The game of Y is a connection game invented by Craige Schenstead and Charles Titus. In its original form, it is played on a triangular grid of hexagons. There are two players, who have one colour each, and a move consists of placing a piece of your colour in one of the hexagons on the board. The winner is the first player to complete a chain connecting all three sides of the board. Y is a kind of generalisation of Hex, perhaps the one the nearest from it, but there are some strategic peculiarities, such as corner templates
Y cannot end in a draw. That is, once the board is complete there must be one and only one winner.
Less than two winners
There cannot be two winners at the same time. If there were, each player would have a region of the board touching all three sides of the triangle as well as the opponent's region. Considering the three sides as regions themselves, this gives a map of five regions, each of which is adjacent to the other four. However, this is impossible, as the graph K5 is non-planar.
At least one winner
It can be proved by an algorithm that once a board is complete there is at least one player linking the 3 sides. Let the "state" of a board refer to the answer to the question "Is there at least one winner?" We want to prove that the state of every board is "Yes".
First step: if there is a pawn group (red for instance) completely surrounded by the opponent (blue for instance) let's consider the board with this pawn group replaced by opponent's pawns (blue ones). The new board has the same status as the older one as the remote group was not winning and the new big (blue) one is winning iff it was in the former board. Also note that there is still at least one group left.
Repeat this step until there is no completely surrounded group more (of either color). The board obtained has the same state as the original.
Second step: if there is a pawn group surrounded by the opponent and a side, removing it does not change the state of the board (for similar reasons as in step 1) and there is still at least one group left.
Repeat this step.
Third step: if there is a pawn group surrounded by the opponent and two sides removing it does not change the state of the board (for similar reasons as in step 1) and there is still at least one group left.
Repeat this step.
It is quite clear that after applying this algorithm there is no group connected to more than 1 opponent's groups. No group is connected to zero sides and one opponent's group, no group is connected to one side and one opponent's group, no group is connected to two sides and one opponent's group. No group can be connected to 0 1 or 2 sides without connecting an opponent group. Moreover there is at least one group left. Hence this group left is connected to 3 sides.
So the state of the board is "yes"; as it is the same as the state of the beginning board, there was a winner to begin with.
Note that this algorithm ends because the number of different groups is finite.
Extension to Hex
The proof above extends to Hex because a Hex game can be seen as a subset of a Y game.
For instance consider the following Y board of size 8. (The star marked hexes are not a part of the board.)
We can simulate a Hex game of size 4 on it.
Now the only way to win for Blue is to cross the board horizontally, whereas the only way for Red to do so is to cross the board verticaly.
Each game of Hex on size n Hex board can be played on a Y board of size 2n or 2n-1 with the rules of Y, player just need to play some stones to "construct" the Hex board.
The first player wins
In Y the strategy-stealing argument can be applied. It proves that the second player has no winning strategy. The argument is that if the second player had a winning strategy, then the first player could chose a random first move and then pretend that she is the second player and apply the stratey. An important point is that an extra pawn is not a disadvantage in Y. Y is a complete and perfect information game in wich no draw can be conceived, so there is a winning strategy for one player. The second player has no winning strategy so the first player has one.
The game is usually played with the swap_rule. Alternatively, one can play Double-move Y, also known as Master Y: The first player places one piece on the board, and each subsequent move consists of placing two pieces on the board. This is a pretty challenging variant, even on small boards.
The inventors tried out a number of alternative playing grids, and eventually concluded that the most suitable one is the following "bent" version. The pieces are placed on the intersections (like in Go).
You can find more boards here: Printable Y boards
Help for programming the bent Y board
Try this Y puzzle.
On the web
- http://www.gamepuzzles.com/revugy.htm (Games magazine reviews)
- http://home.flash.net/~markthom/html/the_game_of_y.html (Dead link?)
- http://www.iggamecenter.com/ (igGameCenter - play "Y" online with other opponents from your iGoogle homepage)