Problem : Two players have in front of them a single pile of objects, say a stack of 7 pennies. The first player divides the original stack into two stacks that must be unequal. Each player alternatively thereafter does the same to some single stack when it is his turn to play. The game proceeds until each stack has either just one penny or two—at which point continuation becomes impossible. The player who first cannot play is the loser. Show, by drawing a game tree, whether any of the players can always win.
Why is the state 6-1 not going to 3-3-1?If we have 6-1 pennies we can remove 3 pennies from the 6 stack and we have 3-3-1 pennies.So why isn't 3-3-1 not a child of 6-1?
Asked By : Sid
Best Answer from StackOverflow
Question Source : http://cs.stackexchange.com/questions/6363
Answered By : Vor
Because - according to the rules above - each player must divide a stack into two stacks that must be unequal.
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