Hex on a Torus

I came up with this today after two years of thinking on and off about the possibility of a hex/y connection-style game where the potential value of the first move was the same no matter where it was played (doing away with the pie rule). I originally thought about playing on the surface of polyhedra, but the platonic solids (the only ones that satisfied my "first move is equal" criteria) didn't seem to work with anything. Then I looked at spherical geometries, with some kind of great arc/circle being the goal, but those and other possible winning conditions I thought of seemed forced. However, inspiration came to me today after playing the classic arcade game Asteroids: In the game, when you move off one edge of the screen, you reappear on the other; it wraps around. What if Hex was played on that kind of field? A bit of research later, I found out that that kind of geometry is called a torus. So here is what I came up with.

First, an 8x8 example board, with the pieces played on the vertices:

http://img819.imageshack.us/img819/8483/torus.png

Notice that the "9th" column and row are actually the 1st column and row! Like Asteroids, the playingfield "wraps around". When a piece is played on the edge, a "hologram" of it is placed on the opposite side, and when a piece is played on a corner, the "holograms" are places on the other three.

Now, the rule difference: instead of connecting one side to another, the goal is to make a connection "ring" around the board. One player plays vertical, and the other horizontal. The only catch (which I will explain in detail in the next paragraph) is that if the vertical player's path crosses the "wrap-around" boundary on the left or right "n" times, it must cross back over "n" times for the ring to count, and the same goes for the horizontal player and the top or bottom boundary. Like in Hex, a board can only have one winner. Some examples of a win for Red (the vertical player):

http://img96.imageshack.us/img96/9692/win1sf.png

(Completely straight ring)

http://img204.imageshack.us/img204/8247/win2g.png

(Wavy ring)

http://img806.imageshack.us/img806/4804/win3.png

(Crosses boundary once, and crosses back once)

However, here is the case that is NOT a win, and which allows for ties:

http://img138.imageshack.us/img138/6418/notawin.png

Topologically, this is not a ring around the torus, but actually a "curly-q". Visualize it like this: on a torus (donut) shape, you can draw a circle that goes through the "loop" (a win for vertical), or a circle that goes around the whole thing, like if you drew along the outside only or inside only, or in-between only (a win for blue). However, you can also have one that wraps through the "loop" as it is making its way around. That is what happens in the third image. Notice that unlike the circles, both players can have multiple connected "curly-qs" at the same time, which leads to the possibility (though unlikely except upon perfect play) of a tie.

Does anyone understand what I'm talking about here, and if so, does anyone know if this game would actually be playable, without say, a game-breaking win/tie-forcing strategy? Thanks for any responses!

EDIT: In the paragraph about the "curly-q", I meant to refer to the fourth, or last image, not the third.

I found a neat graphic that sort of displays what a win for vertical would look like vs. horizontal on the wikipedia page for "torus":

http://en.wikipedia.org/wiki/File:Torus_cycles.png

I was thinking about this to but I didn't realize about the issue with "curly-qs" I thought that there would be always a win for one player. You propose that such a curly q would be a draw. It would be fine If it really can occur rarely in game play. But I think we can distinguish right q and left q On your Image we see the left q it is the same as going from the top right corner to the bottom left. The right q would be going from the top left corner to the bottom right. They cant occur simultaneously so we could make a rule that the right q is a win for first player and the left q is a win for second player.

I feel that with a fully filed board under thous rules there always have to be a winner. But I am not certain yet.

I thought about that a bit, but quickly realized that it would not work, as the curly-qs will always slant the same direction using Hexagons on a torus. However, I just thought of something that would make your solution work: instead of playing on a torus, I'm going to see how a Klein bottle works out. The principle is the same, only instead of emerging directly opposite from your position, you would emerge as if the board was rotated 180 degrees.

Actually I'm not sure it will work, but I am testing it right now.

I've done it! Behold, the first hexagon connection game (to my knowledge) where:

1. All spaces are "equal" at the start of play,

2. It is impossible for there to be two winners, regardless of how the board fills up, and

3. No ties are possible.

http://img20.imageshack.us/img20/3313/92694394.png

It turns out, a "square" torus wasn't what was needed, but a hexagonal one.

Here is how the wraparound works:

http://img8.imageshack.us/img8/6707/24303612.png

"Pieces" of the same color are actually the same exact piece.

