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# Pour over this

Go to solution Solved by k-man,

## Question

You have two 10-liter beakers filled with water, and two empty beakers, of capacity 4 and 5 liters, respectively. You may pour from one beaker to another, in such a way that either the source beaker is emptied, or the receiving beaker is filled. The object is for the two smaller beakers each to contain 2 liters of water. No water is to be wasted, and no more than nine pourings is allowed.

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• Solution

Oops, you're right. I skipped a pour.

Initial State: 10 10 0 0
1: 10 6 0 4
2: 10 6 4 0
3: 10 2 4 4
4: 10 2 5 3
5: 10 7 0 3
6: 10 7 3 0
7: 6 7 3 4
8: 6 7 5 2
9: 6 10 2 2
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10-10-5-4 ---> beakers

10-10-0-0 ---> initial state

10-5-5-0

10-5-1-4

10-9-1-0

6-9-5-0

2-9-5-4

7-9-0-4

7-4-5-4

10-4-2-4

10-8-2-0

6-8-2-4

6-10-2-2 ---->finish 11 moves

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Barcallica - nice work.

OK we have a solution for 11 pourings.

Can anyone get it down to 10?

Or 9?

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Initial State: 10 10 0 0
1: 10 6 0 4
2: 10 2 4 4
3: 10 2 5 3
4: 10 7 0 3
5: 10 7 3 0
6:  6 7 3 4
7:  6 7 5 2
8:  6 10 2 2
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I think your second pour is not right.

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Bravo. Not an easy solve.

Yes, that is the coveted bonanova gold star.

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