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4 unknown coins - another weighing puzzle
Posted 28 November 2012 - 06:28 PM
Posted 28 November 2012 - 07:00 PM
Posted 28 November 2012 - 08:11 PM
I think I see it more clearly now--any number of coins always weighs strictly more than a smaller number of coins.
Aye, Captain. You can interpret it that way. I put this condition there to exclude the possibility of one coin weighing as much as 2 others.
Posted 28 November 2012 - 09:30 PM
Edited by phil1882, 28 November 2012 - 09:31 PM.
Posted 22 December 2012 - 04:06 AM
This is Sort problem. Computer scientists have been tackling this very puzzle for the past 60 years or so, producing a great variety of algorithms.
Spoiler for One less
What you have suggested here seems like “Tree Sort” with self-balancing binary tree. The "worst case" order for such algorithm is O(n log n). That's as fast as it gets. Or, at least, that’s the present day theory.
Past prime, actually.
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