Question

B-1 Graph coloring Given an undirected graph G (V. E), a k-coloring of G is a function c : V → {0, 1, . . . ,k-1} such that c(u)≠c(v) for every edge (u, v) ∈ E. In other words, the numbers 0.1,... k-1 represent the k colors, and adjacent vertices different colors. must have

c. Let d be the maximum degree of any vertex in a graph G. Prove that we can color G with d +1 colors.

Answer 1

We use induction on the number of vertices in the graph, which we denote by n. Let P(n) be the proposition that an n-vertex graph with maximum degree at most d is (d + 1)-colorable.

Base case (n = 1):

1. A 1-vertex graph has maximum degree 0 and is 1-colorable, so P (1) is true.

Inductive step:

1. Now assume that P (n) is true, and let G be an (n + 1)-vertex graph with maximum degree at most d.
2. Remove a vertex v (and all edges incident to it), leaving an n-vertex subgraph, H. The maximum degree of H is at most d, and so H is (d + 1)-colorable by our assumption P (n).
3. Now add back vertex v. We can assign v a color (from the set of d + 1 colors) that is different from all its adjacent vertices, since there are at most d vertices adjacent to v and so at least one of the d + 1 colors is still available.
4. Therefore, G is (d + 1)-colorable. This completes the inductive step, and the theorem follows by induction.

Answer 2

Solution:-------------------

We do induction on the number of nodes.
If n ≤ d + 1, this is trivial.
Suppose the result holds for all graphs with ≤ n vertices.
Then given a graph G on n + 1 vertices and maximum degree d, remove some vertex v to obtain G".
G" has n vertices, and maximum degree at most d, and thus has a d + 1 coloring by our hypothesis. Now simply assign v some color that is not used by its neighbors (such a color exists as deg(v) ≤ d).