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Expert-verifiedShow that Cn is chromatically 3-critical whenever n is an odd positive integer, \(n \ge 3\)
Cn is chromatically 3-critical whenever n is an odd positive integer, \(n \ge 3\)
Given Cn
We need to show that Cn is chromatically 3-critical whenever n is an odd positive integer, \(n \ge 3\)
The chromatic number of a graph is the smallest number of colors that can be used in an edge coloring of the graph. The edge chromatic number of a graph is denoted by \(\chi \left( G \right)\)
A connected graph G is called Chromatically k-critical if the chromatic number of G is k , but for every edge of G , the chromatic number of the graph obtained by deleting this edge from G is k -1.
Let Cn represents a cycle of edges of size n
i.e, graph with n vertices \({v_1},{v_2},{v_3},......{v_n}\) and edges \(\left\{ {{v_1},{v_2}} \right\},\left\{ {{v_2},{v_3}} \right\},.........\left\{ {{v_n},{v_1}} \right\}\)
as a circle is not possible with 2 points so n must be 3 or greater than 3
ie \(n \ge 3\)
let n =5 and proceed as follows
Therefore, the chromatic number of C5 is 3
The graph is
Similarly ,we will get of graph of C3 which is chromatically 3- critical as follows.
This is true for every odd number of vertices.
If we delete any edges of C5 and C3, they become chromatically 2- critical
These figures show that Cn is chromatically 3-critical whenever n is an odd positive integer, \(n \ge 3\)
Thus,
Cn is chromatically 3-critical whenever n is an odd positive integer, \(n \ge 3\)
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