Mohamed G. Gouda						     CS 311
Fall 2013							     Midterm 3

1) Let f:A->B be a function.
   a) Prove by direct inference that (f(A) sub B)
   b) Prove by direct inference that ((f is surjective) => (B sub f(A)))

Sol.:
a) 
	y in f(A)
=> {definition of f(A)}
	(y in B) and 
	(Exist x in A, f(x) = y)
=> {property of "and"}
	y in B

b)
	y in B
=> {f is surjective}
	(Exist x in A, f(x) = y)
=> {definition of f(A)}
	y in f(A)

2) Let f:A->B be a function where set B is finite. Use direct inference
and the two predicates
	(f(A) sub B)
	((f is surjective) => (B sub f(A)))
to prove the predicate
	((f is surjective) => (|B| = |f(A)|))

Sol.:
	(f is surjective)
=> {the two given predicates}
	(f(A) sub B) and (B sub f(A))
=> {definition of set equality}
	B = f(A)
=> {B is finite and so f(A) is finite and both |B| and |f(A)| are 
    well-defined}	
	(|B| = |f(A)|)

3) Let f:A->B be a function, where A is the set {x, x', x"}, B is the
set {y, y"}. Also let A1 be the set {x}.
   a) Define function f such that 
	f(A-A1) != (f(A) - f(A1))
   b) Define function f such that
	f(A-A1)  = (f(A) - f(A1))

Sol.:
a) f(x) = y, f(x') = y, f(x") = y"
   f(A-A1) =	f({x', x"})
	   =    {y, y"}
	  !=	f(A) - f(A1)
	   =	({y, y"} - {y})
	   =    {y"}

b) f(x) = y, f(x') = y", f(x") = y"
   f(A-A1) =	f({x', x"})
	   =    {y"}
	   =	f(A) - f(A1)
	   =	({y, y"} - {y})
	   =    {y"} 

4) Use the characteristic polynomial method to derive the closed equation
for the following recurrence equation:
	R(0) = 2				(1)
	R(1) = 0				(2)
	R(n+2) = 4*R(n)		      n >= 0	(3)

Sol.:
From (3), the characteristic polynomial (r^2 - 4 = 0) has two roots:
r1 = 2, r2 = -2, and the closed equation is
	R(n) = x*(2)^n + y*(-2)^n     n >= 0  	(4)
From (1), (2), and (4) we get
	x = 1 and y = 1
The closed equation is 
	R(n) = 2^n + (-2)^n

5) Show by direct inference that the function f(n) = (n^3 - n^2) is 
Omega(g(n)) where g(n) = n^3.

Sol.:
|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