Chinese remainder theorem in java

In this tutorial, we will learn how to calculate chinese remainder theorem in java with some cool examples and easy examples. In many situations, you might have to come up with type of  requirements.

I know you are here to learn the easy process on chinese remainder theorem in java. First we have to do some operations such as taking an input, Error handling, calculate moduli, calculate modula and inverse modular.

            Solving remainder theorem in java

1. First we have to take two integer arrays(1. remainders ,2.moduli), representing the remainders and moduli.
2. next after taking two integer arrays we have to check if the lengths of the arrays are equal and throws an exception if not equal
3. After checking the lengths of the arrays we have to calculate M, the product of all moduli, using Big Integer  for potential large values
4. After calculating the product of all moduli using Big Integer we have calculate mi=M/mi and yi, the modular inverse of mi modulo mi, using Big Integer for modular arithmetic
5. Then it calculates the solution x using chinese remainder theorem , ensuring correct modular arithmetic with big integer

here is the example code :

import java. math .Big Integer;

public class Chinese Remainder Theorem {
    public static Big Integer  solve CRT(int[] remainders, int[] moduli) {
        if (remainders. length != moduli .length) {
            throw new Illegal Argument Exception("Lengths of remainders and moduli must be equal");
        }

        int n = remainders .length;
        Big Integer Mk = BigInteger.ONE;
        for (int i = 0; i < n; i++) {
            Mk = M. multiply(Big Integer .value Of(moduli[i]));
        }

        Big Integer[] Mi = new Big Integer[n];
        Big Integer[] yi = new Big Integer[n];
        for (int i = 0; i < n; i++) {
            Mi[i] = M. divide(Big Integer. value Of(moduli[i]));
            yi[i] = Mi[i].mod Inverse(Big Integer .value Of(moduli[i]));
        }

        Big Integer x = Big Integer .ZERO;
        for (int i = 0; i < n; i++) {
            x = x. add(Big Integer .value Of(remainders[i]).multiply(Mi[i]).multiply(yi[i]));
        }

        return x.mod(M);
    }

    public static void main(String[] args) {
        int[] remainders = {2, 3, 2};
        int[] moduli = {3, 5, 7};
        Big Integer solution = solve CRT(remainders, moduli);
        System .out. print ln("Solution: " + solution);
    }
}

we also can improve efficiency, Error handling and Flexibility, Testing so on etc.