Re: dictionary

[Mark Ellzey <mark () rdi st>]

I have written a quick perl program that will generate all possible
'words' for X number of chars, where X = a number supplied at the
arg, this is good for making dictionary lists of great size.

For other char sets just add to the @chars array.

#!/usr/bin/perl -w
# Word generator
# Mark Ellzey (mark () rdi st) - 9/18/02

use strict;

my $how_many_words = $ARGV[0];
my @chars = qw/a b c d e f g h i j k l m n o p q r s t u v w x y z/;
my %hash;
my $num_of_chars = @chars;

foreach my $char (@chars) {
        for(my $i = 1; $i <= $how_many_words; $i++) {
                push @{$hash{$char}}, $char;
        }
}

foreach my $entry (keys %hash) {
        my $word;
        foreach my $thing(@{$hash{$entry}}) {

                $word .= $thing;
        }
        print "$word\n";
        for(my $i = 1; $i <= $how_many_words-1;$i++) {
                change_array($i,$entry,$hash{$entry});
        }
}

sub change_array {
        my $offset = shift;
        my $letter = shift;
        my $array  = shift;


        my @arrayz = @$array;


        foreach my $char (@chars) {
                next if $char eq $letter;
                #print "$letter - $char $offset\n";
                $arrayz[$offset] = $char;
                my $word;
                foreach my $thing (@arrayz) {
                        $word .= $thing;
                }
                print "$word\n";
                if($offset+1 != $how_many_words) {
                        change_array($offset+1,$char,\@arrayz);
                }
        }
}

On Tue, Sep 17, 2002 at 08:28:51AM +0000, alex hajii wrote:
> 70% of passwords "break" under a good dictionary attack.
> is anyone here familiar with a good dictionary database ?
> is there any already written software for this reason that is downloadable?
> thak U.
>
>
>
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