Difference between revisions of "Y"
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| − | The game of Y is a [[connection game]] invented by | + | The game of Y is a [[connection game]] first invented in the early 1950s by John Milnor, and independently discovered in 1953 by Craige Schensted (later known as Ea Ea) and Charles Titus. In its original form, it is played on a triangular grid of hexagons, but Schensted and Titus preferred grids the include some pentagons, see [[#Variations|Variations]] below. There are two [[player]]s, who have one colour each, and a move consists of placing a stone 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 closest to it, but there are some strategic peculiarities, such as [[corner template]]s. |
<hexboard size="7x7" | <hexboard size="7x7" | ||
| Line 9: | Line 9: | ||
== No draws == | == No draws == | ||
| − | Y cannot end in a draw. That is, once the board is | + | Y cannot end in a draw. That is, once the board is completely filled, there must be one and only one winner. |
| + | |||
| + | === At most one winner === | ||
| − | |||
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. | 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 === | === At least one winner === | ||
| − | It can be proved by an algorithm that once | + | It can be proved by an algorithm that once the board is completely filled, there is at least one player linking the 3 sides. Let the "status" of a board refer to the answer to the question "Is there at least one winner?" We want to prove that the status of every board is "yes". |
| − | ''First step'': if there is a | + | ''First step'': if there is a group of stones that is completely surrounded by the opponent, let's consider the board with the surrounded group replaced by opponent's stones. The new board has the same status as the old one, as the surrounded group was not winning and the new big group is winning if and only if it was winning on the old board. Also note that there is still at least one group left. |
| − | Repeat this step until there | + | Repeat this step until there are no more completely surrounded groups of either color. The resulting board has the same status as the original. |
| − | ''Second step'': if there is a | + | ''Second step'': if there is a group of stones surrounded by the opponent and an edge, removing it does not change the status of the board (for similar reasons as in step 1), and there is still at least one group left. |
Repeat this step. | Repeat this step. | ||
| − | ''Third step'': if there is a | + | ''Third step'': if there is a group of stones surrounded by the opponent and two edges, removing it does not change the status of the board (for similar reasons as in step 1), and there is still at least one group left. |
Repeat this step. | Repeat this step. | ||
| − | It is quite clear that after applying this algorithm there is no group connected to more than | + | It is quite clear that after applying this algorithm there is no group connected to more than one of the opponent's groups. No group is connected to zero edges and one opponent's group, no group is connected to one egdes and one opponent's group, no group is connected to two edges and one opponent's group. No group can be connected to 0, 1 or 2 edges without connecting to an opponent group. Moreover there is at least one group left. Hence this group left is connected to 3 edges. |
| − | So the | + | So the status of the board is "yes"; as it is the same as the status of the original board, there was a winner to begin with. |
Note that this algorithm ends because the number of different groups is finite. | Note that this algorithm ends because the number of different groups is finite. | ||
| − | + | For another proof, see [[#Y-Reduction|Y-Reduction]] below. | |
| − | The proof above extends to Hex because a Hex game can be seen as a | + | == 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 strategy. An important point is that an extra stone is not a disadvantage in Y. Since Y is a perfect information game without draws, there is a winning strategy for one player. The second player has no winning strategy so the first player has one. | ||
| + | |||
| + | == Extension of [[Hex]] == | ||
| + | |||
| + | The proof above extends to Hex because a Hex game can be seen as a special case of a Y game. | ||
For instance consider the following Y board of size 7. | For instance consider the following Y board of size 7. | ||
| Line 60: | Line 67: | ||
== Y-Reduction == | == Y-Reduction == | ||
| − | Given a Y board of size n filled with red or blue stones, there is an operation of replacing each of the upper | + | Given a Y board of size ''n'' filled with red or blue stones, there is an operation of replacing each of the upper triangles of size 2 with a hex at its center, and with a stone of the color representing the majority of the three hexes being replaced. The result is a Y board of size ''n''−1. This operation is called the '''Y-reduction''', introduced by Craige Schensted. |
<hexboard size="5x24" | <hexboard size="5x24" | ||
coords="none" | coords="none" | ||
edges="none" | edges="none" | ||
| − | visible="area(e1,a5,e5) area(j2,g5,j5) area(n3,l5,n5) q4 p5 q5 s5" | + | visible="area(e1,a5,e5) area(j2,g5,j5) area(n3,l5,n5) area(q4,p5,q5) area(s5)" |
contents="R b4 c3 c4 d4 d5 e2 e5 h4 i4 j5 i5 m4 m5 n5 p5 q5 s5 B a5 b5 c5 d2 d3 e1 e3 e4 j2 i3 j3 g5 h5 j4 n3 n4 l5 q4" | contents="R b4 c3 c4 d4 d5 e2 e5 h4 i4 j5 i5 m4 m5 n5 p5 q5 s5 B a5 b5 c5 d2 d3 e1 e3 e4 j2 i3 j3 g5 h5 j4 n3 n4 l5 q4" | ||
/> | /> | ||
| Line 71: | Line 78: | ||
For example, the above Y board of size 5 can be reduced to size 4, and so on. | For example, the above Y board of size 5 can be reduced to size 4, and so on. | ||
| − | An important property of this operation is that | + | An important property of this operation is that one color has a winning [[chain]] if and only if the color has a winning chain for its Y-reduction. As a consequence, one can repeatedly reduce the board until size 1 to determine the winner. This can also be seen as a proof that there is exactly one winner in Y. |
| − | + | ||
| − | + | ||
| − | + | ||
== Swap == | == Swap == | ||
| − | The [[ | + | The [[swap rule]] can be used in Y. Opening moves in the center are good, and opening moves in the corners are bad, so there may well exist average opening moves. Further information, see [[Where to swap (y)]]. |
