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
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