unforum

[Stratman]

Sudden panic for a moment there a lot of effort had been wasted...
But a closer read of http://www.schneier.com/solitaire.html
This is how I understand it...I think!
The jokers have to be individually identifiable (hence the star on one) and can be numbered 53 and and 54 (A joker and B joker) for the initial setting up of the cipher.
It is at step 4 (the count cut) when you value either joker as 53 - I assume so that the last card is not disturbed (if the last card was 54 then you would cut all the cards)
Quote 'The reason the last card is left in place is to make the step reversible. This is important for mathematical analysis of its security.'
I am sure the Java applet we have all been using (and the windows Solitaire program if you have a PC) take this into account.
Incidentally a flash of inspiration led me to try the 108 character string split in two as the passkey and the ciphertext and ran the applet with every possible split - gibberish as usual!

[locqust]

Im still fairly sure that the 108 characters are two messages encoded with the same key.

Call it a gut feeling

[Guin]

[Ashin]

Hey locqust, could you elaborate on your theory? Sounds interesting, but I'm not quite sure I see what you're proposing.

[cheesey]

[locqust]

Ok this is a theory I came up with as we didnt seem to be getting anywhere with the straightforward approach. So I thought, ok MC must of been sneaky somehow. The phrase we got isn't long enough to be a passphrase, its supposed to be longer according to Schneier, who invented this I believe. Also it's far too cryptic (or else were dumber than MC think we are) to be useful as a clue to deck ordering. So I do think it is to do with the heat sensitive letters and nothing else. (I may be wrong so don't start shouting me down if you think I am, because I already know I may be.)

Anyway to the theory.

If, according to Schneier's site, we have two messages encoded with the same deck ordering/key, whatever you want to call it. Something we're never supposed to do in real life. Then it is possible, and easy, for a cryptologist to break both messages without ever using or knowing the deck ordering/keystream.

Now I'm not a cryptologist so I'm not entirely sure this will work at all. But if we assume each line, minus the heat sensitive letters are two different messages not one, both encoded using the same deck order/keystream. It should be possible to decode both without using a keystream. By comparing one set of characters with the other set.And either one should be our answer. As for the shorter second line, ummm well thats the hole in the idea really

The Schneier site does briefly give a clue how to do this but I tried and couldn't make heads or tails of how to go about it. A + K = Q sorta thing, just look for that on his site then you'll know what I'm on about.

Hope that gives people some ideas!

Oh also I don't know if this will work either but maybe if it is one message then we could split it in half and treat it as two messages to get the same result as having two original messages? (And that bit just proves my amateur experience in code breaking! )

Edit: Italics are added as I dont think I explained myself properly the first time. [/i]

[AngusA]

[UKver2.0]

Although I don't think it's the right path, I threw together a crap example spreadsheet of what locqust is talking about... It's not meant to be used for the solve, but just to show how to do it. The example assumes that the first 10 characters and last 10 characters have been encrypted with the same keystream. What you do is type in something that might show up in the message and if the output side starts to look like real words there's a good chance you are on to something. You don't need a full message to play with it. If you think "kiteway" might be in there, replace "ITWASSENTE" with "KITEWAYNTE". Nothing. Copy/Paste it down a line "KKITEWAYTE". Nothing. Move It Again "KKKITEWAYE". Nothing. If you think "KITEWAY" would show up in the second group, you can flip "EOD..." with "FHU..." and work it that way. To do the first line against the second just grab the first cell of the output and fill down 55 cells. Replace "EOD..." with line 1 etc. Like I said, it's just an example. A real cryptanalyst would have written a program to do the same, but automatically run through dictionary words, names, etc and parse the outputs. They might even use word parts vs. whole words. I suppose I could work on that, but my skill level would make it a much bigger task than for someone who does that kind of thing in their sleep.

[ramsfan]

Light relief. Read Handful of Dust by Evelyn Waugh on holiday. Cracking. Stephen Fry should direct it now. Don't know why it's not better known. Definitely not a comedy, but a great story. Anyway, the sister's Pekingese (made in china?) is called Djinn. I'm sure it's of no relevance, but hey, those Oxford types like their Waugh, you never know.

[maniacfive]

Hello Y'all.

First post actually contributing to trying to solve a card, so please be gentle, I've just read through 42 pages of forum, making notes the whole time, but I may well say stuff that has already been said. I apologise, it 2:49am, i won't feel the trouting, trout away.

My first thoughts are regarding the disappearing texts. Now I'm not very good with this sort of cipher, but is it possible, that to understand the whole message you first need to decode the string WITH the text that disppears, then decode the string WITHOUT the disappearing text? Making a double length message? Perhaps the doubles in the 108 characters supports that the same string can be decoded in 2 diffrent manners?

or 2nd, Mind Candy went to a lot of effort printing the card with heat responsive ink that disappears? What significance lies in the disappering letters? myabe those letters shuld disappear from the rest of the text as well?

