Question

Answer the following true or false questions with a brief justification.

A) There exists an undirected graph on 6 vertices whose degrees are 4, 5, 8, 9, 3, 6.

B) Every undirected graph with n vertices and n − 1 edges is a tree.

C) Let G be an undirected graph. Suppose u and v are the only vertices of odd degree in G. Then G contains a u-v path.

Answer 1

ans(a)true,Degree of vertex in undirected graph is the number of edges incident to it.

V = {1, 2, 3, 4, 5, 6}

E = {{1,2}, {1,5}, {2,5}, {3,6}}

• Edge (u, v) is incident to u and v

: For undirected graphs, every edge is stored twice. – If graph is weighted, a weight is stored with each edge.

ans (b) false,not every undirected graph is strongly connected. We know that the minimum number of edges required to make a graph of n vertices connected is (n-1) edges. We can observe that removal of one edge from the graph G will make it disconnected. Thus a connected graph of n vertices and (n-1) edges cannot have a circuit. Hence a graph G is a tree.

ans (c)true, We will prove the statement by contradiction. That is, we first assume that G is a graph with exactly two vertices of odd degree, u and v, and that there is no u, v-path in G.

By definition, G is a disconnected graph and the vertices u and v must lie in separate connected components of G. Thus the vertex u lies in H, a connected subgraph of G, and all other vertices in H (which, of course, are also vertices in G) have even degree. Therefore, the sum of the degrees of the vertices in H is an odd number.

But by the Hand-Shaking Theorem, the sum of the degrees of the vertices in H (as in any graph) must be an even number. So we have a contradiction and our assumption, that there is no u, v-path in G, must be false.