CME 305: Discrete Mathematics and Algorithms

“⇒”: assume G has an Eulerian circuit. Clearly, G must be connected, otherwise we will be unable to traverse all the edges in a closed walk. Suppose that an Eulerian circuit starts at v1; note that every time the walk enters vertex v 6= v1, it must leave it by traversing a new edge. Hence, every time that the walk visits v it traverses two edges of v. Thus dv is even. The same is true for v1 except for the first step of the walk that leaves v1 and the last step that it enters v1. Again, we can pair the first and last edges in the circuit to conclude that dv1 is also even. “⇐”: we prove this by strong induction on the number of nodes. For n = 1, the statement is trivial. Assume G has k + 1 vertices, it is connected, and all the nodes have even degrees. Pick any vertex v ∈ V and start walking through the edges of the graph, traversing each edge at most once. Given that all the degrees of the nodes in G are even, we can only get stuck at v. Remove the circuit of visited edges, c1, from G; Call this new graph G′. Node v is an isolated vertex in G′ therefore any connected component of G′ has at most k vertices. Moreover, it is easy to see that the degree of nodes in G′ are even (we removed an even number of edges from each vertex). Thus by the induction assumption, all the connected components of G′ are Eulerain. Now, we have a collection of circuits c1, . . . , cl. We can merge them into one large circuit in the following way: