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So i have a problem regarding proving this statements:

1) for any $n\in\mathbb{N}$ tripartite graph $K_{n,2n,3n}$ is Hamiltonian.

2) tripartite graph $K_{n,2n,3n+1}$ is not Hamiltonian

Any help would be really helpful. I am having trouble even with understanding the concept behind tripartite graph, since i only had experience with birtatite until this very moment. Thank you in advance.

1 Answers 1

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A complete tripartite graph is like a fully connected 3 layered network constrained such that no connections exists in the nodes of same layer. Take $n=1$ as an example and $1,2,3$ as number of nodes in consecutive layers. $1$ is connected to all nodes in layer $2$ and layer $3$ and this goes same to all layers. So a total of $11n^2$ edges exist in a $K_{n,2n,3n}$.

Use Ore's or Dirac theorem for proving whether a graph can be Hamiltonian or not. By Ore's theorem, sum of the degrees of any two non-adjacent vertices should be $>=n$ ($n$ is the number of vertices). By this theorem, in $K_{n,2n,3n}$, take any two nodes in the same layer. It has at least 6n degrees. So $K_{n,2n,3n}$ is Hamiltonian.

But in a $K_{n,2n,3n+1}$ graph, take two nodes from $3n+1$ layer and you got $6n$ degrees as sum which is less than total number of vertices $6n+1$.

Thanks @Joffan for correcting.

Hope this helps.

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    Not quite. A complete tripartite graph has an edge from every vertex to every other vertex **except** those in its own part - including edges between the top and bottom layers in your description. So there are $11n^2$ edges in $K_{n,2n,3n}$ – 2017-01-13
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    So it is not like a Feed forward network? If I not I am sorry, I will edit it properly. – 2017-01-13
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    A feed forward network as you describe is effectively bipartite. The top and bottom layers are in one part, and the middle layer is the other part. – 2017-01-13
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    Thanks @Joffan. Got it after looking at an image. – 2017-01-13
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    In your second and third paragraphs, I think you ought to emphasize that the condition of Ore's theorem is sufficient for a graph to be Hamiltonian, but not necessary. Non-existence of a Hamiltonian cycle in $K_{n,2n,3n+1}$ will have to be proved some other way. – 2017-01-14