/*! 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 3 Does there exist' a graph of ord... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 3

Does there exist' a graph of order 5 whose degree sequence equals \((4,4,3,2,2)\) ?

Short Answer

Expert verified
No, there does not exist a graph of order 5 whose degree sequence equals (4,4,3,2,2).

Step by step solution

01

Verify Handshaking Lemma

Calculate the sum of the sequences. If it is even, then the Handshaking lemma is satisfied. The sum of the given sequence (4,4,3,2,2) is 15, which is odd, so the Handshaking lemma is not satisfied.
02

Analyse Havel-Hakimi Theorem

Since the Handshaking lemma is not satisfied, there is no need to proceed with the Havel-Hakimi theorem, which is a technique for testing whether a finite list of integers can be the degree sequence of a finite simple graph. The theorem states that firstly, the degree sequence should be in non-increasing order. Secondly, if the first degree 'd' is removed from the sequence and the next 'd' degrees are reduced by 1, the resultant sequence should also be graphic. But since the Handshaking lemma is already violated, it does not need to be tested.
03

Conclusion

Since the Handshaking lemma is not satisfied, we conclude that there is no graph of order 5 with the degree sequence (4,4,3,2,2).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Degree Sequence
The degree sequence of a graph is a list of its vertices' degrees. When you list the degrees in non-increasing order, each number in the sequence represents the number of edges connected to a particular vertex. In graph theory, understanding the degree sequence is crucial as it determines whether a graph can be constructed from these degrees or not. When trying to figure out if a graph exists for a given degree sequence, the sequence must satisfy certain conditions:
  • All vertex degrees are non-negative integers.
  • The sequence should sum to an even number (as per the Handshaking Lemma).
  • The sequence should be sorted in non-increasing order when applying certain theorems, like the Havel-Hakimi theorem.
The question of graph existence often hinges on these conditions being met.
Handshaking Lemma
The Handshaking Lemma is a fundamental principle in graph theory. It states that in any graph, the sum of all the vertex degrees is twice the number of edges. This means that the sum of the degrees must be even, because each edge adds two to the total degree sum (one for each endpoint). The lemma is beneficial for checking the feasibility of a degree sequence.

To apply this to the problem, we sum the degrees \[ 4 + 4 + 3 + 2 + 2 = 15 \].Since 15 is odd, the Handshaking Lemma tells us that these degrees cannot represent any graph. Thus, immediately answering the question of whether such a degree sequence is possible.
Havel-Hakimi Theorem
The Havel-Hakimi theorem is a technique used to determine if a given degree sequence can form a simple graph. When applying this theorem, you should have your sequence sorted in non-increasing order. The process involves:
  • Remove the first degree \(d\) in your sequence.
  • Subtract 1 from the next \(d\) degrees in the sequence.
  • Check if the modified sequence can still represent a degree sequence using the same theorem.
For our sequence, due to the Handshaking Lemma being violated already, the exercise doesn't proceed with the Havel-Hakimi theorem. However, if the lemma had been satisfied, using the Havel-Hakimi theorem would verify if the remaining degrees can actually form a valid graph. This theorem ensures that after reducing the degrees, the result still qualifies as a potential graph's degree sequence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove that \(K_{m, n}\) is isomorphic to \(K_{n, m}\).

Let \(G\) be a connected multigraph having a vertex of degree \(k\). Prove that \(G\) has a minimal edge-cut \(F\) with \(|F| \leq k\).

Use the pigeonhole principle to prove that a graph of order \(n \geq 2\) always has two vertices of the same degree. Does the same conclusion hold for multigraphs?

Let \(G\) be a forest of \(k\) trees. What is the fewest number of edges that can be inserted in \(G\) in order to obtain a tree?

Use the algorithm for growing a spanning tree in order to grow a spanning tree of the graph of the regular dodecahedron.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.