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[locard]

Major tangent alert!

I have read through the many posts for this card and to be honest am somewhat confused with the solitaire cipher. Doing an internet search for joker and djinn etc, DC Comics kept springing up.

It seems that both the Joker and Ra's al Ghul (The Demons Head, a type of Ghul / Djinn) are foes of Batman. A bit of further reading on Wiki reveals that Ra's al Ghul likes to partake in the odd bit of alchemy. Having clicked on alchemy it mentions something about cold fussion, a la disappearing text on the card.

I apologise for the rant but this card is driving me up the wall.

[TekkiBreki]

I've been playing around with the decryption function of Bruce Schneier's Go go gadget routine and have come across some quirky little things.

I applied cipher streams and keystreams of the same length and would get "plaintexts" of a different length - it didn't matter if the streams were very short (10 characters) or fairly long (27 to 54 characters), the "plaintext" would always be longer. The "plaintext" was always junk, too, just longer junk than I expected.

I don't understand the algorithm to be generative (or consumptive), so I'm wondering if some bug in it is causing the extra characters.

I've tested it by passing the same string as ciphertext and keytext expecting a string of zz...zz and discovered a sensitivity - if the cursor is not at the end of the stream, whatever nulls, returns, blank spaces seem to get processed as well. The "plaintext" will be a string of zz...zz with some non-zee letters at the end.

The impact to solvers is not to conclude that junk returned by this tool is junk.

[Stratman]

I am certainly not expert enough to tell if Leland Aldridge's Javascript Solitaire implementation (Go Go Gadget) has a bug, but I tested it on examples from other sites and always found it to work.
For example
CLEPK HHNIY CFPWH FDFEH
(from http://www.rubyquiz.com/quiz1.html)

Put the above in the Ciphertext field (remove spaces!)
There is no passkey so just hit Key
Message length is 20 so put 20 in the box and hit keystream (you can put a bigger number in here if you are unsure of the message length)
Hit decrypt.
You should get YOURCIPHERISWORKINGX in the Cleartext box.

If you hunt around you can probably find others to test it on. Not exhaustive proof that it works, but I am am on a Mac and it is the only software I can find that appears to work. I have tried hundreds of possibilities - timewise it is not realistic to do that by hand with a pack of cards.

[TekkiBreki]

In order to decrypt, the go go gadget routine requires a keystream - the same DWJXHYRFDGTMSHPUURXJ string used in the Ruby example would go in the keystream field. It just sits there if no keystream value is supplied.

The Ruby example can't be done by hand without using the keystream either.

It correctly returns YOURCIPHERISWORKINGX.

I'll keep playing with it and see if I can figure out why it returns arbitrary lengths.

[Stratman]

Sorry if I am misunderstanding you, I dont see the problem.
The keystream is the way Solitaire (and many other ciphers) works.
It is produced from the deck order, which in this case we do not know, but Garnet must have deduced it (or he used the pack in the order he was given it) Either way he may have left us a clue we at present are not understanding. You have to have a keystream to decrypt (by hand or otherwise) - you get it from the deck order.
In the earlier example I posted, the program assumes a deck order of...
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,
35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54...if no passkey is entered.

Another example... Use the passkey
THISISHOWITWORKSWITHAPASSKEY
hit key and you will get a deck order of
50,39,15,31,4,27,26,47,16,43,17,14,37,10,42,45,54,20,51,38,41,30,19,53,25,34,35,
44,23,24,8,11,32,33,21,1,3,6,7,9,52,29,48,2,12,13,18,5,49,40,46,28,22,36
Message length 19
Hit generate keystream and you get
VLXXWGSFEIBYVYCMSST
Enter the ciphertext
PTGQFZCAXCCMAWDZIEY
Hit decrypt. You should get the text
THISISJUSTANEXAMPLE
All the above could be done by hand but would take some time for long ciphertexts

[BrianEnigma]

[Stratman]

Hmm what's a troll?
I made it up just to explain how it works
(It is Garnet's email encrypted with Solitaire using If entropy...etc as a passkey)
Sorry I thought that was obvious

[locqust]

big ugly brute that tends to hide under bridges and eats goats, has an annoying tendency to regrow limbs as well. OR its a web-bot that spams forums like this

so basically your saying that the cyphertext isnt a name or phrase its an email address??

[TekkiBreki]

I understand how to generate the keystream, given a pass phrase, but I was under the impression from the earlier posts that the 108-bit stream is a solitaire ciphertext with no pass phrase for generating a key.

I have been exploring the possibility that the key is given.

These next bits are purely speculative, since they hang on the assumption that we do have the key:

There are 27 lines of text and if the 108-bit stream is the ciphertext, we'd need a source of 27*4 pieces of keystream data.

A "difficult" puzzle can include the classic misdirection of having a complex thing work once (Garnet's email) and then not requiring it a second time.

If the key was extracted from the text on the card, it will support modulo-26 addition and subtraction, like a solitaire cipher, but it does not have to conform to solitaire generation on a newly opened deck of cards. This means that working backwards once we know the key to see what the deck order must have been, we find the key generator to have been either a shuffled deck or a corrupt deck.

The key being embedded in the text is congruent with Von's clue and with Mind Candy's assurance that no other cards are needed.

[Stratman]

Sorry if I have confused anyone here.
All I was trying to do was post an example of how the cipher operates with a passkey. To do this I made it up by encrypting Garnet's email using Solitaire so that it can show that the Leland Aldridge implementation of the cipher does work.
Sorry, I thought I explained all that and it was obvious. Apparently not.
Strat
I will edit my explanatory post to make it more obvious

Garnet sent the email to bertyb (original thread p22) it said
Hello -
You're quite right, Professor Kiteway was involved.
If you can tell me what the message you deciphered was, I can shed some light on the mystery for you.
Best Wishes,
Garnet.

[arnezami]

[locqust]

Right I understand now Strat! Cheers for clearing that up, I thought that perhaps I'd missed something!

[TekkiBreki]

Working back to find out the original deck order is something a purist might do.

My thought was that if the pointers to a solitaire cipher were only to get us to perform modulo-26 arithmetic, then working out the deck order would just be an academic exercise. If the key is text-based, that academic exercise might lead to a keyed deck whose order we have insufficient information to discern or worse a "corrupt" deck.

One example of a corrupt deck would be 54 cards containing 20 face cards and no low hearts or spades. For every 52 bits of keystream generated, A, B, C, D, N, O, P and Q should average one instance, while K, L, M, X, Y, Z could appear three or more times each.

My mention above about Garnet's email was assuming it was one of the cipher streams I'd seen in various places and not something that had been delivered in cleartext and encrypted as a demonstration. This weakens my case for the classic misdirection I described.

[TekkiBreki]

Here are some of the things I've explored so far.

I tried the initial and terminal letters of each of the first four words in each line (8 candidate keystreams) acrostic-style as a keystream against the 108-character presumed ciphertext. I used 27 letters of ciphertext at a time to minimize the possible noise effects I posted about originally.

I shuffled the text block by alphabetizing the lines (line 1 numbered above becomes line 14) and applied columns of 27 letters to a depth of 17 letters (the length of the shortest line). I did not try this working from the right-hand side of the text block.

At least one line has 29 letters in it. I cut off the first and last (outward looks) and tried the remaining 27 as a keystream.

[Sh1ft]

I have been trying to get a keyed deck using a strategy of looking for the instances of "S", "D", "C", and "H" in the sentences maybe thinking that they represent spades, diamonds, clubs, and hearts respectively. Note that I wrote a quick perl script to pull the letters out so I do know they are accurate. Not that I have had any luck, but here it is: