Chinese Remainder Theorem | Set 1 (Introduction)

We are given two arrays num[0..k-1] and rem[0..k-1]. In num[0..k-1], every pair is coprime (gcd for every pair is 1). We need to find minimum positive number x such that:

x % num[0] = rem[0],

x % num[1] = rem[1],

.......................

x % num[k-1] = rem[k-1]

Basically, we are given k numbers which are pairwise coprime, and given remainders of these numbers when an unknown number x is divided by them. We need to find the minimum possible value of x that produces given remainders.

Examples :

Input: num[] = {5, 7}, rem[] = {1, 3}

Output: 31

Explanation:

31 is the smallest number such that:

(1) When we divide it by 5, we get remainder 1.

(2) When we divide it by 7, we get remainder 3.

Input: num[] = {3, 4, 5}, rem[] = {2, 3, 1}

Output: 11

Explanation:

11 is the smallest number such that:

(1) When we divide it by 3, we get remainder 2.

(2) When we divide it by 4, we get remainder 3.

(3) When we divide it by 5, we get remainder 1.

Chinese Remainder Theorem states that there always exists an x that satisfies given congruences. Below is theorem statement adapted from wikipedia.

Let num[0], num[1], …num[k-1] be positive integers that are pairwise coprime. Then, for any given sequence of integers rem[0], rem[1], … rem[k-1], there exists an integer x solving the following system of simultaneous congruences.

## Code in JAVA

// A Java program to demonstrate the working of Chinese remainder

// Theorem

import java.io.*;

class Hactoberfest {

// k is size of num[] and rem[]. Returns the smallest

// number x such that:

// x % num[0] = rem[0],

// x % num[1] = rem[1],

// ..................

// x % num[k-2] = rem[k-1]

// Assumption: Numbers in num[] are pairwise coprime

// (gcd for every pair is 1)

static int findMinX(int num[], int rem[], int k)

{

int x = 1; // Initialize result

// As per the Chinese remainder theorem,

// this loop will always break.

while (true)

{

// Check if remainder of x % num[j] is

// rem[j] or not (for all j from 0 to k-1)

int j;

for (j=0; j<k; j++ )

if (x%num[j] != rem[j])

break;

// If all remainders matched, we found x

if (j == k)

return x;

// Else try next number

x++;

}

}

// Driver method

public static void main(String args[])

{

int num[] = {3, 4, 5};

int rem[] = {2, 3, 1};

int k = num.length;

System.out.println("x is " + findMinX(num, rem, k));

}

}