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### Re: Puzzle for the puzzle: Re: Puzzle...

 on Jul 09, 2002 at 23:36 UTC ( #180654=note: print w/replies, xml ) Need Help??

in reply to Puzzle for the puzzle: Re: Puzzle...
in thread Puzzle: need a more general algorithm

And here is an optimized version. Add in the question of what optimizations are present, and why they work. Some are more obvious than others. The thoughtful hacker might also wonder at why the \$max_mid, \$min_mid logic switched.

Oh right. And this adds the restriction that the heights should be integers. I think that is a reasonable rule for the speedup...

use strict; { use integer; my @sizes; my %ans; my @align; sub group_cats { (my \$num, @sizes) = @_; %ans = (); my \$a = 1; for my \$i (1..\$#sizes) { if (\$a + \$a < \$i) { \$a += \$a; } \$align[\$i] = \$a; } \$align[0] = 0; my (\$size, @partition) = _group_cats(\$num, 0, \$#sizes); return @partition; } sub _group_cats { my \$key = join ":", @_; my (\$num, \$start, \$end) = @_; if (not exists \$ans{\$key}) { if (\$num < 1) { \$ans{\$key} = [0]; } elsif (1 == \$num) { my @part = @sizes[\$start..\$end]; my \$sum = 0; \$sum += \$_ for @part; \$ans{\$key} = [\$sum, \@part]; } else { my \$num_a = \$align[\$num]; my \$num_b = \$num - \$num_a; my \$min_mid = \$start + \$num_a - 1; my \$max_mid = \$end - \$num_b; my \$mid = \$min_mid + \$align[\$max_mid - \$min_mid]; my (\$last_a, @part_a) = _group_cats(\$num_a, \$start, \$mid); my (\$last_b, @part_b) = _group_cats(\$num_b, \$mid + 1, \$end); my \$best = \$last_a < \$last_b ? \$last_b : \$last_a; my @best_part = (@part_a, @part_b); while (\$min_mid < \$max_mid) { if (\$last_b <= \$last_a) { if (\$max_mid > \$mid) { \$max_mid = \$mid; } else { \$max_mid--; } } else { \$min_mid = \$mid; } \$mid = \$min_mid + \$align[\$max_mid - \$min_mid]; (\$last_a, @part_a) = _group_cats(\$num_a, \$start, \$mid); (\$last_b, @part_b) = _group_cats(\$num_b, \$mid + 1, \$end); my \$max = \$last_a < \$last_b ? \$last_b : \$last_a; if (\$max < \$best) { \$best = \$max; @best_part = (@part_a, @part_b); } } \$ans{\$key} = [\$best, @best_part]; } } return @{\$ans{\$key}}; } }
BTW the original question had an straightforward recursive solution that scaled exponentially. I found that too easy, hence the faster solution.

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