Question

Discrete Math: Please help with all parts of question 5. I have included problem 3 to help answer part (a) but I only need help with question 5!

5.

(a) (4 points) Prove that a graph is bipartite if and only if there is a 2-coloring (see problem 3) of its vertices. (b) (4 points) Prove that if a graph is a tree with at least two vertices, then there is a 2-coloring of its vertices. (Hint: Here are two strategies: (1) use induction on the number of vertices, using the fact that a tree always has a vertez of degree 1, or (2) make the tree rooted, and consider the levels of each verter.)

3.

Answer 1

K-Colorable Graph : A graph G is said to be K-Colorable if G has K-coloring of a graph.

5) (a) Prove that a graph G is bipartite if and only if there is a 2-coloring of its vertices.

Answer of 5 (a):

Suppose G is 2-Colorable.

First to prove that G is bipartite.

i.e. we have to partition the vertex set $V(G)$ into two sets $V_{1}$ and $V_{2}$ such that $G<V_{1}> = \phi$ and $G<V_{2}> = \phi$

Let $C_{1}$ and $C_{2}$ be two colors used to color graph G.

Let $V_{1}$ be the set of all the vertices of G which receives the color $C_{1}$ and $V_{2}$ be the set of all the vertices of G which receives the color $C_{2}$.

$\therefore$$V_{1}$ and $V_{2}$ is a bi-partition of the vertex set of G also there is no edge between any two vertices of $V_{1}$ and $V_{2}$ since they are of same color.

$\therefore V(G)=V_{1}\cup V_{2}$ and $\therefore V_{1}\cap V_{2}=\phi$ also  $G<V_{1}> = \phi$ and $G<V_{2}> = \phi$

$\Rightarrow$ G is bipartite graph.

Now conversely suppose G is bipartite graph.

Let $V_{1}$ and $V_{2}$ be the vertex set of G such that $V(G)=V_{1}\cup V_{2}$   and $V_{1}\cap V_{2}=\phi$    also  $G<V_{1}> = \phi$ and $G<V_{2}> = \phi$

Now assign the color $C_{1}$ to all the vertices of the set $V_{1}$ and color $C_{2}$ to all the vertices of set $V_{2}$

Thus all the vertices of graph G are colored using only two colors.

$\Rightarrow$ G is 2-colorable.

Thus a graph G is bipartite if and only if there is a 2-coloring of its vertices.

5) (b) Prove that if a graph is a tree with at least two vertices, then there is a 2-coloring of its vertices.

Answer of 5 (b):

I will prove this using the 2nd Hint given in the problem.

We will make the tree as rooted tree.

Let T(V,E) be tree with at least two vertices.

V be the vertex as V = {v₁, v₂, v₃, . . . v_p} and E be the set of edges as E ={e₁, e₂, e₃, . . . e_q}

The image of the rooted tree as follows:

Let v₁ be the root of the tree which is at the level 0. This vertex v₁ can be coloured by using color C₁. Now in the next level, vertices are adjacent to vertex v1. Hence we cannot color them using color C₁. But the vertices at level 1 are not adjacent to themselves. Hence we can color them using a single color say C₂. So suppose all the vertices at level 1 are colored using color C₂.

Similarly the vertices at the level 2 are adjacent to vertices at level 1. Thus they cannot be coloured using colour C₂. But those are not adjacent with the vertices at level 2 and also with the vertices at level 0. Thus the vertices at level 2 can be colored using the color C₁.

Thus alternatively we can color the vertices of a tree by color C₁ or color C₂.

Thus if a graph is a tree with at least two vertices, then there is a 2-coloring of its vertices.

​​​​​​​5) (c) Parts (a) and (b) together imply that every tree is bipartite. Show that the converse is false. i.e. draw a bipartite graph that is not a tree.

Answer of 5 (c):

Yes parts (a) and (b) together imply that every tree is bipartite graph. To show that the converse is false we have to give an example of a bipartite graph which is not a tree.

The example is as follows.

Graph G is bipartite graph because we can bipartite its vertex set V = {v₁, v₂, v₃, v₄} in two sets $V_{1}$ and $V_{2}$ as follows.

$V_{1} =\left \{ v_{1}, v_{2} \right \}$ and $V_{2} =\left \{ v_{3}, v_{4} \right \}$ also $G<V_{1}> = \phi$ and $G<V_{2}> = \phi$

but it is not a tree because it contains a cycle.