While wailing away the hours I've put my mind to a problem posed by newly minted Mauritian FM Roy Phillips (7/10 at the Olympiad!): Out of 12 balls, identical in appearance, one is of a slightly different weight to others. You have a set of balancing scales. How do you find the (a) odd ball out and (b) determine it is either heavier or lighter than the others, using the scales only 3 time?
He says he eventually solved it but for now I don't have an answer.
3 comments:
Hello Shaun.
For the 12 balls and a balance beam.
Measure 6 a side, then retain the heavier 6.
Do the same for 3 balls each.
Out of the remaining 3, pick 2 at random. If they are equal then the heavier is the remaining ball, otherwise you know the heavier ball with the one ball on easch side.
But heavier isn't given, the different ball may be lighter.
Thinking a long way back, I think the solution begins with dividing into three groups of four balls each.
Weigh one group against another, if equal, then the different ball is one of the unweighed balls. Which should be easy to distinguish with the remaining weighings.
The difficult part is finding the odd ball when it is among the two original groups. This involves some switching of balls between the two groups in the remaining weighings, but I don't recall the exact procedure nor can work it out at the moment. So, I had a won game, but resign here :)
OK. I've seen a solution to this problem (as I eventually gave up) and while it is along the lines of what I was thinking (and what has been discussed here), it is probably 1 level of complexity beyond what I was looking at. Here is a tip: It helps if you label the balls.
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