A wise man has three vessels of 12, 8 and 5 liters. The 12 is full; the other two, empty. You must divide the contents into two equal parts, using only transfers between vessels and without intermediate marks. How do you get it?
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The three vessels of the wise man (India antigua)
The Three Pots of the Sage (Ancient India) belongs to the great tradition of problems that seem narrative and end up being exact. It retains that old-fashioned flavor in the wording, but the solution is still surprisingly modern.
Hints
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- Don't try to get to 6 directly; Use the 5 pot to craft useful scraps.
- It is advisable to first produce a situation in which a vessel is almost full or almost empty by necessity.
- The goal is to end up with 6 and 6, so each transfer should bring you closer to a symmetrical distribution.
Solution
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Answer: Yes. It is available in 7 transfers and ends with two 6 liter containers. Sequence of states $(12L,8L,5L)$:
- $(12,0,0) \to (4,8,0)$
- $(4,8,0) \to (4,3,5)$
- $(4,3,5) \to (9,3,0)$
- $(9,3,0) \to (9,0,3)$
- $(9,0,3) \to (1,8,3)$
- $(1,8,3) \to (1,6,5)$
- $(1,6,5) \to (6,6,0)$ Explanation: The sequence seeks to first manufacture useful intermediate quantities and, from them, force an exact partition into two halves. The final state $(6,6,0)$ solves the problem perfectly.
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