Modular multiplicative inverse from 1 to n (original) (raw)

Last Updated : 23 Nov, 2023

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 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.

JavaScript

PHP

a,inta, int a,intprime) { a=a = a=a % $prime; for ( x=1;x = 1; x=1;x < prime;prime; prime;x++) if (($a * x)x) % x)prime == 1) return $x; return -1; } function printModIverses( n,n, n,prime) { for ( i=1;i = 1; i=1;i <= n;n; n;i++) echo modInverse($i, $prime) , " "; } // Driver Program n=10;n = 10; n=10;prime = 17; printModIverses($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+primey1 = gcd(prime%a, prime) => (prime%a)x1+primey1 = 1 -----(ii) =>(prime - (prime/a)a)x1 + primey1 = 1 =>-(prime/a)x1a+(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 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.

JavaScript

PHP

dp[0]=dp[0] = dp[0]=dp[1] = 1; for ($i = 2; i<=i <= i<=n; $i++) dp[dp[dp[i] = dp[dp[dp[prime % $i] * ($prime - intval($prime / $i)) % $prime; for ($i = 1; i<=i <= i<=n; $i++) echo ($dp[$i].' '); } // Driver code n=10;n = 10; n=10;prime = 17; modularInverse($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)