FOUR WEIGHINGS
Taking this now to the point of obsessive stupidity, what if you had four weighings? What is the maximum number of balls out of which you would be able to detect a fake of different weight, and whether it is heavier or lighter?
These are my workings.
(1) Over four weighings there are 3 x 3 x 3 x 3, i.e. 81 outcomes. Write them down in their mirror-image pairs, and assign them to balls (ignore last two columns for now):
(2) Put the balls on the balance as indicated by the 'Heavy' column (see final table).
This puts 27 balls in each weighing, although not in equal numbers at each side.
(3) The numbers of balls on each side are now 20/7 21/6 19/8 20/7 i.e. 27
balls on each weighing, an odd number. Remove b06 to give even numbers
19/7 20/6 19/7 20/6 i.e. 26 balls on each balance.
(4)
Move others across so that there are 13 balls on each side. For those
balls that are moved across, the heavy and light indicators in the main
table need to be switched, e.g LLLL to RRRR for b01
I made the following moves:
Starting with (after removing b06) 19/7 20/6 19/7 20/6
move b01 (LLLL to RRRR heavy) to give 18/8 19/7 18/8 19/7
move b02 (LLLR to RRRL heavy) to give 17/9 18/8 17/9 20/6
move b03 (LLRL to RRLR heavy) to give 16/10 17/9 18/8 19/7
move b04 (LRLL to RLRR heavy) to give 15/11 18/8 17/9 18/8
move b09 (LLLB to RRRB heavy)to give 14/12 17/9 16/10 18/8
move b10 (LLBL to RRBR heavy) to give 13/13 16/10 16/10 17/9
move b11 (LBLL to RBRR heavy) to give 12/14 16/10 15/11 16/10
move b12 (BLLL to BRRR heavy) to give 12/14 15/11 14/12 15/11
move b28 (BLLB to BRRB heavy) to give 12/14 14/12 13/13 15/11
move b29 (BLBL to BRBR heavy) to give 12/14 13/13 13/13 14/12
move b34 (RBBL to LBBR heavy) to give 13/13 13/13 13/13 13/13
It seems clear that there are a whole load of different possible moves that would give the same end result, such is the magnitude of combinatorial mathematics. Also, in order to achieve an even number of balls in each weighing I could have taken out any one of b01 to b08.
We now have the following items at each weighing:
So four weighings would be able to detect a fake in 39 balls.
Outcomes BBBB, LLRR and RRLL are not used.
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