On Extremal Graphs With No Long Paths

Connected graphs with minimum degree δ and at least 2δ+ 1 vertices have paths with at least 2δ + 1 vertices. We provide a characterization of all such graphs which have no longer paths. Extremal problems involving paths and cycles have been considered since the infancy of graph theory. The question which interests us here is the question of what minimum degree condition guarantees a path of a preassigned length. This question was answered by Erdös and Gallai [4] and again by Andrasfai [1]. Our formulation of their answer is Theorem 1 Let G be a connected graph with minimum degree δ and at least 2δ + 1 vertices. Then G contains a path of at least 2δ + 1 vertices. The complete bipartite graphs Kδ,n−δ with n ≥ 2δ+ 1 show that the theorem is best possible in the sense that there exist graphs of minimum degree δ with no longer paths. Our purpose in this article is to characterize all such extremal graphs. For a graph G with n vertices we define f(G) to be the number of vertices in a longest path of G, and for a vertex v of G we define f(v) to be the number of vertices in a longest path of G with initial vertex v. A set of vertices is called independent if no two of them are adjacent. By the sum H ⊕K of two graphs we mean the graph obtained from the disjoint union of H and K by adding edges joining every vertex of the electronic journal of combinatorics 3 (1996), #R20 2 H to every vertex of K. All graphs considered here are simple. Definitions of terms not explicitly given here can be found in [2]. Theorem 2 Let G be connected and v ∈ G. Suppose δ ≥ 2. Then f(v) ≥ 1 + δ. If f(v) = 1 + δ then every component of G− v is a Kδ. Proof: To show f(v) ≥ 1 + δ, first note that it is trivial for δ = 2. Suppose the result is true for graphs with δ = k and let G have δ = k + 1. Let v be a vertex of G and note that G− v has minimum degree at least k. Let w be a neighbor of v. Then there is a path of k + 1 vertices in G− v beginning at w. Hence there is a path in G of k+ 1 vertices in G− v beginning at w. Hence there is a path in G of k+ 2 vertices beginning with v followed by w. Now suppose that f(v) = 1+δ. By Theorem 1, there are paths of 2δ+1 vertices in G. Now v must lie in the center of every such path P, or else a path from v to P followed by the longer subpath of P would result in a path of more than 1 + δ vertices beginning at v. Now consider the component or components of G− v. Each such component has at least δ vertices or else the minimum degree of G would be too small. We now claim that each component C of G− v has exactly δ vertices. Let Q be a longest path in C ∪ v starting at v. Certainly the final vertex of Q has all its neighbors on Q, forcing Q to have exactly δ+ 1 vertices. Hence the final vertex of Q is adjacent to all vertices of Q. Say Q: v = v1, v2, · · · , v1+δ. Since v1+δ is adjacent to every vi, it follows that every vi, i = 2, · · · , δ + 1, is the final vertex of a longest path in C ∪ v starting at v, say v1, v2, · · · , vi−1, vδ+1, vδ, · · · , vi. Hence vi has all its neighbors on Q, so the component C consists only of the vertices v2, · · · , vδ+1 and these are all adjacent, proving the theorem. 2 Theorem 3 If G is connected and f(G) = 2δ + 1, δ ≥ 2 then i) If G has no cut vertex and n ≥ 2δ + 2 then G = H ⊕ I where H has δ vertices and I is an independent set. ii) If G has a cut vertex v, then G is the union of copies of Kδ+1 with vertex v in common.