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++
#include<iostream>
using namespace std;
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) << " ";
}
int main()
{
int n = 10, prime = 17;
printModIverses(n, prime);
return 0;
}
|
Java
import java.io.*;
class GFG {
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) + " ");
}
public static void main(String args[])
{
int n = 10, prime = 17;
printModIverses(n, prime);
}
}
|
Python3
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= " ")
n = 10
prime = 17
printModIverses(n, prime)
|
C#
using System;
class GFG {
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) + " ");
}
public static void Main()
{
int n = 10, prime = 17;
printModIverses(n, prime);
}
}
|
PHP
<?php
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) , " ";
}
$n = 10; $prime = 17;
printModIverses($n, $prime);
?>
|
Javascript
<script>
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) + " ");
}
let n = 10;
let prime = 17;
printModIverses(n, prime);
</script>
|
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
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)) % primeWe 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++
#include <iostream>
using namespace std;
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] << ' ';
}
int main()
{
int n = 10, prime = 17;
modularInverse(n, prime);
return 0;
}
|
Java
import java.io.*;
class GFG {
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] + " ");
}
public static void main(String args[])
{
int n = 10, prime = 17;
modularInverse(n, prime);
}
}
|
Python3
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=" ")
n = 10
prime = 17
modularInverse(n, prime)
|
C#
using System;
class GFG {
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] + " ");
}
public static void Main()
{
int n = 10, prime = 17;
modularInverse(n, prime);
}
}
|
PHP
<?php
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].' ');
}
$n = 10; $prime = 17;
modularInverse($n, $prime);
?>
|
Javascript
<script>
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] + ' ');
}
let n = 10;
let prime = 17;
modularInverse(n, prime);
</script>
|
Output:
1 9 6 13 7 3 5 15 2 12
Time Complexity: O(n)
Auxiliary Space: O(n)