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Mohamed G. Gouda						     CS 311
Fall 2015							     Midterm 3
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1. (5 points)
Prove by contradiction that the graph K3,4 is nonplanar.
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2. (5 points)
Let A and B be two sets and let v and ^ denote the set operators of union and 
intersection. Prove the following predicate using a single 2-sided inference 
proof:
	(A = B) => (AvB) = (A^B)
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3. (5 points)
Let A be the set {1, 2, 3, 4} and f:A->A be a function where f(1)=1 and 
f(2)=3. What are all the possible values of the pair f(3) and f(4) such that
function f has an inverse.
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4. (5 points)
Show by direct inference that the function f(n) = (n^3 - n^2) is Omega(g(n))
where g(n) = n^3.
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Solutions
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1.
	(K3,4 = (V, E) is planar
=> {definition of K3,4 and region degrees of K3,4}
	(|V|=7 and |E|=12 and for every region r in K3,4, rdeg(r) = 4)
=> {Euler's Formula}
	|R| = (|E| - |V| + 2) = 7
=> {Region Handshake Theorem}
	28 = 4*|R| =< (Sum of rdeg(r)) =< 2*|E| =24
=> {(28 =< 24) is false}
	F
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2.
	x in (AvB)
<=> {definition of v}
	(x in A) or (x in B)
<=> {definition of A=B}
	(x in A) or (x in A)
<=> {(P or P) <=> P}
	(x in A)
<=> {P <=> (P and P)}
	(x in A) and (x in A)
<=> {definition of A=B}
	(x in A) and (x in B)
<=> {definition of ^}
	x in (A^B)
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3. 
First possible value is f(3)=2 and f(4)=4.
Second possible value is f(3)=4 and f(4)=2.

Explanation: f needs to be both injective and surjective. Since f is 
injective, f(3) can't be 1 and can't be 3. Thus, f(3)=2 or f(3)=4.
Since f is surjective and f(3)=2, we have f(4)=4. Since f is surjective 
and f(3)=4, we have f(4)=2.
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4.
|f(n)|	=	|n^3 - n^2|
	=	n^3 - n^2		for n > 1
	=	n^3/2 + n^3/2 - n^2
	>=	n^3/2			for n > 2
	=	(1/2)*|n^3|
	=	C*|n^3|			for C = 1/2
	=	C*|g(n)|		for K=2 and C=(1/2)
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Mohamed G. Gouda						     CS 311
Fall 2015							     Midterm 3
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1. (5 points)
Prove by contradiction that the graph K3,5 is nonplanar.
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2. (5 points)
Let A and B be two sets and let v and ^ denote the set operators of union and 
intersection. Prove the following predicate using a single 2-sided inference 
proof:
	(A = B) => (AvB) = (A^B)
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3. (5 points)
Let A be the set {1, 2, 3, 4} and f:A->A be a function where f(1)=1 and 
f(2)=3. What are all the possible values of the pair f(3) and f(4) such that
function f has an inverse.
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4. (5 points)
Show by direct inference that the function f(n) = (n^3 - n^2) is Omega(g(n))
where g(n) = n^3.
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Solutions
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1.
	(K3,5 = (V, E) is planar
=> {definition of K3,5 and region degrees of K3,5}
	(|V|=8 and |E|=15 and for every region r in K3,5, rdeg(r) = 4)
=> {Euler's Formula}
	|R| = (|E| - |V| + 2) = 9
=> {Region Handshake Theorem}
	36 = 4*|R| =< (Sum of rdeg(r)) =< 2*|E| =30
=> {(36 =< 30) is false}
	F
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2.
	x in (AvB)
<=> {definition of v}
	(x in A) or (x in B)
<=> {definition of A=B}
	(x in A) or (x in A)
<=> {(P or P) <=> P}
	(x in A)
<=> {P <=> (P and P)}
	(x in A) and (x in A)
<=> {definition of A=B}
	(x in A) and (x in B)
<=> {definition of ^}
	x in (A^B)
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3. 
First possible value is f(3)=2 and f(4)=4.
Second possible value is f(3)=4 and f(4)=2.

Explanation: f needs to be both injective and surjective. Since f is 
injective, f(3) can't be 1 and can't be 3. Thus, f(3)=2 or f(3)=4.
Since f is surjective and f(3)=2, we have f(4)=4. Since f is surjective 
and f(3)=4, we have f(4)=2.
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4.
|f(n)|	=	|n^3 - n^2|
	=	n^3 - n^2		for n > 1
	=	n^3/2 + n^3/2 - n^2
	>=	n^3/2			for n > 2
	=	(1/2)*|n^3|
	=	C*|n^3|			for C = 1/2
	=	C*|g(n)|		for K=2 and C=(1/2)
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