To win, one must make a loop through the barrier of your color. Red's path must travel NW/SE and Blue's NE/SW for it to count as a loop. The paths can wrap through the top and bottom of the board as much as needed.

I am extremely excited to try this game out! It is perfect for IgGameCenter's sandbox. I just really hope there is no "game-breaking" tactic of play, because this is one of the most elegant things I have ever seen.

*sigh*

Never mind my last post. I just realized something...

http://img25.imageshack.us/img25/8304/faileh.png

Back to the drawing board :(

The game works well on your previous board (squares witch one diagonal)

First player(red) wins when he will make a "vertical circle" or a "right-q"

and second(blue) wins when he will make a "horizontal circle" or a "left-q"

There is lack of symetry between the right-q and the left-q. The left q is shorter require only n pices on nxn board and the right-q require 2n of them (exept 2x2 board where 3 is the number). Look on the simplest board 2x2 there are only 3 unequivalent endgame positions:

1.Two vertical circles of different collour

2.Two horizontal circles of different collour

3.Two left-qs of different collour

With perfect play second player wins with the 3-rd ending position

I wonder how it will look on different boards, on odd size boards first player have one more pice at the end of the game I wonder if this allows him to reduce the advantage of the easier winning condition

This is an interesting discussion.

I have some objections regarding the topology of the board. To me it is not a torus. The connectedness of the edges is like P1^3 (P1=projective line), but the inside is something like R^2 (R=real line).

I suggest to use a usual hex board. There you can identity the first and the last row resp. column for the red and the blue player. This leads to the toric geometry you wanted.

When you are now dissatisfied with the rhombic form of the board, you can cut off the sharp corners and attach them at the obtuse corners. This is permitted, because the board is a torus.

Copy the following text to an editor with fixed witdh text to see what I mean...

A1 A2 A3 A4 A5 A1 . A2 A3 A4 A5 A1 A2

B1 B2 B3 B4 B5 B1 . B2 B3 B4 B5 B1 B2

C1 C2 C3 C4 C5 C1 . C1 C2 C3 C4 C5 C1

D1 D2 D3 D4 D5 D1 . D1 D2 D3 D4 D5 D1

E1 E2 E3 E4 E5 E1 . E5 E1 E2 E3 E4 E5

A1 A2 A3 A4 A5 F1 . A5 A1 A2 A3 A4 A5

Sorry, forgot that double spaces are ignored in HTML...

.A1..A2..A3..A4..A5..A1.............:...A2..A3..A4..A5..A1..A2

...B1..B2..B3..B4..B5..B1...........:.....B2..B3..B4..B5..B1..B2

.....C1..C2..C3..C4..C5..C1.........:...C1..C2..C3..C4..C5..C1

.......D1..D2..D3..D4..D5..D1.......:.....D1..D2..D3..D4..D5..D1

.........E1..E2..E3..E4..E5..E1.....:...E5..E1..E2..E3..E4..E5

...........A1..A2..A3..A4..A5..F1...:.....A5..A1..A2..A3..A4..A5

Your way of describing the torus board with hexagonal fields is the same. You identify the same fields Aj with Fj and X1 with X6 where j ranges over 1,2,3,4,5,6 and X over A,B,C,D,E,F

The square board is the same as the rombus one, the vertices are identified with hexagonal fields (you place your stones there) and each vertex is conected with 6 others by an edge, which is identified with one side of a hexagon.

In hex the first player have a winning strategy on the nxn board. I don't remember if the second player have a winning strategy on (n+1)xn board or if this is an open question? And how the sytuation would look in torus hex (the version without a draw), what would be the board sizes with winning strategy for red?

I don't have any reference at hand, but if I remember well, there is a constructive winning strategy for the player with long edges on (n+k)xn. Philip's handicap hex is based on this property.

http://img138.imageshack.us/img138/6418/notawin.png this is in my eyes a win for whichever player reached this formation: It is a combination of a circle that wins for player 1 and also a circle that wins for player 2 (the 2 circles seen in the wiki pic).

http://img806.imageshack.us/img806/4804/win3.png this would be a win for the player playing 'up-down'

I also don't get how you unfold the thorus to get this hexanol 2D representation: http://img25.imageshack.us/img25/8304/faileh.png

Just woke up and I think I found a solution. Same board as my last post:

http://img20.imageshack.us/img20/3313/92694394.png

Only instead of being directly mapped to the "mirror" opposite hex, the edges are mapped to the "rotated" opposite hex:

http://img15.imageshack.us/img15/5348/hextor.png

So A1=O15, B1=N15 ... H1=H15, I2=G14

It is not a torus anymore, but if there is nothing I overlooked, it is impossible to tie and all initial spaces are equal.

