## Tantrix from combinatorial viewpoint

This question is about the popular logic game called Tantrix. I would like to collect combinatorial theorems about it, eg. necessary conditions for making a cycle of one color from a given set of tiles that passes through all the tiles and the union of tiles has no holes in it. On the web I could only find theorems about the complexity of the game which is a completely different question. To show what kind of theorems I want, here are three easy observations.

Theorem Trivial. If there is a red cycle using all the tiles, then red must appear on all the tiles.

Theorem Crossing Parity. If there is a red cycle using all the tiles, then the number of red-blue crossings must be even.

Theorem Winding. Count straight red tiles 0 (can be denoted by I), big turns (which are almost straight, can be denoted by L) as 1 and small turns (when red touches adjacent sides, can be denoted by V) as 2. If there is a red cycle using all the tiles, then it must be possible to assign a sign to each tile such that the sum is 6.

I would be also interested in configurations that satisfy these theorems but it is still impossible to make a cycle, like 3 Vs and a non-zero number of Is.

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 I assume we're talking puzzles using general multi-subsets of the sixteen possible tiles, and not just the particular set of 10 tiles given, right? If so, from the rules linked above (as well as the ones on my copy of the game), the crossing parity theorem is false, because there's nothing stopping the configuration from having holes. For instance, you can make a teardrop-shaped red loop with tiles 2, 4, 5, 6, 8, 9, and 10 which is has three red-blue crossings. – Jonah Ostroff Jan 5 2010 at 16:03 That said, some versions of the rules I can find online do say "no holes", and with that stipulation the theorem is true. – Jonah Ostroff Jan 5 2010 at 16:04 Yes, no holes, I added it. But what 16 and 10 tiles? There are 56 tiles in the game. – domotorp Jan 5 2010 at 18:06 Sorry, I think we have different versions; mine is the solitaire puzzle with three colors. There are sixteen possible pieces with three colors (up to rotation, but not caring about under/over crossings), but the game only comes with ten. With four colors, I suppose there would be 64 possibilities. Judging by the picture in your profile, it looks like the eight that are missing are the ones where all three strands are straight and cross in the middle. That's good, because it means we can actually look at these things from a knot theory perspective. Not my area of expertise, though. – Jonah Ostroff Jan 5 2010 at 20:09 Yeah, exactly tose are the ones in the standard tantrix set. – domotorp Jan 6 2010 at 6:00

## 2 Answers

Hi. I am trying to figure out the combination of pieces needs to form a loop so that solutions are easier to conceptualize before trying to play with the pieces. The tight turns (V) are 120 degrees and the less tight turns (L) are 60 degrees and a loop must contain a total of 360 degrees so only specific i.e. 3V facing inwards (plus any even number of Vs or Ls used in opposite orientations) 2V plus 2L facing inwards (plus any even number of Vs or Ls used in opposite orientations) 1V plus 4L facing inwards (plus any even number of Vs or Ls used in opposite orientations) There must be similar logic for the lengths in each direction that would indicate the need oriention of the straight pieces and ratio of Vs and Ls on all sides Any thoughts? Moksha

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You might wish to look at some papers by Paul Zinn-Justin. For example "Littlewood--Richardson coefficients and integrable tilings" defines a model of random tilings which count the Littlewood-Richardson coefficients. As early as 2001, puzzles were being used to compute the cohomology of complex Grassmanians. See "The honeycomb model of GL(n) tensor products II: Puzzles determine facets of the Littlewood-Richardson cone" by Allen Knutson and Terry Tao.

Another might try looking in relation to the Temperley Lieb-Algebra or other Planar Algebras.

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