Scale Lighted

12 balls, 1 level, 3 weights, 1 lighter or heavier ball, find and tell if it is heavy or light. How?

Yanked the network in some party Date: 12/09/97 at 14:49:06 From: Doctor Tom Subject: Re: 3 pesos problem Hi Paul, is a difficult problem. Here is how to solve the general case (ie 4 pesos, 5 pesos, etc.): Here is the algorithm for the resolution of the 12 balls into the problem of three dollars. It also happens to solve the 3 balls in two heavy or 39 balls in 4 pesos, or 120 balls in 5 pesos or … problems. First, the list of all 3-digit combinations of 0, 1 and 2: 000 100 200 001 101 201 002 102 202 010 110 210 011 111 211 012 112 212 020 120 220 021 121 221 022 122 222 For the 4 weight problems, all combinations of 4 digit list, and so on. Then delete all sets with identical numbers (000, 111, 222), and also to expel all those whose first digit (change of reading left to right) is not 01, 12, or 20. In other words, it stays at 112, because the first change is 1 to 2, but 212 is eliminated, since the former change is 2 to 1. This leaves (a) 001 (b) 010 (c) 011 (d) 012 (E) 112 (f) 120 (g) 121 (h) 122 (i) 200 (j) 201 (k) 202 (L) 220, where a, b, …, l are the balls. Now for the first weigh all the balls with a 0 in the first place against all the balls with a 2 in the first position, in other words, (a, b, c, d) are compared with (i, j, k, l). If the balls with 0 are heavier, type 0, and if the balance of the balls, type 1, otherwise, type 2. Then, weigh all the balls with a 0 in the second position, against all the balls with a 2 in the second position (a, i, j, k) against (f, g, h, l), and enter the digit corresponding to the result of weighing, and they do again in the third position in the third weighing. As an example, suppose f ball (120) is heavy. In the first weighing, f is not involved, so write 1. Then (f, g, h, l) – the "2" balls are heavier than (a, i, j, k) to score 2. Finally, in weighing (b, f, i, l) versus (d, e, h, k), the "0" side (b, f, i, l) is heavier, so write a 0. You have written "120" – the code f ball – so f is the heavy ball. If you find the combination that you enter is not on the list, say 211, then change the 2 to 0s, 0 to 2s, and leave the 1s only, giving 011. C This is the ball, but since we had to flip the digits, c ball is lighter. With heavy N, there are 3 ^ combinations of N, which removes between 3 and cut the rest in the middle, so for $ 4, one can distinguish between (3 * 3 * 3 * 3-3) / 2 = 39 balls, etc-Doctor Tom, The Math Forum

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