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I am teaching this semester graph theory for undergraduate students. Now, I am discussing with them about Hamilton Paths in finite graphs. Last time we meet, I presented the following theorem:

Theorem. For $n\geq 3$ the complete graph $K_n$ is decomposable into edge disjoint Hamilton cycles iff n is odd. For $n\geq 2$ the complete graph $K_n$ is decomposable into edge disjoint Hamiltonian paths iff $n$ is even.

During the class I noted that my argument to prove this theorem was not complete. I started proving that the second statement implies the first one, which is ok. But I had not a correct argument to show that there exist an edge disjoint decomposition of $K_n$ in $n/2$ Hamilton paths if $n$ is even.

Can we explicitly construct such decomposition or just present an existence argument ?

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up vote 8 down vote accepted

We can explicitly construct such a decomposition.

Label the vertices of the graph with $\{0,1,...,n-1\}$, take the first path to be $0, n-1, 1, n-2, 2,... ,n/2$ and generate the other paths by addition modulo $n$ (the $n$ paths come in pairs in which one is the reverse of the other).

More generally, a symmetric sequencing in a group with a single involution is sufficient to construct the decomposition.

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Thank you Matt. – Leandro Apr 6 '11 at 23:51

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