Two positive integers 'a' and 'b' have a sum of 's' and a bitwise XOR of 'x'. How many possible values are there for the ordered pair (a, b)?

Question

We need to find ordered pairs (a,b).

(2<=s<=10^12) and (0<=x<=10^12)

For example -

s=9 x=5 We have number of ordered pairs = 4{(2,7),(7,2)(3,6)(6,3)}

Can someone please provide me a method to solve this question !!!

Answer 1

Here is the logic let the numbers be a and b, we know

s = a + b
x = a ^ b

therefore

x = (s-b) ^ b

Since we know x and we know s, so for all ints going from 0 to s - just check if this last equation is satisfied

here is the code for this

public List<Pair<Integer>> pairs(int s, int x) {
    List<Pair<Integer>> pairs = new ArrayList<Pair<Integer>>();
    for (int i = 0; i <= s / 2; i++) {
        int calc = (s - i) ^ i;
        if (calc == x) {
            pairs.add(new Pair<Integer>(i, s - i));
            pairs.add(new Pair<Integer>(s - i, i));
        }
    }
    return pairs;
}

Class pair is defined as

class Pair<T> {
    T a;
    T b;

    public String toString() {
        return a.toString() + "," + b.toString();
    }

    public Pair(T a, T b) {
        this.a = a;
        this.b = b;
    }
}

Code to test this:

public static void main(String[] args) {
    List<Pair<Integer>> pairs = new Test().pairs(9,5);
    for (Pair<Integer> p : pairs) {
        System.out.println(p);
    }
}

Output:

2,7
7,2
3,6
6,3