How many different Tsuro tiles can exist?

Question

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.

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 ;)

Answer 4

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