Find two numbers from their sum and XOR

Given the sum and xor of two numbers X and Y s.t. sum and xor , we need to find the numbers minimizing the value of X. Examples :

Input : S = 17
        X = 13
Output : a = 2
         b = 15

Input : S = 1870807699 
        X = 259801747
Output : a = 805502976
         b = 1065304723

Input : S = 1639
        X = 1176
Output : No such numbers exist

Variables Used: X ==> XOR of two numbers S ==> Sum of two numbers X[i] ==> Value of i-th bit in X S[i] ==> Value of i-th bit in S

A simple solution is to generate all possible pairs with given XOR. To generate all pairs, we can follow below rules.

1. If X[i] is 1, then both a[i] and b[i] should be different, we have two cases.
2. If X[i] is 0, then both a[i] and b[i] should be same. we have two cases.
   1. If X[i] = 0 and A[i] = 0, then a[i] = b[i] = 0. Only one possibility for this bit.
   2. If X[i] = 0 and A[i] = 1, then a[i] = b[i] = 1. Only one possibility for this bit.
   3. If X[i] = 1 and A[i] = 0, then (a[i] = 1 and b[i] = 0) or (a[i] = 0 and b[i] = 1), we can pick any of the two.
   4. If X[i] = 1 and A[i] = 1, result not possible (Note X[i] = 1 means different bits)

Let the summation be S and XOR be X.

Below is the implementation of above approach:

C++

// CPP program to find two numbers with
// given Sum and XOR such that value of
// first number is minimum.
#include <iostream>
using namespace std;

// Function that takes in the sum and XOR
// of two numbers and generates the two
// numbers such that the value of X is
// minimized
void compute(unsigned long int S,
            unsigned long int X)
{
    unsigned long int A = (S - X)/2;

    int a = 0, b = 0;

    // Traverse through all bits
    for (int i=0; i<8*sizeof(S); i++)
    {
        unsigned long int Xi = (X & (1 << i));
        unsigned long int Ai = (A & (1 << i));
        if (Xi == 0 && Ai == 0)
        {
            // Let us leave bits as 0.
        }
        else if (Xi == 0 && Ai > 0)
        {
            a = ((1 << i) | a);
            b = ((1 << i) | b);
        }
        else if (Xi > 0 && Ai == 0)
        {
            a = ((1 << i) | a);

            // We leave i-th bit of b as 0.
        }
        else // (Xi == 1 && Ai == 1)
        {
            cout << "Not Possible";
            return;
        }
    }

    cout << "a = " << a << endl << "b = " << b;
}

// Driver function
int main()
{
    unsigned long int S = 17, X = 13;
    compute(S, X);
    return 0;
}

Java

// Java program to find two numbers with
// given Sum and XOR such that value of
// first number is minimum.
class GFG {

// Function that takes in the sum and XOR
// of two numbers and generates the two
// numbers such that the value of X is
// minimized
static void compute(long S, long X)
{
    long A = (S - X)/2;
    int a = 0, b = 0;
        final int LONG_FIELD_SIZE     = 8;

    // Traverse through all bits
    for (int i=0; i<8*LONG_FIELD_SIZE; i++)
    {
        long Xi = (X & (1 << i));
        long Ai = (A & (1 << i));
        if (Xi == 0 && Ai == 0)
        {
            // Let us leave bits as 0.
        }
        else if (Xi == 0 && Ai > 0)
        {
            a = ((1 << i) | a);
            b = ((1 << i) | b);
        }
        else if (Xi > 0 && Ai == 0)
        {
            a = ((1 << i) | a);

            // We leave i-th bit of b as 0.
        }
        else // (Xi == 1 && Ai == 1)
        {
            System.out.println("Not Possible");
            return;
        }
    }

    System.out.println("a = " + a +"\nb = " + b);
}

// Driver function
    public static void main(String[] args) {
        long S = 17, X = 13;
    compute(S, X);

    }
}
// This code is contributed by RAJPUT-JI

Python3

# Python program to find two numbers with
# given Sum and XOR such that value of
# first number is minimum.


# Function that takes in the sum and XOR
# of two numbers and generates the two
# numbers such that the value of X is
# minimized
def compute(S, X):
    A = (S - X)//2
    a = 0
    b = 0

    # Traverse through all bits
    for i in range(64):
        Xi = (X & (1 << i))
        Ai = (A & (1 << i))
        if (Xi == 0 and Ai == 0):
            # Let us leave bits as 0.
            pass

        elif (Xi == 0 and Ai > 0):
            a = ((1 << i) | a)
            b = ((1 << i) | b)

        elif (Xi > 0 and Ai == 0):
            a = ((1 << i) | a)
            # We leave i-th bit of b as 0.

        else: # (Xi == 1 and Ai == 1)
            print("Not Possible")
            return

    print("a = ",a)
    print("b =", b)


# Driver function
S = 17
X = 13
compute(S, X)

# This code is contributed by ankush_953

C#

// C# program to find two numbers with
// given Sum and XOR such that value of
// first number is minimum.
using System;

public class GFG {

// Function that takes in the sum and XOR
// of two numbers and generates the two
// numbers such that the value of X is
// minimized
static void compute(long S, long X)
{
    long A = (S - X)/2;
    int a = 0, b = 0;


    // Traverse through all bits
    for (int i=0; i<8*sizeof(long); i++)
    {
        long Xi = (X & (1 << i));
        long Ai = (A & (1 << i));
        if (Xi == 0 && Ai == 0)
        {
            // Let us leave bits as 0.
        }
        else if (Xi == 0 && Ai > 0)
        {
            a = ((1 << i) | a);
            b = ((1 << i) | b);
        }
        else if (Xi > 0 && Ai == 0)
        {
            a = ((1 << i) | a);

            // We leave i-th bit of b as 0.
        }
        else // (Xi == 1 && Ai == 1)
        {
            Console.WriteLine("Not Possible");
            return;
        }
    }

    Console.WriteLine("a = " + a +"\nb = " + b);
}

// Driver function
    public static void Main() {
        long S = 17, X = 13;
        compute(S, X);
    }
}
// This code is contributed by RAJPUT-JI

Javascript

// JavaScript program to find two numbers with
// given Sum and XOR such that value of
// first number is minimum.


// Function that takes in the sum and XOR
// of two numbers and generates the two
// numbers such that the value of X is
// minimized
function compute(S, X)
{
    var A = Math.floor((S - X) / 2);
    var a = 0;
    var b = 0;

    // Traverse through all bits
    for (var i = 0; i < 64; i++)
    {
        var Xi = (X & (1 << i));
        var Ai = (A & (1 << i));
        if (Xi == 0 && Ai == 0)
        {
            // Let us leave bits as 0.
            continue;
        }

        else if (Xi == 0 && Ai > 0)
        {
            a = ((1 << i) | a);
            b = ((1 << i) | b);
        }

        else if (Xi > 0 && Ai == 0)
        {
            a = ((1 << i) | a);
            // We leave i-th bit of b as 0.
        }

        else
        {
        // (Xi == 1 and Ai == 1)
            console.log("Not Possible");
            return;
        }
    }

    console.log("a = " + a);
    console.log("b = " + b);
}

// Driver function
var S = 17;
var X = 13;
compute(S, X);

// This code is contributed by phasing17

Output
a = 15
b = 2

Time Complexity: O(logn)

Auxiliary Space: O(1)