On Mar 11, 2013, at 5:26 PM, Joe Niederberger <firstname.lastname@example.org> wrote:
> R Hansen says: >> No, I meant period. You cannot take 4 coins, even if you know that the counterfeit coin is heavy and find it in one weighing. > && >>> You cannot end up with a group of 4 coins in step 3 > > Now I have no idea what you are claiming. Look at, say, the Frank Cole solution again: > > http://www.iwriteiam.nl/Ha12coins.html
Frank couldn't do it either. If I gave Frank 4 coins (or more) and told him that the counterfeit coin is heaver (or lighter) he would not be able to determine the coin in one weighing.:) Obviously, I am talking about the group strategy I posted earlier.
But Frank's strategy operates the same way. After 2 weighings, Frank's method eliminates 9 coins (their score/value is unreachable).