Question

(4) Let G be a nontrivial tree. Prove that there exists a cut vertex \textit{v} whose neighbors, with an exception of at most one, are all leaves.

          (4) Let G be a nontrivial tree. Prove that there exists a cut vertex \textit{v} whose neighbors, with an exception of at most one, are all leaves.
        
(4) Let G be a nontrivial tree. Prove that there exists a cut vertex v whose neighbors, with an exception of at most one, are all leaves.

Added by Daniela S.

Close

Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Assume the tree T has at least three vertices. (4) Let G be a nontrivial tree. Prove that there exists a cut vertex v whose neighbors, with an exception of at most one, are all leaves
Close icon
Play audio
Feedback
Powered by NumerAI
Ivan Kochetkov Kathleen Carty
Jennifer Stoner verified

William Casper and 95 other subject Algebra educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
let-t-be-a-tree-on-n-vertices-prove-that-if-t-contains-no-vertices-of-degree-two-then-at-least-half-of-its-vertices-are-leaves-30499

Let T be a tree on n vertices. Prove that if T contains no vertices of degree two, then at least half of its vertices are leaves.

William C.

prove-that-if-g-is-a-tree-with-an-even-number-of-edges-then-g-must-contain-at-least-one-vertex-of-even-degree-19966

Prove that if G is a tree with an even number of edges, then G must contain at least one vertex of even degree

Muhammad A.

let-t-v-e-be-a-rooted-tree-with-the-property-that-all-internal-vertices-all-non-leaf-vertices-have-exactly-3-children-if-there-are-k-internal-vertices-prove-by-induction-on-k-that-v-3k-1-86319

Let T = (V, E) be a rooted tree with the property that all internal vertices (all non leaf vertices) have exactly 3 children. If there are k internal vertices, prove, by induction on k, that |V | = 3k + 1.

Shaiju T.


*

Recommended Textbooks

-
Elementary and Intermediate Algebra

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition
achievement 1,983 solutions
Elementary and Intermediate Algebra

Elementary and Intermediate Algebra

Marvin L. Bittinger, David J. Ellenbogen,Barbara L. Johnson 4th Edition
achievement 1,245 solutions
Algebra and Trigonometry

Algebra and Trigonometry

James Stewart, Lothar Redlin, Saleem Watson 4th Edition
achievement 1,991 solutions

*

Transcript

-
00:01 Hey there, in this video will prove that a tree with no vertices of degree 2 must have at least half of its vertices be leaves.
00:07 To prove this, we'll use two facts.
00:09 First of all, if n is the number of edges, or sorry, number of vertices, and e is the number of edges, then in a tree, the number of edges has to be equal to the number of vertices minus 1.
00:28 The second fact is true for any graph.
00:31 Let's let v be the set of vertices.
00:34 Then the sum over all vertices and v of the degree of the vertex is going to be equal to twice the number of edges.
00:48 Where we remember the degree of a vertex is exactly the number of edges that are coming out of that particular vertex.
00:54 Putting these together, we can say that for a tree, the sum over all the vertices of the degree of the vertex has to be equal to 2 times n minus 2, where n here is the total number of vertices.
01:09 We can rewrite the term on the right hand side, actually, as the sum over all the vertices of 2 minus 2.
01:20 Because it just add 2 to itself the number of times i have vertices, i'm going to get 2 times the number of vertices.
01:27 So what i've got is the sum of the degrees is equal to the sum over 2 minus 2.
01:35 And we can rewrite this a little bit.
01:38 Further, just by bringing all the sums to the left -hand side, and i can say that the sum over all the vertices of the degree of the vertex minus 2 is equal to negative 2.
01:48 Now let's think about that sum on the left -hand side.
01:53 When v is a leaf, the degree of v is going to be exactly 1, because leaves are the terminal parts of the tree.
02:01 The degree of v can never be 2 by assumption.
02:05 So the terms on the right -hand side are the left -hand side of the sum are always going to negative 1 if i've got a leaf or greater than 1 if i don't have a leaf.
02:16 So if i let n l be the number of leaves, i can re -express the sum as the sum over all vertices, which are leaves, plus the sum over all vertices which are not leaves...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever