If you don’t know what the Water Jug Problem is, click on the link first before you go on. For those who’ve read it, and consequently tried to answer it, I’ll refresh you on the problem.
We have three water jugs, and each can hold 8oz., 5oz., and 3oz. of water, respectively. Without the possibility of water spilling when poured from one jug to another, and given that the jugs have no calibration, how do we divide the 8oz. of water equally among two jugs?
To illustrate, from:
What I’m going to do is to draw these jugs into:
I said that the jugs aren’t calibrated, and they’re not calibrated here either. It’s easier to understand when we can see what’s going on. We have to remember though, that when we pour from one jug to another, it must either fill the jug where water is being poured into, or until the other jug becomes empty.
I will show the solution I have here. These are all trial-and-error stuff, so I guess with a little persistence, you’ll be able to get the solution. Read the captions so you know what I did in that step.
. . .
There you go. Fun, right? What did I tell you?
I want to share this website with you. It’s got a simple program where you can try out the solution for the three jugs problem, as well as taking a look at how the Barycentric Coordinate System can be used to solve it.