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Chapter 10: Problem 12

Let \(G\) be an undirected graph with a loop at every vertex. Show that the relation \(R\) on the set of vertices of \(G\) such that \(u R v\) if and only if there is an edge associated to \(\\{u, v\\}\) is a symmetric, reflexive relation on \(G .\)

Short Answer

Expert verified
The relation R is both reflexive and symmetric because each vertex has a loop (uRu) and undirected edges imply uRv -> vRu.

Step by step solution

01

Define the Relation

Let relation R be defined on the set of vertices in graph G such that for any vertices u and v, uRv if and only if there is an edge between u and v.
02

Prove Reflexivity

To show reflexivity, we need to prove that every vertex u is related to itself in relation R. Given that there is a loop at every vertex in G, by definition, there is an edge from u to itself for any vertex u. Therefore, uRu holds for every vertex u, proving that R is reflexive.
03

Prove Symmetry

To demonstrate symmetry, we need to show that if uRv then vRu. If uRv, by definition, there is an edge between u and v. In an undirected graph, edges do not have a direction, meaning if there is an edge between u and v, there is also an edge between v and u. Therefore, if uRv, then vRu, proving that R is symmetric.

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Key Concepts

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

undirected graphs
In graph theory, an undirected graph is a type of graph in which edges have no direction. This means that the connection between two vertices is bidirectional.
For example, if there is an edge between vertices u and v in an undirected graph, you can travel from u to v and from v to u.
Key points about undirected graphs include:
  • No direction: Edges do not have directions, so the connection is mutual.
  • Loop edges: It's common for vertices to have edges that loop back to themselves, indicating a loop.
  • Simplicity: Many undirected graphs are simple, meaning they do not have multiple edges between the same vertices.
Understanding undirected graphs helps to comprehend complex relationships and connectivity in various systems, such as social networks or computer networks.
reflexive relation
A reflexive relation is a fundamental concept in set theory and graph theory. For a relation R on a set to be reflexive, every element must relate to itself.
Mathematically, a relation R on a set S is reflexive if, for every element u in S, the statement uRu holds.
For instance, if we have a graph with vertices representing people, a reflexive relation would mean everyone has a connection to themselves, like knowing oneself.
In the context of the given exercise:
  • Each vertex has a loop: Because of this loop, every vertex u relates to itself.
  • The presence of loops confirms the reflexive property of the relation R defined.
  • Reflexivity ensures every element in the set maintains a self-connection.
This characteristic is essential in various areas, including computer science and logic, to ensure complete connectivity within a set.
symmetric relation
A symmetric relation means that if one element is related to another, the second element is also related to the first. Formally, a relation R on a set S is symmetric if, for all elements u and v in S, if uRv then vRu.
This property is visually and conceptually easy to understand in undirected graphs.
Key points about symmetric relations include:
  • Bidirectionality: If there is a connection from u to v, there must be a connection from v to u.
  • Undirected edges: In undirected graphs, every edge naturally represents a symmetric relationship.
  • Real-world analogy: Think of friendships where if person A is friends with person B, it implies person B is also friends with person A.
In the given exercise, the symmetry is evident because:
  • Edges in the undirected graph ensure a two-way relationship.
  • The relation R is defined by these connections, making R symmetric by nature.
Understanding symmetric relations helps in analyzing social networks, relationships, and various bidirectional systems efficiently.

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Most popular questions from this chapter

The thickness of a simple graph \(G\) is the smallest number of planar subgraphs of \(G\) that have \(G\) as their union. $$ \begin{array}{l}{\text { Show that if } G \text { is a connected simple graph with } v \text { ver- }} \\ {\text { tices and } e \text { edges, where } v \geq 3, \text { and no circuits of length }} \\ {\text { three, then the thickness of } G \text { is at least }[e /(2 v-4)]}\end{array} $$

The thickness of a simple graph \(G\) is the smallest number of planar subgraphs of \(G\) that have \(G\) as their union. $$ \text { Show that } K_{3,3} \text { has } 2 \text { as its thickness. } $$

Show that an edge in a simple graph is a cut edge if and only if this edge is not part of any simple circuit in the graph.

Show that a bipartite graph with an odd number of vertices does not have a Hamilton circuit.

The parts of this exercise outline a proof of Ore's theorem. Suppose that \(G\) is a simple graph with \(n\) vertices, \(n \geq 3,\) and \(\operatorname{deg}(x)+\operatorname{deg}(y) \geq n\) whenever \(x\) and \(y\) are non-adjacent vertices in \(G .\) Ore's theorem states that under these conditions, \(G\) has a Hamilton circuit. a) Show that if \(G\) does not have a Hamilton circuit, then there exists another graph \(H\) with the same vertices as \(G,\) which can be constructed by adding edges to \(G,\) such that the addition of a single edge would produce a Hamilton circuit in \(H .[\text {Hint} : \text { Add as many edges as }\) possible at each successive vertex of \(G\) without producing a Hamilton circuit.] b) Show that there is a Hamilton path in \(H\) c) Let \(v_{1}, v_{2}, \ldots, v_{n}\) be a Hamilton path in \(H .\) Show that \(\operatorname{deg}\left(v_{1}\right)+\operatorname{deg}\left(v_{n}\right) \geq n\) and that there are at most \(\operatorname{deg}\left(v_{1}\right)\) vertices not adjacent to \(v_{n}\) (including \(v_{n}\) itself). d) Let \(S\) be the set of vertices preceding each vertex adjacent to \(v_{1}\) in the Hamilton path. Show that \(S\) contains \(\operatorname{deg}\left(v_{1}\right)\) vertices and \(v_{n} \notin S .\) e) Show that \(S\) contains a vertex \(v_{k}\) that is adjacent to \(v_{n}\) implying that there are edges connecting \(v_{1}\) and \(v_{k+1}\) and \(v_{k}\) and \(v_{n} .\) f) Show that part (e) implies that \(v_{1}, v_{2}, \ldots, v_{k-1}\) \(v_{k}, v_{n}, v_{n-1}, \ldots, v_{k+1}, v_{1}\) is a Hamilton circuit in \(G\) . Conclude from this contradiction that Ore's theorem holds.
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