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Mohamed G. Gouda						     CS 311
Summer 2014							     Midterm 2
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1. (6 points)
(a) Draw graphs G1 and G2 such that the following three conditions hold:
	1. G1 is 2-colorable and has at least two vertices u1 and v1
	2. G2 is 2-colorable and has at least two vertices u2 and v2
	3. If u1 and u2 are connected by an edge and
	   if v1 and v2 are connected by an edge
 	   then the resulting graph G is 2-colorable

(b) Same as (a) except that the resulting graph is not 2-colorable.
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2. (7 points)
Prove by contradiction that the complete graph K6,6 is not planar.
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3. (7 points)
Let G = (V, E) be a planar graph with 4 vertices where the degree of each 
vertex is 3. Assume that the two sides of each edge in G belong to distinct 
regions. Compute the sum of the region degrees of G. Explain your answer.
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Solutions:
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1.
(a) 	G1 = (V1, E1) where V1={u1, v1} and E1={(u1, v1)}
	G2 = (V2, E2) where V2={u2, v2} and E2={(u2, v2)}
	Note that both G1 and G2 are trees and so are 2-colorable
	Also G consists of one cycle length 4; thus it is 2-colorable

(b) 	G1 = (V1, E1) where V1={u1, v1, w} and E1={(u1, w), (w, v1)}
	G2 = (V2, E2) where V2={u2, v2} and E2={(u2, v2)}
	Note that both G1 and G2 are trees and so are 2-colorable
	Also G consists of one cycle length 5; thus it is not 2-colorable
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2.
The proof is by-contradiction

	(K6,6 is planar)
=> {definition of K6,6}
	(K6,6 is planar and connected and where |V|=12 and |E|= 36)
=> {Euler's corollary}
	(30 = 3*|V|-6 >= |E| = 36)
=> {arithmentics}
	F
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3.
	Sum of rdeg(r)'s
= {Region Handshake Theorem; two sides of each edge belong to distinct regions}
	2*|E|
= {Handshake Theorem}
	Sum of deg(u)'s
= {|V| = 4 and deg(u) = 3 for each vertex u}
	4*3
= {arithmetics}
	12
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