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Dec 9, 2016 at 7:43 vote accept Vepir
May 7, 2016 at 17:11 comment added David G. Stork I would be interested in the answer to a modified question in which each weighing operation (on a pan balance, say) returns just either heavier, lighter or equal of the candidate sets--not their weight difference.
May 7, 2016 at 6:59 comment added Vepir That is great! So overall this solves the $2n$ cases in either $n+1$ or $n$ weightings. So far in my original question I've posted some "manual methods" for specific cases which seem to need exactly $n$ weightings or exactly the average of $n$ weightings up to the case of $10$ balls which I've found to do in either $4$ or $5$ weightings I believe, which averages out at $\approx 4.86$ weightings thus is even less than $n$, on average. But I think I need to double check and revise all my methods so far, and hopefully come up with something to follow the pattern of exactly $n$ weightings or less.
May 7, 2016 at 6:45 history edited Tony Huynh CC BY-SA 3.0
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May 6, 2016 at 23:12 history edited Tony Huynh CC BY-SA 3.0
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May 6, 2016 at 23:05 history edited Tony Huynh CC BY-SA 3.0
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May 6, 2016 at 21:23 history answered Tony Huynh CC BY-SA 3.0