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

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# 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.

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>

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

Time Complexity: O(n*prime)
Auxiliary Space: O(1)

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
one of the numbers is prime. So we
know
  ax + prime * y = 1 ---- (i)

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).

if we take for gcd(prime%a,prime) it'll be 1 

so (prime%a)*x1+prime*y1 = gcd(prime%a, prime)
=> (prime%a)*x1+prime*y1  = 1 -----(ii)
=>(prime - (prime/a)*a)x1 + prime*y1  = 1
=>-(prime/a)*x1*a+(x1+y1)*prime
using eq(i) and eq(ii) comparing the co-eeficient of a & prime we get

x = -(prime/a)*x1, & y = (x1+y1)

 x = inv(a)  & x1 = inv(prime%a)

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

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

(-x % m = (m-x) % m)

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;
}