== Variations == | == Variations == | ||
| − | + | As an alternative to the [[swap rule]]. Alternatively, one can play Double-Move Y, also known as Master Y: The first player places one stone on the board, and each subsequent move consists of placing two stones 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 | + | The inventors tried out a number of alternative playing grids, and eventually concluded that the most suitable one is the following "bent" version. The stones are placed on the intersections (like in [[Go]]). |
[[Image:Y93_bent.gif]] | [[Image:Y93_bent.gif]] | ||
| Line 104: | Line 108: | ||
* http://www.iggamecenter.com/ ('''igGameCenter''' - play "Y" online with other opponents from your iGoogle homepage) | * http://www.iggamecenter.com/ ('''igGameCenter''' - play "Y" online with other opponents from your iGoogle homepage) | ||
| + | == References == | ||
| + | |||
| + | * John F. Nash. Some games and machines for playing them. RAND Corporation Report D-1164, February 2, 1952. https://www.rand.org/pubs/documents/D1164.html Credits John Milnor as the inventor of the game. | ||
[[Category: Y]] | [[Category: Y]] | ||
[[Category: Theory]] | [[Category: Theory]] | ||
Latest revision as of 21:57, 20 June 2024
The game of Y is a connection game first invented in the early 1950s by John Milnor, and independently discovered in 1953 by Craige Schensted (later known as Ea Ea) and Charles Titus. In its original form, it is played on a triangular grid of hexagons, but Schensted and Titus preferred grids the include some pentagons, see Variations below. There are two players, who have one colour each, and a move consists of placing a stone 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 closest to it, but there are some strategic peculiarities, such as corner templates.
Contents
No draws
Y cannot end in a draw. That is, once the board is completely filled, there must be one and only one winner.
At most one winner
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 the board is completely filled, there is at least one player linking the 3 sides. Let the "status" of a board refer to the answer to the question "Is there at least one winner?" We want to prove that the status of every board is "yes".
First step: if there is a group of stones that is completely surrounded by the opponent, let's consider the board with the surrounded group replaced by opponent's stones. The new board has the same status as the old one, as the surrounded group was not winning and the new big group is winning if and only if it was winning on the old board. Also note that there is still at least one group left.
Repeat this step until there are no more completely surrounded groups of either color. The resulting board has the same status as the original.
Second step: if there is a group of stones surrounded by the opponent and an edge, removing it does not change the status 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 group of stones surrounded by the opponent and two edges, removing it does not change the status 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 one of the opponent's groups. No group is connected to zero edges and one opponent's group, no group is connected to one egdes and one opponent's group, no group is connected to two edges and one opponent's group. No group can be connected to 0, 1 or 2 edges without connecting to an opponent group. Moreover there is at least one group left. Hence this group left is connected to 3 edges.
So the status of the board is "yes"; as it is the same as the status of the original board, there was a winner to begin with.
Note that this algorithm ends because the number of different groups is finite.
For another proof, see Y-Reduction below.
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 strategy. An important point is that an extra stone is not a disadvantage in Y. Since Y is a perfect information game without draws, there is a winning strategy for one player. The second player has no winning strategy so the first player has one.
Extension of Hex
The proof above extends to Hex because a Hex game can be seen as a special case of a Y game.
For instance consider the following Y board of size 7.
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 vertically.
Each game of Hex on a board of size n can be played on a Y board of size 2n−1 with the rules of Y. The players just need to place some stones to "construct" the Hex board.
Y-Reduction
Given a Y board of size n filled with red or blue stones, there is an operation of replacing each of the upper triangles of size 2 with a hex at its center, and with a stone of the color representing the majority of the three hexes being replaced. The result is a Y board of size n−1. This operation is called the Y-reduction, introduced by Craige Schensted.
For example, the above Y board of size 5 can be reduced to size 4, and so on.
An important property of this operation is that one color has a winning chain if and only if the color has a winning chain for its Y-reduction. As a consequence, one can repeatedly reduce the board until size 1 to determine the winner. This can also be seen as a proof that there is exactly one winner in Y.
Swap
The swap rule can be used in Y. Opening moves in the center are good, and opening moves in the corners are bad, so there may well exist average opening moves. Further information, see Where to swap (y).
Variations
As an alternative to the swap rule. Alternatively, one can play Double-Move Y, also known as Master Y: The first player places one stone on the board, and each subsequent move consists of placing two stones 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 stones are placed on the intersections (like in Go).
See also
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/gameofy.htm
- http://www.gamepuzzles.com/revugy.htm (Games magazine reviews)
- http://home.flash.net/~markthom/html/the_game_of_y.html (Dead link?)
- http://www.neutreeko.net/y.htm
- http://www.iggamecenter.com/ (igGameCenter - play "Y" online with other opponents from your iGoogle homepage)
References
- John F. Nash. Some games and machines for playing them. RAND Corporation Report D-1164, February 2, 1952. https://www.rand.org/pubs/documents/D1164.html Credits John Milnor as the inventor of the game.
