Question

Show that Kn has a Hamiltonian circuit whenever n >= 3.

with a proof please

Answer 1

To show that the complete graph Kn has a Hamiltonian circuit whenever n >= 3, we first need to understand what a Hamiltonian circuit is and then proceed with the proof.

Definition: Hamiltonian CircuitA Hamiltonian circuit in a graph is a closed walk that visits every vertex exactly once and returns to the starting vertex.

Proof:We will proceed with a constructive proof by demonstrating a method to construct a Hamiltonian circuit in the complete graph Kn for n >= 3.

Step 1: Setting up the Complete Graph KnThe complete graph Kn has n vertices, where every vertex is connected to every other vertex by an edge. In other words, there are (n-1) edges emanating from each vertex, leading to the other (n-1) vertices.

Step 2: Labeling the VerticesLabel the vertices of Kn as v1, v2, v3, ..., vn.

Step 3: Constructing the Hamiltonian Circuit

Start at any arbitrary vertex, say v1.

Move to an adjacent vertex, v2, as there is an edge connecting v1 and v2.

Continue to the next adjacent vertex, v3, as there is an edge connecting v2 and v3.

Repeat this process, visiting all the vertices in the order v1, v2, v3, ..., vn-1.

At this point, we have visited n-1 vertices, and we haven't visited vn yet. We need to complete the Hamiltonian circuit by returning to v1. Since every vertex in Kn is connected to every other vertex, there is an edge connecting vn and v1. Hence, we can simply move from vn to v1, completing the Hamiltonian circuit.

Step 4: ConclusionIn this construction, we have visited all n vertices of Kn exactly once and returned to the starting vertex, v1. Therefore, we have successfully constructed a Hamiltonian circuit in the complete graph Kn for n >= 3.

As a result, we have proven that the complete graph Kn has a Hamiltonian circuit whenever n >= 3.

Answer 2

K_n has a Hamiltonian circuit when n >= 3:

Hamiltonian circuit:

• It is a circuit in graph where each vertex is presented one and only one time.

• The circuit at the end will back to the starting node or vertex.

Proof:

• Let K_n be the connected graph with n number of vertices .

• Assume that vertices ordered in a graph as v₁, v₂, ... v_n, where v₁ considered as a starting vertex.

• Hamilton circuit is presented as v1 — v2 — .... — vn — v1 such that all the edges are present and it will back to starting vertex.

• Hence, having more numbers of vertices result in more number of edges that forms Hamiltonian circuit.

• In order to form a Hamiltonian circuit in graph which returns back to starting vertex, there must be n >= 3.

• Example: Suppose you have 3 vertices A, B and C then Hamiltonian circuit for K₃ is represented as A — B — C — A.