A Proof of the Two-path Conjecture

Let G be a connected graph that is the edge-disjoint union of two paths of length n, where n ≥ 2. Using a result of Thomason on decompositions of 4-regular graphs into pairs of Hamiltonian cycles, we prove that G has a third path of length n. The “two-path conjecture” states that if a graph G is the edge-disjoint union of two paths of length n with at least one common vertex, then the graph has a third subgraph the electronic journal of combinatorics 8 (2001), #R00 1 that is also a path of length n. For example, the complete graph K4 is an edge-disjoint union of two paths of length 3, each path meeting the other in four vertices. The cycle C6 is the edge-disjoint union of two paths of length 3 with common endpoints. In the first case, the graph has twelve paths of length 3; in the second there are six such paths. The two-path conjecture arose in a problem on randomly decomposable graphs. An H-decomposition of a graph G is a family of edge disjoint H–subgraphs of G whose union is G. An H-decomposable graph G is randomly H–decomposable if any edge disjoint family of H–subgraphs of G can be extended to an H–decomposition of G. (This concept was introduced by Ruiz in [7].) Randomly Pn-decomposable graphs were studied in [1, 5, 6, 4]. In attempting to classify randomly Pn-decomposable graphs, in [5] and [6] it was necessary to know whether the edge-disjoint union of two copies of Pn could have a unique Pn-decomposition. The two-path conjecture is stated as an unproved lemma in [3]. Our notation follows [2]. A path of length n is a trail with distinct vertices x0, . . . , xn, ([2], p. 5). We say that G decomposes into subgraphs X and Y when G is the edge-disjoint union of X and Y . Theorem. If G decomposes into two paths X and Y , each of length n with n ≥ 2, and X and Y have least one common vertex, then G has a path of length n distinct from X and Y . Proof. Label the vertices of X as x0, x1, . . . , xn, with xi−1 adjacent to xi for 1 ≤ i ≤ n. Similarly, label the vertices of Y as y0, y1, . . . , yn. Let s be the number of common vertices; thus G has 2n+ 2− s vertices. If s = 1, then we may assume by symmetry that xi = yj with i ≥ j and i ≥ 1 and j < n. In this case, the vertices x0, . . . , xi, yj+1, . . . yn form a path of length at least n having a subpath of length n different from X and Y . Similarly, if s = 2, then we may let the common vertices be xi1 , xi2 and yi1 , yi2 with xi1 = yj1 and xi2 = yj2 . Using symmetry again, we may assume that i1 < i2, j1 < j2, and i1 ≥ j1. With this labeling, again the vertices x0, . . . , xi1 , yj1+1, . . . yn form a path with a subpath of length n different from X and Y . Hence we may assume that s ≥ 3. The approach above no longer works, since now the points of intersection need not occur in the same order on X and Y . Suppose first that the intersection contains an endpoint of one of the paths. We may assume that x0 = yk for some k with k < n. Now we consider two cases. If yk+1 is not a vertex of X, then we replace the edge xn−1xn with the edge yk+1x0 to create a third path of length n. If yk+1 = xi for some i, then we replace the edge xixi−1 with the edge yk+1x0 to create a new path of length n. Therefore, we may assume that s ≥ 3 and that none of {x0, xn, y0, yn} is among the s shared vertices. We apply a result of Thomason ([8], Theorem 2.1, pages 263-4): If H is a regular multigraph of degree 4 with at least 3 vertices, then for any two edges e and f there are an even number of decompositions of H into two Hamiltonian cycles C1 and C2 with e in C1 and f in C2. From the given graph G, we construct a 4-regular multigraph H. We first add the edges e0 = x0xn and f0 = y0yn. We then “smooth out” all vertices of degree 2; that is, we the electronic journal of combinatorics 8 (2001), #R00 2 iteratively contract edges incident to vertices of degree 2 until no such vertices remain. Since every vertex of G∪ {e0, f0} has degree 2 or degree 4, the resulting multigraph H is regular of degree 4. Since s ≥ 3, H has at least three vertices. In H, the edge e0 is absorbed into an edge e, and f0 is absorbed into an edge f . The cycles X ∪ {e0} and Y ∪ {f0} have been contracted to become Hamiltonian cycles in H. Together they decompose H. By the theorem of Thomason, there is another Hamiltonian decomposition C1, C2 of H with e in C1 and f in C2. Now we reverse our steps. Restore the vertices of degree 2 and remove the edges e0 and f0. The cycle C1 becomes a path from x0 to xn, and C2 becomes a path from y0 to yn. Neither of these paths is the original X or Y . Since G has 2n edges and is the edge-disjoint union of these two paths, one of the paths has length at least n. It contains a new path of length n.