in reply to
Keys beside keys on keyboards

An interesting problem, even just for English keyboards. The good news is that there's somewhat less than 35,000 possible 3-letter sequences, so the easiest method might just be to generate a hash of those sequences and test every 3-letter sequence in the input string, preferably with a routine that stays up permanently so you don't have to regenerate the hash every time someone enters a new password.

Of course, the interesting part is writing a routine to generate the sequences for you. The following perhaps isn't the cleanest piece of code in the world, but it works as proof of concept:

`use strict;
use warnings;
print find('mskrtgdiwpa'), "\n";
print find('mwslkdftghm'), "\n";
sub find {
my $s = $_[0];
my $key = init();
for (0..(length($s)-2)) {
return substr($s, $_, 3)
if exists $key->{substr($s, $_, 3)};
}
}
BEGIN {
my (@keys, $rows, $cols, %seq, $run);
@keys = ('1 2 3 4 5 6 7 8 9 0',
'q w e r t y u i o p',
'a s d f g h j k l ;',
'z x c v b n m , . /');
$_ = [split / /, $_] for @keys;
$rows = $#keys;
$cols = $#{$keys[0]};
sub init {
if (!$run) {
my ($x, $y);
for $y (0..$rows) {
for $x (0..$cols) {
spider($x, $y, 3, '');
}
}
$run = 1;
}
return \%seq;
}
sub spider {
my ($x, $y, $depth, $s) = @_;
$s .= $keys[$y][$x];
if (!--$depth) {
$seq{$s} = ();
return;
}
spider($x, $y-1, $depth, $s) if $y > 0;
spider($x+1, $y-1, $depth, $s) if $y > 0 && $x < $cols;
spider($x-1, $y, $depth, $s) if $x > 0;
spider($x, $y, $depth, $s);
spider($x+1, $y, $depth, $s) if $x < $cols;
spider($x-1, $y+1, $depth, $s) if $y < $rows && $x > 0;
spider($x, $y+1, $depth, $s) if $y < $rows;
}
}
`

As it turns out, I miscalculated. There are probably less than 1500 sequences.