Here are examples:

Win for Red:

http://img140.imageshack.us/img140/270/redwin1.png

Another win for Red:

http://img828.imageshack.us/img828/4386/redwin2.png

NOT a win for Red:

http://img253.imageshack.us/img253/5543/notredwin.png

I think this works!

Actually that doesn't work either. However, I think the Klein Bottle on the nxn grid may have potential:

http://en.wikipedia.org/wiki/File:Klein_Bottle_Folding_1.svg

@maraca

As Mjungo said the q formations are not a circles on torus (You cant continuisly trasform it to a real circle in the R^3 space where the torus is embedded).

So in the second variant the rules would be:

2)First player(red) wins when he will make a "vertical circle" or a "left\right-q"

and second(blue) wins when he will make a "horizontal circle" or a "left\right-q"

instad of:

1)First player(red) wins when he will make a "vertical circle" or a "right-q"

and second(blue) wins when he will make a "horizontal circle" or a "left-q"

The second variant have simpler rules, there is clear first player advantage on a square board (winning strategy by the same reasoning as in hex i think). The difference would be that if the board is completely filled both players can have a winning formations which is impossible in hex or in the 1) variant of rules.

And the swap rule if implemented would have to apply at lest after 3-rd move since on torus every first move is essentially the same.

Maybe I did not make my point clear enough. The board layout of MJongo (see http://img140.imageshack.us/img140/270/redwin1.png) is not a torus in the desired sense.

Just look at the edges -- in a toric design it should be possible to shift them arbirarily (because they are only aids for the players). This means that you should be able to move both red edges one row to the upper left side without changing the board. This is not possible. One other way to see this is by displaying more than one hexagonal elementary cell of the board. Here you can clearly see that not all fields are equivalent. Some are better for red, some are better for blue.

The original graphical representation (http://img806.imageshack.us/img806/4804/win3.png) is correct. Here you can shift the edges and all fields are essentially the same.

Can you spot the difference?

On the hexagonal board you have three pairs of edges, leading to three different fundamental directions. On the rhombic board you have two pairs of edges. This is probably what you want. Both boards are regular tilings of the plane, but only the rhombic one has infinite edges (when you put more than one copies of it together). The end of an edge does not continue at its starting point (but you might need more than one board to see this).

I'm sorry I didn't realize that you ware talking about the hexagonal version of the board. I agree that this is not the correct implementation of a torus board.

I still don't see it, but if it's only a problem of transformation, why don't you just ignore this and say the one who reaches a curly-q first wins?

Another thought: if all spots are equal, how you wanna equalize the first move, I guess the pie rule is useless there... You could use a similar system than the one proposed for gomoku in the forum somewhere.

The curly-q it self is not enough. If both players would form the same type of circles there is no way to form a curly-q anymore.

So I think The following rule would be ok.

First player(red) wins when he will make a "vertical circle" or a "left\right-q" (or simply curly-q since there is no point to distinguish them)

and second(blue) wins when he will make a "horizontal circle" or a "left\right-q"

I must say that I like this version better. There is no problem in choosing the relatively fair shape of the board which would be a square I think. In my first version as I pointed out there was a lack of symmetry in the winning conditions and the square board might be a big advantage for the second player on big boards.

The classic pie rule (possible to swap just after first move) would not work since the swap will be always a good thing. I have no idea what the best variation of it would be.

1. Possible to swap after 3-rd move

2. Possible to swap after 5-th move

3. First player places 3 red and 2 blue stones and second decides which he wonts to play.

Or maybe completely reduce the obvious first player advantage and play the game similarly to connect6 first player play one stone in the center and then both players take turns placing two of their own stones. It would be a completely different game then hex maybe it would be nice to try the classic hex under this rule. But there is no place on this discussion on this treed I think.

kfiecio, this is actually what I meant in both points ;-) so I agree with you. The gomoku rule I was talking about was something like 1st player makes 3 moves, then the second player can take the color with more stones on the board or play another peg, player 1 can then swap or make a move and the game continues normally.

