Minus' Solution to

The Jeweler's Jumble
Posted February 17 - 23, 1997

1) If the two diamonds weigh the same, these two are the ones in the middle. Jeffery sets these aside, and weighs the other two diamonds. The lighter of these two is the lightest of the four, and the heavier of the two is the heaviest of the four.

2) If one diamond is lighter, and one diamond is heavier, Jeffery takes a third diamond to weigh against the lighter diamond. Again, there are several possibilities.

a) If this third diamond is lighter, then this one is the lightest of the four. The lighter diamond from the first weighing is one of the middle diamonds, as is the unweighed diamond. The heavier of the initial two is the heaviest of the four.

b) If the third diamond is heavier, then Jeffery will know which is lightest, but not which are in the middle, and which is heaviest. Jeffery then weighs the third diamond against the second diamond (the heavier of the first two). The heaviest of these two will be the heaviest of all. The lighter of those two will be in the middle.

c) If the third diamond is the same as the lighter diamond from the first weighing, then these two are the ones in the middle. The heavier diamond from the first weighing is the heaviest of the four, and the unweighed diamond is the lightest.

Mr. T & his Little Texan Einsteins had to go into over-drive to come up with this week's solution. But the DEN Keepers and Minus unanimously agreed that they still came up with the best solution to this week's open-ended problem. Hope you enjoyed the challenge, Little Einsteins!

Because of the amount of unknown information concerning the ratio of weight between the various stones, we concluded that we should put one diamond on each side of the scale and the other two on the counter. From this weighing, one of two things must happen:

Option A:
The two sides are equal. If the two sides are equal, then they must be the medium stones; that means that in one more weighing we can determine which of the remaining two weights is the heaviest, thus being the heaviest of the four. Likewise, the lighter one will be the lightest of the four. Two weighings, all info known, no prob!!!

Option B:
The two sides are unequal. If we allow one stone to be "A", one to be "B", one to be "C", and the other "D", then we weigh A & B and A is heavier, then one of three situations exist:

If A > B, then:
1) A=HEAVIEST & B=LIGHTEST (M,M)
2) A=HEAVIEST & B=MIDDLE (M,L)
3) A=MIDDLE & B=LIGHTEST (H,M)

You then must weigh C & D. If they are equal, then they are the two middle diamonds, and you have situation #1. If they are uneven and C > D, then you simply take D from that weighing, and weigh it against B from the previous weighing. If B > D, then B is one of the middle ones, D is the light one, leaving A as the heavy one and C as the other middle one. Conversely, if D > B, then D must be the middle one, B the light one, C the heavy one and A the other middle one.

So if all goes well, you could determine all the weights in two weighings (Option A), or if things take a turn for the worst (Option B), three will still do the trick.

Jeffrey can easily find these by putting any two diamonds on the scale. The heavier one should be compared with another diamond and if in the two times you got the two that weigh the same you'll know which are the middle and the lightest and the heaviest. Otherwise you'll know which is the heaviest and which is the lightest the remaining two are the middle. That does it in two tries!

Dear Minus,

First, Jeffery should weigh the diamond which he believes to be the lightest, because it will give him a variable to work with. Next, he should put the diamond that he just weighed and put on one try. of the balance beam. Put another diamond on the other tray and compair the weights. If he uses this method with each of the diamonds' weights with all of the other diamonds weights, he will get his answer.