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Mohamed G. Gouda						     CS 311
Fall 2014							     Midterm 2
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1. (4 points)
Let TR be a tree with 104 vertices.
(a) What is the largest possible value of max-deg(TR). Explain your answer.
(b) What is the smalles possible value of max-deg(TR). Explain your answer.

2. (4 points)
Define a graph G whose chromatic number is 4 and whose max-deg(G) is 6.
Explain your answer.

3. (4 points)
Prove by contradiction that every planner connected graph has a vertex whose 
degree is at most 9.

4. (4 points)
Let G be a graph that can be constructed in two steps as follows:
	a. Construct a tree TR that has at least 3 vertices.
	b. Connect any two leaves of TR by one edge.
Show that G is 3-colorable.

5. (4 points)
Define a planner graph G = (V, E) where V has 5 vertices and E has 5 edges 
such that (1) G can be viewed as having two regions of equal degrees and (2)
G can be viewed as having two regions of unequal degrees.

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Solution:
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1.
(a) 103. The number of edges in TR is 103 and the largest possible value of
max-deg(TR) is achieved when all the edges of TR are incident at one vertex.

(b) 2. TR has a vertex of degree at least 2. Thus, the smallest possible value
of max-deg(TR) is achieved when the degree of each vertex in TR is at most 2.

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2.
G = (V, E) where 
V = {1, 2, 3, 4, 5, 6, 7} and
E = {(1,2), (1,3), (1,4), (1,5), (1,6), (1,7), (2,3), (2,4), (3,4)}

Vertex 1 is colored Black, each of the vertices 2, 5, 6, and 7 are colored
White, vertex 3 is colored Red, and vertex 4 is colored Yellow. Thus G
is 4 colorable. Moreover G has a complete subgraph with 4 vertices. Thus the
chromatic number of G is 4.

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3.
	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at least 10)

=> {Handshake Theorem}
	2*|E| = Sum of deg(u)'s >= 10*|V|

=> {Euler's Corollary}
	(|E| >= 5*|V|)  and  (|E| =< (3*|V| - 6)

=> {P and not P = F}
	F 

(Another Proof):

	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at least 10)

=> {Theorem of Small Vertex Degrees}

	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at least 10) and
	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at most 5)

=> {P and not P = F}
	F

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4.
The initial tree TR is 2 colorable. Thus, color the vertices of TR using 
Black and White. If the 2 leaves that are connected by an edge in Step b are
are colored using the same color (both Black or both White), then change the
color of one leaf to Gray. The reulting coloring of G uses at most three
colors: Black, White, and Gray. 

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5.
G = (V, E) where
V = {1, 2, 3, 4, 5}
E = {(1,2), (1,3), (1,4), (1,5), (2,3)}

G can be viewed as having two regions of degree 4 each: the first region is
bordered by the edges (1,2), (1,3), (1,4) and (2,3), and the second region
is bordered by the edges (1,2), (1,3), (1,5), and (2,3).

G can be viewed as having two regions of degree 3 and 5: the first region is
bordered by the edges (1,2), (1,3), and (2,3), and the second region is
bordered by the edges (1,2), (1,3), (1,4), (1,5), and (2,3).
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Mohamed G. Gouda						     CS 311
Fall 2014							     Midterm 2
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1. (4 points)
Let TR be a tree with 104 vertices.
(a) What is the largest possible value of max-deg(TR). Explain your answer.
(b) What is the smalles possible value of max-deg(TR). Explain your answer.

2. (4 points)
Define a graph G whose chromatic number is 3 and whose max-deg(G) is 8. 
Explain your answer.

3. (4 points)
Prove by contradiction that every planner connected graph has a vertex whose 
degree is at most 9.

4. (4 points)
Let G be a graph that can be constructed in two steps as follows:
	a. Construct a tree TR that has at least 3 vertices.
	b. Connect any two leaves of TR by one edge.
Show that G is 3-colorable.

5. (4 points)
Define a planner graph G = (V, E) where V has 5 vertices and E has 5 edges 
such that (1) G can be viewed as having two regions of equal degrees and (2)
G can be viewed as having two regions of unequal degrees.

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Solution:
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1.
(a) 103. The number of edges in TR is 103 and the largest possible value of
max-deg(TR) is achieved when all the edges of TR are incident at one vertex.

(b) 2. TR has a vertex of degree at least 2. Thus, the smallest possible value
of max-deg(TR) is achieved when the degree of each vertex in TR is at most 2.

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2.
G = (V, E) where 
V = {1, 2, 3, 4, 5, 6, 7, 8, 9} and
E = {(1,2), (1,3), (1,4), (1,5), (1,6), (1,7), (1,8), (1,9), (2,3)}

Vertex 1 is colored Black, each of the vertices 2, 4, 5, 6, 7, 8, and 9 are 
colored White, and vertex 3 is colored Red. Thus G is 3 colorable. Moreover 
G has a complete subgraph with 3 vertices. Thus chromatic number of G is 3.

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3.
	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at least 10)

=> {Handshake Theorem}
	2*|E| = Sum of deg(u)'s >= 10*|V|

=> {Euler's Corollary}
	(|E| >= 5*|V|)  and  (|E| =< (3*|V| - 6)

=> {P and not P = F}
	F 

(Another Proof):

	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at least 10)

=> {Theorem of Small Vertex Degrees}

	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at least 10) and
	(Exist a planar connected graph G = (V, E) where the degree of
	each vertex is at most 5)

=> {P and not P = F}
	F

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4.
The initial tree TR is 2 colorable. Thus, color the vertices of TR using 
Black and White. If the 2 leaves that are connected by an edge in Step b are
are colored using the same color (both Black or both White), then change the
color of one leaf to Gray. The reulting coloring of G uses at most three
colors: Black, White, and Gray. 

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5.
G = (V, E) where
V = {1, 2, 3, 4, 5}
E = {(1,2), (1,3), (1,4), (1,5), (2,3)}

G can be viewed as having two regions of degree 4 each: the first region is
bordered by the edges (1,2), (1,3), (1,4) and (2,3), and the second region
is bordered by the edges (1,2), (1,3), (1,5), and (2,3).

G can be viewed as having two regions of degree 3 and 5: the first region is
bordered by the edges (1,2), (1,3), and (2,3), and the second region is
bordered by the edges (1,2), (1,3), (1,4), (1,5), and (2,3).

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