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Mohamed G. Gouda						CS 311 
Spring 2015							Homework 3
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1. (2.5 points) 
Use a by-contradiction proof to prove the predicate:
	(A sub c(B)) => ((A^B) = { })   
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2. (2.5 points)
Let A and B be two finite and nonempty sets and let f:A->B be a function.

(a) Given that f is injective, which of the two integers |A| and |B| is larger.
Explain your answer.

(b) Given that f is surjective, which of the two integers |A| and |B| is larger.
Explain your answer.

(c) Given that f is bijective, which of the two integers |A| and |B| is larger.
Explain your answer.
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3. (2.5 points)
Derive the closed equation for the following recurrence equation:
	R(0) 	= 3
	R(1) 	= 5
	R(n+2)	= -R(n+1) + 6R(n)
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4. (2.5 points)
Show by direct inference that (f(n) = n^3 - 5((n)^2)) is Omega(g(n)) where
(g(n) = n^3).
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Solutions
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1.
	(not((A^B) = { }))
=> {definition of { }}
	(Exist x, x in (A^B))
=> {definition of ^}
	(Exist x, (x in A) and (x in B))
=> {(A sub c(B))}
	(Exist x, (x in c(B)) and (x in B))
=> {definition of c}
	(Exist x, not(x in B) and (x in B))
=> {(not(P) and P => F}
	F
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2. 
(a) |A| =< |B|. Each element in A has a unique and distinct image in B

(b) |A| >= |B|. Each element in A has a unique image in B and each element in B has
at least one pre-image in A.

(c) |A| = |B|. This case follows from (a) and (b).

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3.
The recurrence equation is 
	R(0)		= 3					(1)
	R(1)		= 5					(2)
	R(n+2)		= -R(n+1) + 6R(n)			(3)

From (3), the characteristic polynomial is
	r^2		= -r + 6
	(r-2)(r+3)	= 0					(4)

From (4), the characteristic polynomial has two roots:
	r1		= 2					(5)
	r2		= -3					(6)

The closed equation is
	R(n)		= x*(r1)^n + y*(r2)^n			(7)

From (1), (2), (5), (6), and (7), we get
	x = 14/5 and y =1/5					(8)
 
Thus, the closed equation is
	R(n)		= (14/5)*(2^n) + (1/5)*((-3)^n)		(9)
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4.
	|f(n)|	=  |n^3 - 5*(n^2)|
		=  (n^3 - 5*(n^2))		for n > 5
		=  (n^3)/2 + (n^3)/2 - 5*(n^2)
		>= (n^3)/2			for n >10
		=  (1/2) * (n^3)
		=  C * |g(n)|			for K=10 and C=(1/2)
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