Find two numbers from their sum and XOR

Difficulty Level : Hard
Last Updated : 09 Nov, 2020

Given the sum and xor of two numbers X and Y s.t. sum and xor $\in [0, 2^{64}-1]$ , 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

If X[i] is 1, then both a[i] and b[i] should be different, we have two cases.

If X[i] is 0, then both a[i] and b[i] should be same. we have two cases.

So we generate 2^n possible pairs where n is number of bits in X. Then for every pair, we check if its sum is S or not.

An efficient solution is based on below fact.

S = X + 2*A
where A = a AND b
We can verify above fact using the sum process. In sum, whenever we see both bits 1 (i.e., AND is 1), we make resultant bit 0 and add 1 as carry, which means every bit in AND is left shifted by 1 OR value of AND is multiplied by 2 and added.

So we can find A = (S – X)/2.

Once we find A, we can find all bits of ‘a’ and ‘b’ using below rules.

If X[i] = 0 and A[i] = 0, then a[i] = b[i] = 0. Only one possibility for this bit.
If X[i] = 0 and A[i] = 1, then a[i] = b[i] = 1. Only one possibility for this bit.
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.
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