Minus' Solution to

King Harold's Jewels
Posted March 25-31, 1996

Harold is about to be crowned king of MathLand. He is selecting the jewels for his crown, and wants only the finest, heaviest stones. He has only a finely-calibrated balance scale to assist him in his task.

a) Harold has three blue sapphires, b1; b2; and b3. The weight of
b1 < b2 < b3. How will Harold determine which sapphire is which, and what is the minimum number of weighings you will need?

Harold begins by weighing two of the sapphires. One will be lighter than the other.

Harold takes out the lighter one, and sets it aside. He then puts in the last jewel.

If the last jewel is heavier than the other jewel, the last jewel is the heaviest. The other jewel in the tray is the middle weight, and the lightest one has been set aside. With this scenario, Harold needs only two weighings.

If the last jewel is lighter than the other jewel in the tray, Harold now knows which jewel is the heaviest, but he does not know which is the lightest. He has to weigh the jewel that he set aside with the last jewel to determine that. In this case, Harold needs three weighings.

b) Harold figured that out, and is now on to selecting his rubies and emeralds. He has two rubies, (r1 and r2) and two green emeralds (g1 and g2). The weight of r1 = the weight of g1. The weight of r2 = the weight of g2. The weight of r1 < r2. How will Harold determine what r1, r2, g1, and g2 are, and what is the minimum number of weighings he will need to do so?

Harold begins by weighing one red against one green.

If the two are not equal, Harold can tell which jewel is the heavier, and will therefore know that its unweighed mate is the lighter. In that case, Harold needs only one weighing.

If the red and green are equal, Harold can take out the red, and put the other green in. Then Harold can tell which green is lighter and which is heavier. By remembering the first weighing, Harold will know which red jewel equals which green jewel, and he will know which jewels are heavier and which are lighter. (Of course, Harold could also take out the green, and put the other red jewel in. Or he could take out either one, and put the opposite colour in, and get the same result.) Harold would need two weighings in this instance.

c) Harold has four amethysts and is hoping that their reputed good-luck qualities will help him. Two of the amethysts are equally light. The other two amethysts are the same weight compared to each other, but are a little heavier than the two lighter amethysts. Harold does not know the weight of any of the amethysts. How will Harold determine which amethysts are the light ones and which are the heavy ones using the minimum number of weighings?

Harold begins by weighing two amethysts. If the two are unequal, Harold knows which is which, and sets them aside. He then weighs the other two, which will also be of different weights.

If the two amethysts are of equal weight, Harold takes one out of the tray, and substitutes one of the unweighed ones. Now Harold can tell which are the light amethysts, and which are the heavy amethysts.

In either case, Harold needed two weighings.