Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 6

1. How to prove that a graph G=(V,E) is bipartite? And how to prove that it
is not bipartite? 

Sol:
To prove that G=(V,E) is bipartite, exhibit a partitioning V1 and V2 of V
such that for every edge (u1,u2) in E, u1 is in V1 and u2 is in V2.

To prove that G=(V,E) is not bipartite, exhibit a cycle of odd length in G

2. Which of the following two graphs is bipartite and which of them is not.

(a) G1 = (V1,E1) where 
    V1 = {1,2,3,4}, 
    E1 = {(1,2),(2,3),(2,4),(3,4),(4,1)}

(b) G2 = (V2,E2) where
    V2 = {1,2,3,4,5}
    E2 = {(1,2),(2,3),(2,5),(3,4),(4,1),(5,4)}

Sol:
(a) G1 is not bipartite because it has a cycle (1,2,4,1) of odd length

(b) G2 is bipartitle because its vertices can be partitioned into two 
partitions {1,3,5} and {2,4}

3. Use a direct inference proof to show that 
   (TR is tree) => (TR is bipartite)

Sol:
  (TR is tree)
=> {definition of Tree}
  (TR is connected and has no simple circuits)
=> {Every cycle is a simple circuit}
  (TR has no cycles)
=> {prperty of All with empty domains}
  (All cycles in TR are of even length)
=> {Theorem of Bipartite Graphs}
  (TR is bipartite)

4. Give an example graph that is a bipartite but not a tree.

Sol:
An example graph is G=(V,E) where 
V={1,2,3,4}
E={(1,2),(2,3),(3,4),(4,1)}

G consists of a single cycle of even length. Thus, G is a bipartite and not 
a tree.

5. Let G be a graph that consists of two connected components G1 and G2. Give 
a two-sided inference proof that shows 
  (G is bipartite) <=> (G1 is bipartite) and (G2 is bipartite)

Sol:
  (G is bipartite)
<=> {Theorem of Bipartite Graphs}
  (every cycle in G is of even length)
<=> {definition of components}
  (every cycle in G1 is of even length) and 
  (every cycle in G2 is of even length)
<=> {Theorem of Bipartite Graphs}
  (G1 is bipartite) and (G2 is bipartite)

6. Let G be a graph that consists of k connected components and let these 
components have n.1, n.2, ..., n.k vertices. Show that the number of edges in 
G is at most 
  (sum over i, 1 =< i =< k, n.i(n.i-1)/2)

Sol:
The number of edges in G is maximum iff for every i, 1 =< i =< k, the i-th 
component in G is a complete graph with n.i vertices. Thus, the number of 
edges in G is maximum iff for every i, 1 =< i =< k, the i-th 
component in G has n.i(n.i - 1)/2 edges. Thus, the number of edges in G is at 
most 
  (sum over i, 1 =< i =< k, n.i(n.i-1)/2)

7. Write an algorithm that takes a graph G as input and computes a 2-coloring 
of G such that the computed 2-coloring is valid iff input G is 2-colorable.

Sol:
The required algorithm consists of three steps.

Step 1: Color Black any vertex v in G

Step 2: Color Black every vertex whose shortest distance from v is even

Step 3: Color White every vertex whose shortest distance from v is odd

8. Let G be a graph. Show, by contradiction, that if G has exactly two 
vertices u1 and u2 whose degrees are odd, then G has a path between u1 and 
u2.

Sol:
  (G has exactly two vertices u1 and u2 whose degrees are odd) and
  (G has no path between u1 and u2)
=> {definition of components}
  (G has exactly two vertices u1 and u2 whose degrees are odd) and
  (G has two components G1 and G2 such that u1 is in G1 and u2 is in G2)
=> {only u1 and u2 have odd degrees in G}
  (u1 is the only vertex whose degree is odd in G1)
=> {Corollary 2 of Handshake Theorem}
   F

9. Let G be a graph where max-deg(G) =< 3. Show, by contradiction, that 
every simple circuit in G is a cycle.

Sol:
  (max-deg(G) =< 3) and
  (G has a simple circuit that is not a cycle)
=> {no cycle has repeated vertices}
  (this simple circuit has a repeated vertex v and it is of the form 
  (u, ..., a, v, b, ..., x, v, y, ..., u))
=> {every simple circuit has no repeated edges}
  (in this simple circuit, the four edges (a,v), (v,b), (x,v), (v,y) are 
  all distinct)
=> {definition of deg(v)}
  (deg(v)=4)
=> {definition of max-deg(G)}
  (max-deg(G) >= 4 and max-deg(G) =< 3)
=> F