From edge-disjoint paths to independent paths

Let f(k) denote the maximum such that every simple undirected graph containing two vertices s, t and k edge-disjoint s–t paths, also contains two vertices u, v and f(k) independent u–v paths. Here, a set of paths is independent if none of them contains an interior vertex of another. We prove that f(k) = ( k if k ≤ 2, and 3 otherwise. Since independent paths are edge-disjoint, it is clear that f(k) ≤ k for every positive integer k. Let P be a set of edge-disjoint s–t paths in a graph G. Clearly, if |P| ≤ 1, then the paths in P are independent. If P = {P1, P2}, a set of two independent u–v paths can easily be obtained as follows. Set u := s and let v be the vertex that belongs to both P1 and P2 and is closest to s on P1. Then, the u–v subpaths of P1 and P2 are independent. This proves that f(k) = k if k ≤ 2. The lower bound for f(k), k ≥ 3, is provided by the following lemma. Lemma 1. Let G = (V,E) be a graph. If there are two vertices s, t ∈ V with 3 edge-disjoint s–t paths in G, then there are two vertices u, v ∈ V with 3 independent u–v paths in G. Proof. Let P1, P2, P3 denote 3 edge-disjoint s–t paths, and let S = {s1, s2, s3}, where si neighbors s on Pi, 1 ≤ i ≤ 3. Consider the connected component G′ of G \ {s} containing t. Then, G′ contains all vertices from S. Let T be a spanning tree of G′. Select v such that the si–v subpaths of T , 1 ≤ i ≤ 3, are independent. This vertex v belongs to every subpath of T that has two vertices from S as endpoints. To see that this vertex exists, consider the s1–s3 subpath P1,3 of T and the s2–s3 subpath P2,3 of T . Set v to be the vertex that belongs to both P1,3 and P2,3 and is closest to s2 on P2,3 (if P1,3 contains s2, then v = s2). Set u := s, and obtain 3 independent u–v paths in G by moving from u to si, and then along the si–v subpath of T to v, 1 ≤ i ≤ 3. For the upper bound, consider the following family of graphs, the recursive diamond graphs [4]. The recursive diamond graph of order 0 is G0 = ({s, t}, {st}), and the diamond graph Gp of order p ≥ 1 is obtained from Gp−1 by replacing each edge e = xy by the set of edges {xpe, pey, xqe, qey}, where pe and qe are new vertices. See Figure 1 for an illustration. The following lemma entails the upper bound for f(k), k ≥ 3. Lemma 2. For every k ≥ 3, there is a graph G = (V,E) containing two vertices s, t ∈ V with k edge-disjoint s–t paths, but no two vertices u, v ∈ V with 4 independent u–v paths. Proof. Consider the diamond graph G = Gp of order p = dlog ke. G has 2 ≥ k edge-disjoint s–t paths. Let u, v be any two vertices in G. We will show that there are at most 3 independent u–v paths. Observe that each recursive diamond graph Gr contains 4 edge-disjoint copies of Gr−1. The extremities of Gr are the vertices s and t, and the extremities of a subgraph H of Gr that is isomorphic to Gr′ , r′ < r, are the two vertices from H whose neighborhoods in Gr are not a subset of V (H). Let Q be the smallest vertex set containing u and v such that G[Q] is a recursive diamond graph. Let q be the order of the recursive diamond graph G[Q]. If q = 0, then uv is an edge in G, and either u or v has degree 2. But then, the number of independent u–v paths in G is at most 2 since independent paths pass through distinct neighbors of u and v. If q > 0, then uv is not an edge in G. Decompose G[Q] into 4 edge-disjoint graphs H1, . . . ,H4 isomorphic to Gq−1 such that u ∈ V (H1) and the Hi are ordered cyclically by their index (i.e., V (H1) ∩ ∗Institute of Information Systems, Vienna University of Technology, [email protected] Research supported by the European Research Council (COMPLEX REASON, 239962).