Question

To prove if G is a chromatically k-critical graph, then the degree of every vertex of G is at least k-1.

Short answer

K-tuple coloring of graphs.

Step-by-step solution

Given information

K-tuple coloring of graphs.

Definition and Concept

A k-tuple coloring of a graph G is the assignment of a set of k different colors to each of the graph's vertices, with no two adjacent vertices having the same color. The lowest positive integer such that G has a k-tuple coloring with n colors is denoted by (G). For instance,\({X_2}\)(C4) 4 To see this, note that we may assign two colors to each vertex of \({C_4}\)using only four colors, as shown, ensuring that no two neighboring vertices are allocated the same color. Furthermore, four colors are required because the vertices\({V_1}\)and\({V_2}\)each require two colors, and a single color cannot be assigned to both\({V_1}\)and\({V_2}\)

Solution

Let G be a chromatically k-critical graph with a vertex v of degree-2 or fewer. One of the edges adjacent to vis has now been eliminated. The resulting graph can be colored using &- 1 colors, as defined by the k-critical definition. Then, with the exception of v, repair the missing edge and use this coloring for all Vertices. We've completed proper coloring of the smaller graph; as a result, adjacent vertices must be colored differently. Also, because G contains at most k-2, we can simply A-1 color it, i.e. the final color is not next to. We have colored Gink-1, which contradicts the definition that G has chromatic number k. As a result, our hypothesis was incorrect. As a result, each vertex of G has a degree of at least & - I.

The final conclusion is that the degree of every vertex of G is at least k-1.