K5 Graph: Hamiltonian Circuits & Analysis

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Discussion Overview

The discussion centers on the presence of Hamiltonian circuits in the complete graph with 5 vertices, denoted as K5. Participants explore the definition of Hamiltonian circuits, the conditions under which a graph may possess them, and the implications of K5's structure in relation to Hamiltonian properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that K5 does not contain Hamiltonian circuits due to a misunderstanding related to the properties of connected graphs with two nodes.
  • Another participant references a source that suggests the number of Hamiltonian circuits in complete graphs can be calculated using the formula (n-1)!, indicating a potential for K5 to have Hamiltonian circuits.
  • A participant clarifies that any complete graph with more than two vertices, including K5, inherently has a Hamiltonian cycle by traversing the vertices in order.
  • Further, a participant questions the sufficiency of having more than two vertices for guaranteeing Hamiltonian circuits, noting that while it is necessary, K5's complete connectivity facilitates the construction of such circuits.

Areas of Agreement / Disagreement

Participants express differing views on the conditions necessary for a graph to contain Hamiltonian circuits. While some assert that K5 does contain Hamiltonian circuits due to its structure, others emphasize that merely having more than two vertices is not a sufficient condition without considering the graph's connectivity.

Contextual Notes

Some statements reflect misunderstandings about graph theory concepts, particularly regarding the implications of vertex count and connectivity on Hamiltonian properties. The discussion does not resolve these misunderstandings or clarify the definitions fully.

Joystar77
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Consider the complete graph with 5 vertices, denoted by K5.

E.) Does K5 contain Hamiltonian circuits? If yes, draw them.

I know that a Hamiltonian circuit is a graph cycle through a graph that visits each node exactly once. However, the trivial graph on a single node is considered to possesses a Hamiltonian cycle, but the connected graph on two nodes is not. A graph possessing a Hamiltonian circuit is said to be a Hamiltonian graph.

Is it correct that K5 doesn't contain Hamiltonian circuits because this is a connected graph on two nodes?
 
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I have never done anything related with Graph Theory but with a google search for your problem I found this pdf, where it states that the number of Hamilton circuits in K(n) is
(n-1)! [factorial]

Hope the link will help you with the knowledge you have on graph theory!

http://www.math.ku.edu/~jmartin/courses/math105-F11/Lectures/chapter6-part2.pdf
 
It does not have two nodes, but five nodes. Any complete graph with more than two vertices has a Hamiltonian cycle: just go around the graph in order.
 
Consider the complete graph with 5 vertices, denoted by K5.

Does K5 contain Hamiltonian circuits? If yes, draw them.

Is it correct to say that K5 does contain Hamiltonian circuits because it has more than two vertices?
 
Joystar1977 said:
Is it correct to say that K5 does contain Hamiltonian circuits because it has more than two vertices?

Well, having more than two vertices is not sufficient, by itself, to ensure that any particular graph contains a Hamiltonian circuit. It is necessary. However, because $K_{5}$, in addition to having more than two vertices, contains an edge from any vertex to any other vertex, it is quite straight-forward to construct a Hamiltonian circuit.
 

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