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Alexander Harrowell's avatar

that's a really neat solution; my own was to do a binary search (split the bottles into two equally sized groups; test one pooled sample at random; if it's positive, that's the group, if negative, it's the other group; repeat) which will also ID the poison bottle out of 100 in 7 rounds providing you know there's one and one only.

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Alan Morgan's avatar

Brilliant - simple but clever! Thanks for posting…

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