A graph is G is semi-Eulerian if there are distinct vertices u, v ∈ V (G),...

A graph is G is semi-Eulerian if there are distinct vertices u, v ∈ V (G), u =v such
that there is a trail from u to v which goes over every edge of G. The following
sequence of questions is towards a proof of the following:
Theorem 1. A connected graph G is semi-Eulerian (but not Eulerian) if and only
if it has exactly two vertices of odd degree.
Let G be semi-Eulerian with a trail t starting at a vertex u₀ and ending at a vertex
v₀ Let G ⁰ be a graph obtained by adding an edge e₀ joining u₀ and v₀ , so that
G ⁰ − e = G.
(a) Prove that given a semi-Eulerian trail t in G from v₀ to u₀ , it is possible to
construct a Eulerian trail in G₀.
(b) Prove that given an Eulerian trail in G 0 it is possible to construct a semi-Eulerian
trail in G.
(c) Prove Theorem 1.

Expert Solution

a) Let be a semi-Eulerian trail in G from v₀ to u_0. Now, let us add the edge e in G. Then, we can see that this edges makes this trail in a Eulerian trail. Hence, we get a Eulerian trail in G₀

b) Now, let us take an Eulerian trail in G₀. Then we will write the trail as , since it passes each edge once (basically I reset the starting parameter of the trail such that it starts from v_0 and it traverses the edges e at the veyr end). Then, if we take out the edge e from the graph G₀ . Then, we can get a semi Eulerian trail in the graph in G.

c) From part a) and b) we get that a Graph can be converted from Semi Eulerian graph to a Eulerian graph by adding an edge between the endpoints of the semi Eulerian trail. And we can remove an edge from Eulerian graph to make it semi Eulerian. Now, we also know that a Graph is Eulerian if and only if every vertex has even degree. Then, from these information we get that a graph is Semi Eulerian if and only if it has exactly two odd degree vertices. This can be concluded because we are only removing one edge from a Eulerian graph to make it semi Eulerian.

Colby Messinger answered 7 months ago

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