Quote:
 Originally Posted by ZorbaTehZ There is a famous solution to that problem, published in Eureka, by the Blanche Descartes collective: (F refers to Professor Felix Fiddlesticks)
The solution is distinct from mine in the sense that it cannot be transformed into my solution by combinations of (a) renumbering the coins, (b) reordering the three weighings, and (c) reversing the sides of the scale in any particular weighings.

This can be seen by looking at the three coins that are in only one weighing and the three coins that appear in all three weighings.

In my solution, the three coins in only one weighing are 4, 10 and 12, and the three coins in all three weighings are 2, 6 and 8. Using the letter X to signify a coin in only one weighing, Y to signify a coin in exactly two weighings, and Z to signify a coin in all three weighings, my three weighings are:

(1) XYYY against YZZZ
(2) YYZZ against XYYZ
(3) YYZZ against XYYZ

In ZorbaTehZ's solution, the three coins in only one weighing are T, L and C, and the three coins in all three weighings are O, I and D. Using the same notation, Zorba's weighings can be represented as:

(1) YYYZ against XYZZ
(2) XYZZ against YYYZ
(3) YYYZ against XYZZ

There is no way of rearranging Zorba's weighings to give the same pattern as my weighings, so the two solutions must be distinct.

This raises the question: how many distinct solutions of the 12 coin problem are there? Note that we need consider only those solutions where there are four coins on both sides in all three weighings - we can convert any solution involving fewer than four coins on both sides by adding coins known to be true to make up the numbers. So the decision tree solution of Iancar29 can be converted into a "four against four" solution quite easily. Though is there a way of doing so that gives a solution isomorphic to any of the published "four against four" solutions?