Eric Purdy

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Among a set of <i>n</i> coins of two weights (good and bad), and using a balance, we wish to determine the number of bad coins using as few measurements as possible. There is a known adaptive decision tree that answers this question in <i>O</i>((log(<i>n</i>))<sup>2</sup>) measurements, and a slight modification of this decision tree determines the parity(More)
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