Graph Theory In Details Prove that for K fixed almost every
Graph Theory
In Details
Prove that for K fixed, almost every graph is k - connected.Solution
Let Hn,k denote the class of all connected graphs of order n with k pendant vertices. We may assume that 1kn1 and hence n3. We first consider the question of maximizing the algebraic connectivity over Hn,k We now define a collection of graphs, denoted Pkn, which have exactly k pendant vertices and hence are members of Hn,k. For k not equal to n 2, the graph Pkn, is obtained by adding k pendant vertices adjacent to a single vertex of the complete graph Knk. There is no graph of order n= 3 with exactly one pendant vertex. For n4 and k=n2, the graph Pn2n is obtained by adding n 3 pendant vertices adjacent to a pendant vertex of the path P3.
