Here’s a little doodle on a grid based on a standard six-sided die:

I started by deciding that the pip positions should all connect North to south and East to West. It followed logically that I could have a puzzle where the solver could choose one of two possibilities for each empty cell: connecting North to West and South to East, or North to East and South to West. Because there are 33 non-pip squares, there would therefore be 2^{33}=8,589,934,592 ways to fill the grid. The lines on the outside of the grid show how the squares would connect when folded into a cube.

As an exercise, I found a way to make a single circuit, which is shown above. While that turned out to be about the difficulty of puzzle I can handle in something I am solving by hand, I’m sure there are more interesting specimens to be found.

Unfortunately, because the number of pips is odd, it’s impossible to have two circuits where each go through all of the pips. The circuits would have to cross each other an even number of times. But we could have one of the circuits cross itself once, and then have both circuits go through the remaining 20 pips. (Let’s call that problem number, oh what are we on, **#12**. By the way, the problem numbers are so that I can keep track of solvers of numbered problems and give them the fame they deserve. Nobody has solved any yet. You can be the first!)

Another possibility would be to use three circuits, each crossing itself once, and each visiting 12 pips in addition to the self-crossing. That’s **#13**.

I like big, wide ranging circuits here, so a constraint I like is to have circuits that visit all 6 sides. So bonus points on the preceding problem for having all three circuits do that. And that suggests **#14**: Maximize the number of circuits in a solution where all circuits present visit all 6 sides.

I thought early on that I could take this in the direction of a knot theoretical puzzle, but then one would have to keep track of which thread went under which in the crossings, which seemed like an unnecessary complication. I also think it would be interesting to make a multi-state maze (See Robert Abbott’s site for some good examples) using this template, but I haven’t yet had any good ideas for how that would work. If you have a good idea for a variation on this puzzle, I’d love to hear it. (This is of course true for all of my puzzles.)

For your solving convenience, I have an empty grid image here, and the Inkscape SVG file I used to produce it and the image above is here.

One glance at the diagram suggests #14 has an upper bound of 6 circuits, with each circuit entering and exiting each face once. A little trial and error shows this doesn’t work, so the upper bound is 5 circuits.

I have some preliminary considerations for problem #12. In regards to notation, I’ll use the following diagram, best viewed in a fixed-width font:

Consider the corners DFG and PST, each of which has three segments going through two pips each. With two circuits, one of each of these triplets must be visited twice, parity problem or no.

So according to the problem statement, no other pip is revisited, and also no disjoint triplets can exist in the final layout. Thus, 5f must connect up to right and down to left to keep KNO from being a triplet, which implies in turn that P is not revisited.

I might make more progress later.

I put pre tags in your comment to force it to be fixed width. Actually, according to my problem statement, at least as I meant it, only one self crossing should occur. Since you showed that there must be at least 2, with the parity problem, the lowest possible number of self crossings becomes 3. Ah, well. I liked #13 better anyway.

…Ooohh. Well, three self-crossings is only a minimum when there are two paths, because three paths could cover those corners with no self-crossings. But while I think one total self-crossing is possible with three circuits (one pip is AA, six are AB, six are AC, eight are BC), your #13 is more symmetrical.

The idea of three self crossings with two paths brought me back around to knot theory again, since that could give you one trefoil knot and one unknot.

If we have that, the solved puzzle itself could become a puzzle, to determine which path is the knot, and which the unknot. But then, to make the second puzzle more interesting, there should be enough self-crossings on each path to obfuscate the matter. A solution where crossings alternate between over- and underpasses, and where the two loops aren’t linked, would be especially nice.