March 8th, 2003, 11:06 pm
I give a different solution in which weighings are unconditional (i.e. don't depend on previous weighings) inhttp://wilmott.com/310/messageview.cfm?catid=15&threadid=5185&highlight_key=yIn that solution you weigh 4 against 4 each time. However this solution is a sort of "guess". I have no idea how to approach the general problem. Does anyone know by the way if it's the famous Besikovich solution the general coins problem (which was according to some source taking so many mathematicians man hours during the war that it was suggested to drop it on germans?)You see, in the "true" version those are neither pills nor balls, but coins By the way, it is easy to get a lower bound of log(2N)/log(3) by primitive "information theory" arguments, which I describe in my solution for N=12. That's close to the actual answer log(2N+3)/log(3) stated by Chukcha.