Graphs and Algorithms Exercise 1 (small and Regular) Exercise 2 (combining K-connected Graphs)

Consider an arbitrary 3-regular graph G = (V,E) with connectivity 1. By the definition it has a cut vertex v ∈ V , and since its degree is three, there are either 2 or 3 connected components in G \ v. Moreover, in every connected component of G \ v there exist either one or two vertices with degree 2, and all others have degree 3. On the other hand, there has to be an even number of vertices of degree 3 in every component of G \ v, and further there has to be at least 4 vertices in each component. Since the graph on Figure 1 satisfies all these minimal constraints, we conclude that there doesn’t exist a 3-regular graph with less vertices which has connectivity 1.