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A sticky problem

Posted By: Stein Kulseth
Date: Friday, 21 August 2009, at 2:54 p.m.

In Response To: A sticky problem (Joe Russell)

If he is required to partition all the tokens into exactly two groups, he will have to flip one of them to make an even number, 86 or 88. He won't know which, but it does hardly matter. He will then have to partition the tokens, best in two equal groups, and hope that he lucks out to have respectively 43 or 44 in each.

But, if he is allowed to partition the tokens into more than two subgroups, he can partition all of them into groups of one token each. Then any two groups made from the 87 tokens that has the 'sucks' face up have exactly one 'sucks' face up, and any two of the remaining groups have exactly zero. (Or in general he can make do with 3 groups of 1 token out of which at leat two must match - or 4 groups of 2, 5 groups of 3 ... but partitioning into N groups of 1 does seem cooler, somehow)

Now, if it is not enough that he is able to make the subgroups, but must also while blindfolded be able to point to which two groups contain the equal number, his best bet seems to be to make two empty groups, each containing no token, and each containing zero 'sucks' facing up.

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