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Chapter 11: Problem 17

Determine the value(s) of \(n\) for which the complete graph \(K_{n}\) has an Euler circuit. For which \(n\) does \(K_{n}\) have an Euler trail but not an Euler circuit?

Short Answer

Expert verified
A complete graph \(K_{n}\) will have an Euler circuit when \(n\) is odd. However, there isn't a value of \(n\) such that the complete graph \(K_{n}\) would have an Euler trail but not an Euler circuit.

Step by step solution

01

Determine \(n\) for Euler Circuit in Complete Graph \(K_{n}\)

In a complete graph \(K_{n}\), each vertex is connected to \(n-1\) other vertices. That indicates that the degree of each vertex is \(n-1\). Therefore, for \(K_{n}\) to have an Euler circuit, each vertex must have even degree, meaning \(n-1\) must be even. As such, \(n\) should be odd for \(K_{n}\) to have an Euler circuit.
02

Determine \(n\) for Euler Trail But Not Euler Circuit in Complete Graph \(K_{n}\)

For \(K_{n}\) to have an Euler trail but not an Euler circuit, exactly two vertices must have odd degree. However, in a complete graph \(K_{n}\), either all vertices have even degree (when \(n\) is odd) or all vertices have odd degree (when \(n\) is even). As such, there is no value of \(n\) for which \(K_{n}\) would have an Euler trail but not an Euler circuit.

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