I'm new here, so please tell me if I'm doing anything wrong.
Here is a puzzle that I came up with, which I believe falls someone in the realm of graph theory or topology (I could be very wrong, please correct me!):
Given a standard six-sided die (with opposite faces adding to seven), is it possible to draw a single non-crossing closed loop on those faces such that the loop crosses through face $n$ exactly $n$ times?
The answer to this question is yes: I found a solution with code and a friend.
However, if I add one more restriction ...
The loop can't return to the face it was at one face before (eg. the loop can't go through face 1 and then face 5 and then immediately back to face 1),
... it becomes very difficult to find a solution. So difficult, in fact,
I started wondering if it is possible at all. However, I can't prove one way or the other whether there is a solution or not.
So:
Is this version of my problem impossible? Or is there a solution?
Also, if you know, what is this type of problem called? It certainly seems related to graph theory and topology, but I don't know if there are any keywords that describe this problem.
Thanks!
Added by @Blue. Here's a different view of the sample path. (Circled nodes indicate turn-arounds that the revised rules ask to avoid.)
Here's the path from @DanielMathias' comment to @JimFerry's answer, having just one turn-around:
I did some computations with strings such as $126453654623654635645$ that represent a cycle of visits to die faces, without regard to whether the path crosses itself. The number of such cycles without the doubling-back pattern "aba" is $43399$. Each cycle may be represented as a string that begins with $1$ and continues with the smaller of its two adjacent faces.
We whittle down these $43399$ cycles with planarity restrictions.
First we filter out cycles that connect both pairs of opposite edges of any face: e.g., $265$ and $364$. Eliminating strings with these crossing patterns leaves $6565$.
A more powerful pattern to exclude is "abcab". With this spiral pattern, planarity + no-doubling-back forces "c" to appear again on one end or the other. The spiral then continues forever and the loop cannot close. By itself this pattern reduces the $43399$ cycles to $1228$.
One might hope that, together, the two filters eliminate all cases. Not quite. There are $11$ surviving cycles: $$ 126453654623654635645\\ 126453654635642635645\\ 126456326453654635645\\ 126456326456354635645\\ 126456326456354653645\\ 126456354623654635645\\ 126456354635642635645\\ 126456354653624635645\\ 126456354653624653645\\ 126456354653642635645\\ 126456354653645632645 $$ These can be ruled out by inspection. They all contain the substring $64512645$, so the $645$ is "stacked" on top of itself. Regardless of which one is "on top," the pattern cannot be continued as required in both directions according to one of the $11$ strings above without violating planarity.
Because computations like this are opaque (and, umm, dicey?), I'd appreciate it if someone could reproduce the numbers of remaining cycles after the various filters. Even better, of course, would be a direct argument, or at least a set of filters that eliminate all cases instead of leaving a small number of them.