New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Graph Theory

by: Mr. Eladio Murphy
Mr. Eladio Murphy
GPA 3.59


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Applied Math And Statistics

This 2 page Class Notes was uploaded by Mr. Eladio Murphy on Monday October 19, 2015. The Class Notes belongs to 1016 467 at Rochester Institute of Technology taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/225062/1016-467-rochester-institute-of-technology in Applied Math And Statistics at Rochester Institute of Technology.

Popular in Applied Math And Statistics


Reviews for Graph Theory


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/19/15
122 124 126 128 Graph Theory 1016 467 pring 005 Set 1 Remarks and Solutions The graph is K 5 This can be shown by the brute force method of simply checking all of the vertex connections directly However to reduce the work you might consider making use of the symmetries in this problem S itself is symmetric about 0 that is S S This together with the given rules for joining vertices allows us to prove that v w is an edge if and only if v w is an edge Some direct checking is still needed but it is much less than if no notice is taken of the symmetries Let G be a disconnected graph Then VG l2 2 If IV G 2 then G E2 so G K2 which has diameter 1 Therefore suppose IV G I2 3 Claim If v and w are vertices in then dvw s 2 Proof If v w E then dvw 1 and our claim is satisfied Suppose therefore that v w E Then v w E As G is disconnected there must be a vertex 2 not joined to v by any path in G In particular vz E and so vz E Also were zw E then zwv would be a path joining z and v in G which is forbidden Therefore zw E and so zw E With this vzw is now a path in Gjoining v and w That is dvw 2 and the claim is proved As dvw s 2 for all vertices v and win we have diamG s 2 i The graph G1 is bipartite with bipartition A r1u1w1t1y1q1 and B x1v1s1zl The graph G2 is not bipartite Suppose it were with bipartition A B and v2 E A Now vzwzrzxzu2 v2 is a cycle oflength 5 in G2 So we have v2 r2u2E A and wz x2v2 E B As v2 E Am B we havea H i contradiction and so no bipartition of G2 is possible For a given nonempty set of integers V we construct a graph G as follows Put VG V and join vertices v and w with an edge if and only if v w E 1mod 2 Claim G is bipartite Proof Put AvvE0mod2andBwwE1mod2 If v wEAthen vwE0 1mod2 If v wE Bthen vwE2 1mod2 If vEA and wEB then v w E 1 mod2 Therefore a pair of vertices are joined by an edge in G if and only if one vertex is in A and the other in B If either A Q or B Q then G is edgeless If both A Q and B Q then G is the complete bipartite graph KIAHBI In this problem recall the usual labeling of Q n where adjacent bit strings differ in exactly one place Parts a and b can then be regarded as asking for special subgraphs of Q n Note A union of graphs is taken as a disjoint union In particular the expression G U G denotes the graph made up of two disjoint copies of G a i R2 Q 2K2UK2 ii Adjacent vertices in Q 3 differ in 2 or 3 places Removing from each vertex the edge that joins it to its binary complement gives R3 When you ve done this you see that R3 K 4 U K 4 shown 22 24 O 9quot 0 P below 000 1 1 1 110 100 101 010 If two 3 bit binary strings differ in exactly 3 places then each is the binary complement of the other So in S3 each vertex is joined only to its binary complement Therefore S3 K2 U K2 U K2 U K2 In the graphs C for n 2 3 every vertex has degree 2 and so no vertex has odd degree The graph K2 X C3 has six vertices all of degree 3 It is not a complete graph since it has just 9 edges while K 6 has 15 Think of the given condition this way Every edge joins vertices of different degrees 1 1 The easiest way is to take the complement of your answer in part c Vertices that are not adjacent will then have different degrees This complement looks a bit busy if you simply draw it using the above configuration It looks nicer if you move things about so that there are no unsightly crossings 3 3 Let G be a graph with 6 vertices and 10 edges If 6G 3 and AG 4 then every vertex has degree 3 or 4 Let x denote the number of degree 3 vertices y the number of degree 4 vertices Then x y 6 and 3x 4y 2 10 20 Solving this system we get x 4 and y 2 To get the desired example start with K2 XC3 a graph with six vertices all of degree 3 and add a single edge joining some pair of vertices that are not already adjacent


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Janice Dongeun University of Washington

"I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.