Problem 1. Let G(V,E) be a connected, undirected graph and let sV be a source node (you may assume that the graph is given in an adjacency list format). For any vV, we call path from s to v shortish if it is either a shortest path or it has one more edge than a shortest path. (a) Consider the BFS algorithm. Suppose that u is the next node that the algorithm explores its neighbors. Prove that for any neighboring node vG[u] that has already been explored by the algorithm we have dists(v)=dists(u)1ordists(v)=dists(u), where dists(a) denotes the distance between s and some node aV (this is, the length of the shortest path between s and a )