It’s an interesting question, but I think that the answer is that it’s too hard to say anything at all. There does not seem to be any common structure to the valid solutions, and this makes it seem unlikely to me that they will enumerate nicely — indeed, even recognizing that a set of weighings is a solution may be nontrivial. It is possible that a question of the form, “How many solutions are there that can be proven to work via the rearrangement inequality method?” has an answer, but one should probably first solve the problem, “For which n does there exist a solution using the rearrangement inequality?”

]]>1+2+4 < 8

1+3 = 4

(after this we knew coins 1,2,3,4 and 8)

1+5 < 7

You are right Markus. The case n = 8 is all wrong. I will look into it, or, better, you can solve it.

]]>interesting problem, but I’m a bit confused by the case n=8. There seems to be a mistake there. First 1 + 2 + 3 + 4 + 5 < 7 + 8 is false, it’s 1 + 2 + 3 + 4 + 5 = 7 + 8. Then 1 + 2 + 5 < 4 + 6 does not provide any useful information I can see. And after 2 + 4 = 6 it’s impossible to distinguish 7 from 8.

I’m a bit quizzed whether I’m missing an obvious point or there is a snafu.

]]>Thanks, I made a close-parenthesis an italic one. It is ugly, but no smiley faces.

]]>Thanks for writing this up! I’m still in awe of these solutions with the rearrangement inequality. One small complaint: it seems that your blog software replaces an eight followed by a close-parenthesis with a sunglasses-wearing face 8) — this makes a few places a little difficult to decipher. Any way to fix this? ]]>