<nitpick>

If GCD(X, Y) <> 1, then any value of T can be obtained provided that T > X*Y - X - Y

I think your condition is not stated correctly. If X = 2 and Y = 4, then no odd T is achievable. Perhaps the condition should be any T where T is a multiple of GCD(X, Y)? </nitpick> And on to the list.

It seems like dynamic programming is the way to go for either optimization. Essentially I see it in two stages: determining algorithms for obtaining all water amounts less than the bigger jug and then application of the Knapsack problem. I'm gonna go off with a pen and paper to think about how to achieve that first part. Don't tell my boss.


In reply to Re: Challenge: N Jugs Problem by kennethk
in thread Challenge: N Jugs Problem by Limbic~Region

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