pset3soln

# pset3soln - EECS 310, Fall 2011 Instructor: Nicole...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #3 Due: October 13, 2011 1. (25 points) This question explores some basic properties of graphs. (a) (13 points) Prove the following claim using induction . Consider carefully whether it is easier to do induction on the number of vertices or the number of edges. The number of odd degree vertices in a graph is even. (b) (12 points) There is an EECS faculty party with N professors going. At this party, certain pairs of professors are already friends (friendships are mutual i.e. if Alice is friends with Bob then Bob must be friends with Alice), while other professors are not friends (for this example, people are not friends with themselves). For a particular guest g i at this party, let F ( g i ) be the number of friends that g i has at this party. Prove that there are at least two professors that have the same number of friends at this party. In notation, prove that F ( g i ) = F ( g j ) for some i 6 = j . Solution: (a) By induction on the number of edges in a graph. Let m be the number of edges in any graph. Base Case m = 0. When there are no edges in a graph, then every vertex has degree 0 and thus there are 0 vertices of odd degree, which is even. So the base case is proved. Inductive Step . Assume this property holds for any graph with m edges. Now we show that it holds for any graph with m + 1 edges. Let G = ( V,E ) be any graph on m + 1 edges (so | E | = m + 1). Now remove any edge e E and we get the resulting graph G = ( V,E \ e ). Now G is a graph on m edges, so by our induction hypothesis, there are an even number of odd-degree vertices in G . Let u,v be the vertices in V that e connected. Now we have three cases Case 1 : u and v are both odd degree vertices in G . Then adding back in edge e will result in two less odd-degree vertices in G than in G . Since there was an even number of odd-degree vertices in G , and we subtracted 2 odd- degree vertices when we added edge e back in, there is still an even number of odd-degree vertices in G . Case 2 : u and v are both even degree vertices in G . Then adding back in edge e will result in two more odd-degree vertices in G than in G . Since there was an even number of odd-degree vertices in G , there will be an even number of odd-degree vertices in G . Case 3 : u is an odd degree vertex and v is an even degree vertex in G (notice that we do not have to worry about the case where u is even and v is odd since we can simply rename the vertices and use an identical argument). Then adding back in edge e will cause u to go from an odd-degree vertex to an even-degree vertex and will cause v to go from an even-degree vertex to an odd-degree vertex....
View Full Document

## This note was uploaded on 01/15/2012 for the course ECON 101 taught by Professor N/a during the Fall '11 term at Middlesex CC.

### Page1 / 6

pset3soln - EECS 310, Fall 2011 Instructor: Nicole...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online