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| Two players, two pots (Posted on 2009-02-21) |
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A and B are going to play a game: two pots are filled with stones, one with 57 stones and the other with 62 stones.
A legal move consists of emptying a pot - throwing away all the stones it contains - and splitting the stones of the other pot among the two pots, so that each pot contains at least one stone.
Players A and B alternate their moves, and player A is the first to make a move.
If a player is unable to make a legal move, then he loses.
Who wins this game and what is his optimal play strategy?
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Submitted by pcbouhid
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Rating: 2.0000 (1 votes)
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Solution:
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(Hide)
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A wins.
Strategy: A throws away all the stones of the pot with the odd number (57) and distributes the 62 stones (there are at least two stones) among the two pots, an odd number of stones in each.
For a strategy, which forces as fast as possible the victory, A should distribute the stones as evenly as possible.
B must throw away all the stones in a pot, and split the stones of the second pot (an odd number) among the two pots, leaving one pot with an even number of stones and the other with and odd number.
Following this strategy, at some time, after Bīs move, one of the pots will contain only 2 stones, and the other pot an odd number of stones.
A empties the odd pot and distributes the two stones among the two pots, which contain now one stone each.
B cannot play any legally move, now: He can empty one of the pots, but he canīt split the other stone among the two pots.
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