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Modular multiplicative inverse from 1 to n

  • Difficulty Level : Hard
  • Last Updated : 31 Mar, 2021

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

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.
?>

Javascript




<script>
 
// Javascript 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(a, prime)
{
    a = a % prime;
     
    for(let x = 1; x < prime; x++)
        if ((a * x) % prime == 1)
            return x;
     
    return -1;
}
 
function printModIverses( n, prime)
{
    for(let i = 1; i <= n; i++)
        document.write(modInverse(i, prime) + " ");
}
 
// Driver code
let n = 10;
let prime = 17;
 
printModIverses(n, prime);
 
// This code is contributed by _saurabh_jaiswal
 
</script>

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 xprev and yprev. x and y are updated using below expressions. 

x = yprev - ⌊prime/a⌋ * xprev
y = xprev

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)
?>

Javascript




<script>
 
// Javascript 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)
{
    let dp = [];
    dp[0] = dp[1] = 1;
     
    for(let i = 2; i <= n; i++)
        dp[i] = dp[prime % i] * (prime -
         parseInt(prime / i)) % prime;
 
    for(let i = 1; i <= n; i++)
        document.write(dp[i] + ' ');
}
 
// Driver code
let n = 10;
let prime = 17;
 
modularInverse(n, prime);
 
// This code is contributed by _saurabh_jaiswal
 
</script>

Output: 

1 9 6 13 7 3 5 15 2 12

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

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