$\begingroup$

Question 208 on Project Euler describes walks of "robots" the move in parts of a circular arc:

A [$5$-]robot moves in a series of one-fifth circular arcs (72°), with a free choice of a clockwise or an anticlockwise arc for each step, but no turning on the spot.

Let an $n$-robot be a robot that moves in $1/n$ of a circular arc.

Let an $(i, j)$-path be a path that consists of $i$ clockwise steps, followed by $j$ anticlockwise steps, followed by $i$ clockwise steps, and so on.

The following picture shows $(i,j)$-paths for all $0 < i < j < 5$ of a $5$-robot. In order, these are: a $(1, 2)$-path, $(1, 3)$-path, $(1, 4)$-path, $(2, 3)$-path, $(2, 4)$-path, and a $(3, 4)$-path.

It is clear from the picture that the $(1, 2)$-path, $(2, 3)$-path, and $(3,4)$-path are non-self-intersecting.

If you want to play around for yourself, you can do so on this web app. In particular, you can change the n=5 and w=1,4 in the URL to whatever value of $n>2$ that you want.

Data

Here's some data for $(i,j)$-paths for an $n$-robot with $3 \leq n \leq 12$ and $1 \leq i < j < n$.

Question

In general, is there a combinatorial rule that will determine whether an $(i, j)$-path is non-self-intersecting for an $n$-robot? If so, does it predict how many intersections there will be?

Conjecture

Suppose that $i < j < n$. Then an $(i,j)$-path is non-self-intersecting if and only if $(j-i) \mid n$ and $6j < 5n$.