Hi, 


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 ? 



Hamilton Paths in $K_{2n}$
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. 