Btw. on iggamecenter are games on the hex board with more than one move per player, the idea is nice, but I have to say that it changes the game completely (the knight moves became more or less useless, they can be broken in 1 move).

See Bill Taylor's posts to rec.games.abstract. The exactly and always one winner property can be maintained simply by making corkscrews of different chirality winning positions, IIRC.

I just read some posts about the torus-hex rec.games.abstract. including Bill Taylor's posts. The conclusion is basically the same as I mentioned earlier except that he was certain that the first set of rules give second player an easy win on big enough boards and he also gave an example of that game on 2x2 board.

The second version of rules also appear In some other topic But Bill didn't like It because like he said it was more racing game rather then a connection game.

You said ¨The exactly and always one winner property can be maintained simply by making corkscrews of different chirality winning positions¨

Maybe I am wrong once again assuming that, but it sounds like you are saying that the different corkscrews condition is sufficient. But If both players would form the same type of circles there is no way to form a corkscrew anymore. In some of the Bills posts he also seems to omit the horizontal/vertical circles but In the 2x2 game explanation he uses the terms vertical/horizontal player.

So at this point we have two games

The first one where if one player will make a winning connection the second cant do that anymore. The down side of this game is a big advantage for the second player so a special Equalization rule would be needed.

The second one where players share common goal which make this a racing game and a simple pie rule has to be replaced by more complex one.

About the 2 moves rule. I found on igameceter rules of Master Y. which rules was presented in 1950. I thought that the 2 move rule was rather new thing that come with c6. But now I think that I sow a chess variant with this rule also.

I am a big fan of this rule. It create a great balanced game with simple rules like c6 from an unbalanced with a rather complex rule set game like gomoku.

The other nice thing is that this rule make games harder for computers to master. Like in hex without a symmetry there are n^2-1 possible second moves with 2 move rule there would be (n^2-1)(n^2-2)/2. Breakthrough with 2 move rule could be nice game (I am not saying that it isn't now. I like Breakthrough).

The other thing is that this additional rule create a completely new game, where the developed strategy for the original one most likely fails like maraca pointed out about hex.

I am completely out of subject now but we might also consider 2n move rule:

we take a base game except that first player in his first turn make n moves then both make 2n moves in their turns. Sometimes you have to change the winning conditions also like in c6. I thought about connect8 (gomoku with 4 move rule) But there was not much interest on gomoku forum and I am not certain if the game would have a lot of draws or not.

I apologize for going out of the subject but I think that there is not much to say abut it anymore. We have two concepts for torus hex both have some weak sides.

Forgive me for not reading the whole post in detail, but what's wrong with one side working to complete a vertical ring while the other side works to complete a horizontal ring on the normal rhombus made into a torus by wrapping in both dimensions?

You have tie situations. There will always be the problem of defining who will be the winner in a case such as alternating black and white rows of stones in the diagonal direction.

@Dvd Avins: Yes, concluding from the discussion we had above, I say this is possible, except that the curly-q is no circle, but if you say this is a winning configuration too, then there should be no problems.

@Jonathan Rydh: I think this is wrong, because alternating diagonal pieces would give you automatically horizontal or vertical rings... It is also not possible to have several curly-qs if the first one wins.

N-move rule:

I have mixed feelings about this rule, it has advantages and disadvantages like everything... With N bigger it becomes harder to look moves ahead. I even think, if you can look 10 moves ahead with N=1, then with N=2 you often can only look 4 or less moves ahead instead of 5. So the bigger N the more the game becomes a puzzle, where you almost can't look any moves ahead and just try to play the best move available for each situation... so with increasing Ns the focus goes from planning (strategy) to 'balance'. If this is a good thing depends on the game and personal taste.

Yes if you introduce sloppy "first to reach goal"-kind of rules you can avoid ties. But it gives a whole other kind of game where you race to the goal like Havannah. In my opinion not very interesting games.

