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Hex tile design help?

11 replies [Last post]
Joined: 09/12/2013
Hex example

I'm not really even sure what to call the help I need.... I'll do my best to explain.

I want to design hex tiles. Tsuro style, where the tile has connections that lead to other tiles (see attached photo). In the photo, I have each connection named A, B, C, etc.. to L. What I am wondering is, is there a mathematical, or Excel, way of generating all of the possible combinations using these connections? I could go through and draw out each and every one, but I'm worried I will duplicate one and not notice, or skip a lot of them.

In a perfect world, Excel would spit out a list of letters, which would give me a list of all possible connections. For the image attached, it would read "AG BH CF DE IK JL" or something along those lines. Basically telling me which two connections need to be connected.

Any and all help would be appreciated. Thanks!

Joined: 06/14/2014
If I understand what you're

If I understand what you're asking correctly, then the simplest way is just to list them off.

This method assumes that the direction of the connection is not significant, i.e. a-b is the same as b-a. It also assumes that a connection cannot connect to itself.

If you think of each letter as an ascending number, so A =1, B = 2, etc. Then if you follow the formula of Starting with the lowest number, and listing it paired with each subsequent number, i.e.
1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9 etc...
Then you know you've paired 1 with all of the numbers, so all possible connections using 1 has been accounted for. You can then advance to connections starting with 2, however as you've accounted for 1,2 and because 2 cannot connect to itself then you can begin with the next highest number, 3. Producing:
2,3 2,4 2,5 2,6 2,7 2,8 2,9
And the same with connections beginning with 3
3,4 3,5 3,6 3,7 3,8 3,9
Note how the new sets of connections gets shorter for each step up, it will eventually lead to
7,8 7,9
and that'll be all of the possible connections.

With your example it will look like this:

a,b a,c a,d a,e a,f a,g a,h a,i a,j a,k a,l
b,c b,d b,e b,f b,g b,h b,i b,j b,k b,l
c,d c,e c,f c,g c,h c,i c,j c,k c,l
d,e d,f d,g d,h d,i d,j d,k d,l
e,f e,g e,h e,i e,j e,k e,l
f,g f,h f,i f,j f,k f,l
g,h g,i g,j g,k g,l
h,i h,j h,k h,l
i,j i,k i,l
j,k j,l

I'd be inclined to use numbers though, as it's far more obvious (to me at least) if one has been missed out of the sequence.

If you're really set on an excel solution this works:
The duplicate connections are in red, the self connections in orange.

let-off studios
let-off studios's picture
Joined: 02/07/2011
Kinda Like Dice Outcomes, But...

Yeah, I would have simply listed all the outcomes just as if I was doing a probability outcome matrix (or whatever it's called), and - just like Fwyver had done.

However, keep in mind that the player will be able to rotate the tiles, so some of the combinations will be redundant. To put it another way: connection A will not always be in the same place when a tile is laid down on the table. This makes it a bit different from dice, and more like a deck of cards, with certainty - as opposed to probability - being your driving metric.

Joined: 09/12/2013
Thanks for the replies. I

Thanks for the replies. I had done something similar already, but seeing it written out your way definitely helped.

I assume there is no way to "automatically" create all of the possible combinations though. So, I'll just do it manually, and carefully.

Thanks again!

seo's picture
Joined: 07/21/2008
Are you planning on having

Are you planning on having all those possible tiles in your game?

It's 10,395 (11 * 9 * 7 * 5 *3) tiles. That is going to require some huge game box! ;)

Joined: 09/12/2013
That was part of what I

That was part of what I wanted to come up with... the total number of combinations. I figured it was a lot, didn't think it was that many, haha.

I've re-thought the design and decided to go with only 10 designs with 4 color variations of each.

Thanks for the help though... again, very appreciated.

seo's picture
Joined: 07/21/2008
I've kept thinking a bit

I've kept thinking a bit about this problem (just because I enjoy this kind of mental excercise), and it is actually much more complex than that, and a lot less tiles than that, if rotation of the tiles is possible, as already pointed out.

Still, I think it is probably quite a lot of tiles, but not over 10,000.

To explain the issue a bit better, a quick analysis of a much more simpler case: hex tiles with just one anchor per side.

