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Re: Golf: Grocery Bagging

by grinder (Bishop)
on May 23, 2001 at 20:04 UTC ( [id://82599] : note . print w/replies, xml ) Need Help??

in reply to (Golf) Grocery Bagging

I'm not even going to attempt to write any code for this because it's pointless. For positive numbers, the problem is already NP-complete. When you mix in negative numbers, it becomes worse. The only feasible way to proceed is to generate all possible groupings and brute-force your way through them.

For three elements, it's easy: abc, ab c, ac b, bc a.

For four elements there is already an explosion of combinations: abcd, abc d, abd c, acd b, bcd a, ab cd, ac db, ad cb, ab c d, ac b d, ad b c, bc a d, bd a c, cd a b, a b c d.

I won't even post what the results of 5 gives, and I doubt this server has enough disk space available to store all the possible enumerations of 10 elements.

One could start to develop heuristics that paid special attention to groups of elements whose sigma matches the container size, but then that's going against the spirit of golf.

Noodling around some code here, it looks like the complexity of the algorithm is O(n!). Which means that it's going to be quite... slow.

g r i n d e r

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Re^2: Golf: Grocery Bagging
by tadman (Prior) on May 23, 2001 at 20:18 UTC
    It is NP-complete, so there's no way a super-efficient general solution is going to emerge here. However, the evaluation of any particular ordering doesn't require evaluating all possible groupings of that ordering. You can merely start adding stuff until an overflow occurs, switching to the next bag as necessary. The "best" solution out of the possibilities evaluated is the one that uses the least containers.

    Perl is buff enough to spin through several million possibilities in realistic time, unless you're using a version of Perl for Palm Pilot. I would submit that as long as a Perl Golf program runs in finite time (even extremely, exponentially long), the point is to make a minimally sized program that produces the correct solution.
      You can merely start adding stuff until an overflow occurs, switching to the next bag as necessary.
      No, I thought that too, at first, but you've got the problem of having permitted negative elements, so adding the next element might overflow, but the element after that (if negative) might bring it back. Argh!

      So while there might be a solution other than generating all possible partitions and seeing which ones have acceptable weights, it's not along the line of "fill until it won't fit". Sorry.

      -- Randal L. Schwartz, Perl hacker

        It might, but if you're going to evaluate all possibilities, who cares? As in, if you are worried about a scenario such as the following, where the 10 gets bagged "solo" despite there being a -5 farther down the pipeline:      ( [ 9 ], [ 10, -5], [ 11 ] ) Then later on you will inevitably evaluate a scenario where the -5 is inserted earlier.      ( [ 9, -5, 10 ], [11] ) So, taking the "brute force" approach, you might lose points for style, but you get the job done, no?

        Has anyone ever pointed out to a grocery checker that the bagging problem was NP-complete? The result might be similar to explaining that dogs can solve quadratic equations (i.e. capturing a frisbee in a parabolic arc while in linear motion).