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Chapter 11: Problem 9

Let \(G=(V, E)\) be a loop-free undirected graph. Prove that if \(G\) contains no cycle of odd length, then \(G\) is bipartite.

Short Answer

Expert verified
The contradiction that was deduced from assuming that the graph \(G\) is not bipartite, which means \(G\) contains an odd cycle, contradicts our original statement that the graph contains no odd cycle. Thus \(G\) must be bipartite.

Step by step solution

01

Understanding Key Definitions

First understand the key terms - a 'bipartite graph' is a graph whose vertices can be divided into two disjoint sets such that each edge connects a vertex in one set to a vertex in the other set. An 'odd cycle' refers to a closed path in a graph that includes an odd number of edges.
02

Assume Graph Is Not Bipartite

Next, let's suppose for contradiction that \(G\) is not bipartite. It means there exists an odd cycle in the graph, because an undirected graph is bipartite if and only if it has no odd cycles.
03

Prove That Graph Is Bipartite

Now, prove if there is no odd cycle in the graph \(G\), it must be bipartite. In a bipartite graph, all cycles are of even length because the cycle alternates between the two vertex sets, adding two edges each time it switches from one set to the other. Thus, no odd cycle means the graph is bipartite.

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Most popular questions from this chapter

Seven towns \(a, b, c, d, e, f\), and \(g\) are connected by a system of highways as follows: (1) I-22 goes from \(a\) to \(c\), passing through \(b ;(2)\) I-33 goes from \(c\) to \(d\) and then passes through\(b\) as it continues to \(f ;(3)\) I-44 goes from \(d\) through \(e\) to \(a ;(4)\) I- 55 goes from \(f\) to \(b\), passing through \(g\); and, (5) I-66 goes from \(g\) to \(d\). a) Using vertices for towns and directed edges for segments of highways between towns, draw a directed graph that models this situation. b) List the paths from \(g\) to \(a\). c) What is the smallest number of highway segments that would have to be closed down in order for travel from \(b\) to \(d\) to be disrupted? d) Is it possible to leave town \(c\) and return there, visiting each of the other towns only once? e) What is the answer to part (d) if we are not required to return to \(c\) ? f) Is it possible to start at some town and drive over each of these highways exactly once? (You are allowed to visit a town more than once, and you need not return to the town from which you started.)

As the chair for church committees, Mrs. Blasi is faced with scheduling the meeting times for 15 committees. Each committee meets for one hour each week. Two committees having a common member must be scheduled at different times. Model this problem as a graph-coloring problem, and tell how to determine the least number of meeting times Mrs. Blasi has to consider for scheduling the 15 committee meetings.

For \(n \in \mathrm{Z}^{+}\)where \(n \geq 4\), let \(V^{\prime}=\left\\{v_{1}, v_{2}, v_{3}, \ldots, v_{n-1}\right\\}\) be the vertex set for the complete graph \(K_{n-1}\). Construct the loop-free undirected graph \(H_{n}=(V, E)\) from \(K_{n-1}\) as follows: \(V=V^{\prime} \cup\\{v\\}\), and \(E\) consists of all the edges in \(K_{n-1}\) together with the new edge \(\left\\{v, v_{1}\right\\}\). a) Show that \(H_{n}\) has a Hamilton path but no Hamilton cycle. b) How large is the edge set \(E\) ?

Let \(G=(V, E)\) be a loop-free undirected graph. We call \(G\) color-critical if \(\chi(G)>\chi(G-v)\) for all \(v \in V\). a) Explain why cycles with an odd number of vertices are color-critical while cycles with an even number of vertices are not color-critical. b) For \(n \in \mathbf{Z}^{+}, n \geq 2\), which of the complete graphs \(K_{n}\) are color-critical? c) Prove that a color-critical graph must be connected. d) Prove that if \(G\) is color-critical with \(\chi(G)=k\), then \(\operatorname{deg}(v) \geq k-1\) for all \(v \in V\).

Give an example of a loop-free connected undirected multigraph \(G=(V, E)\) such that \(|V|=n\) and \(\operatorname{deg}(x)+\operatorname{deg}(y) \geq n-1\) for all \(x, y \in V\), but \(G\) has no Hamilton path.
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