[Math] Edge-coloring of the complete graph without any rainbow paths

co.combinatoricsgraph theorygraph-colorings

For a given $2k-1$ edge coloring of the complete graph $K_{2k}$,
say a Hamiltonian path $P$ is a rainbow path if every color appears exactly once in $P$.

My question is

"For each $2k (k \geq 2)$, is there a proper $2k-1$ edge coloring of $K_{2k}$ with no rainbow paths?"

It is easy to see that
every proper edge-coloring of $K_4$ or $K_6$ does not contain any rainbow paths.

For $K_8$, the statement is still true but some proper 7-edge-coloring contains rainbow paths.

When the number of vertices is a power of 2, then the statement is true, but I do not know if it is true for every even number (I don't even know for $2k = 10$).

Here's why the statement is true for $2^k$.

Label the vertices of $K_{2^k}$ by the elements of the group $((Z_2)^k, +)$
and label each edge by the sum(or difference) between two ends. Then the edge-labels give us a proper $2^k-1$ edge coloring, and this coloring does not have any rainbow paths.

Best Answer

This is an open problem, see the intro of this paper for more details: http://www.renyi.hu/~gyarfas/Cikkek/136_orthogonal.pdf

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