Find two numbers from their sum and XOR

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 

Output

a = 15
b = 2

Time complexity of the above approach $O(b)$ where b is number of bits in S.

My Personal Notes

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

C++

Java

Python3

C#

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