Graphs and Algorithms Exercise 1

(a) (i) Since T is a connected graph with ≥ 2 vertices, there are edges in T . Hence, any maximal path in T has length ≥ 1, and thus two distinct end vertices. Let v0, vk be the end vertices of such a path P = (v0, v1, . . . , vk). As T is acyclic, the only neighbor of v0 on P is v1 and the only neighbor of vk on P is vk−1. Because P was chosen to be maximal, both v0 and vk have no neighbors outside of P . This shows that v0 and vk are two distinct leaves. (ii) Let v0 be a leaf of T and T ′ := T −v0. Clearly, deleting v0 cannot create a cycle and T ′ is thus cycle-free. It remains to show that T ′ is connected. Let u, v ∈ V (T ′). Since T is connected there exists a u, v-path P = (u, v1, v2, . . . , vs, v) in T . Clearly, deg(vi) ≥ 2 for all 1 ≤ i ≤ s, hence v0 cannot be in P and P is also a u, v-path in T ′. (b) (i)⇒ (ii) We need to show that G has n − 1 edges. For n = 1 the statement is trivial. We proceed by induction on n. Since G is a tree we know from (a)(i) that G contains a leaf v. From (a)(ii) we know that G− v is also connected and has no cycles. Hence by induction, G− v has |V (G− v)| − 1 = n− 2 edges, and thus G has n− 1 edges. (ii)⇒ (iii) We need to show that G has no cycles. Assume that this was not true and let e ∈ E(G) be an edge contained in a cycle in G. Then the graph G − e, in which e is removed, is still connected. (If it wasn’t, the end vertices u and v of e would have to lie in different components of G− e, but they are still connected by a path.)