5. The purpose of this question is to prove the following statement, which is one half of Theorem 4.5.2 ( so you can't use it in answering your questions!) Statement. Let G be a connected graph. If G has at most two vertices with odd degree, then G has an Euler trail. In each of the following questions, assume G is a connected graph with at most two vertices with odd degree (note that we are NOT saying that G has at most two vertices! In all the following, G can have millions of vertices, but only at most two of them can have odd degree!). You may also assume and use the fact that "if every vertex of G has even degree then G has an Euler ciruit." Since G has at most two vertices with odd degree, there are three cases depending on the number of vertices of G that have odd degree (i.e., 0, 1, or 2). (a) Our first case is when G has no vertices of odd degree. Explain why G has an Euler trail in this case. (b) Our second case is when G has exactly one vertex with odd degree. Explain why this is actually impossible! (Hint: it has NOTHING to do with any Eulerian stuff) (c) Our last case is when G has exactly two vertices with odd degree. Suppose u and u are those two vertices (again G can have millions of vertices!). Create a new graph G' from G by adding a new vertex z that is adjacent to only u and v. What does G' contain? Explain why. (d) Using G' and what you learned about G', explain why G contains an Euler trail (where do you start? where do you go? etc.)

5. The purpose of this question is to prove the following statement, which is one half of Theorem 4.5.2 ( so you can't use it in answering your questions!) Statement. Let G be a connected graph. If G has at most two vertices with odd degree, then G has an Euler trail. In each of the following questions, assume G is a connected graph with at most two vertices with odd degree (note that we are NOT saying that G has at most two vertices! In all the following, G can have millions of vertices, but only at most two of them can have odd degree!). You may also assume and use the fact that "if every vertex of G has even degree then G has an Euler ciruit." Since G has at most two vertices with odd degree, there are three cases depending on the number of vertices of G that have odd degree (i.e., 0, 1, or 2). (a) Our first case is when G has no vertices of odd degree. Explain why G has an Euler trail in this case. (b) Our second case is when G has exactly one vertex with odd degree. Explain why this is actually impossible! (Hint: it has NOTHING to do with any Eulerian stuff) (c) Our last case is when G has exactly two vertices with odd degree. Suppose u and u are those two vertices (again G can have millions of vertices!). Create a new graph G' from G by adding a new vertex z that is adjacent to only u and v. What does G' contain? Explain why. (d) Using G' and what you learned about G', explain why G contains an Euler trail (where do you start? where do you go? etc.)

Linear Algebra: A Modern Introduction

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter3: Matrices

Section3.7: Applications

Problem 80EQ

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5. The purpose of this question is to prove the following statement, which is one half of Theorem 4.5.2 (
so you can't use it in answering your questions!)
Statement. Let G be a connected graph. If G has at most two vertices with odd degree, then G has
an Euler trail.
In each of the following questions, assume G is a connected graph with at most two vertices with odd
degree (note that we are NOT saying that G has at most two vertices! In all the following, G can have
millions of vertices, but only at most two of them can have odd degree!). You may also assume and
use the fact that "if every vertex of G has even degree then G has an Euler ciruit."
Since G has at most two vertices with odd degree, there are three cases depending on the number of
vertices of G that have odd degree (i.e., 0, 1, or 2).
(a) Our first case is when G has no vertices of odd degree. Explain why G has an Euler trail in this
case.
(b) Our second case is when G has exactly one vertex with odd degree. Explain why this is actually
impossible! (Hint: it has NOTHING to do with any Eulerian stuff)
(c) Our last case is when G has exactly two vertices with odd degree. Suppose u and u are those two
vertices (again G can have millions of vertices!). Create a new graph G' from G by adding a new
vertex z that is adjacent to only u and v. What does G' contain? Explain why.
(d) Using G' and what you learned about G', explain why G contains an Euler trail (where do you
start? where do you go? etc.)

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