Mohamed G. Gouda						     CS 313K
Fall 2012							     Midterm 5

1. Consider the recurrence equation: 
   T(1) = 1
   T(n+1) = 2*T(n) + 1
Prove, by induction, that the closed equation for T(n) is:
   T(n) = 2^n - 1    	      for n >= 1

Sol:
Let P(n) be the predicate T(n) = 2^n - 1.

Base Case: n=1:
P(1) <=> T(1) = 2-1 = 1 <=> T

Induction Step:
For n >= 1, P(n) => P(n+1)

 T
=> {Definition of T(n+1)}
 T(n+1) = 2*T(n) + 1
=> {Induction Hypothesis P(n)}
 T(n+1) = 2*(2^n-1) + 1
 	= 2^(n+1) - 2 + 1
	= 2^(n+1) - 1
=> {Definition of P(n+1)}
 P(n+1)

2. Consider the recurrence T(1), T(2), ... where T(n) = number of slices that 
can be obtained from a pizza by making n straight cuts (along some diameters of
the pizza) using a pizza knife. Define the recurrence equation and derive the 
closed equation for this recurrence.

Sol:
	T(1) = 2,
	T(2) = 4,
	T(3) = 6, ...

Each cut makes four slices out of two slices. Thus, the recurrence equation can 
be defined as:

   	T(1) = 2
	T(n+1) = T.n + 2		for n >= 1

A derivation of the closed equation can proceed as follows:

  	T(n) = T(n-1) + 2
	     = T(n-2) + 2 + 2
	     ...
	     = T(n-i) + 2^i
	     = {choosing i to be n-1}
	       T(1) + 2^(n-1)
	     = 2 + 2^(n-1)


3. Use the Characteristic Polynomial method to compute the closed equation for 
the following recurrence equation:
      T(0) = 2						(1)
      T(1) = 8						(2)
      T(n+2) = (-4)*T(n+1) + 5*T(n)	for n >= 0	(3)

Sol:
The Characteristic Polynomial from (3) is:
t^2 = (-4)*t + 5
t^2 + 4*t - 5 = 0
(t+5)(t-1) = 0
t1 = -5 and t2 = 1

Expression for T(n):
T(n) = x*(-5)^n + y*(1)^n				(4)

From (1) and (4):
2 = x + y						(5)

From (2) and (4):
8 = (-5)*x + y						(6)

From (5) and (6),
x = -1 and y = 3					(7)

The closed equation from (4) and (7):
T(n) = (-1)*(-5)^n  + 3

4. Show, by direct inference, that f(x) = floor(x) is Theta(g(x)) where 
g(x)=x.

Sol:
Proving f(x) is O(g(x)):
|f(x)| = |floor(x)|
       = floor(x)	for x>1
       =< x
       = C*|x|		for C=1
       = C*|g(x)|	for K=1 and C=1

Proving f(x) is Omega(g(x)):
|f(x)| = |floor(x)|
       = floor(x)	for x>1
       >= (x/2)
       = C*|x|		for C=(1/2)
       = C*|g(x)|	for K=1 and C=(1/2)

5. Let d be any positive number and m be any positive integer. Show, by direct 
inference, that f(x) = (x+d)^m is O(g(x)) where g(x) = x^m.

Sol:
|f(x)| = |(x+d)^m|
       = (x+d)^m	for x>0
       =< (2*x)^m	for x>d
       = 2^m * x^m
       = C*|x^m|	for C=2^m
       = C*|g(x)|	for K=d and C=2^m