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Mohamed G. Gouda						    CS 311
Fall 2014							    Homework 2
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1. (2 points)
Let G be a disconnected graph that has 13 edges. G consists of two connected 
components G1 and G2. G1 has 4 vertices whose degrees are 2, 3, 3, x. G2 has
8 edges and 5 vertices whose degrees are 3, 3, 3, 3, y. Compute the values of 
x and y.
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2. (2 points)
A tree is defined as a connected graph that has no cycle of length at least 3.
(a) Can a tree be also defined as a connected graph that has no cycle? Explain
your answer.
(b) Can a tree be also defined as a connected graph that has no cycle of length
at least 1? Explain your answer.
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3. (2 points)
Prove by induction the predicate (All n, n >= 1, any tree with n vertices has 
(n-1) edges).
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4. (2 points)
Let G = (V, E) be a planar connected graph that has 6 vertices whose degrees
are 1, 1, 2, 2, 3, 3. Compute the number of regions in G.
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5. (2 points)
Let G be a planar disconnected graph that consists of 2 connected componenets
G1 and G2. G1 has 3 edges and the two sides of each edge in G1 belong to
distinct regions. G2 has 7 edges and the two sides of each edge in G2 belong to
the same region. Compute the sum of region degrees in G.
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Solutions:
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1. 
G1 = (V1, E1) has 13-8 = 5 edges

Applying Handshake Theorem to G1,
	2*|E1| = 2*5 = 10 = Sum of deg(u)'s in G1 = 2+3+3+x = 8+x
Thus x = 2.

Applying Handshake Theorem to G2,
	2*|E2| = 2*8 = 16 = Sum of deg(u)'s in G2 = 3+3+3+3+y = 12+y
Thus y = 4.
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2.
(a) No because each graph (whether a tree or not) has at least one vertex and so
has a cycle of length 0.

(b) Yes because no cycle is of length 1 or of length 2. Thus, stating that a tree
has no cycle of length at least 3 is the same as stating that a tree has no cycle 
of length at least 1.
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3.
Let P(n) be the predicate "any tree with n vertices has (n-1) edges"

Base case: n = 1:
P(1) <=> {any tree with 1 vertex has no edges} T

Induction step:
A direct inference proof of (All n, n >= 1, P(n) => P(n+1)) proceeds as follows

		P(n)
=> {big hint: Let TR be any tree with (n+1) vertices; from the Theorem of Tree
Leaves, TR has a leaf u that is connected with an edge (u,v) to a tree TR' with
n vertices; from P(n), tree TR' has (n-1) edges; thus, TR has n edges}
		P(n+1)
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4.
From Handshake Theorem, 
2*|E| = Sum of deg(u)'s = 1+1+2+2+3+3 = 12
Thus, |E| = 6

From Euler Formula,
(|V| - |E| + |R|) = 2

Thus from |V|=6 and |E|=6,
|R| = 2
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5.
Applying the Region Handshake Theorem to G1,
Sum of region degrees in G1 = 2*|E1| = 6

Applying the Region Handshake Theorem to G2,
Sum of region degrees in G2 = 1*|E2| = 7

Sum of region degrees in G = 6+7 = 13
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