Find the values of X and Y in the Given Equations

Given two numbers A and B. Find the values of X and Y in the equations.

  1. A = X + Y
  2. B = X xor Y

The task is to make X as minimum as possible. If it is not possible to find any valid values for X and Y then print -1.

Examples:



Input : A = 12, B = 8
Output : X = 2, Y = 10

Input : A = 12, B = 9
Output : -1

Let’s take a look at some bit in X, which is equal to 1. If the respective bit in Y is equal to 0, then one can swap these two bits, thus reducing X and increasing Y without changing their sum and xor. We can conclude that if some bit in X is equal to 1 then the respective bit in Y is also equal to 1. Thus, Y = X + B. Taking into account that X + Y = X + X + B = A, one can obtain the following formulas for finding X and Y:

  • X = (A – B) / 2
  • Y = X + B = (A + B) / 2

One should also notice that if A < B or A and B have different parity, then the answer doesn’t exist and output is -1. If X and (A – X) not equal to X then the answer is also -1.

Below is the implementation of the above approach :

C++

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// CPP program to find the values of
// X and Y using the given equations
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the
// values of X and Y
void findValues(int a, int b)
{
    // base condition
    if ((a - b) % 2 == 1) {
        cout << "-1";
        return;
    }
  
    // required answer
    cout << (a - b) / 2 << " " << (a + b) / 2;
}
  
// Driver Code
int main()
{
    int a = 12, b = 8;
  
    findValues(a, b);
  
    return 0;
}

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Java














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// Java program to find the values of 


// X and Y using the given equations 


import java.io.*; 


  


class GFG  


{ 


      


// Function to find the 


// values of X and Y 


static void findValues(int a, int b) 


{ 


    // base condition 


    if ((a - b) % 2 == 1)  


    { 


            System.out.println ("-1"); 


        return; 


    } 


  


    // required answer 


    System.out.println (((a - b) / 2)+ " " + 


                            ((a + b) / 2)); 


} 


  


    // Driver Code 


    public static void main (String[] args) 


    { 


        int a = 12, b = 8; 


        findValues(a, b); 


    } 


} 


  


// This code is contributed by ajit...  



















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Python3














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# Python3 program to find the values of 


# X and Y using the given equations 


  


# Function to find the values of X and Y 


def findValues(a, b): 


  


    # base condition 


    if ((a - b) % 2 == 1): 


        print("-1"); 


        return; 


  


    # required answer 


    print((a - b) // 2, (a + b) // 2); 


  


# Driver Code 


a = 12; b = 8; 


  


findValues(a, b); 


  


# This code is contributed 


# by Akanksha Rai 



















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C#














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// C# program to find the values of 


// X and Y using the given equations 


using System; 


  


class GFG 


{ 


          


// Function to find the 


// values of X and Y 


static void findValues(int a, int b) 


{ 


    // base condition 


    if ((a - b) % 2 == 1)  


    { 


            Console.Write ("-1"); 


        return; 


    } 


  


    // required answer 


    Console.WriteLine(((a - b) / 2)+ " " + 


                        ((a + b) / 2)); 


} 


  


// Driver Code 


static public void Main () 


{ 


    int a = 12, b = 8; 


    findValues(a, b); 


} 


} 


  


// This code is contributed by @Tushil..  



















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PHP














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<?php 


// PHP program to find the values of 


// X and Y using the given equations 


  


// Function to find the values  


// of X and Y 


function findValues($a, $b) 


{ 


    // base condition 


    if (($a - $b) % 2 == 1) 


    { 


        echo "-1"; 


        return; 


    } 


  


    // required answer 


    echo ($a - $b) / 2, " " ,  


         ($a + $b) / 2; 


} 


  


// Driver Code 


$a = 12; 


$b = 8; 


findValues($a, $b); 


  


// This code is contributed by jit_t 


?> 



















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Output:


2 10





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