Problem 50E Show that a graph is bipartite if, and only if, it does not have a circuit with an odd number of edges. (See exercise for the definition of bipartite graph.) bipartite graph A bipartite graph G is a simple graph whose vertex set can be partitioned into two disjoint nonempty subsets V1 and V2 such that vertices in V1 may be connected to vertices in V2, but no vertices in V1 are connected to other vertices in V1 and no vertices in V2 are connected to other vertices in V2.For example, the graph G illustrated in (i) can be redrawn as shown in (ii). From the drawing in (ii), you can see that G is bipartite with mutually disjoint vertex sets V1 = {v1, v3, v5} and V2 = {v2, v4, v6}. (i) ________________ (ii) Find which of the following graphs are bipartite. Redraw the bipartite graphs so that their bipartite nature is evident a. ________________ b. ________________ c. ________________ d. ________________ e. ________________ f.
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Textbook Solutions for Discrete Mathematics with Applications
Question
The solution for Example shows a graph for which every vertex has even degree but which does not have an Euler circuit. Give another example of a graph satisfying these properties.ExampleShowing That a Graph Does Not Have an Euler CircuitShow that the graph below does not have an Euler circuit. Solution Vertices v1 and v3 both have degree 3, which is odd. Hence by (the contrapositive form of) Theorem, this graph does not have an Euler circuit.TheoremIf a graph has an Euler circuit, then every vertex of the graph has positive even degree.Proof:Suppose G is a graph that has an Euler circuit. [We must show that given any vertex v of G, the degree of v is even.] Let v be any particular but arbitrarily chosen vertex of G. Since the Euler circuit contains every edge of G, it contains all edges incident on v. Now imagine taking a journey that begins in the middle of one of the edges adjacent to the start of the Euler circuit and continues around the Euler circuit to end in the middle of the starting edge. (See Figure. There is such a starting edge because the Euler circuit has at least one edge.) Each time v is entered by traveling along one edge, it is immediately exited by traveling along another edge (since the journey ends in the middle of an edge).Figure Example for the Proof of Theorem In this example, the Euler circuit is v0v1v2v3v4v2v5v0, and v is v2. Each time v2 is entered by one edge, it is exited by another edge.Because the Euler circuit uses every edge of G exactly once, every edge incident on v is traversed exactly once in this process. Hence the edges incident on v occur in entry/exit pairs, and consequently the degree of v must be a positive multiple of 2. But that means that v has positive even degree [as was to be shown].
Solution
The first step in solving 10.2 problem number 10 trying to solve the problem we have to refer to the textbook question: The solution for Example shows a graph for which every vertex has even degree but which does not have an Euler circuit. Give another example of a graph satisfying these properties.ExampleShowing That a Graph Does Not Have an Euler CircuitShow that the graph below does not have an Euler circuit. Solution Vertices v1 and v3 both have degree 3, which is odd. Hence by (the contrapositive form of) Theorem, this graph does not have an Euler circuit.TheoremIf a graph has an Euler circuit, then every vertex of the graph has positive even degree.Proof:Suppose G is a graph that has an Euler circuit. [We must show that given any vertex v of G, the degree of v is even.] Let v be any particular but arbitrarily chosen vertex of G. Since the Euler circuit contains every edge of G, it contains all edges incident on v. Now imagine taking a journey that begins in the middle of one of the edges adjacent to the start of the Euler circuit and continues around the Euler circuit to end in the middle of the starting edge. (See Figure. There is such a starting edge because the Euler circuit has at least one edge.) Each time v is entered by traveling along one edge, it is immediately exited by traveling along another edge (since the journey ends in the middle of an edge).Figure Example for the Proof of Theorem In this example, the Euler circuit is v0v1v2v3v4v2v5v0, and v is v2. Each time v2 is entered by one edge, it is exited by another edge.Because the Euler circuit uses every edge of G exactly once, every edge incident on v is traversed exactly once in this process. Hence the edges incident on v occur in entry/exit pairs, and consequently the degree of v must be a positive multiple of 2. But that means that v has positive even degree [as was to be shown].
From the textbook chapter Trails, Paths, and Circuits you will find a few key concepts needed to solve this.
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