lukestein’s avatarlukestein’s Twitter Archive—№ 6,128

  1. …in reply to @iamlemec
    @iamlemec Thanks to you and @ben_golub for a terrible night of sleep tossing and turning and eventually convincing myself that (allowing all equal), • 12 is in fact the *maximum* number of balls possible with 3 weighings • (3ⁿ-3)/2 is max number of balls possible with n weighings 😴
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    1. …in reply to @lukestein
      @iamlemec @ben_golub I have an intuitive argument though not a proof, and it “works” for 1 weighing → 0 balls 2 → 3 balls 3 → 12 balls I think using Doug’s heuristic it would be pretty straightforward to gen the algorithm to do 39 balls in 4 weighings, and I don’t think you could do more 🤷🏻‍♂️
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