That's actually not that far fetched.

use Digest::MD5 'md5_hex';

my $x = 'a';
my %found;
my $key;
while (1) {
    $key = substr md5_hex($x), 0, 8;
    if ( exists $found{$key} ) {
        my $first_md5 = md5_hex( $found{$key} );
        my $second_md5 = md5_hex( $x );
        die "found $key at $x ($second_md5) and $found{$key} ($first_m
+d5)\n";
    }
    else {
        $found{$key} = $x;
    }
    $x++;
}
__END__
found a986d9ee at bwma (a986d9ee140c5acbf0d51c00bc5a7810) and kot (a98
+6d9ee785f7b5fdd68bb5b86ee70e0)
[download]

With only eight hex digits, there's only 4_294_967_296 different hashes. You can exhaust that pretty quck.

An md5 hash of a large dataset is always a reduction of information. Different sets of information can result in the same reduction, which is actually the case with md5. There's a paper about md5 collisions.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

(Also in theory two MD5s from two different inputs can be identical, but this is very unlikely)

You are right. I didn't realise he would take a substring from the result.