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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
Based on the proof by contradiction, it's concluded that the loop-free undirected graph \(G=\(V, E)\) with no odd cycle is bipartite.

Step by step solution

01

- Assumption

Assume that the graph \(G\) is not bipartite. An uncomplicated way to define a bipartite graph is that you can color a graph in two colors such that no two adjacent vertices have the same color. So if \(G\) isn't bipartite, then it can't be colored using 2 colors, implying that there exists at least one cycle of odd length.
02

- Contradiction of Assumption

Since it's given that the graph \(G\) doesn't contain any cycles of odd length, this is a contradiction to our Step 1 statement that there is at least one cycle of odd length because the graph can't be colored using 2 colors. Hence the assumption that \(G\) is not bipartite is false.
03

- Proving Bipartite

So, the original graph \(G\) must be bipartite. That means you can color the graph in such a way that no two adjacent vertices share the same color and hence no cycle of odd length.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graph Theory
Graph theory is a significant area of discrete mathematics that studies the properties of graphs - a collection of nodes, called vertices, connected by edges. This concept is fundamental to several fields, including computer science, networking, and even biology.

Graphs can be undirected, with no indicated direction in their connections, or directed, often illustrated with arrows. Loop-free graphs, which are our focus, do not have edges that connect a vertex to itself. Within graph theory, a graph's ability to be divided into two distinct sets, where no two adjacent vertices are in the same set, is particularly captivating and leads us to the concept of bipartite graphs.
Discrete Mathematics
Discrete mathematics encompasses a broad array of topics that involve discrete elements, that is, those which are countable and not continuous. It includes concepts like set theory, combinatorics, logic, and graph theory. One of the key appeals of discrete mathematics is its applicability to real-world problems and its foundation for algorithms and computational processes.

In terms of graph theory within discrete mathematics, problems often involve proving certain properties about graphs, such as the existence of a path, connectivity, or as in our exercise, the graph's colorability and its implications on the graph being bipartite.
Odd Cycle
An odd cycle in a graph is a closed path with an odd number of edges, where the starting and ending vertex are the same. Odd cycles play a significant role in graph theory, particularly when discussing graph coloring.

Understanding odd cycles is crucial in graph coloring as they influence whether a graph can be considered bipartite. In general, the existence of an odd cycle in a graph indicates that it's not possible to divide the graph into two parts without breaking the rule that each part should only contain non-adjacent vertices. Therefore, proving the absence of odd cycles is a strong starting point to demonstrate a graph's bipartiteness.
Graph Coloring
Graph coloring involves assigning colors to the vertices of a graph so that no two adjacent vertices share the same color. It's a way of distinguishing different vertices and can have practical applications, like scheduling to avoid conflicts.

In our graph coloring context, a graph that can be colored using just two colors, adhering to the mentioned rule, is referred to as a bipartite graph. The connection between graph coloring and bipartiteness is fundamental: a bipartite graph has no odd cycles, and conversely, a graph with no odd cycles must be bipartite, precisely due to such a two-coloring being possible.

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Most popular questions from this chapter

a) Let \(G=(V, E)\) be a connected bipartite undirected graph with \(V\) partitioned as \(V_{1} \cup V_{2}\). Prove that if \(\left|V_{1}\right| \neq\) \(\left|V_{2}\right|\), then \(G\) cannot have a Hamilton cycle. b) Prove that if the graph \(G\) in part (a) has a Hamilton path, then \(\left|V_{1}\right|-\left|V_{2}\right|=\pm 1\) c) Give an example of a connected bipartite undirected graph \(G=(V, E)\), where \(V\) is partitioned as \(V_{1} \cup V_{2}\) and \(\left|V_{1}\right|=\left|V_{2}\right|-1\), but \(G\) has no Hamilton path.

For \(n \geq 3\), let \(C_{n}\) denote the undirected cycle on \(n\) vertices. The graph \(\bar{C}_{n}\), the complement of \(C_{n}\), is often called the cocycle on \(n\) vertices. Prove that for \(n \geq 5\) the cocycle \(\bar{C}_{n}\) has a Hamilton cycle.

a) How many vertices and how many edges are there in the complete bipartite graphs \(K_{4,7}, K_{7,11}\), and \(K_{m, n}\), where \(m, n, \in \mathbf{Z}^{+} ?\) b) If the graph \(K_{m, 12}\) has 72 edges, what is \(m\) ?

Are any of the planar graphs for the five Platonic solids bipartite?

Consider the complete graph \(K_{n}\) for \(n \geq 3\). Color \(r\) of the vertices in \(K_{n}\) red and the remaining \(n-r(=g)\) vertices green. For any two vertices \(v, w\) in \(K_{n}\) color the edge \(\\{v, w\\}\) (1) red if \(v, w\) are both red; (2) green if \(v, w\) are both green; or (3) blue if \(v, w\) have different colors. Assume that \(r \geq g\). a) Show that for \(r=6\) and \(g=3\) (and \(n=9\) ) the total number of red and green edges in \(K_{9}\) equals the number of blue edges in \(K_{9}\). b) Show that the total number of red and green edges in \(K_{n}\) equals the number of blue edges in \(K_{n}\) if and only if \(n=r+g\), where \(g, r\) are consecutive triangular numbers. [The triangular numbers are defined recursively by \(t_{1}=\) \(1, t_{n+1}=t_{n}+(n+1), n \geq 1 ;\) so \(t_{n}=n(n+1) / 2\). Hence \(\left.t_{1}=1, t_{2}=3, t_{3}=6, \ldots\right]\)
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