When I started thinking about it though I realised that while both Jeff and Atacrawl had come up with a different way to do the second part of the problem (4 unknown), my way was in fact different again.
Now I've looked at the three alternatives to the second part and they all seem to be both correct and distinct solutions. If you think they're not then speak now or forever hold your peace.
A different method for the 8-unknown
What I've suddenly realised is that there's also at least one other solution for the 8-unkown section of the problem and I'm starting to suspect there may be more. Is there perhaps even another alternative to the initial 8/4 split?If you want a clue as to how the alternative solution to the 8-unknown works then have a look at the second comment I wrote on the previous post which describes an alternative to the 4-unknown. First person to this or any other solution will join the ranks below.
Current winners:
Jeff Koke - first to a solution on both the 8-unknown and 4-unknownAtacrawl - a different solution to the 4-unknown
??
3 comments:
Let's label the five balls 1, 2, 3, 4, 5, and we'll use C for a control.
First weigh 1,2 vs. 3,C.
If those are even, then weigh 4 vs. C and you can determine whether 4 or 5 is odd.
If 1,2 is heavy, then weigh 1 vs. 2. The heavy side is odd, or if they're balanced then 3 is light.
If 1,2 is light, then again weigh 1 vs. 2. The light side is odd, or if they're balanced then 3 is heavy.
What is additionally interesting is that should you determine whether 1-4 are odd, you'll also know whether it's heavy or light. But if 5 is odd, you won't know whether it's heavy or light. Therefore you're getting some additional information beyond which one is odd. Wouldn't it be cool if there were a way to find the odd among even more balls by somehow giving up on the heavy/light information....
Eric