A1 opening - Revision history

Selinger: added a link

2023-05-25T16:28:43Z

added a link

Selinger: Fixed syntax.

2020-09-08T12:53:12Z

Fixed <hex> syntax.

Selinger: Fixed typo.

2020-06-14T02:19:22Z

Fixed typo.

Selinger: Added a cross-reference

2020-06-14T01:05:44Z

Added a cross-reference

Selinger: Added a cross-reference

2020-06-14T00:46:07Z

Added a cross-reference

Selinger: Added information about which board sizes the statement (and proof) applies to.

2020-06-13T23:40:19Z

Added information about which board sizes the statement (and proof) applies to.

Selinger: Fixed "dominated" -> "captured".

2020-06-13T23:11:26Z

Fixed "dominated" -> "captured".

Selinger: Updated the proof to use terminology like "dominated" and "dead", and made it more general.

2020-06-13T23:10:34Z

Updated the proof to use terminology like "dominated" and "dead", and made it more general.

Selinger: Reformulated the proof sketch, using dead cell and captured cell terminology.

2020-06-13T19:54:23Z

Reformulated the proof sketch, using dead cell and captured cell terminology.

Roland Illig: replaced image with hex markup

2012-11-12T04:30:20Z

replaced image with hex markup

@@ Line 18: / Line 18: @@
 == Sketch of proof ==
-In the following, when we say that a position is "winning" or "losing", we always mean under optimal play by both players.
+In the following, when we say that a position is "winning" or "losing", we always mean under [[optimal play]] by both players.
 Assume a symmetric board of size at least 3×3. To show that each of a1 and b1 is a losing opening move, we will prove a stronger statement: the position with red pieces in ''both'' a1 and b1 (and no other pieces on the board), with Blue to move, is losing for Red. Since an additional red piece can only help Red, it then follows that having just a single piece at a1 or b1 is also losing for Red.

@@ Line 4: / Line 4: @@
 <hex>
-R5 C5 border
+R5 C5 Q1
-play numbered a1
+numbered a1
 </hex>

@@ Line 14: / Line 14: @@
 Due to the symmetry of the Hex board, the same is true of the opposite hexes, but they are not usually referred to explicitly because their coordinate depends on the size of the board.
-The above statements about the a1 and b1 openings hold for board of size ''n''×''n'', where ''n'' ≥ 3. For smaller board sizes, the situation is different but fairly trivial: on the 1×1 board, a1 is (obviously) winning, and on the 2×2 board, a1 is losing and b1 is winning. On [[Hex_theory#Winning_strategy_for_non-square_boards|non-square boards]], the player with the shorter distance to cover has a winning strategy regardless of the opening move.
+The above statements about the a1 and b1 openings hold for boards of size ''n''×''n'', where ''n'' ≥ 3. For smaller board sizes, the situation is different but fairly trivial: on the 1×1 board, a1 is (obviously) winning, and on the 2×2 board, a1 is losing and b1 is winning. On [[Hex_theory#Winning_strategy_for_non-square_boards|non-square boards]], the player with the shorter distance to cover has a winning strategy regardless of the opening move.
 == Sketch of proof ==

@@ Line 10: / Line 10: @@
 (This does not mean that these are the worst possible opening moves. There might be other opening moves that are also losing but for which it is harder for Blue to carry out the win. Compare with the diagrams on [https://web.archive.org/web/20091013152809/http://www.cs.ualberta.ca/~queenbee/openings.html the openings page at the Queenbee site]).
-Like the proof that Hex is a win for the first player, the proof that a1 (or b1) loses is non-constructive: Although we know that it exists, the winning strategy has not been found for regular sized Hex boards.
+Like the [[Hex_theory#No_winning_strategy_for_Blue|proof that Hex is a win for the first player]], the proof that a1 (or b1) loses is non-constructive: Although we know that it exists, the winning strategy has not been found for regular sized Hex boards.
 Due to the symmetry of the Hex board, the same is true of the opposite hexes, but they are not usually referred to explicitly because their coordinate depends on the size of the board.

@@ Line 14: / Line 14: @@
 Due to the symmetry of the Hex board, the same is true of the opposite hexes, but they are not usually referred to explicitly because their coordinate depends on the size of the board.
-The above statements about the a1 and b1 openings hold for board of size ''n''×''n'', where ''n'' ≥ 3. For smaller board sizes, the situation is different but fairly trivial: on the 1×1 board, a1 is (obviously) winning, and on the 2×2 board, a1 is losing and b1 is winning. On non-square boards, the player with the shorter distance to cover has a winning strategy regardless of the opening move.
+The above statements about the a1 and b1 openings hold for board of size ''n''×''n'', where ''n'' ≥ 3. For smaller board sizes, the situation is different but fairly trivial: on the 1×1 board, a1 is (obviously) winning, and on the 2×2 board, a1 is losing and b1 is winning. On [[Hex_theory#Winning_strategy_for_non-square_boards|non-square boards]], the player with the shorter distance to cover has a winning strategy regardless of the opening move.
 == Sketch of proof ==

@@ Line 26: / Line 26: @@
 <hexboard size="5x5" contents="R a1 b1 B b2 E *:a2 *:a3"/>
-The cells marked "*" are [[dominated cell|dominated]], and therefore behave as if they had blue pieces in them. This makes the red pieces at a1 and b1 [[dead cell|dead]]. Therefore, after Blue's move at b2, the position is strategically equivalent to the one where a1, a2, a3, and b1 are blue.
+The cells marked "*" are [[captured cell|captured]], and therefore behave as if they had blue pieces in them. This makes the red pieces at a1 and b1 [[dead cell|dead]]. Therefore, after Blue's move at b2, the position is strategically equivalent to the one where a1, a2, a3, and b1 are blue.
 <hexboard size="5x5" contents="B a1 b1 B b2 B a2 a3"/>

