/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Let \(G\) be a cycle on \(n\) ve... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(G\) be a cycle on \(n\) vertices. Prove that \(G\) is selfcomplementary if and only if \(n=5\).

Short Answer

Expert verified
A cycle graph is selfcomplementary if and only if it has \(n=5\) vertices.

Step by step solution

01

Understanding the properties of a cycle graph

In a cycle graph with \(n\) vertices each vertex is connected to two other vertices, so the number of edges is also \(n\). This rule can be formulated: \( E = n \), where \(E\) represents the number of edges.
02

Understanding the properties of a complementary graph

A complementary graph \(Gc\) of \(G\) covers the same set of vertices, but the edges in \(Gc\) are the edges not present in \(G\). The total number of edges in a graph is \( n*(n-1)/2 \), so in the complementary graph, the number of edges \( Ec \) can be described as follows: \( Ec = n*(n-1)/2 - E \).
03

Set up and solve the equation

Since for a self-complementary graph, the number of edges should be equal in the original graph and its complement: \( E = Ec \). Substituting \( E = n \) and \( Ec = n*(n-1)/2 - n \), we get the equation \( n = n*(n-1)/2 - n \). After simplifying, we get the quadratic equation \( 0 = n^2 - 5n + 4 \). Solving this equation for \(n\), we find that \(n = 1\) or \(n = 4\). But in a cycle graph there cannot be 1 or 4 vertices. Therefore, \(n ≠ 1\) or \(n ≠ 4\).

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Key Concepts

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

Cycle Graph
A cycle graph is a simple circular-shaped graph where every vertex connects to precisely two other vertices. Think of it like forming a closed loop where you visit all points without lifting your pencil. This graph is notable because, regardless of the number of vertices, if you start at one vertex and follow the edges, you'll eventually return to your starting point. This property creates fascinating problems in graph theory.
For any cycle graph with \( n \) vertices, the number of edges is also \( n \). This means, if there are 5 vertices arranged cyclically, you'll also have 5 edges connecting them in a loop. Each vertex connects to its immediate neighbors, ensuring a seamless circular connection.
Self-Complementary Graph
A self-complementary graph is quite special because it means the graph and its complementary graph are the same after rearranging. In simpler terms, if you were to draw both the original graph and its complement, the two would look the same, even though their connections differ.
For a cycle graph to be self-complementary, its structure must be just right for a transformation that results in the same appearance. Typically, not all graphs can have this property; they require a balanced number of edges and vertices. In the specific context of cycle graphs, only particular vertex counts, such as 5, satisfy this condition.
Complementary Graph
The complementary graph of a given graph captures all the connections not present in the original graph. Imagine having a set of vertices, where initially some pairs are connected. By creating a complementary graph, every pair of vertices not originally connected gets an edge, and every original edge disappears.
Mathematically, if a graph \( G \) has \( n \) vertices, the complete graph would have \( n(n-1)/2 \) edges. The complementary graph \( G_c \) would therefore have a number of edges calculated by subtracting the edges of \( G \) from this complete set. Complementary graphs fundamentally change how connections within the vertex set are illustrated.
Number of Edges
Determining the number of edges in any graph involves understanding its precise structure. For a cycle graph with \( n \) vertices, the number of edges also matches \( n \). But, when considering its complement, the calculation involves all possible vertex pairs, minus the edges present in the original.
If a complete graph on \( n \) vertices has all possible connections \( n(n-1)/2 \), then subtracting the original cycle graph's edges provides the edge count of its complementary graph. This relationship is crucial in creating and analyzing complementary and self-complementary graphs.
Quadratic Equation
Quadratic equations come into play when mathematical relationships need resolving, often revealing things like dimensions or counts in graph theory. We use them to solve problems involving self-complementary graphs by balancing properties of the original and its complement.
For example, in proving a graph's self-complementary nature, solving \( E = E_c \) leads to forming and factoring a quadratic equation. In the case of the equivalent cycle and complementary graphs, solving \( n^2 - 5n + 4 = 0 \) shows solutions like \( n = 5 \). Thus, quadratic equations help unlock specific values ensuring that graphs can meet complex conditions, such as being self-complementary.

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Most popular questions from this chapter

If \(G=(V, E)\) is an undirected graph, a subset \(D\) of \(V\) is called a dominating set if for all \(v \in V\), either \(v \in D\) or \(v\) is adjacent to a vertex in \(D\). If \(D\) is a dominating set and no proper subset of \(D\) has this property, then \(D\) is called minimal. The size of any smallest dominating set in \(G\) is denoted by \(\gamma(G)\) and is called the domination number of \(G\). a) If \(G\) has no isolated vertices, prove that if \(D\) is a minimal dominating set, then \(V-D\) is a dominating set. b) If \(I \subseteq V\) is independent, prove that \(I\) is a dominating set if and only if \(I\) is maximal independent. c) Show that \(\gamma(G) \leq \beta(G)\), and that \(|V| \leq \beta(G) \chi(G)\). [Here \(\beta(G)\) is the independence number of \(G-\) first given in Exercise 25 of Section 11.5.]

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.

a) How many paths of length 4 are there in the complete graph \(K_{7}\) ? (Remember that a path such as \(v_{1} \rightarrow v_{2} \rightarrow\) \(v_{3} \rightarrow v_{4} \rightarrow v_{5}\) is considered to be the same as the path \(\left.v_{5} \rightarrow v_{4} \rightarrow v_{3} \rightarrow v_{2} \rightarrow v_{1} \cdot\right)\) b) Let \(m, n \in \mathbf{Z}^{+}\)with \(m

Give an example of a connected graph that has (a) Neither an Euler circuit nor a Hamilton cycle. (b) An Euler circuit but no Hamilton cycle. (c) A Hamilton cycle but no Euler circuit. (d) Both a Hamilton cycle and an Euler circuit.

a) Let \(G\) be an undirected graph with \(n\) vertices. If \(G\) is isomorphic to its own complement \(\bar{G}\), how many edges must \(G\) have? (Such a graph is called self-complementary.) b) Find an example of a self-complementary graph on four vertices and one on five vertices. c) If \(G\) is a self-complementary graph on \(n\) vertices, where \(n>1\), prove that \(n=4 k\) or \(n=4 k+1\), for some \(k \in \mathbf{Z}^{+}\).

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