Unfortunately I don't think it's possible to have the property that every full board has a unique single winner and still have a fully symmetric board where every hex is equivalent (like the torus). But I would be happy to be proven wrong.

In my opinion it is not sloppy. If you forget the whole Torus thing and say you play on a wrap around board, one tries to make a ring horizontal and the other vertical... this sounds like a normal connection game to me. That's why I didn't understand what's the problem about the curly-q.

The problem is that both players can have a connection simultaneously. For me a connection game does not have this property.

Yes, but only without the sloppy rule ;-) It would have to be tested if it is played like a connection game or if it is more or less a race for the first curly-q.

What would be interesting is no compensation for the first move but curly-qs are wins for the 2nd player... then we would have a real connection game. The question is: is this fair?

Ah.. sacrifice the player symmetry to get board symmetry. But isn't it too big sacrifice? The small asymmetry of the first move advantage is handled very well with the pie rule to the point I have no preference of being first or second player in Hex.

Question is: Is it possible to have a fully symmetrical board, player symmetry (except player order) and still have the unique winner property? Unlikely perhaps but I'm still not convinced until it proven impossible.

Wait a minute. I don't understand what curly-Q is, but I don't think I need to for the purpose of the argument I'm about to make. A horizontal ring bisects the playing surface such that no vertical ring of the other color is possible, no? And vice versa, of course. While what's called a ring in Havannah does not do that and would not be a win. So is the problem that ties are possible? I don't see whether they are or not, but if they're not, isn't the simple wrapping-rhombus torus adequate for all the desired conditions?

I've been thinking about this a bit more. Problem is: no matter how the wrap around is done, how do you assign these three special cases of striping to two players in a symmetrical way?

http://img138.imageshack.us/img138/4321/stripesi.png

There three cases correspond to a horizontal, vertical and curly-q. I would say that in a perfect symmetrical board, these three cases will be equally important. That is, making a curly-q connection would be equally difficult to make as a vertical or horizontal one. Simply because they correspond to the three directions of the hex grid.

The big problem is we must decide who of blue or red wins in each of the three cases, but this isn't possible to do symmetrically since three cannot be divided by two.

How is it done in Hex? Well, the edges make it so that there exist two cases for each of A, B and C. (Basically the color of A1 will distinguish between the two cases.) That makes a total of six ways we can put stones in stripes in Hex. Three of these cases are assigned a win to one player and the other three are assign to the second.

But if we have a wrap-around like a torus, we will only have three cases. If we want to keep player symmetry must we not assign one case to be a tie?

The only solution as I see it, is to have a wrap around that does not allow to make stripes. (Simply having an odd length before wrapping around doesn't cut it, since a similar argument can be made.)

Not allowing striping seems to mean that if you make a straight row of stones in this geometry you either visit all hexes or cross yourself before you reach your tail. Maybe such a geometry could work?

Either that or introduce some kind of asymmetry to the board (like edges) or the players themselves. Or a sloppy first-to-rule.. None of which are very appealing options.

@Dvd http://img138.imageshack.us/img138/6418/notawin.png look at this and you will probably see the problem... I guess you would say this is a win for both? In this case it is still possible for both players to make other winning paths.

Yes, I see now. Like the seam on the cardboard at the center of a roll of paper towel or toilet paper. To actually bisect the arena, a win for one dimension would have to have zero net traversals of the other dimension.

Jonatan, interesting thoughts. I agree.

However I think an asymetry wouldn't be such a tragedy. I guess the smaller the board the bigger the advantage of the first move without compensation, so maybe there could be found the right size where the first move is equal to the added possibility of making curly-qs.

Another solution would be to choose a different tesselation of the plane. For example if you take octagons and squares you will have 4 possible line directions which can be divided by 2.

PS: I came to the conclusion that the curly-q is much too strong to assign it to one player. Maybe another assymetry works.

Jonatan Rydh make a good point about 3 equivalent directions on a hexagonal torus board. there is hard to solve this problem. Actually we have many more distinct formations that can be assigned to a certain player as a win so that if he made one there is no way to form another one. Torus knots as explained in wikipedia we can denote by q(p,q) where p is an integer q is a natural number and p, q are coprime

circles in this notation would be denoted as q(0,1) and q(1,0) left corkscrew as q(1,q) including the shortest one q(1,1), right corkscrew as q(-1,q). the shortest right would be twice as long as the shortest left. I still refer to the board presented at the beginning of this topic

http://img204.imageshack.us/img204/8247/win2g.png

Where I defined left and right in my first post here.

