Mohamed G. Gouda						     CS 311
Fall 2013							     Midterm 2

1) Let G be any 2-colorable graph. Show by contradiction that every cycle in G is of even length.

Sol:
	not (every cycle of G is of even length)
=>{De-Morgan's}
	(G has a cycle C of odd length)
=>{cycle C is a subgraph of G and G is 2-colorable}
	(cycle C is of odd length and is 2-colorable)
=>{no cycle of odd length is 2-colorable}
	F

2) Define a graph G such that all the following five conditions hold:
	1. The chromatic number of G is 3
	2. If a certain edge is removed from G, the chromatic number remains 3
	3. If any other edge is removed from G, the chromatic number becomes 2
	4. If a certain vertex and its incident edges are removed from G, the 
	   chromatic number remains 3
	5. If any other vertex and its incident edges are removed from G, the 
	   chromatic number becomes 2

Sol:
	G=(V,E) where V={1,2,3,4} and E={(1,2), (1,3), (2,3), (3,4)}
	1. The chromatic number of G is 3
	2. If edge (3,4) is removed, the chromatic number remains 3
	3. If any other edge is removed, the chromatic number becomes 2
	4. If vertex 4 and its incident edge are removed, the chromatic number 
	   remains 3
	5. If any other vertex and its incident edges are removed, the chromatic
	   number becomes 2

3) Let G be any graph 
	a. Is it possible that the degree of every vertex in G be odd? Explain your answer
	b. Is it possible that the degree of every vertex in G be even? Explain your answer
	
Sol:
	a. Yes, if the number of vertices in G is even, then the degree of every vertex 
	   in G can be odd satisfying Corollary 2 of Handshake Theorem
	b. Yes, the degree of every vertex in G can be even becasue in this case the
	   number of vertices with odd degrees is 0 or even satisfying Corollary 2 of
	   Handshake Theorem 

4) Let G1 be a graph that has a vertex v1 and G2 be another graph has a vertex v2.
Assume that max-deg(G1) = m and max-deg(G2) = p. Let G be the graph that results from combining G1 
and G2 by collapsing the 2 vertices v1 and v2 into one vertex v. Compute an integer k from m and p
such that G can be colored using k colors. Explain your answer

Sol:
From the Graph Coloring Theorem, G can be colored using max-deg(G)+1 colors.
But max-deg(G) is at most max-deg(G1)+max-deg(G2) which equals m+p.
Thus, G can be colored using m+p+1 colors.

A Better Sol:
Without loss of generality, let max(m, p) be m. Therefore, graph G2 can be re-colored 
using the m+1 colors that have been used to color grah G1 such that the two vertices v1 
and v2 end up being colored with the same color. Thus, G ends up being colored using m+1 
colors.

5) Let G=(V,E) be a graph where the degree of every vertex is at least 4. Show by direct
inference that |E| >= 2*|V|. (Hint: Show that T => (|E| >= 2*|V|).)

Sol:
	T
=>{Handshake Theorem}
	2*|E| = sum of deg(u)'s
=>{deg(u) is at least 4 for every u in V}
	2*|E| >= 4*|V|
=>{arithmetics}
	|E| >= 2*|V|