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### Re: "The Skirting Board Problem"

 on Jan 22, 2008 at 15:06 UTC ( #663596=note: print w/ replies, xml ) Need Help??

in reply to "The Skirting Board Problem"

This is obviously an example of the general problem of having a set of numbers and trying to find the minimum number of groups, the sum of whose members does not exceed a certain value.
What you describe is the bin-packing problem. Each length of board represents a bin and the lengths to be cut for each wall represent the set of items to be packed into the fewest number of bins.
I can conceive of a sort of brute force approach of just trying out a whole bunch of combinations, but there must be a more intelligent way.
Not likely, since the problem is NP-complete. On the bright side, there are many simple approximations for this problem which give a guaranteed approximation ratio*. The wikipedia article mentions a few which give quite good ratios (11/9). They use very simple rules. For example: First-fit decreasing would look like this (untested):
```my \$B = ...;           # bucket size
my @items = qw[ ... ]; # items
my @bins;
my @binsizes;

ITEM:
for my \$item (sort { \$b <=> \$a } @items) {
for my \$bin (0 .. \$#binsizes) {
if (\$binsizes[\$bin] + \$item <= \$B) {
push @{ \$bins[\$bin] }, \$item;
\$binsizes[\$bin] += \$item;
next ITEM;
}
}
push @bins, [\$item];
push @binsizes, \$item;
}

print "[@\$_]\n" for @bins;

Other places to look include Algorithm::Bucketizer or Algorithm::BinPack, but I don't know which heuristic algorithm they use under the hood.

BTW, it's likely you have few enough items (with small values) so that the brute-force method will not take too long. But it might take longer to code the brute-force search than to execute it (or install your skirting boards)!

Appendix/Update: A guaranteed approximation ratio of 11/9 means that if the optimal solution to a problem instance uses N bins, then the approximation/heuristic algorithm is guaranteed to return a solution for that instance that uses no more than (11/9)*N bins. (Note that I am ignoring the extra +1 in the guarantee for first-fit decreasing -- it will give you a solution that uses at most (11/9)*N+1 bins)

Also note that this ratio is tight, which means that there are indeed pathological cases where the algorithm gives solutions that really are a (11/9) factor worse than optimal. However, chances are that most inputs will not yield this worst case -- and at least there is a known bound about how bad things can really get.

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