[Guest]

wo players take turns biting off pieces of m × n rectangular chocolate bar marked into unit squares. Each bite consists of selecting a unit square and biting off that square plus all remaining unit squares above and/or to its right. Also, each player wishes to avoid getting stuck with the lowest-left square which is poisonous...

Refer to the attached image.

Assuming the chocolate bar has more than one unit square, which player (first or second) has a winning strategy and why?

[bonanova]

much like the game of Nim, only in two dimensions.

[wolfgang]

Hi...

its no matter howmany squares they are..

in all cases,,I`ll begin with the square which is diagonal to the yellow square,leaving the right side and the squares above( leaving only two raws).

and I´ll win easily after that,, i.e. if he`ll take the right,I`ll take the above,and vice verse.

[araver]

is that the second player does not have a winning strategy.

I got this argument after observing the 2x2 game, where there is the only one good opening move.

If the second player had a strategy against all possible first-moves, then he has a strategy if I remove the single square opposite to the poisonous one - the right-uppermost square.

In this case, whatever he chooses in his winning strategy I could have already chosen instead of choosing the right-uppermost square and beat him.

So there's no "bulletproof" strategy for the second player.

is how I could win as a first player.

All I can see is that if I choose something from the left-most column or the lowest-most row, I enter a new smaller chocolate-bar game as the second player, which means I no longer have a strategy. So whatever my winning strategy could be, I can never leave a rectangular chocolate bar for the second player. This is a necessary but not sufficient condition to win.

[Guest]

Hi...

its no matter howmany squares they are..

in all cases,,I`ll begin with the square which is diagonal to the yellow square,leaving the right side and the squares above( leaving only two raws).

and I´ll win easily after that,, i.e. if he`ll take the right,I`ll take the above,and vice verse.

This strategy only wins if m=n. If both branches are different, the second player only have to do both equal to win.

Edited October 25, 2010 by :-)

[Guest]

I think you should clarify what the contest of the game is; in other words, it is unclear what the players must do to win. Otherwise, it is impossible to figure out the puzzle.

[Guest]

I think you should clarify what the contest of the game is; in other words, it is unclear what the players must do to win. Otherwise, it is impossible to figure out the puzzle.

Not die?

[Guest]

how to find win strategy