You have 4 weights weighing 2,3,5 and 7 pounds. The problem is none of them are marked. What is the fewest number of weighings you need using a balance scale figure out which weights are which?
(In reply to
re: a computerized start by Charlie)
If we start by weighing ab v cd and then ac v bd, the possibilities are:
a 2 2 2 2 2 2 3 3 3 3 3 3 5 5 5 5 5 5 7 7 7 7 7 7
b 3 3 5 5 7 7 2 2 5 5 7 7 2 2 3 3 7 7 2 2 3 3 5 5
c 5 7 3 7 3 5 5 7 2 7 2 5 3 7 2 7 2 3 3 5 2 5 2 3
d 7 5 7 3 5 3 7 5 7 2 5 2 7 3 7 2 3 2 5 3 5 2 3 2
ab v cd \ \ \ \ / / \ \ \ \ / / \ \ \ \ / / / / / / / / 12 0 12
ac v bd \ / \ / \ \ \ / \ / \ \ \ / \ / \ \ / / / / / / 12 0 12
At this point we've narrowed the possibilities down to six, depending on the sequence of results. If, for example, the left side was heavy in both instances, it is the rightmost six permutations that are possible, and weighing a v bc and b v c will settle the matter
a v bc \ \ \ \ \ \ \ \ \ \ \ \  \  \ \ \ /  / \  \ 2 4 18
b v c \ \ / \ / / \ \ / \ / / \ \ / \ / / \ \ / \ / / 12 0 12
as the last four columns are unique sets of (taken vertically) / \,  \, / /, \ \,  /, and \ /.
This is thus done in 4 weighings.
Now appropriate rows have to be found for the other three possibilities of the results of the first two weighings.

Posted by Charlie
on 20040725 12:36:07 