Graph Theory, 02/24/2015; Quiz 3 (three problems;) (Solutions)

Problem 1 Let v be a cut-vertex in a connected graph G and let the complement G be also connected. Prove that v is not a cut-vertex in the complement graph G.

Hint. The complement of a graph G(V,E), denoted G, is a graph with V as the vertex set; two vertices are adjacent in G iff they are not adjacent in G. A vertex v in a connected graph G is called a cut-vertex, if G − v is disconnected.

Proof. Let G − v consist of connected components C1,...,Ck (k ≥ 2). Then the subgraph G − v of G induced on C1 ∪ ... ∪ Ck is connected, since for any i,j ∈ [1,k] (i 6= j) every vertex in Ci is adjacent in G to every vertex in Cj. In addition, for any i ∈ [1,k], ′ ′′ if x and x are two vertices in Ci, then we select y ∈ Cj, where j 6= i and a path x′yx′′ which connects x′ with x′′. Thus, G − v is a connected graph.

Problem 2 Let the length of a shortest in a graph be 5 and the degree of every vertex be exactly k > 0. Prove that the graph has at least k2 +1 vertices. Find one such graph for k =2 and k =3.

Solution. Let x be a vertex in G, and let y1,y2,...,yk be the vertices adjacent to x in G.

1 1 k−1 For every i ∈ [1,k], denote Si = {yi ,...,yi } the set of neigh- bors of yi that are different from x. Then

1. ∀i 6= j,yi 6∈ Sj (since G has no triangles); and

2. ∀i 6= j,Si ∩ Sj = ∅ (since G has no cycle of length 4).

Thus, G contains k disjoint subsets {yi ∪Si} of size k and a vertex x, in total k2 + 1 vertices. For k = 2, the graph satisfying all conditions, is a cycle of length 5. For k = 3, such a graph is the .

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5 6 2

10 7

9 8 4 3

Problem 3 Let G be an undirected graph for which there are exactly two vertices, u and v, of odd degree. Prove that G contains an u,v-path, that is a path connecting u with v.

Proof. If G had no u,v-path, graph G would have two connected components, one containing u and the other containing v. In each of these components, all vertices but one would have even degrees, contrary to the Theorem proved in class, stating that every graph has an even number of vertices of odd degree.

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