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### (Golf) Grocery Bagging

 on May 23, 2001 at 15:19 UTC Need Help??

Objective: Given a "container" size and a list of numbers, return an array of array references ("containers") which contain the supplied number list in groups of for which the sum does notexceed the specified container size. Further, the function should return only the minimum number of containers required to solve the problem.

Any numbers which absolutely cannot be accomodated given the input data and container size should be returned "unbagged", that is, as scalar data trailing the returned array references.

Other notes: Negative, zero, and fractional values are valid input data for both the input list, and the container size. As such, numbers larger than the container size cannot be immediately discarded without first considering other input data.

Sample Input:
```    @o = qw [ 14 6 1 22 4 2 7 6 ];

@b = f(15,@o);    # "Container size" = 15
Sample output:     @b = ( [14,1], [ 7, 6, 2], [4, 6], 22);

Replies are listed 'Best First'.
Re: Golf: Grocery Bagging
by grinder (Bishop) on May 23, 2001 at 20:04 UTC

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
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.

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