Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 7

1. Let G=(V,E) be a connected graph. Show, by contradiction, that 
   (|V|=n and |E|=n-1 and n > 1)
=> (G has at least one leaf)

Sol:

 (|V|=n and |E|=n-1 and n > 1) and (G has no leaves)
=> {G is connected and so has no isolated vertices}
 (|V|=n and |E|=n-1 and n > 1) and (every vertex in G is of degree at least 2)
=> {From Handshake Theorem}
 (|V|=n and |E|=n-1 and n > 1) and (2*|E| >= 2*n)
=> {Arithmetics}
 (|E|=n-1) and (|E| >= n)
=> F

2. Let T be any tree. Show, by contradiction, that every simple path in T has 
no repeated vertices (which means it is a cycle).

Sol:
 (Exist a simple path (u, ..., v) in T has a repeated vertex w)
<=> {Definition of repeated vertices}
 (This simple path in T is of the form (u, ..., w, ..., w, ..., v))
=> {Definition of simple circuit}
 (Tree T has a simple circuit (w, ..., w))
=> {Definition of a Tree}
 (Tree T has a simple circuit) and (Tree T has no simple circuits)
=> F

3. Let T be a tree and u and v be two vertices in T. Show, by contradiction, 
there is a unique simple path connecting u and v.

Sol:
 (Exist two simple paths between u and v in tree T)
=> {From Problem 12 in Exercise 5}
 (Exist a simple circuit in tree T)
=> {Definition of a tree}
 (Exist a simple circuit in tree T) and (T has no simple circuits)
=> F

4. Let T be any tree with at least two vertices. Show, by contradiction, that 
T has a vertex of degree 1.

Sol: 
 (Tree T has no vertex of degree 1)
=> {T is connected and has at least two vertices}
 (Every vertex in T is of degree at least 2)
=> {From Problem 11 in Exercise 5}
 (Tree T has a cycle)
=> {Definition of Tree}
 (Tree T has a cycle) and (Tree T has no cycle)
=> F

5. Let T be a tree with 
       2m vertices of degree 1 each,
       3m vertices of degree 2 each,
       m vertices of degree 3 each.
Compute m and the number of vertices in T and the number of edges in T.

Sol:
Let n be the number of vertices in T and e be the number of edges in T.

From Handshake Theorem, we have
  2e = 2m+6m+3m = 11m        (1)
  
Because T is a tree, we have
  e = n-1 = 2m+3m+m-1 = 6m-1 (2)

From (1) and (2), we have
  2*(6m-1) = 11m
     12m-2 = 11m
         m = 2

  n = 6m = 12
  e = n-1 = 11

6. Let G be a bipartite graph with n vertices and e edges. Show that 
       	    e =< (n^2)/4 

Sol:
Because G is bipartite, its vertices can be partitioned into V1 and V2.
Let the number of vertices in V1 be m and the number of vertices in V2 be n-m.
Therefore, the number e of edges in G is at most m*(n-m). Thus, 
	   e =< m*(n-m)
	     =  m*n - m^2
	     =  -(m^2 - m*n + (n^2)/4) + (n^2)/4
 	     =  -(m - n/2)^2 + (n^2)/4
	     =<  (n^2)/4

7. Let G be a bipartite whose vertices are partitioned in V1 and V2. Show that 
       	 (sum over vertex u in V1, deg(u))
       = (sum over vertex v in V2, deg(v))

Sol:
Let e be the number of edges in G. Each edge in G contributes 1 to the 
(sum over vertex u in V1, deg(u)). Thus, 
     	   (sum over vertex u in V1, deg(u)) = e.
Similarly, (sum over vertex v in V2, deg(v)) = e.

8. Let G=(V,E) be a graph. Show that if |V|=<|E|, then G has a cycle.

Sol:
Execute the following two operations on graph G until neither operation can be 
executed any further:
	 (1) if G has a vertex u of degree 0,
	     then remove u from G
	 (2) if G has a vertex v of degree 1,
	     then remove v and its incident edge from G

The resulting graph G'=(V',E') has the following three properties:
    a) |V'| =< |E'| because each execution of operation (1) or (2) maintains 
       this relation
    b) |V'| > 1 because if |V'|=1 then |E'|=0 violating the relation
       |V'| =< |E'|
    c) Each vertex in G' is of degree at least 2 because if G' has a vertex of 
       degree less than 2 then operation (1) or (2) can still be executed on G'

Because |V'| > 1 and each vertex in G' is of degree at least 2 then, by Problem 
11 in Exercise 5, graph G'=(V',E') has a cycle. Because G' is a subgraph of G 
then graph G has a cycle.