Logisch toch? :-)
Ja dat zei ik! ;-) Vanadium, nu je toch bezig bent, ik zie niet helemaal goed wat er in onderstaande sub
ocr gebeurt, dus als je effe hebt... Komt hier vandaan:
http://www.drdobbs.com/the-fourth-annual-obfuscated-perl-contes/199101795#!/usr/bin/perl -l
print ocr(<<TPJ);
# # # ## ## ## ## # # # # # ## # # # #
### # # # # # # # # # # # # # # # # ## # # # #
# ### ## ## ## ## # # # # # # ## # ## ### #
# # # # # # # # # # # # # # # # # # # # # #
# # # ## # ## # # ## ### # ### # # # # # # ##
TPJ
sub ocr{@{$-[$@++]}=split$,for(split'\n',shift);for$@(0..4){for(0..51){++$_{$_
}if($-[$@][$_]=~$")}}@&
-1);for(sort{$a<=>$b}keys%_){push@&,$_ if($_{$_}>4)
}push@&,52;for$@(0..13){@{$|[$@][$_]}=@{$-[$_]}[$&[$@]+1..$&[$@+1]-1]for(0..
4)}for(@|){**=$_;$w=@{$*[$^=$$=0]}-1;for$@(0..4){for(1..$w){$^++if$*[$@][$_
]ne$*[$@][$_-1]}}for(0..$w){for$@(1..4){$$++ if$*[$@][$_]ne$*[$@-1][$_]}}
for(0..20){push@},chr$_+65if(7*(8,4,2,9,2,3,7,8,1,$@,5,4,9,10,10,6,3,8,4,
8,8)[$_]+(5,8,3,3,4,2,1,2,8,2,7,1,5,4,6,$@,3,6,8,4,1)[$_]==7*$^+$$)}}@}}