Re: cryptquote solver

[blakem]

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