A sultan has 9 seemingly identical pearls, but one weighs a little more than the others. You have a two-pan scale and want to identify the heaviest pearl using the fewest possible weighings. What is the optimal strategy?
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The wise men and the perlas (Persia antigua)
It has the charm of old scale problems: few movements, very expensive information and a solution that must be clean from the beginning. It does not ask for brute force, but for a good distribution of uncertainty.
Hints
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- With a two-pan scale, a single weighing never distinguishes more than three situations.
- To make good use of it, it is advisable to distribute the pearls in balanced groups.
- The first weighing should tell you where to look; the second, what is the pearl.
Solution
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Answer: The minimum is 2 weighings. Explanation: With a single weighing, a pan scale can only give three results: left heavier, right heavier, or balance. This allows us to distinguish at most 3 cases, and here there are 9 possible pearls. With two weighings you can distinguish up to
$$ 3^2=9 $$
cases, just the necessary ones. The specific strategy is to divide the 9 pearls into three groups of 3:
- weighs 3 against 3;
- if they balance, the weighing is in the remaining group;
- if not, it's on the saucer that goes down. Then, in the second weighing, you compare 1 against 1 within the suspected group. If they tie, the third is the heavy one; If not, the one that goes down is the one sought.
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