The idea is to make the q(1,0), q(1,1) a win for second player and all others a win for a firs player.

This would be equivalent with that second player have to make a vertical/horizontal circle and second any other formation q(p,q).

I think that on big enough board that could equalize the chance for both players preserving the ¨unique winning formation condition¨ If not then we could abandon the nxn board and try different rectangles. But there is still a problem to determine which size would be most fair.

So I propose 3-rd version of rules

Second player goal is to form a vertical/horizontal circle

and first player goal is to make second player goal impossible to make.(Sorry for not nice form but i hope that it is clear what i mean)

@ maraca

Regarding the n move rule I agree with your point especially in games where there it is already hard to think many moves ahead like in twixt. But I Think that the first iteration of this rule is worth to explore in any game. Amazons with 2 move rule might be fun and I think there is still not a problem to think couple moves ahead.

If I understand your new version correct kfiecio you are assigning two of the three main directions to one player and one to the other. Note that stopping opponent to make either a vertical or horizontal connection is equivalent of making a curly-q connection.

(When you make a connection in any of the three directions - it automatically stops connections to be made in the other two by your opponent.)

To make up for it, one could make one dimension shorter. But we now talk about a game with serious player asymmetry. Like Hnefatafl.

Another possible solution would be to ban one of the direction from being made. (A stone may not be placed by either player so that an curly-q connection appears) This way we make sure that each full board has either a vertical or horizontal connection. However ties might appear if an hex is banned for both player simultaneously. Can this situation occur??

To answer my own question. Unfortunately yes. Here is an example:

http://img218.imageshack.us/img218/7338/exampleg.png

B2 is banned for both player since playing there would make a connection in all three directions simultaneously. The banning rule would lead to ties.

As a sidenote it seems for any fully played torus hex board, either one player has a connection in all three directions or both players has a connection in the same one direction. If this conjecture is correct and these two cases are equally common - maybe one can assign each case to one player?

Interesting I completely forgot the Torus knots. Other from choosing another tesselation, this would be the rule I would say feels most like a normal connection game:

-horizontal wins if he makes a horizontal circle or a torus knot (or corkscrew) where the number of the horizontal wrap arounds is bigger than the vertical wrap arounds.

-vertical wins if he makes a vertical circle or a torus knot where the number of the vertical wrap arouds is bigger than the horizontal wrap arounds.

-it's a draw if we get a torus knot (or corkscrew) where the number of horizontal and vertical wrap arounds is equal.

Unfortunately it won't work. Considering one third of all layouts will be drawn there should be easy strategies to ensure a drawn game. In fact this is one pairing strategy that works for even sized boards assuming vertical player plays first:

http://img546.imageshack.us/img546/5540/pairing.png

Second player simply plays on corresponding letter. The best vertical can achieve is a wrap around drawn game. Check yourself: unless I'm mistaken there is no way to draw a vertical path without crossing two of the same letter.

For odd sized boards.. well I'm sure there are almost as equally simple strategies.

maybe we can combine both ideas

-horizontal wins if he makes a horizontal circle or a torus knot (or corkscrew) where the number of the horizontal wrap arounds is bigger than the vertical wrap arounds.

-vertical wins if he makes a vertical circle or a torus knot where the number of the vertical wrap arouds is bigger than the horizontal wrap arounds.

-It is forbidden to create a torus knot (or corkscrew) where the number of horizontal and vertical wrap arounds is equal unles player creates winning formation for himself at the same time.

The question is what to do if player have no legal move i see 3 options

1)It is a draw

2)Player to move lose

3)Player to move passes (If both have to pass it is a draw)

@ Jonathan

Your pairing strategy seems to work also under the rules I presented above so there are not good. But what if we switch the second type of winning formation?

-horizontal wins if he makes a horizontal circle or a torus knot (or corkscrew) where the number of the vertical wrap arounds is bigger than the horizontal wrap arounds.

-vertical wins if he makes a vertical circle or a torus knot where the number of the horizontal wrap arouds is bigger than the vertical wrap arounds.

The pairing strategy does not work then.