All partitions of a set - W. Luis Mochán style


W. Luis Mochán explains a much simpler way to generate all partitions of a set.

I am so happy and humbled for having discovered that it’s possible to generate an iterator for all partitions of a set with a quite compact implementation:

sub all_partitions_iterator_wlm_style (@items) {
   return sub { state $r = 0; return $r++ ? () : [] } if @items == 0;
   my $last = pop @items;
   my ($pit, @presets, @postsets);
   return sub {
      $pit //= all_partitions_iterator_wlm_style(@items);
      if (@postsets == 0) { (@postsets, @presets) = $pit->() or return }
      my @pres = @presets;
      push @presets, my $set = shift @postsets;
      return (@pres, [$set->@*, $last], @postsets ? @postsets : []);

This is it. You can throw away pretty much the whole series of posts regarding this topic, and only use the function above.

It’s almost a drop-in replacement for what described in previous post All partitions of a set, with the following changes:

  • I figured that a proper representation of a partition should also include the empty set. Hence, the output always includes the empty set as well.
  • The order in which partitions are emitted is different from the previous one.

For the latter point, I have to admit that I liked the sorting in my original implementation best. On the other hand, this implementation is so much simpler that this aspect goes in second, or maybe third or fourth row.

Regarding the first bullet, this is actually scratching my itch with the seeming difference between $B_0$ and $B_n$ with $n > 0$, because the justification for $B_0 = 1$ is that there is one partition containing the empty set.

Why should the empty set disappear in later partitioning actions?

The obvious reason is that this empty set is probably of little to no benefit for using a partition (most probably it will be ignored), so removing it in the first place can be handy.

On the other hand… I’m still convinced that it belongs (or at least it can belong) to any partition, so why not including it? It also makes the implementation shorter 😄 Besides, if you don’t want to fiddle with the empty set, you can just ignore it because it’s always located at the end of the list.

So… thanks W. Luis Mochán for putting the partitioning problem in such a simple way to understand, and to Colin Crain’s review for making it easy to find interesting stuff!

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