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Re: Crypto Japh

by secret (Beadle)
on Dec 06, 2005 at 17:03 UTC ( #514553=note: print w/replies, xml ) Need Help??

in reply to Crypto Japh

veRy Nic3 !
Here is my solution to this riddle ;)

The core of the job is done by the first map which will crypt the 6 words of the string "writing Library. Building consider TO($FOO) *binary*" (split on space) with the numbers at the begining of the list @:: ( obtained with shift ) .

This step can be rewrited as :

@keys=qw'79 11 34 69 41 27'; @list_to_crypt = split/ /, 'writing Library. Building consider TO($FOO +) *binary*' ; @result=map{ crypt ($_.' ', shift @keys) } @list_to_crypt ;
If you print @result you will see :
79jUStLZ6Ck9c 11anOTHZ33A12 34ERehEMmayDo 69pERLsGUZYU2 41HAcKS9/lrEQ 27eRTYtC6a5HI
Now, in the same map liverpole pops the value at the end of @:: into $x , which is used into a regexp against the crypted values . This can be rewriten as :
@matcher = qw'2 4 4 2 5 4' ; foreach $crypted (@result ) { $x = pop @matcher ; $crypted =~/..(.{$x}).*/g ; push @final, $1 ; }
If you print @final you get :
The match is fairly understandable : it could read as "any character", "any character", "remember those next $x any character", "anything till the end of line" . And as it happens, $x selects the characters we need from @result !

And the next map is giving the proper formating .

But there is one last thing that i didn't know here : in the first map, the value returned by the map function is what was matched in the regexp ! Now that's interesting !!

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Re^2: Crypto Japh
by liverpole (Monsignor) on Dec 06, 2005 at 18:14 UTC
    Very nice analysis++.

    The thing that took the most time was, of course, finding words (and combinations of word fragments) which gave the appropriate crypt results, since crypt is a one-way algorithm.  I was especially surprised when I found the fairly long (5-letter) string "anOTH".

    @ARGV=split//,"/:L"; map{print substr crypt($_,ord pop),2,3}qw"PerlyouC READPIPE provides"

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