Removing EODMF XRUTH from the first line you get

WBBCG IBLCQ YWZIP JLWYL JYYNG YJBIO GI
28 characters, not a magic number to me but still.

Removing EHURG IFULP from the second line you get

XAXNQ JXAWJ TAMSM MOJMV BAAKP GWWXO VMYKZ
35 characters

If you remove all the disappearing characters you are left with

WBBCG BCQYW ZJWYJ YYNYJ BOVAN QJAWJ ASOJV BAAKP GWWOV YKZ
48 characters?


My only other thoughts probably just recap to stuff already mentioned, but i wrote down cause it occoued to me it was important, so it can't hurt to recap right?

The decka are all unmarked EXCEPT for the joker with the star. The star was not on the pack, it was added deliberately to somehow help with the cipher. I'm sure Garnet can tell the difference bewteen a star added to the cards later. The star is very important.

The particular brand of deck is not important, It's made on Earth, so the Perplex City deck was not the exact same deck re: deck order. It couldn't be.


I hope I've thrown at least one fresh idea out there. Shuffled is driving me mad, luckily I have a space just above it in my card folder, and I have placed a picture of kittens in a barrel in that spot. The kitten on the left is called Philip, and when I look at his cute little eyes, I can't stay mad, even at Shuffled. So long as I have my kitten picture, my sanity is intact.

Cheers

Al

EDIT: Clearly I had missed characters I was supposed to be removing and they had been left in. Will go back and try again

[Guin]

good to meet you maniacfive. At least I know I am not going mad as I sat and did something similar last night with the letters. Removing them etc. I questioned my reason, could think of non so drank more and carried on - hey im mad.

One thing I picked up on though is the freqency of certain letters and the abscence of others. No D's for example lots of G's.

i am really trying to sus how either the 108 letters as grouped pairs represent the deck - for example SM appears twice - could this be the jokers? I ve tried letter values - eg if SM = joker (53) is S = 5 and M = 3) but it seems to fall short.

Loq i read that thing on the site too about the A+K - B+K etc and I really cannot get my head rounud what he is saying so am happy to sit back on that one. It seems to be worth investigating but I really do not understand it.

i was wondering - I had pmed Strat already - but would it be worth setting up a room in IRC one night to chat together and bounce ideas? Many hands make light work - or too many cooks (to quote another card!)

well I left the pesky cards at home today and am resting my brain so no new ideas here (well apart from getting a mob and some torches (see my post in Riemann!))

See I am rambling now........la lallla lallal a

la

[locqust]

Sure Guin, As you've noticed Im on most nights, except Tuesdays but Im off to Majorca next Friday so anytime before then, just let me know!

Must admit Im not sure how it works either, it just seems a logical idea considering all the other ideas we've tried, including the basic ones.
Effectively though it would entail doing a dictionary attack of sorts against both sets of characters, I think!

Im going to google this and see if I can find some *helpful* websites to explain this process or even hint better at what I need to do.

On the bright side if this isnt the way to go at least Ill have learnt how to crack solitaire encoded messages (and stegged pictures at this rate too!)

[jamesays]

I think I understand the (A+K)-(B+K) thing.


For example, assume we encode two messages with the same keystream:

Message A PROPS (16,18,15,16,19)

Message B BLEED (2,12,5,5,4)

Keystream (K) JXFDQ (10,24,6,4,17)


The two encoded messages are (A+K) and (B+K), i.e.

(A+K) ZPUTJ (26,16,21,20,10)

(B+K) LJKIU (12,10,11,9,21)


Now by taking the difference between the messages (A+K)-(B+K) the keystream disappears in a puff of logic. It's the same as cancelling like terms on either side of a maths equation, you don't need to know what they are, they just have to be the same.


(A+K)-(B+K) (14,6,10,11,15) NFJKO

At this point the numbers are more useful. We're not looking at a message, but the difference between messages A and B. So the first letter of message A is 14 characters further throught the alphabet than the first letter of message B. The second letter of message A is 6 characters further on etc. That whittles it down to only 26 combinations, which shouldn't take long to search through.

If there are two messages that we can crack simultaneously like this we need to know where they both start. Also, if the messages aren't the same length, the characters at the end of the longer message will be impossible to decode in this fashion, because the keystream falls out as a constant, we never actually discover what it is.

I hope this makes sense, and maybe even helps a little (who knows, stranger things have happened!). Even if nothing comes of it, it's another avenue that we can ignore.

[jamesays]

...oh, hang on. It whittles it down to 26 combinations for the first letters of A and B, 26 combinations for the second letters of A and B, 26 combinations for the third letters of A and B, 26 combinations....

Might take a little bit longer than I thought...oops.

[UKver2.0]

...and, of course, that's exactly what the little spreadsheet I attached a few posts back does. The way I have it set up, you don't even see the keystream - poof (of logic)!