For what value of n is the graph of Kn Eulerian ?
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Kn has an Eulerian trail (or an open Eulerian trail) if there exists exactly two vertices of odd degree. Since each of the n vertices has degree n − 1, we need n = 2.
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Answer: n=2
Step-by-step explanation:
- For the graph of Kn to be Eulerian there must be two vertices of odd degree.
- So the value of n will always be (n-1).
- The minimum value of n will be 2.
- Hence,value of n will be 2 for the graph to be Eulerian.
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