Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 8

1. A connected planar graph G has 6 vertices of degree 4 each. How many regions 
does G have? Explain.

Sol:
Let	n be the number of vertices in G,
	e be the number of edges in G, and
	r be the number of regions in G.

From Handshake Theorem,
     2e = (sum over v in G, deg(v))
        = 6*4
	= 24
      e = 12

From Euler's Formula,
     n-e+r  = 2
     6-12+r = 2
          r = 8

2. Let G be a connected planar graph that has e edges and r regions. Show, by 
direct inference, that 2e >= 3r.

Sol:
 T
=> {Region Handshake Theorem}
 (2e >= sum over r.i in G, rdeg(r.i))
=> {For every r.i, rdeg(r.i) >= 3}
 (2e >= sum over r.i in G, rdeg(r.i)) and 
 (sum over r.i in G, rdeg(r.i) >= 3r)
=> {Arithmetics}
 (2e >= 3r)

3. Let K3,3 be the complete bipartite graph with 6 vertices and 9 edges. Show, 
by contradiction, that K3,3 is non-planar.

Sol:
Let	n be the number of vertices in K3,3,
	e be the number of edges in K3,3, and
	r be the number of regions in K3,3.

 K3,3 is a connected planar graph
=> {n=6 and e=9 and by Euler's Formula, 6-9+r=2}
 (e = 9) and (r = 5)
=> {K3,3 has no cycle of length 3 or less. Thus, rdegree of each region in K3,3
    is at least 4. Then, from the Region Handshake Theorem, 2e >= 4r. Then
    e >= 10}
 (e = 9) and (e >= 10)
=> F

4. Let G be a connected planar graph with n vertices (where n >= 3), e edges 
(where e >= 5), r regions, and no cycle of length 4 or less. Show that 
       e =< (5n - 10)/3

Sol:
Because e >= 5 and G has no cycle of length 4 or less, the rdegrees of each 
region in G is at least 5. 

From the Region Handshake Theorem, 
     2e >= (sum over r.i in G, rdeg(r.i))
        >= 5r	     	    (1)

From Euler's Formula, 
     r = 2-n+e		    (2)

From (1) and (2),
     2e >= 5*(2-n+e)
     3e =< 5n - 10
      e =< (5n - 10)/3

5. Let A, B, and C be three sets. Show, by direct inference, that 
       (A sub B) and (B sub C) => (A sub C)

Sol:
 x in A
=> {A sub B}
 x in B
=> {B sub C}
 x in C

6. Let A and B be two sets. Show, by two direct inference proofs, that 
       (A sub B) <=> (PS(A) sub PS(B))
where PS(A) is the power set of A.

Sol:
Proving =>

 x in PS(A)
=> {Definition of PS(A)}
 x sub A
=> {From A sub B and from the previous Problem}
 x sub B
=> {Definition of PS}
 x in PS(B)

Proving <=

 T
=> {Theorem 1; see below}
 A in PS(A)
=> {From PS(A) sub PS(B)}
 A in PS(B)
=> {Definition of PS(B)}
 A sub B

Proving Theorem 1: A in PS(A)

 T
=> {Theorem 2; see below}
 A sub A
=> {Definition of PS(A)}
 A in PS(A)

Proving Theorem 2: A sub A

 x in A
=> {For every predicate P, P => P}
 x in A

7. Let A and B be two sets. Prove, by two-sided inference, that 
       PS(A^B) = PS(A) ^ PS(B)
where PS(A) is the power set of A.

Sol:
 x in (PS(A) ^ PS(B))
<=> {Definition of ^}
 (x in PS(A)) and (x in PS(B))
<=> {Definition of PS}
 (x sub A) and (x sub B)
<=> {Theorem; see below}
 (x sub A^B)
<=> {Definition of PS}
 (x in PS(A^B))

Proving Theorem ((x sub A) and (x sub B)) => (x sub A^B)

1. Proving =>: 
 y in x
=> {(x sub A) and (x sub B)}
 (y in A) and (y in B)
=> {definition of "^"}
 y in A^B

2. Proving <=:
 y in x
=> {(x sub A^B)}
 y in A^B
=> {definition of A^B}
 (y in A) and (y in B) 

8. Let A be a set and * be the Cartesian Product operator. Show, by 
contradiction, that A*{} = {}.

Sol:
 A*{} != {}
=> {Definition of "!= {}"}
 (Exist (x,y) in A*{})
=> {Definition of *}
 (Exist x in A) and (Exist y in {})
=> {not(Exist y in {})}
 (Exist x in A) and (Exist y in {}) and not(Exist y in {})
=>
 F

9. Let A be a set and c(A) be the complement of set A. Show, by two-sided 
inference, that c(c(A)) = A.

Sol:
 x in c(c(A))
<=> {Definition of the "c" set operator}
 not(x in c(A))
<=> {Definition of the "c" set operator}
 not(not(x in A))
<=> {not(not(Predicate P)) = P}
 x in A

10. Let A and B be two sets, and let "v", "^", and "c" be the union, 
intersection, and complement set operators (respectively). Show, by two-sided 
inference, that 
	   c(c(A) v c(B)) = A ^ B

Sol:
 x in c(c(A) v c(B))
<=> {definition of the "c" set operator}
 not (x in (c(A) v c(B)))
<=> {definition of the "v" set operator}
 not ((x in c(A)) or (x in c(B))
<=> {definition of the "c" set operator}
 not (not(x in A) or not(x in B))
<=> {De Morgan's}
 ((x in A) and (x in B))
<=> {definition of the "^" set operator}
 x in (A ^ B)