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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++; }

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..@_)}); }

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.


In reply to Re: Re: Re: Big Picture by duelafn
in thread constructing large hashes by duelafn

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