Mohamed G. Gouda						     CS 313K
Summer 2013							     Homework 2

1) Let G={V,E} be a graph where
	V={0,1,...,9}
	E={(0,1),(1,2),(2,3),(3,4),(4,0),
	   (5,6),(6,7),(7,8),(8,9),(9,5),
	   (0,5),(1,6),(2,7),(3,8),(4,9)}

Compute the chromatic number of this graph. Explain your answer.

Sol.
The chromatic number of this graph is 3. 
Explanation: This graph can be colored using 3 colors B(Black),
W(White), R(Red) as follows:
Vertex 0 is colored B
Vertex 1 is colored W
Vertex 2 is colored B
Vertex 3 is colored W
Vertex 4 is colored R
Vertex 5 is colored W
Vertex 6 is colored R
Vertex 7 is colored W
Vertex 8 is colored R
Vertex 9 is colored B
Moreover, this graph has 2 cycles of odd length each, and each of these 
cycles needs at least 3 colors. Thus, the chromatic number of this graph
is 3.

2) Consider graph G in Problem 1. Is this graph bipartite? Explain your
answer.

Sol.
This graph is not bipartite because it has a cycle of odd length, and each 
cycle in a bipartite graph needs to be of even length.

3) Prove by contradiction that any connected planar graph has at least
one vertex of degree 5 or smaller.

Sol.
	(Exist a connected planar graph G=(V,E) where every vertex is 
	 of degree 6 or larger)
=> {Handshake Theorem}
	2*|E| = SUM over every vertex u in G of deg(u) >= 6*|V|
=> {Euler's Corollary}
	(|E| >= 3*|V|) and (|E| =< 3*|V|-6)
=> 	F

4) Prove by induction that any graph G can be colored using max-deg(G)+1
colors.

Sol.
In order to prove the theorem by induction, we need to restate the theorem
as follows:
	(All n, n >= 1, any graph G with n vertices can be colored using
	 max-deg(G)+1 colors)

Now, we can prove the theorem by induction on n.
Let P(n) be the predicate: any G with n vertices can be colored using 
max-deg(G)+1 colors.

Base case: (n=1):
P(1) <=> {any G with 1 vertex can be colored using 1 color} T

Induction step: 
	P(n)
=> {Let G be any graph with (n+1) vertices, u be any vertex in G, and
    G' be graph G after removing vertex u and its incident edges from G.
    Note that G' has n vertices and so P(n) holds for G'. Also, 
    max-deg(G') is at most max-deg(G). Thus, G' can be colored using 
    max-deg(G')+1 colors, and can be colored using max-deg(G)+1 colors.
    Also note that vertex u has at most max-deg(G) neighbors}
	(G' can be colored using max-deg(G)+1 colors. And one of those
	 colors can be used to color vertex u in G)
=>	(G can be colored using max-deg(G)+1 colors)
=> {Definition of P(n+1)}
	P(n+1)