Simple solution
my %norm = reverse %dem;
Of course in general there are, so you have to be more careful:
\nmy %norm;\nwhile( my ($key, $val) = each %den) {\npush @{ $norm{$val} }, $key;\n}\nVoila!The problem in your code is that you keep on switching between hashes and arrays, and references to hashes or arrays. You can't do that in Perl. Also you have to always look at how lists expand. So, for instance, you write:
\nmy %norm =\n{\n"a" => (),\n"b" => (),\n"c" => ()\n};\nall of those ()s get expanded to nothing, Perl actually sees,
\nmy %norm =\n{\n"a",\n"b",\n"c"\n};\nwhich is
\nmy %norm =\n{\n"a" => "b",\n"c" => undef,\n};\nand - even worse - {} constructs a reference to a hash so that is assigned as the key of a hash. So you wind up actually with something like:
\nmy %norm = ('HASH(0x815dba8) => undef);\n(The address of the hash will vary - but it doesn't matter since you didn't keep a reference to it so it got unallocated.)
In the same way if you say:
\nmy @list = $norm{$val};\nwhat you get is an array with a reference to an array in it. (Actually with undef rather than a reference to an array since %norm didn't have what you expect.) You then append to @list in the next line, but that doesn't do anything because then @list gets discarded.
So your code is going to wind up with %norm having one unexpected key and no values.
For playing around with this kind of thing, I strongly recommend using Data::Dumper. Something like this would have helped you see what is happening:
\nuse Data::Dumper;\nprint Dumper(\\%norm);\n
Cheers,
Ben
PS For a lot of problems like this, the Perl Cookbook has useful recipies. I highly recommend using it as a reference until your brain starts "getting" how Perl is supposed to work.
EDIT I simplified the real solution slightly by getting rid of a superfluous my.
