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<3, and f(k)=3 otherwise.

💡 Research Summary

The paper investigates the relationship between edge‑disjoint s‑t paths and vertex‑independent u‑v paths in simple undirected graphs. For a given integer k, let f(k) denote the largest integer such that every graph containing two vertices s and t with k edge‑disjoint s‑t paths must also contain two vertices u and v with f(k) independent u‑v paths (a set of paths is independent if no path contains an interior vertex of another). The authors prove that f(k)=k for k≤2 and f(k)=3 for all k≥3.

The easy direction f(k)≤k follows from the fact that independent paths are automatically edge‑disjoint. For k=1 the single path is trivially independent. For k=2 the construction is explicit: take s as the common start, locate the first vertex v where the two edge‑disjoint s‑t paths intersect, and use the s‑v subpaths of both paths as the required independent pair.

The non‑trivial part is establishing the lower bound f(k)≥3 when k≥3. Lemma 1 shows that any three edge‑disjoint s‑t paths can be transformed into three independent u‑v paths. Let P₁,P₂,P₃ be the three edge‑disjoint s‑t paths and let sᵢ be the neighbor of s on Pᵢ. After deleting s, consider the connected component containing t and take a spanning tree T of this component. Within T there exists a vertex v that lies on the three sᵢ‑v subpaths; such a vertex can be found by intersecting the s₁‑s₃ and s₂‑s₃ subpaths of T and choosing the common point closest to s₂. The three paths from s to v obtained by concatenating the edge s‑sᵢ with the sᵢ‑v subpath of T are pairwise independent. This argument essentially uses Menger’s theorem: the three edge‑disjoint s‑t paths guarantee a vertex cut of size at most two separating s from v, which forces the existence of three independent paths.

To prove the upper bound f(k)≤3 for k≥3, Lemma 2 introduces the recursive diamond graphs Dₚ. D₀ consists of a single edge st. For p≥1, each edge of Dₚ₋₁ is replaced by a “diamond” consisting of four parallel two‑edge paths, creating four edge‑disjoint copies of Dₚ₋₁. Choosing p=⌈log₂k⌉ ensures that Dₚ contains at least k edge‑disjoint s‑t paths. The authors then show that for any pair of vertices u, v in Dₚ, there are at most three independent u‑v paths. The proof proceeds by selecting the smallest recursive diamond subgraph Q that contains u and v, analyzing its order q, and observing that Q can be decomposed into four sub‑diamonds H₁,…,H₄ each isomorphic to D_{q‑1}. Depending on which sub‑diamonds u and v belong to, one can always find a u‑v vertex cut of size two (or three when q<p and u is an extremity of another disjoint sub‑diamond). By Menger’s theorem, the size of a minimum u‑v cut bounds the number of independent u‑v paths, establishing the desired upper bound of three.

Combining the two lemmas yields the exact function: f(k)=k for k=1,2 and f(k)=3 for all k≥3.

The paper also discusses an application to SAT solving. In a recent algorithm for detecting strong backdoor sets to the class of nested formulas, Lemma 1 is used to exhibit three independent paths in an auxiliary graph, which after edge expansion yields a K_{2,3} minor—a structure that certifies the absence of a small backdoor set. Lemma 2, however, shows that this technique cannot be extended to guarantee larger independent sets, limiting the approach when trying to broaden the target formula class.

Overall, the work provides a clean characterization of how many vertex‑independent paths are forced by a given number of edge‑disjoint paths, highlights the tightness of the bound via a simple yet powerful family of graphs, and connects these combinatorial insights to practical algorithmic problems in computational complexity.