I was surprised to find that 7 weighings is achievable.
I tried to make sure to split the possible orderings as evenly as possible at each step. The 4th weighing was the crucial step as 15 had to be split into 7 and 8.
label the coins A, B, C, D, E (5!=120 possible orders)
1st weighing: AvsB
(120/2=60 possible orders)
2nd weighing CvsD
(60/2=30 possible orders)
3rd weighing BvsD
(30/2=15 possible orders)
No symmetries have been broken at this point so we may assume
A<B<D and C<D. The possible orders are
times 5 possibilities for E.
4th weighing BvsE (the lighter of the 3rd weighing)
CASE 1, on the 4th weighing B<E
(7 possible orders)
The 3 cases marked with a * can be separated from the 4 without by 5th weighing DvsE.
If D<E the final two weighings are simple and have a few choices.
IF E<D the 6th weighing is BvsC and the 7th will be obvious.
CASE 2, on the 4th weighing E<B
(8 possible orders)
The 4 cases marked with a * can be separated from the 4 without by 5th weighing EvsA
If E<A the 6th weighing is AvsC and the 7th will be obvious.
If A<E the 6th weighing is CvsE and the 7th will be obvious.
Posted by Jer
on 2016-07-14 13:47:58