teest.md

To prove that every ( k )-regular graph on ( 2k + 1 ) vertices is Hamiltonian, we’ll use some concepts and arguments from graph theory. A ( k )-regular graph means that every vertex has exactly ( k ) edges.

Here's how we can approach the proof step-by-step:

Key Definitions and Theorem

1. ( k )-regular Graph: A graph where every vertex has exactly ( k ) edges.
2. Number of Vertices: ( 2k + 1 ) vertices.
3. Hamiltonian Cycle: A cycle that visits each vertex exactly once and returns to the starting vertex.

Proof Steps:

1. Graph Basics:

   • Let ( G ) be a ( k )-regular graph with ( n = 2k + 1 ) vertices.
   • Since ( G ) is ( k )-regular, every vertex has exactly ( k ) neighbors.

2. Considering the Degrees and Neighbors:

   • Since ( G ) is ( k )-regular, every vertex is connected to ( k ) other vertices.
   • The total number of edges in ( G ) can be computed as ( \frac{k \times (2k + 1)}{2} ) because each edge is counted twice when summing the degrees of all vertices.

3. Using the Theorem by Dirac:

   • Dirac’s theorem states that a graph with ( n ) vertices (where ( n \geq 3 )) is Hamiltonian if every vertex has a degree of at least ( \frac{n}{2} ).
   • In our case, every vertex has degree ( k ). To apply Dirac’s theorem, we need ( k \geq \frac{n}{2} ).

4. Applying Dirac’s Condition:

   • Substitute ( n = 2k + 1 ) into Dirac's condition: ( \frac{n}{2} = \frac{2k + 1}{2} ).
   • Since ( k ) is always an integer, the smallest possible value for ( k ) is ( \frac{2k + 1}{2} - \frac{1}{2} ), which simplifies to ( k \geq \frac{2k}{2} = k ).
   • This simplifies to ( k \geq k ), which is always true.

5. Constructing the Hamiltonian Cycle:

   • Because every vertex has a degree ( k ), and this satisfies Dirac's condition, ( G ) must have a Hamiltonian cycle.
   • Thus, by Dirac’s theorem, the graph ( G ) is Hamiltonian.

Conclusion:

By applying Dirac's theorem and considering that in a ( k )-regular graph on ( 2k + 1 ) vertices every vertex has a degree ( k ), which is sufficient for the graph to be Hamiltonian, we conclude that every ( k )-regular graph on ( 2k + 1 ) vertices is Hamiltonian.

This completes the proof.