Eulerian trails in complete graphs

Question

Assume that for some odd $n$ we have a complete graph $K_n$. Is it possible to always find Eluerian trail on $K_n$ such that if we have visited vertices $axb$ then we never before nor after visit $ayb$ nor $bya$ for some $y$. Basically if I split my Eulerian trail into paths of length two I want each two points to be endpoints of a single such path.

For prime $n$ I can find such Eulerian trail by just labelling vertices 1,2,...,n and then doing hamiltonian cycles starting from 1 where first cycle goes 1,2,...,n,1 ; second 1,3,5,7,...,n-1,1 and so on. But I am not sure how to do it for composite $n$ since in this case this isn't an Eulerian trail. Does someone have a reference or an idea how this would work?

Edit: Thanks to the comment of Ilya Bogdanov below I see that my attempt was very obviously incorrect. The question of weather such an eulerian trail exists is still interesting to me,

Answer 1

In Ref 1 we called these things semi-Eulerian circular designs or semi-Eulerian quasigroups. They exist for all odd $n\ge 7$.

I'll state the problem again. Find an Eulerian circuit in $K_n$ which (considered as a circular sequence of $\binom n2$ vertices) has every unordered pair of vertices exactly once at distance two. An example is $$(0, 1, 2, 0, 3, 4, 1, 5, 6, 0, 5, 3, 1, 6, 2, 4, 5, 2, 3, 6, 4).$$

If the sequence is $(a_i)$, define the binary operation $\circ$ by $a_i\circ a_{i+1}=a_{i+2}$ for all $i$ and $x\circ x=x$ for all $x$. Then the multiplication table of $\circ$ is a Latin square, i.e. it defines a quasigroup. However not all quasigroups are suitable as usually the sequence starting $(0,1)$ repeats before exhausting all pairs.

The relevant theorem is Theorem 2.

Incidentally, even though we only give one example for each $n$, computational experiments suggest that there are very many.

Max asked whether it is possible for each unordered pair to also occur exactly once at distance 3. We investigated that but as far as I can determine we never found an example or proved it impossible. We did find the similar example below, though. It is an Eulerian circuit of a directed complete graph with 17 vertices. For each unordered pair $\{x,y\}$, $y$ appears exactly once 3 steps, 2 steps and 1 step before $x$, as well as 1 step, 2 steps and 3 steps after $x$.

$(0, 1, 14, 4, 11, 0, 3, 1, 5, 0, 10, 6, 14, 13, 16, 6, 8, 1, 2, 15, 5,\\  12, 1, 4, 2, 6, 1, 11, 7, 15, 14, 16, 7, 9, 2, 3, 0, 6, 13, 2, 5, 3, 7,\\  2, 12, 8, 0, 15, 16, 8, 10, 3, 4, 1, 7, 14, 3, 6, 4, 8, 3, 13, 9, 1, 0,\\  16, 9, 11, 4, 5, 2, 8, 15, 4, 7, 5, 9, 4, 14, 10, 2, 1, 16, 10, 12, 5,\\  6, 3, 9, 0, 5, 8, 6, 10, 5, 15, 11, 3, 2, 16, 11, 13, 6, 7, 4, 10, 1, 6,\\  9, 7, 11, 6, 0, 12, 4, 3, 16, 12, 14, 7, 8, 5, 11, 2, 7, 10, 8, 12, 7,\\  1, 13, 5, 4, 16, 13, 15, 8, 9, 6, 12, 3, 8, 11, 9, 13, 8, 2, 14, 6, 5,\\  16, 14, 0, 9, 10, 7, 13, 4, 9, 12, 10, 14, 9, 3, 15, 7, 6, 16, 15, 1,\\  10, 11, 8, 14, 5, 10, 13, 11, 15, 10, 4, 0, 8, 7, 16, 0, 2, 11, 12, 9,\\  15, 6, 11, 14, 12, 0, 11, 5, 1, 9, 8, 16, 1, 3, 12, 13, 10, 0, 7, 12,\\  15, 13, 1, 12, 6, 2, 10, 9, 16, 2, 4, 13, 14, 11, 1, 8, 13, 0, 14, 2,\\  13, 7, 3, 11, 10, 16, 3, 5, 14, 15, 12, 2, 9, 14, 1, 15, 3, 14, 8, 4,\\  12, 11, 16, 4, 6, 15, 0, 13, 3, 10, 15, 2, 0, 4, 15, 9, 5, 13, 12, 16, 5, 7)$

1 R. E. L. Aldred, R. A. Bailey, B. D. McKay and I. M. Wanless, Circular designs balanced for neighbours at distances one and two, Biometrika, 101 (2014) 943-956.