIverses(n, prime) : 
    for i in range(1,n+1) : 
        print( modInverse(i, prime) ,end= " ") 
    
# Driver Program 
n = 10
prime = 17
  
printModIverses(n, prime) 
  
# This code is contributed 
# by Nikita Tiwari. 

C#

// C# program to find modular inverse of  
// all numbers from 1 to n using naive 
// method 
using System; 
  
class GFG { 
      
    // A naive method to find modular 
    // multiplicative inverse of 'a' 
    // under modulo 'prime' 
    static int modInverse(int a, int prime) 
    { 
        a = a % prime; 
        for (int x = 1; x <prime; x++) 
            if ((a * x) % prime == 1) 
                return x; 
          
        return -1; 
    } 
      
    static void printModIverses(int n, int prime) 
    { 
        for (int i = 1; i <= n; i++) 
            Console.Write(modInverse(i, prime) + " "); 
    } 
      
    // Driver Program 
    public static void Main() 
    { 
        int n = 10, prime = 17; 
          
        printModIverses(n, prime); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find modular inverse of  
// all numbers from 1 to n using naive 
// method 
  
// A naive method to find modular 
// multiplicative inverse of 'a' 
// under modulo 'prime' 
function modInverse(int $a, int $prime) 
{ 
    $a = $a % $prime; 
    for ( $x = 1; $x < $prime; $x++) 
    if (($a * $x) % $prime == 1) 
        return $x; 
      
    return -1; 
} 
  
function printModIverses( $n, $prime) 
{ 
    for ( $i = 1; $i <= $n; $i++) 
    echo modInverse($i, $prime) , " "; 
} 
  
// Driver Program 
  
    $n = 10; $prime = 17; 
    printModIverses($n, $prime); 
  
// This code is contributed by anuj_67. 
?> 

Output:

1 9 6 13 7 3 5 15 2 12

An efficient solution is based on extended Euclid algorithm.

Extended Euclidean algorithm finds integer coefficients x and y such that:

  ax + by = gcd(a, b)

Let us put b = prime, we get
  ax + prime * y = gcd(a, prime)

We know gcd(a, prime) = 1 because
on of the numbers is prime. So we
know
  ax + prime * y = 1

Since prime * y is a multiple of prime,
x is modular multiplicative inverse of a.

 ax  ≡ 1 (mod prime)

We can recursively find x using below expression (see extended Euclid algorithm for details).

The extended Euclidean algorithm updates results of gcd(a, b) using the results calculated by recursive call gcd(b%a, a). Let values of x and y calculated by the recursive call be x_prev and y_prev. x and y are updated using below expressions.

x = y_prev - ⌊prime/a⌋ * x_prev
y = x_prev

We use above relation to compute inverse using previously computed values.

inverse(a) = (inverse(prime % a) *
              (prime - prime/a)) % prime

We use Dynamic Programming approach that uses above recursive structure.

Dynamic Approach :
dp[1] = 1,
dp[2] = dp[17%2]*(17-17/2)%17 = 9
dp[3] = dp[17%3]*(17-17/3)%17 = 6

and so on..

C++

// CPP code to find modular inverse 
// from 1 to n w.r.t a big prime number 
#include <iostream> 
using namespace std; 
  
// Function to calculate modular 
// inverse using D.P 
void modularInverse(int n, int prime) 
{ 
    int dp[n + 1]; 
    dp[0] = dp[1] = 1; 
    for (int i = 2; i <= n; i++)  
        dp[i] = dp[prime % i] *  
               (prime - prime / i) % prime;     
  
    for (int i = 1; i <= n; i++)  
        cout << dp[i] << ' ';     
} 
  
// Driver code 
int main() 
{ 
    int n = 10, prime = 17; 
    modularInverse(n, prime); 
    return 0; 
} 

Java

// Java code to find modular inverse 
// from 1 to n w.r.t a big prime number 
import java.io.*; 
  
class GFG { 
  
    // Function to calculate modular 
    // inverse using D.P 
    static void modularInverse(int n, int prime) 
    { 
        int dp[]=new int[n + 1]; 
        dp[0] = dp[1] = 1; 
        for (int i = 2; i <= n; i++)  
            dp[i] = dp[prime % i] *  
                (prime - prime / i) % prime;  
      
        for (int i = 1; i <= n; i++)  
            System.out.print(dp[i] + " ");  
    } 
  
    // Driver Program 
    public static void main(String args[]) 
    { 
        int n = 10, prime = 17; 
        modularInverse(n, prime); 
    } 
} 
  
  
// This code is contributed by Nikita Tiwari. 

Python3

# Python 3 code to find 
# modular inverse 
# from 1 to n w.r.t a 
# big prime number 
  
# Function to calculate modular 
# inverse using D.P 
def modularInverse( n, prime) : 
  
    dp =[0]*(n+1) 
    dp[0] = dp[1] = 1
    for i in range( 2, n+1) : 
        dp[i] = dp[prime % i] *(prime - prime // i) % prime  
   
    for i in range( 1, n+1) : 
        print(dp[i] ,end=" ") 
          
   
# Driver code 
n = 10
prime = 17
  
modularInverse(n, prime) 
  
# This code is contributed 
# by Nikita Tiwari. 

C#

// C# code to find modular inverse 
// from 1 to n w.r.t a big prime number 
using System; 
  
class GFG { 
  
    // Function to calculate modular 
    // inverse using D.P 
    static void modularInverse(int n, int prime) 
    { 
        int []dp=new int[n + 1]; 
        dp[0] = dp[1] = 1; 
          
        for (int i = 2; i <= n; i++)  
            dp[i] = dp[prime % i] *  
                (prime - prime / i) % prime;  
      
        for (int i = 1; i <= n; i++)  
            Console.Write(dp[i] + " ");  
    } 
  
    // Driver Program 
    public static void Main() 
    { 
        int n = 10, prime = 17; 
          
        modularInverse(n, prime); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP code to find modular  
// inverse from 1 to n w.r.t 
// a big prime number 
  
// Function to calculate  
// modular inverse using D.P 
function modularInverse($n, $prime) 
{ 
    $dp = array(); 
    $dp[0] = $dp[1] = 1; 
    for ($i = 2; $i <= $n; $i++)  
        $dp[$i] = $dp[$prime % $i] *  
                     ($prime -  
               intval($prime / $i)) %  
                      $prime;  
  
    for ($i = 1; $i <= $n; $i++)  
        echo ($dp[$i].' ');  
} 
  
// Driver code 
$n = 10; $prime = 17; 
modularInverse($n, $prime); 
  
// This code is contributed by 
// Manish Shaw(manishshaw1) 
?> 

Output:

1 9 6 13 7 3 5 15 2 12

Time Complexity : O(n)
Auxiliary Space : O(n)

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My Personal Notes

Aug	SEP	Oct
	26
2019	2020	2021

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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