[yousuffauzan]

Under 1 sec is amazing. When you say brute force, do you mean that for the first step (TS 1), you took all the three routes available, [0,0], [1,0], [2,0] for jar 1 and 2. Cause that just makes the number of branches too many to be solved in a reasonable amount of time. I would love to have a peek at you code.

[daivd]

It is real brute force, but I do not investigate some stupid nodes. If I choose to not use an available jar during a timestep, I disqualify that jar, until another jar has been used, since there is no point in waiting to use a jar unless you want to save your balls for another jar.

The code is ugly and undocumented. I think the same could be accomplished in less than 10 lines of Haskell :). http://pastebin.com/MfXK9fwS

An implementation detail is that my branches are actually "jar1" "jar2" and "stepforward". To use a jar many times, you do jar1, jar1, stepforward.

[yousuffauzan]

As my first erlang program, I tried to solve this problem. The code is very ugly, and rather long for my taste, but it solves the thing in 917 steps. I also took the case where if no output has been gotten in a particular timestep, then do not consider any jars.