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 Four Weights (Posted on 2004-07-25)
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?

 See The Solution Submitted by Brian Smith Rating: 3.5000 (8 votes)

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 All the examples | Comment 9 of 33 |
(In reply to An example by Charlie)

`            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 2ab  v cd      \ \ \ \ / / \ \ \ \ / / \ \ \ \ / / / / / / / / 12  0 12ac  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 18b   v c       \ \ / \ / / \ \ / \ / / \ \ / \ / / \ \ / \ / / 12  0 12`

Suppose on the other hand that the first two weighings resulted in a heavy right pan.  Then we could use (with first two weighings shown for clarity):

`ab  v cd      \   \       \   \       \   \                   12  0 12ac  v bd      \   \       \   \       \   \                   12  0 12            a 2   2       3   3       5   5            b 3   5       2   5       2   3            c 5   3       5   2       3   2            d 7   7       7   7       7   7ab  v d       \   -       \   /       -   /                   18  4  2ac  v d       -   \       /   \       /   -                   18  4  2`

Or, if the first two weighings had been / \, then:

`ab  v cd              / /         / /         / /             12  0 12ac  v bd              \ \         \ \         \ \             12  0 12            a         2 2         3 3         5 5            b         7 7         7 7         7 7            c         3 5         2 5         2 3            d         5 3         5 2         3 2ac  v b               \ -         \ /         - /             18  4  2ad  v b               - \         / \         / -             18  4  2`

Or if the first two weighings had been \ /, then:

`ab  v cd        \   \       \   \       \   \                 12  0 12ac  v bd        /   /       /   /       /   /                 12  0 12            a   2   2       3   3       5   5            b   3   5       2   5       2   3            c   7   7       7   7       7   7            d   5   3       5   2       3   2ab  v c         \   -       \   /       -   /                 18  4  2ad  v c         -   \       /   \       /   -                 18  4  2`

Thus in all cases we can provide an additional two weighings after the first two, that will uniquely identify the order of the weights.  Thus it is found in 4 weighings.

 Posted by Charlie on 2004-07-25 14:12:08

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