Difference between revisions of "A1 opening"
Roland Illig (Talk | contribs) m (replaced image with hex markup) |
(Reformulated the proof sketch, using dead cell and captured cell terminology.) |
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| Line 10: | Line 10: | ||
<hex> | <hex> | ||
| − | + | R5 C5 border | |
play numbered a1 | play numbered a1 | ||
</hex> | </hex> | ||
| Line 16: | Line 16: | ||
== Sketch for proof == | == Sketch for proof == | ||
| − | Let's | + | Let's assume a1 is a winning first move for Red. There exists a [[winning strategy]] for Red beginning with 1.a1. |
| − | Blue answers a2 | + | Blue answers a2. |
| + | <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]]. | |
| − | + | Consequently, the resulting position is strategically equivalent to the position where a1 and a2 are occupied by blue pieces. | |
| − | + | ||
<hex> | <hex> | ||
| − | + | R5 C5 border | |
| − | + | Ha1 Ha2 | |
</hex> | </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 1.a1 is not a winning move for Red. Since a draw is not possible, 1.a1 must therefore be a losing move. | ||
== a2 has been proved a losing answer == | == a2 has been proved a losing answer == | ||
Revision as of 19:54, 13 June 2020
- The title given to this article is incorrect due to technical limitations. The correct title is a1 opening.
The a1 opening (in the acute corner) is one of only two openings known to be defeatable. The other is b1.
(This does not mean that these are the worst possible opening moves. Compare with the diagrams on 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.
WARNING: Unrecognized token: borderSketch for proof
Let's assume a1 is a winning first move for Red. There exists a winning strategy for Red beginning with 1.a1.
Blue answers a2.
WARNING: Unrecognized token: borderThis kills the piece at a1, i.e., makes it into a dead cell. Consequently, the resulting position is strategically equivalent to the position where a1 and a2 are occupied by blue pieces.
WARNING: Unrecognized token: borderFrom 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 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 1.a1 is not a winning move for Red. Since a draw is not possible, 1.a1 must therefore be a losing move.
a2 has been proved a losing answer
A sketch of the proof on kosmanor page
References
References for proofs can be found on the Hex Theory page.
