Solution: x + y = x xor y as long as the addition involves no carries, i.e. the bits that are set in a should be disjoint from the ones in b. Multiples of powers of two are always disjoint from numbers less than that power of two. if x ≠ y, then without loss of generality, one of x and y will be bigger than the other, so choose the k that makes the the bigger one a power of two. If the numbers x and y given are equal to each other, then (x + k) + (y + k) ≠ (x + k) xor (y + k) for all non-negative numbers k (unless x = y = 0).
Solution: This is just a connected components problem. Once you identify the connected components, connect them minimally with # components - 1 edges: connect one vertex in each connected component to one vertex in a single chosen component of your choice. Here's my union find solution. You could also identify and connect the components using a depth first search and no union find structure, but then you need to store the adjacency matrix.