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The boardgame Tsuro consists of tiles, which each have 8 entry points. Each tile connects each point to exactly one other point. The game manual claims every tile is unique. The game consists of 35 such tiles.

A Tsuro tile

How many unique such tiles could possibly exist?

My reasoning:

  • You start with 8 free points. Choose any point, you now have 7 possibilities to connect.
  • You now have 6 free points. Choose any point, you now have 5 possibilities to connect.
  • You now have 4 free points. Choose any point, you now have 3 possibilities to connect.
  • You now have 2 free points. Connect them (no choice possible).

This would lead to 7*5*3 : 105 possibilities. But I wouldn't know how to eliminate "doubles" caused by rotating a tile. Should I divide by 4, since 4 rotations are possible? That would be 26 tiles... but the game itself contains 35 and they are unique.

How should I reason?

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  • 1
    $\begingroup$ Well, here's why you can't just divide by $4$: Not all the rotations are unique - consider the tile in which each point is connected to the one directly opposite it on the other side - then this tile is rotationally symmetric. Another example in which not all rotations are unique is the bottom left tile in the picture you provided. $\endgroup$ – Peter Woolfitt Jan 2 '15 at 10:31
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You can use Burnside's Lemma. I'll go through the calculation here, though you should read that page for this to make sense.

The set X is all 105 possibilities. The group G that acts on X is $\langle r | r^4 = 1 \rangle$, where r is a 90 degree rotation. Burnside's Lemma is applicable since each tile corresponds exactly to one orbit in X under G. The number of elements of X fixed by each element of G is

  • $1: 105$
  • $r: 5$
  • $r^2: 25$
  • $r^3: 5$

So the number of orbits (tiles) is $\frac{1}{4}(105+5+25+5)=35$. The hardest part is computing the $r^2: 25$ entry; here's how I did it:

  • Case 1: No pair of antipodal points is connected. Choose any point: you have 6 legal possibilities to connect. Whatever you choose, $r^2$ fixes a second symmetrical connection, so you have 4 points left with 2 legal ways to connect them, for a total of 6*2=12 possibilities.
  • Case 2: 2 pairs of antipodal points are connected. There are $\binom{4}{2}=6$ ways to pick the pairs and 2 ways to connect the reamining 4 points, for another 12 possibilities.
  • Case 3: All 4 antipodal pairs are connected. There is only 1 such possibility.
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  • $\begingroup$ You'd probably also find this page interesting. $\endgroup$ – Benjamin Cosman Aug 28 '16 at 21:14
  • $\begingroup$ $<abc>$ looks like this: $<abc>$, and $\langle abc\rangle$ looks like this $\langle abc\rangle$. #FriendsDoNotLetFriendsUseTheWrongAnglyThings $\endgroup$ – Mariano Suárez-Álvarez Aug 28 '16 at 21:19
  • $\begingroup$ Interesting approach, thank you! $\endgroup$ – Konerak Aug 29 '16 at 7:07
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One can reduce further the $105$ possible tiles, down to the minimum possible of $35$ topologically unique tiles via $mod\, 8$ arithmetic, $1+7=(8)=0$.

Modulo $8$ arithmetic (each tile has $8$ ports numbered $0123457$) allows $CW$ and $CCW 2$-dimensional rotations (can not flip a tile, as a flip would represent a $3D$ rotation). To keep this simple, there are no negative integers here, $0-1=7$ is being interpreted as $7+1=0$.

A $45^{\circ}$ rotation is represented by adding/subtracting 1 to/from the port #.

Rotations of $90^{\circ}, 180^{\circ}, 270^{\circ}$ are represented by adding/subtracting $2, 4, 6$ respectively. I just added $2, 4, 6$ used only addition.

Represent a pair of two connected ports of a tile by a $2$-digit number. For example $15$ means port#1 is connected to port#5, i.e. not decimal fifteen.

Represent each tile via a quad set of number pairs, non repeating digits,for example $\{01, 23, 45, 67\}$.

Note $\{01, 14, 26, 57\}$ is not a valid tile as digit "1" -port#1- appears twice.

Example of $180^{\circ}$ rotation:

Tile $\{04, 12, 36, 57\}$ is rotated $180^{\circ}$ by adding (modulo 8) $+4$ to each of its digits.

$$0+4=4$$

$$4+4=(8)=0$$

$$1+4=5$$

$$2+4=6$$

$$3+4=7$$

$$6+4=(10)=2$$

$$5+4=(9)=1$$

$$7+4=(11)=3$$

Thus rotated tile becomes $\{40, 56, 72, 13\}$ and will order/describe/ its ports from low-to-high (as each numbers-pair is commutative, $40=04$ and $72=27$).

Therefore this $180^{\circ}$ rotated tile becomes $\{04, 13, 27, 56\}$.

Rotate all $105$ tiles by $0^{\circ}, 90^{\circ}, 180^{\circ}, 270d^{\circ}$, (find & write all quad sets).

Use Excel, reorder the port# numbers from low-to-high, and eliminate all duplicate tiles.

End result = $35$ unique quad sets representing $35$ unique tiles.

May follow my analysis summary at:

https://boardgamegeek.com/thread/1398585/tsuro-35-unique-tiles-math-proof-summary-and-tsuro

A fun game. The tile analysis & methodology was ++fun to perform.

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There are 5 ways to make a tile that is invariant under $90^\circ$ rotations: Just connect one pint to any other point except its rotational predecessor or successor, then the rest is fixed. This would correct your count to $\frac{105-5}4+5=29$, which is more, but still too few. You should also count, how many tiles are possible with a $180^\circ$ symmetry. Of course, the $90^\circ$ symmetric tiles are among them, but there are also some "new" such tiles. If there are $k$ such tiles, then the total count gets corrected to $\frac{105-5-k}{4}+5+\frac k2$, so you can make an educated guess what $k$ is ;)

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Check this link! Tsuro tiles are addressed by the group theory via the Burnside lemma : https://oeis.org/A132102

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  • $\begingroup$ Which, for n=4, also states 35, indeed! $\endgroup$ – Konerak Nov 21 '16 at 15:08

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