Tight bounds for finding degrees from the adjacency matrix

Abstract

King and Smith-Thomas[5] have shown that to find a sink (a vertex with outdegree 0 and indegree n−1) in an n vertex directed graph, 3n-⌊log n⌋-3 probes to its adjacency matrix are necessary and sufficient in the worst case. We first generalize this result to show that for any integer k between 0 and n−1, to find a vertex of outdegree k in a simple n vertex directed graph, Θ(n(n−k)) probes to its adjacency matrix are necessary and sufficient in the worst case, and to find a vertex of degree k in an n vertex undirected graph, Θ(n 2) probes to its adjacency matrix are necessary and sufficient in the worst case. Then we study this problem on a special class of directed graphs called tournaments (between every pair of vertices, there is exactly one directed edge). Recently[1] it has been shown that nk/2 probes are necessary and (4k+2)n probes are sufficient to test whether a tournament has a vertex with outdegree k≤(n−1)/2 (The case when k>(n−1)/2 is symmetric since we can find a vertex of indegree n−1−k). We improve the lower bound, and the upper bound for k=0 and k=1, all by a constant factor to show that if T n is a tournament on n vertices, to test whether T n has a vertex of outdegree k≤(n−1)/2, (2k+2)n+O(k log k) probes to the adjacency matrix are necessary in the worst case, and to test whether T n has a vertex of outdegree k, for k=0 and 1, (2k+2)n+o(n) probes to the adjacency matrix are sufficient, by exhibiting two different algorithms, one for k=0 and k=1. We conjecture that this lower bound for tournaments is optimal (up to lower order terms) for most k≤(n−1)/2.