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Re: Keys beside keys on keyboards

by TedPride (Priest)
on Nov 06, 2006 at 06:41 UTC ( #582387=note: print w/replies, xml ) Need Help??

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.

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