Two players face off in a new way on a chessboard. There is a shot glass with vodka on each of the 64 squares. The vodka in the glass in square H8 is poisoned. The "white" player starts, as usual. On each turn, the players line a rectangle on the board that includes the square A1 and at least one square with a shot glass in it, and drinks everything in that rectangle.
For example, white could start by drinking the six shots in A1..C2. Then black the four glasses in A1..B4, that is, A3, B3, A4, B4. Eventually the only shot glass remaining is the poisoned one in H8, and whoever is next to take a turn ends up poisoned to death.
If both players play a perfect game, despite inebriation, which will die, white or black?
Originally posted by crazyblue It is questionable though, if White would survive having 49 shots of vodka in his first move. Poor Amy (may she rest in peace) didn't. 🙁
"I've had eighteen straight whiskies - I think that's the record."
Dylan Thomas' reputed (and disputed) last words.
An even better solution than that I had in mind as it also tells how. Mine only showed that white wins.
Either there is a way for the player starting with an untouched board to win by doing a move other than drinking A1 only, or there isn't. If there is, white does that move, and wins. If there isn't, white only drinks glass A1, and white is in a position where there is no winning move => either way, white wins.
If you introduce a constraint that you cannot drink more than x drinks without being afflicted by alcohol poisoning, that would make the problem more difficult.
And another variant - say that the players meet once a week for the game, so they have the opportunity to recover between shots, so it is not n total, but a maximum of n at a time that limits choices.
@iamatiger : I believe the player must drink all shots contained within the rectangle with one corner at A1 and the other at x; picking D4 means White must drink the shots from A1 : D4.
Picking anything less than A1:G7 means Black can choose that leaving White in a position where his moves can be mirrored, unless White's starting move is asymmetrical including at least one shot on either the H file, or the 8th rank.
A good point.. are the ways for white to start and win, other than G7? If there are, obviously it would involve choosing H1 .. H6 (or A8.. F8) as first move.
Originally posted by andrew93 @iamatiger : I believe the player must drink all shots contained within the rectangle with one corner at A1 and the other at x; picking D4 means White must drink the shots from A1 : D4.
Picking anything less than A1:G7 means Black can choose that leaving White in a position where his moves can be mirrored, unless White's starting move is asymmetrical including at least one shot on either the H file, or the 8th rank.
could be wrong but the idea is that the diagonal squares a1..g7 are dangerous. If a player can get rid of all of them in one move AND leave a symmetrical position, then the game is won. I cannot prove white can win with d4 (only). But as the diagonal is spit into two equal groups that cannot be simultaneously taken (and the board is symmetrical) so it seems as though black has no easy win either...
Originally posted by talzamir A good point.. are the ways for white to start and win, other than G7? If there are, obviously it would involve choosing H1 .. H6 (or A8.. F8) as first move.
I believe there are. If you don't go for a symmetrical solution (like A1:G7) then an asymmetrical solution works like so : White starts with a move that ends on at least the 8th rank or H file (I'm thinking something like A1:H1) and then White mirrors whatever Black does. In the event Black tries to mirror white, then White creates another asymmetrical position (like A1:H2). While this continues White always has the option of G7 and we are back to the original solution. In the event Black chooses G7 at any point, then all White needs to do is to create a symmetrical position, which is another way of achieving the same outcome.