91Ó°ÊÓ

Before starting, it's important to understand what each term in the exercise means. Here, \( G=(V, E) \) is an n-regular graph. This means that every vertex in the graph \( |V| \) has exactly \( n \) edges connected to it. \( G \) is loop-free means that no vertex has an edge to itself. The size condition \( |V| \geq 2n+2 \) needs to be met. The bar over \( G \), denotes the complement of \( G \) and a Hamilton cycle is a closed loop that visits every vertex in the graph exactly once.

Start by considering the original graph \( G \). Then, conceive its complement graph \( \bar{G} \), where any two vertices will be adjacent in \( \bar{G} \) if and only if they are not adjacent in \( G \). Verify that since \( G \) is \( n \)-regular, \( \bar{G} \) will be \( (|V|-n-1) \)-regular, where \( |V| \) is the number of vertices.

A graph has a Hamilton cycle if for every pair of vertices, the sum of their degrees is at least the total number of vertices in the graph. So, we aim to show that in \( \bar{G} \), the degree sum of any two non-adjacent vertices is at least \( |V| \). All pairs of non-adjacent vertices in \( \bar{G} \) are adjacent in \( G \). Given that \( G \) is an \( n \)-regular graph, the degree of each vertex in \( G \) is exactly \( n \). Consequently, the degree of each vertex in \( \bar{G} \) is \( |V|-n-1 \). Hence, the sum of the degrees of any two non-adjacent vertices in \( \bar{G} \) is \( 2(|V|-n-1) \). Since \( |V|\geq 2n+2 \), we have \( 2(|V|-n-1) \geq |V| \), proving that \( \bar{G} \) has a Hamilton cycle according to Dirac’s Theorem.