# What is the simplest way to represent this situation mathematically?

Sitting in my room with two large jars full of green tea and black coffee, I suddenly realized that I would not be able to drink the coffee first. That is because I only regularly drink out of one of the jars. Yet I also realized that, without a third jar, I would not be able to move all of the contents of jar 1 into jar 2. As such, it seemed like there was something universal about the fact that, given two full jars, the contents of one cannot be poured into the other without the existence of a third jar. I suddenly found myself wanting a mathematical abstraction that could represent this situation. But I have no idea how a set could be called 'full.' I suppose there is some set-theoretic mathematical structure that can have this property though, right?

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Sound like Towers of Hanoi. Or variable swap; too bad you can't xor your mugs! – Raphael Sep 25 '11 at 12:47
Also sounds Pigeonhole principle-ish. – Adam Saltz Sep 25 '11 at 13:10
I'm not really sure about set theoretic, more of a combinatorical nature perhaps. In set theory it is the content of the jar that matters; not its name. So without loss of generality just swap the labels and get it over with. – Asaf Karagila Sep 25 '11 at 13:21
Why is this question down voted? Seems to be a sensible question. – Dinesh Sep 25 '11 at 16:14
If it were a jar of water and a jar of sugar, you might be able to pour the contents of one into the other... – Rahul Sep 25 '11 at 20:14

If you have just two jars, then you can move all the contents of jar 1 into jar 2 - it's just that you either then have a mixture you don't want, or the contents of jar 1 "on top" so that when you transfer back to jar 1 it is the stuff which was already in jar 1 which goes in first. So Jar 1 is always empty or contains some of its original contents, and similarly with jar 2 - so you can't do the swap.

If the contents of the jars mix, then you need one empty jar (however many jars you have which are originally full), and it is then easy to achieve any permutation of contents you want across any finite number of jars.

If the contents of the jars don't mix, but the "poured in" contents stay "on top" and come out first, then you can achieve any permutation you wish provided there are at least three jars. To transpose the contents of A and B, pour the contents of A into a third jar C (which exists by hypothesis), then the contents of B into A, then the top half of C (original contents of A) into B. Since any permutation is a product of transpositions, any permutation can be achieved.

It was not quite clear which case you were considering, but I hope this helps.

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It may be different with liquids which don't mix - (light) oil and water, for example. Pour the water carefully into the oil, and then decant the oil off the top. – Mark Bennet Sep 25 '11 at 17:55