Mohamed G. Gouda						     CS 313K
Summer 2013							     Midterm 1

1) Simplify the formula 
	f(x,y) = (x and y) or (y and not x) or (not (x or not y))

Sol:
	(x and y) or (y and not x) or (not (x or not y))
= {symmetry}
	(y and x) or (y and not x) or (not (x or not y))
= {distribution}
	(y and (x or not x)) or (not (x or not y))
= {x or not x is T and y and T is y}
	y or (not (x or not y))
= {De Morgan's}
	y or (not x and not (not y))
= {not (not y) is y and symmetry}
	y or (y and not x)
= {absorption}
	y

2) Let Dx and Dy be the set of all integers. Compute the boolean values
of the following predicates:
	All x,   Exist y, (x < y+x)
	Exist x, All y,   (x < y+x)
	All x,   All y,   (x < y+x)
	Exist x, Exist y, (x < y+x)
	Exist y, All x,   (x < y+x)
	All y,   Exist x, (x < y+x)

Sol:
	All x,   Exist y, (x < y+x) = T
	Exist x, All y,   (x < y+x) = F
	All x,   All y,   (x < y+x) = F
	Exist x, Exist y, (x < y+x) = T
	Exist y, All x,   (x < y+x) = T
	All y,   Exist x, (x < y+x) = F
	
3) Give an inducton proof to prove that, for every positive integer n,
(5^n - 1) is divisible by 2.

Sol:
Let P(n) be the predicate: (5^n - 1) is divisible by 2

Base Case: n = 1: 
P(1) <=> (5 - 1 is divisible by 2) <=> T

Induction Step: 
for every n > 0, P(n) => P(n+1)

  P(n)
<=> {Definitions of P(n) and of divisible by 2}
  5^n - 1 = 2k, for some integer k
<=> {Arithmetic}
  5^n = 2k+1, for some integer k
<=> {Multiply 5 to both sides of "="}
  5^(n+1) = 5*(2k+1), for some integer k
<=> {Arithmetic}
  (5^(n+1) - 1) = (5*(2k+1) - 1), for some integer k
<=> {Arithmetic}
  (5^(n+1) - 1) = 2*(5k+2), for some integer k
=> {Definition of divisible by 2}
  (5^(n+1) - 1) is divisible by 2
<=> {Definition of P(n+1)}
  P(n+1)

4) Let G be a graph with 6 vertices and 8 edges.
(a) What is the maximum possible value x for max-deg(G)?
(b) Draw a graph G where max-deg(G) is x (from part a).
(c) What is the minimum possible value y for max-deg(G)?
(d) Draw a graph G where max-deg(G) is y (from part c). 

Sol:
(a) 5
(b) G = (V,E) where
    V = {1, .., 6}
    E = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (3,4), (4,5)}
(c) 3
(d) G = (V,E) where
    V = {1, .., 6}
    E = {(1,2), (1,5), (1,6), (2,3), (2,6), (3,4), (4,5), (5,6)}