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I figure you'll always combine the smallest category with the one before or after it (Update: realized that in general, this is not a valid assumption; finding example left as excercise, though maybe finding smallest consecutive two column sum, then generating possibilities summing of columns before and after that might work...), so here's my fairly (in)efficient (O(N**2)) answer:
#!/usr/bin/perl use strict; use warnings; my $num_columns = 4; my @cat = (15, 15, 10, 10, 15, 15); my @ans = squish(@cat); print "@ans\n"; sub squish { my @arr = @_; my @aoa = (\@arr); for ($num_columns..(@arr-1)) { my @tmp_aoa; push @tmp_aoa, squisher(@$_) for @aoa; @aoa = @tmp_aoa; } my $best_ans; my $best_value; for my $aref (@aoa) { my $max_value; for my $value (@$aref) { $max_value = $value if !defined $max_value or $value > $max_valu +e; } $best_ans = $aref, $best_value = $max_value if !defined $best_value or $max_value < $best_value; } @$best_ans; } sub squisher { my @arr = @_; my $min_col; my $min_value; for (0..$#arr) { $min_value = $arr[$_], $min_col = $_ if !defined $min_value or $arr[$_] < $min_value; } my @arr1 = ($min_col > 0) ? @arr : (); my @arr2 = ($min_col < $#arr) ? @arr : (); splice(@arr1, $min_col-1, 2, $arr1[$min_col-1] + $arr1[$min_col]) if @arr1; splice(@arr2, $min_col, 2, $arr2[$min_col] + $arr2[$min_col+1]) if @arr2; return ((@arr1 ? \@arr1 : ()), (@arr2 ? \@arr2 : ())); }
Update: mildly tested...

In reply to Re: Puzzle: need a more general algorithm by runrig
in thread Puzzle: need a more general algorithm by Ovid

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