Why, I'm classifying codimension two hyperplane arrangements in C². What else would I be doing?

Basically, I need to create all of the permutations of 1 to n, then for each one, apply a set of three rules which will give me a list of different permutations which shall be deemed "equivalent" to the current one. I give each of these a common tag and then move on.

I do perform quite a few lookups to determine whether I have already applied the set of rules to a particular permutation through one of its equivalent representations. This is important since each of the rules can give me many equivalent permutations.

At the end I perform one final clean up since sets of equivalent permutations may have been labeled with different tags. I make note of these occurrances during the first pass through though, so the final cleanup is trivial.

Thanks for asking! (but I bet you're sorry you did)
    Dean

Update: Sorry, we're in C² not C⁴

If we didn't reinvent the wheel, we wouldn't have rollerblades.

Thanks for asking! (but I bet you're sorry you did)

You're not wrong there :) :)

Its situations like this where MathML or equivalent would be really nice, since you could just give me the forumlae :)

In essence, what I understood was the following:

For every permutation 1->n check every permutation 1->n to see if it is equivalent according to the rules. If it is, mark it as such.

Results: a set of groups of permutations who are deemed equivalent.

Are the rules transitive?

My head generated psuedocode something like this (no efficiency or anything, just a general algorithm):

foreach $p_1 (permutation(1..n)) {
   if ($tags{$p_1}) { next; }
   $tags{$p_1} = $p_1;
   foreach $p_2 (permutation(1..n)) {
       if (rule1($p_1, $p_2) && rule2($p_1,$p_2) && rule3($p_1, $p_2))
+ {
           $tags{$p_2} = $p_1;
       }
   }
}
[download]

But that only works if the rules are both transitive and symmetric (you hint that they are but I may be reading it wrong)

I would be interested in the contents of the rules 1->3 if they're not going to make my head ache :)

Actually, I (fortunately) do not need two for loops. The main loop of my code is this:

  $i = 'A'; # the class label
  for $n (@temp) {
    next if defined $arrangements{$n}; # already classified it.

    @n = split /,/, $n;
    @classmates = ( $n, join(',', reverse @n) ); # n and its mirror
    @classmates = (@classmates, &M1(@n)); # Maz 1
    @classmates = (@classmates, &M2(@n)); # Maz 2
    @classmates = (@classmates, &M3(@n)); # Maz 3

    &ClassReunion( $i, \@classmates );
    $i++;
  }
[download]

M1, M2, and M3 are the three moves (literally, the permutations represent lines in space) which return a new list of permutations. &ClassReunion just does some magic with the label $i.

M1 is easy, it is just the cyclic permutations of the list Ex: M1(12345) -> 51234, 45123, 34512, 23451. sub M1 { map join(',', @_[$_..$#_], @_[0..$_-1]), 1..$#_ }

M2 is the inverse permutation. The permutation 4132 is equivalent to tr/1234/4132/. The inverse would then be tr/4132/1234/.

sub M2 {
  my %temp;
  @temp{(1..@_)} = @_;
  %temp = reverse %temp;
  return join(',', @temp{(1..@_)});
}
[download]

M3 is, unfortunately much harder to explain, and has much longer code. I've posted a working version of the code on my sratchpad. If you /msg me I will post a better explanation of M3 there as well.

Good Day,
    Dean

---

If we didn't reinvent the wheel, we wouldn't have rollerblades.