Let \(G = (V, E)\) be a simple undirected graph, and \(s\) be a particular…

Key idea: For any edge (u,v) in the graph, a breadth-first search rooted at s guarantees that the distances differ by at most 1, i.e. |d(u) - d(v)| ≤ 1.

Proof that a difference of 2 cannot occur:

• Assume for contradiction that d(u) = d(v) + 2.

• Let k = d(v). Then there is a shortest path from s to v of length k.

• Because (u,v) is an edge, following that shortest path to v and then taking the edge to u gives a path from s to u of length k + 1 = d(v) + 1.

• This implies d(u) ≤ d(v) + 1, contradicting the assumption d(u) = d(v) + 2. Hence a difference of 2 is impossible.

Why the values -1, 0, and 1 are possible:

• -1: This indicates u is one level closer to the source than v. An edge between consecutive levels (other than the tree parent edge) is allowed and does not contradict shortest paths.

• 0: This indicates u and v are at the same distance from s. Such cross edges between vertices on the same BFS level may exist and are simply not part of the BFS tree.

• 1: This indicates u is one level farther from the source than v. As with -1, edges between adjacent levels are possible.

Conclusion: The only value among the choices that cannot occur for d(u) - d(v) is 2.