master-algorithms-py/book/ebook_src/builtin_structures/balance.txt
Mia von Steinkirch 41756cb10c 🏣 Clean up for arxiv
2020-03-04 17:47:53 -08:00

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__author__ = "bt3"
This is the classic "you have 8 balls/coins, which are the same weight, except for one which is slightly heavier than the others. You also have an old-style balance. What is the fewest number of weighings to find the heavy coin/ball?
Answer: 2! You need to use every information available:
Weight 3 x 3 balls/coins.
If they weight the same: weight the 2 balls/coins left outside.
Else, measure 2 of the 3 heavier balls/coins.