@@ Line 1: / Line 1: @@
 {{wrongtitle|title=a1 opening}}
-The '''a1 opening''' (in the acute corner) is one of only two openings known to be defeatable. The other is [[B1 opening|b1]].
+The '''a1 opening''' (in the acute corner) is one of only two openings known to be defeatable. The other is the '''b1 opening'''.
-−
-(This does not mean that these are the worst possible opening moves. Compare with the diagrams on [http://www.cs.ualberta.ca/~queenbee/openings.html the openings page at the queenbee site])
-−
-Like the proof that Hex is a win for the first player, the proof that A1 loses is non-constructive: Although we know that it exists, the winning strategy has not been found for regular sized Hex boards.
-−
-Due to the symmetry of the Hex board, the same is true of the opposite hexes, but they are not usually referred to explicitly because their coordinate depends on the size of the board.
 <hex>
@@ Line 14: / Line 8: @@
 </hex>
-== Sketch for proof ==
+(This does not mean that these are the worst possible opening moves. There might be other opening moves that are also losing but for which it is harder for Blue to carry out the win. Compare with the diagrams on [https://web.archive.org/web/20091013152809/http://www.cs.ualberta.ca/~queenbee/openings.html the openings page at the Queenbee site]).
-Let's assume a1 is a winning first move for Red. There exists a [[winning strategy]] for Red beginning with 1.a1.
+Like the proof that Hex is a win for the first player, the proof that a1 (or b1) loses is non-constructive: Although we know that it exists, the winning strategy has not been found for regular sized Hex boards.
-Blue answers a2.
+Due to the symmetry of the Hex board, the same is true of the opposite hexes, but they are not usually referred to explicitly because their coordinate depends on the size of the board.
-<hex>
-R5 C5 border
-play numbered a1 a2
-</hex>
-This [[Dead cell|kills]] the piece at a1, i.e., makes it into a [[dead cell]].
+== Sketch of proof ==
-Consequently, the resulting position is strategically equivalent to the position where a1 and a2 are occupied by blue pieces.
-<hex>
+In the following, when we say that a position is "winning" or "losing", we always mean under optimal play by both players.
-R5 C5 border
-Ha1 Ha2
+To show that each of a1 and b1 is a losing opening move, we will prove a stronger statement: the position with red pieces in ''both'' a1 and b1 (and no other pieces on the board), with Blue to move, is losing for Red. Since an additional red piece can only help Red, it then follows that having just a single piece at a1 or b1 is also losing for Red.
-</hex>
-From Blue's point of view, this position is at least as good as 1.a1 was from Red's point of view, because Blue has [[captured cell|captured]] a1, and the additional piece at a2 can only help Blue. Since we assumed that Red had a winning strategy from 1.a1, it follows that Blue has a winning strategy from 1.a1, 2.a2. This is a contradiction, since it is not possible for both players to win.
+Therefore, the initial assumption was wrong, and Red does not have a winning strategy for the position with red pieces in a1 and b1. Since a draw is not possible, the position is losing for Red.
-Therefore, the initial assumption was wrong, and 1.a1 is not a winning move for Red. Since a draw is not possible, 1.a1 must therefore be a losing move.
+== a2 is a losing answer to a1 ==
-== a2 has been proved a losing answer ==
+If Red opens at a1, then Blue a2 is a losing response for Blue. To see why, note that we proved above that the position with blue pieces at a1 and a2, with Red to move, is losing for Blue. Replacing the blue piece at a1 by a red piece can only help Red, so is still losing for Blue.
-[http://hex.kosmanor.com/hex/b1loses.html A sketch of the proof on kosmanor page]
 == References ==

@@ Line 1: / Line 1: @@
 {{wrongtitle|title=a1 opening}}
 The '''a1 opening''' (in the acute corner) is one of only two openings known to be defeatable. The other is [[B1 opening|b1]].
-<!-- <hex>R5, C5, Va1</hex> -->
-−
-The '''a1 opening''' (in the [[acute corner]]) is one of only two openings known to be defeatable. The other is [[B1 opening|b1]].
 (This does not mean that these are the worst possible opening moves. Compare with the diagrams on [http://www.cs.ualberta.ca/~queenbee/openings.html the openings page at the queenbee site])
@@ Line 10: / Line 7: @@
 Like the proof that Hex is a win for the first player, the proof that A1 loses is non-constructive: Although we know that it exists, the winning strategy has not been found for regular sized Hex boards.
-Due to the [[Symmetries (board)|symmetry]] of the Hex board, the same is true of the [[opposite hex]]es, but they are not usually referred to explicitly because their [[Coordinates (board)|designation]] depends on the size of the board.
+Due to the symmetry of the Hex board, the same is true of the opposite hexes, but they are not usually referred to explicitly because their coordinate depends on the size of the board.
-[[image:acuteopening.png]]
+<hex>
 == Sketch for proof ==
@@ Line 20: / Line 20: @@
 Blue answers a2, this move makes Red's move useless as b1 does not need a1 to connect to top edge. Blue can now pretend to be the first player.
-Every winning connection that would have used the hexe a1 is a winning connection here too thanks to   a2. Hence Blue can use Red's winning strategy and win too.
+Every winning connection that would have used the hex a1 is a winning connection here too, thanks to a2. Hence Blue can use Red's winning strategy and win too.
-This is absurd and so there is no winning strategy beginning by a1. Therefor 1.a1 is a losing move (because there is no draw possible).
+<hex>
-<hex>R7 C7
+R7 C7 border
-Va1 Ha2</hex>
+play numbered a1 a2
 == a2 has been proved a losing answer ==