On a first approach (what I did on my first post on this thread), it seems you have 15 possible ways to connect the six anchors as three pairs (the general formula would be, for n anchors, n-1 * n-3 etc. until you reach 1, so in this particular case, 5 * 3 = 15).

hex tile connections

BUT if rotation is allowed, there are only five unique tile designs:

unique tiles

The number of tiles of each type is written inside. As you might notice, it corresponds to how many 60 degree rotations you need to apply for the design to match itself. Why is that? Because in the chart above you have one of each design in one of each possible distinct rotations. I feel that might be an important clue to solve the general problem.

I am not sure how to do this automatically, as one would like for more complex cases, like yours, though. If I find a solution, I'll post it here.

Joined: 04/30/2013
It is possible to count all

It is possible to count all possible combinations, but there is no easy way to generate them

Based on your layout in the original post, and the little of what i remember from probability class, it should be:

6sides X 2connectors per side = 12connectors
12connectors X 11connections = 144 possibilities

Now, that should be deterrent enough, but it doesn't end there. Out of these 144 possibilities, we have to choose 12 to fit into 1 tile. Counting possibilities is done with a function called "Choose" represented as "C" on a scientific calculator, so...

144 C 12 = 103,619,293,800,000,000 combinations

there is still one more step to take out certain duplicates, which should chop the number down by about 100 times, but it is still a staggering number.

Using hexes for connection games is usually a bad idea due to the number of possibilities.

BUT this is only assuming you want to make a tile for every possible combination.


I made some severe calculation mistakes.
1) 12 X 11 = 132
2) every possibility chosen closes another opening, only 6 choices happen, which also takes away another possibility, so 2 possibilities are lost with every choice

the proper calculation would be closer to this:

132 + 130 + 128 + 126 + 124 + 122 = 636

much lesser than my previous calculation, but still a substantial number.

Joined: 09/12/2013
Holy crap seo, thank you!

Holy crap seo, thank you!

To clarify, rotation is definitely allowed. The basic idea here is you'll be working towards getting to certain "special" tiles that allow you to collect resources in order to fulfill goals. Hoping I'll have some time this weekend to create a prototype / post some images to get some feedback on the gameplay.

Thanks guys!

seo's picture
Joined: 07/21/2008
Laperen, I am not sure I


I am not sure I follow your reasoning. What do you mean by

12connectors X 11connections = 144 possibilities

Now, that should be deterrent enough, but it doesn't end there. Out of these 144 possibilities, we have to choose 12 to fit into 1 tile.

As I understand it, it is:

12 connectors:

For the first connector, I have 11 possible connections
then, for the next, I have 9 (12 minus two already used, minus the one I'm trying to connect now)
then, for the next, 7, etc.
And by the time I get to the sixth connection, I have no option but to connect the two remaining connectors.

That led me to my first calculation of 11 * 9 * 7 *5 * 3, which I afterwards explain is way too high in terms of actual tile designs required, given that the same tile, when rotated, produces a different combination of connections.

seo's picture
Joined: 07/21/2008
bmrust, One thing to ponder,


One thing to ponder, I think, is, in my one-connector-per-side example, whether to consider that the same tile design, when rotated, results in a different combination of connections (thus reducing the amount of required tiles to just five) or focus on the actual connections, and have the 15 combinations (i.e. several copies of some tile designs, as indicated by the numbers in the bottom image of my post).

I suspect you might want to go for the second option to have a more balanced result.

If you need the twelve connectors on each tile, I guess you will just have to decide which connections to favor and which ones to make harder to complete, to reduce the amount of tiles to a manageable number, but still keep an eye on making this in a methodical manner, so you don't upset the game balance.

Good luck with the prototyping!

Joined: 07/11/2014
Couting tiles


about the tiles you were asking first, their exact number is 1799. It decomposes into 7 tiles that are unchanged by a rotation of 60°, 10 tiles that are unchanged by a rotation of 120° (and not by 60°), 108 tiles that are unchanged by a half turn (and not by a 60° turn), and 1674 other tiles.

You can find other numbers of tiles and general formulae here.

This doesn't tell you how to list them all, but at least you know that there are a large number of them, and probably too many for a game.

This is unfortunate, because the number 1799 has a nice feature: if you fit hexagonal tiles together so that they take the shape of a larger hexagon, with 25 tiles on its side, then you would need a total of 1801 small hexagons. So, if you forbid the central hexagon, you get a (huge) hexagonal board that looks like Tsuro's: it has the same overall shape as one single tile, it fits every existing tile and leaves one empty space. However, it would take forever to play on such a board ;) (up to 5 hours if each player takes only 10s to play!) So you should follow Seo's idea of selecting certain tiles.

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