Mohamed G. Gouda						     CS 313K
Fall 2012							     Homework 3

1. A connected planar graph G has 20 vertices. What is the maximum number of 
edges that G can have? What is the maximum number of regions that G can have?

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 Euler's Corollary,
     e =< 3n - 6
       = 3*20 - 6
       = 54
Thus, emax = 54

From Euler's Formula,
     r = 2 - n + e
       =< 2 - n + emax
       = 2 - 20 + 54
       = 36
Thus, rmax = 36

2. Let G be a planar graph that has n vertices, e edges, r regions, and 
k (connected) components. Show that the Euler's Formula for G can be written 
as: n-e+r = k+1. (Note that if G is connected, then k = 1 and Euler's Formula 
is degenerated to n-e+r = 2.)

Sol: 
The i-th component in G, i = 1, ..., k, is a connected planar graph. And so it 
satisfies Euler's Formula:
	  n.i - e.i + r.i = 2
where n.i is the number of vertices in the i-th component,
      e.i is the number of edges in the i-th component, and
      r.i is the number of edges in the i-th component.

Summing these k formulas together, we get:
	(sum over i, n.i) - (sum over i, e.i) + (sum over i, r.i) = 2*k    (1)

But, we have:
        (sum over i, n.i) = n						   (2)
	(sum over i, e.i) = e						   (3)
	(sum over i, r.i) = r + (k-1)					   (4)
Note that the (sum over i, r.i) counts the outside region k times whereas this 
region is counted only once in r. Therefore, (sum over i, r.i) = r + (k-1) in 
(4). 

Substituting (2), (3), and (4) in (1), we get:
	n - e + r = k+1

Another Sol:
Let the k components of G be G.1, G.2, ..., G.k. Form a new connected planar 
graph G' by adding one edge between vertices in components G.i and G.(i+1) 
for every i, i = 1, 2, ..., k-1. Graph G' is connected and planar and has n 
vertices, (e+k-1) edges, and r regions, where r is also the number of regions 
in the original graph G. Euler's Formula for G' is as follows: 
	  n-(e+k-1)+r = 2
Then, 	  n-e+r = k+1
This formula can be regarded as Euler's Formula for graph G. 

Third Sol:
Prove the formula (n-e+r = k+1) by induction on k.

Let P(k) be the predicate (n-e+r = k+1).

Base Case: k=1: 
P(1) <=> n-e+r = 2 <=> {Euler's Formula} T

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

 T
=> {Let G be a graph that has n vertices, e edges, r regions, and k+1 
components. This G can be viewed as composed of two subgraphs, G' and G''. 
Subgraph G' has n' vertices, e' edges, r' regions, and k components. 
Subgraph G'' has n'' vertices, e'' edges, r'' regions, and 1 component.}
 (n=n'+n'') and (e=e'+e'') and (r=r'+r''-1) and 
 (n''-e''+r'' = 2)
=> {Induction Hypothesis: Applying P(k) on G'}
 (n=n'+n'') and (e=e'+e'') and (r=r'+r''-1) and 
 (n''-e''+r'' = 2) and (n'-e'+r'= k+1)
=> {Arithmetics}
 (n-e+r = k+2)
=> {Definition of P(k+1)}
 P(k+1)

3. Let G be a connected planar graph with n vertices (where n >= 4), e edges 
(where e >= 4), r regions, and no cycle of length 3 or less. Show that 
       	 (e =< 2n - 4)

Sol:
Because e >= 4 and G has no cycle of length 3 or less, the rdegree of each 
region in G is at least 4. 

From the Region Handshake Theorem,
     	 2e >= (sum over r.i, rdeg(r.i))
	    >= 4r
	  e >= 2r			(1)

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

From (1) and (2),
     	  e >= 2*(2-n+e)
	    >= 4 - 2n + 2e
	  e =< 2n - 4
  
4. Let A, B, and C be arbitrary sets, and let ^ denote the set intersection 
operator and * denote the Cartesian product operator. Show, by contradiction, 
that (A^B = {}) => ((A*C)^(B*C) = {}).

Sol:
 (A^B = {}) and (Exist (x,y) in ((A*C)^(B*C)))
=> {Definition of ^}
 (A^B = {}) and ((x,y) in (A*C)) and ((x,y) in (B*C))
=> {Definition of *}
 (A^B = {}) and (x in A) and (y in C) and (x in B) and (y in C)
=> {Definition of ^}
 (A^B = {}) and (x in A^B)
=> F

5. Let A, B, and C be arbitrary sets and let (A sub B) denote the predicate 
(A is a subset of B). Show, by two direct inference proofs, that 
     (C sub A^B) <=> (C sub A) and (C sub B)

Sol:
Proving =>:
 (x in C)
=> {Definition of (C sub A^B)}
 (x in A^B)
=> {Definition of ^}
 (x in A) and (x in B)

Proving <=:
 (x in C)
=> {Definition of (C sub A) and (C sub B)}
 (x in A) and (x in B)
=> {Definition of ^}
 x in (A^B)

6. Let A and B be two arbitrary sets and let ^ denote the set intersection 
operator and - denote the set difference operator. Show, by a two-sided proof, 
that (A-(A-B)) = (A^B)

Sol:
 (x in (A-(A-B)))
<=> {Definition of -}
 (x in A ^ not(A-B))
<=> {Definition of -}
 (x in A ^ not(A ^ not(B))) 
<=> {De-Morgan's law}
 (x in A ^ (not(A) v B))
<=> {Distribution of ^ over v}
 (x in (A ^ not(A)) v (A ^ B))
<=> {A ^ not(A) = {}}
 (x in {} v (A ^ B))
<=> {{} v A = A}
 (x in (A^B))