Chinese Remainder Theorem

If m₁, m₂, .., m_k are pairwise relatively prime positive integers, and if a₁, a₂, .., a_k are any integers, then the simultaneous congruences x ≡ a₁ (mod m₁), x ≡ a₂ (mod m₂), ..., x ≡ a_k (mod m_k) have a solution, and the solution is unique modulo m, where m = m₁m₂⋅⋅⋅m_k .

Problem Statement :

We are given two arrays num[0..n-1] and rem[0..n-1]. In num[0..n-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[2]  =  rem[2],
      .......................
      x % num[n-1]  =  rem[n-1]

Algorithm

The solution is based on the below given formula :

    x =  sum of { (rem[i]* pp[i]* inv[i]) % prod } where 0 <= i <= n-1

    where,
    
    rem[i] is given array of remainders

    prod is product of all given numbers
    prod = num[0] * num[1] * ... * num[k-1]

    pp[i] is product of all divided by num[i]
    pp[i] = prod / num[i]

    inv[i] = Modular Multiplicative Inverse of pp[i] with respect to num[i] 

Example

Consider the above given example for understanding Chinese Remainder Theorem.

    Input : num[] = {3, 5, 7}, rem[] = {1, 2, 3}
    output : x = 52

    Explanation: 52 is the smallest number such that:
    (1) When we divide it by 3, we get remainder 1.
    (2) When we divide it by 5, we get remainder 2
    (3) When we divide it by 7, we get remainder 3.

Let us consider an example for a better understanding of above given formula.

    Input:  num[] = {3, 5, 7}, rem[] = {1, 2, 3}
    Output: x = 52

    Explanation: num[] = {3, 5, 7}, rem[] = {1, 2, 3}
    prod  = 3*5*7 = 105
    pp[]  = {35, 21, 15}
    inv[] = {2,  1,  1}  // (35*2)%3 = 1, (21*1)%5 = 1, // (15*1)%7 = 1

     x = (rem[0]* pp[0]* inv[0] + rem[1]* pp[1]* inv[1] + rem[2]* pp[2]* inv[2]) % prod
       = (1*35*2 + 2*21*1 + 3*15*1) % 105
       = (70 + 42 + 45) % 105
       = 52

Pseudo Code

The code is a simple 4-step process followed to find the solution :

    STEP 1 : Find the Product of all the numbers given in num[].
    STEP 2 : Find the number which is equal to Product divided by num[i] and store it in pp[i].
    STEP 3 : Find the Modular Multiplicative Inverse of pp[i] with respect to num[i] and store it
             in inv[i].
    STEP 4 : Multiply rem[i] with the product of pp[i] and inv[i] for all 0 <= i <= n-1 and
             add them. This is our required Output.

Complexity

Time complexity : O(n*log(max(a,b)) , where

• n is the total number of elements in the array.
• a is the maximum value in num[ ]
• b is the minimum value in rem[ ]

Implementation

• C Code

• CoffeeScript code

• C++ Code

• C# Code

• Dart Code

• Go Code

• Java Code

• JavaScript Code

• PHP Code

• Python Code

Refer this link to know more!

References

Image source: https://www.storyofmathematics.com/chinese.html

Websites:

1. http://homepages.math.uic.edu/~leon/mcs425-s08/handouts/chinese_remainder.pdf
2. https://www.geeksforgeeks.org/chinese-remainder-theorem-set-2-implementation/
3. https://medium.com/free-code-camp/how-to-implement-the-chinese-remainder-theorem-in-java-db88a3f1ffe0