The three jugs problem

Date: Wed, 25 Aug 1999 18:40:03 EDT

Dear Math Central:
I wonder if you could help me figure out this secondary Discrete Math Problem. I cannot figure out where this comes from.

Here's the question:

You have three containers. Container 1 is a three (3) liter container. Container 2 is a five (5) liter container. Container 3 is an eight (8) liter container that is full of liquid. The goal is to get 4 liters in one of the containers.

Stipulations: There are no marks on the containers to measure out the liquid.

All you know is that 1 is 3 l., 2 is 5 l., and 3 is 8 l. full of liquid.

I started out with a chart like this

3	5	8
A	B	C

0	0	8	This is what you have to start.
3	0	5	1st pour
0	3	5	2nd pour
3	3	2	3rd
1	5	2	4th
1	0	7	5th
0	1	7	6th
3	1	4	The goal is reached at this point...4 liters!

Can you describe this method in a way that can be generalized and give me any information that you may have regarding this problem? Can you give me an answer please? Is this something that Karl Gauss did involving algorithms? PLEASE HELP!

Kent Lane

Hi Kent

A great reference for this problem is

H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited, Random House, 1967.

The authors use geometry, in particular reflections, to study this and similar problems. They include a solution to your particular problem that involves only 6 pours. Using your notation the solution is:

3	5	8
A	B	C

0	0	8	This is what you have to start.
0	5	3	1st pour
3	2	3	2nd pour
0	2	6	3rd
2	0	6	4th
2	5	1	5th
3	4	1	The goal is reached at this point...4 liters!

Cheers
Chris

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