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I did a similar thing once upon a time. Instead of writing one regex to match the wordmask, I calculated the wordmask for each dictionary word and compared it to the given word. Its probably less efficient than your approach, but it was easier to write.

Since wordmask() is a non-trivial subroutine I've included a few tests for it...

Run it like so:

% match_word.pl abcdee
% match_word.pl elephant
% match_word.pl run_tests
#!/usr/bin/perl -wT use strict; my $word = shift || 'abcdee'; my $dict = '/usr/dict/words'; run_tests() if $word eq 'run_tests'; my $mask = wordmask($word); open(my $in, '<', $dict) or die "cant open $dict : $!"; while(<$in>) { chomp; next unless length($_) eq length($mask); my $wordmask = wordmask($_); print "$_\n" if $wordmask eq $mask; } sub wordmask { my $word = shift; $word =~ tr/a-z/A-Z/; return 'ERROR' unless $word =~ /^[A-Z]+$/; my $letter = 'a'; while ($word =~ /([A-Z])/) { $word =~ s/$1/$letter/g; $letter++; } return $word; } sub run_tests { eval "use Test::More tests => 4"; is( wordmask('abc'), 'abc' ); is( wordmask('ally'), 'abbc' ); is( wordmask('ggl'), 'aab' ); is( wordmask('*()'), 'ERROR' ); exit; }

-Blake


In reply to Re: cryptquote solver by blakem
in thread cryptquote solver by xChauncey

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