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;
[download]

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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