even harder coin problem err try.. 1-6 vs 7-12 1,7,2,8,3,9 vs 4,10,5,11,6,12 1,4,7,10 vs 8,11,3,6    (2,5,9,12 left out) this right? Cheradenine Zakalwe Thursday, June 06, 2002 No. If the left hand side stays down every time, you don't know whether it's 1 or 11 Paul Paul Viney Sunday, June 09, 2002 ok lets see.. there are 12 coins, each of which can be heavier or lighter, giving a total of 24 possible results.. this means that at each weigh, there must be coins off the balance, since in this way it is possible to represent all the information "encoded" in the results of the balance ie, if L R and B are possible at each weigh, there are 3x3x3 = 27 possible results which can "encode" 24 combinations this leaves us with 3 "extra" combinations. we also note that each combination has an inverse, implying the opposite result.. since BBB is its own inverse, BBB is chosen as one of the extra combinations. we can also choose LLL and RRR as meaningless, being inverses (LLL and RRR are easier to verify) ok, so there are 12 coins at each weigh, but we must always leave coins out, the natural thing is to divide all the coins in 3 with 4 coins on the left, 4 on the right, 4 off. we construct a grid representing this and then enumerate all the combinations, assigning to each one a result eg, LLB implies coin 1 is heavier, LLB => 1H (RRB automatically inverse, 1L) LLB => 1H LLR => 2L LBL => 3H... this process (with one conflict) fills the grid giving 1,3,5,7  vs  2,4,6,8    off 9,10,11,12 1,6,8,11 vs  2,7,9,10  off 3,4,5,12 2,3,8,12 vs  5,6,9,11  off 1,4,7,10 hopefully this is closer to correct than my first cheap attempt? Cheradenine Zakalwe Tuesday, June 11, 2002 Yep, this work work. Checking it took me some time. :-) Paul Paul Viney Wednesday, June 19, 2002   Fog Creek Home