
Re: coins problem
Posted:
Feb 1, 1999 5:09 AM


John Conway wrote: > > Here's the solution to the nweighing version of this problem that
...
> Neat, isn't it?
Yes, it's very nice, thank you.
By the way, you said that an information count would make you think you could do (3^n1)/2  I suppose you mean that there are 3 possibilities for weighing, so n weighing gives 3^n outcomes, and k coins means 2k possibilitis, (anyone light or heavy), so, max k has 2k=3^n, and since that k is not an integer, you get (3^n1)/2.... so how do you prove you can't do this many? And why is it, that if you just have the extra information that you know the odd coin is light, then the 'information count' does work, and you can do 3^n in n weightings? what's the difference?

