(Wikipedia) The four-square cipher is a manual symmetric encryption technique. It was invented by famous French cryptographer Felix Delastelle.
The technique encrypts pairs of letters (digraphs), and thus falls into a category of ciphers known as polygraphic substitution ciphers. This adds significant strength to the encryption when compared with monographic substitution ciphers which operate on single characters. The use of digraphs makes the four-square technique less susceptible to frequency analysis attacks, as the analysis must be done on 676 possible digraphs rather than just 26 for monographic substitution. The frequency analysis of digraphs is possible, but considerably more difficult – and it generally requires a much larger ciphertext in order to be useful.
So although we know K4 has been masked in some way that renders it impossible to use the English language to translate it, who can resist the lure of attempting a foursquare solution. It is a matrix system like Sanborn likes. It’s a different enciphering method than K1-3. It distorts the plaintext letter frequency. In a lot of ways, it seems like the perfect solution.
Unfortunately we know that it doesn’t take into account the masking technique which will interfere with any cryptoanalytic method. It also has 97 letters and foursquare works on a series of digraphs. Foursquare also allows the use of keywords in the Ciphertext quadrants. Even if we were 100% sure a foursquare would take us directly to a solution, we couldn’t be sure of the keywords.
I tried it anyway.
I was hoping that it might force out some plaintext which would then help isolate some of the masking technique. The best I got were some two letter words and it’s inevitable you’d get a few of those.
For my keywords I used Kryptos and Quagmire.
abcde KRYPT
fghi/jk OSABC
lmnop DEFGH
qrstu I/JLMNQ
vwxyz UVWXZ
QUAGM abcde
I/JREBC fghi/jk
DFHKL lmnop
NOPST qrstu
VWXYZ vwxyz
Each ciphertext is located in their quadrants and a line drawn horizontally and vertically. The intersections that occur in the plaintext are naturally the plaintext letter. By convention it would appear that plaintext 1 and ciphertext 1 are the top two sections and ciphertext 2 and plaintext 2 are the bottom.
Since there are 97 letters, I tried it in the original order and then shifted one letter to the right to cover the possibility that the masking technique involved removing some of the letters.
I didn’t find much, just to warn you.
Regular series with the extra R on the end:
IF-BF-QW-TW-ON-LZ-IT-LK-LR-II-NW-GR-BV-TS-RA-OO-RJ-QK-SZ
UT-AQ-RI-MO-ZZ-CX-KA-LE-LV-AS-HV-OR-TF-SL-RZ-EE-PW-OD-CY-LO
YX-KA-PF-AT-AB-EM-CG-MO-EH
Offset series missing the O at the beginning:
OI-BB-VH-AJ-EM-GT-RG-GQ-HO-YG-NP-WB-EQ-IB-OT-QD-RG-ZB-EA
GF-UT-GH-VE-ZW-SK-LT-BR-FL-XG-QQ-IO-YT-YA-UE-XU-SN-VB-OM-VE
TZ-LU-FL-OO-LX-AX-FX-FE-HG
So, not much. It did become obvious that the end of the alphabet for ciphertext would remain the same no matter what the keyword was. It’s likely that v, w, x, y, z would remain in the same position. The largest distortion would be in the first 5-10 letters of the alphabet.
Otherwise we don’t have enough ciphertext to make any conclusions about the digraph frequency.
Maybe it would work after the masking was removed.
But it looked like a failure.
Kind of hard to move past that…
Kryptos - Beyond K4 
This must have been before you found out you can do these online
I’ve been keeping this to myself unless I find someone that tried the same thing I did (or rather “am trying”).
Try keeping the “r” at the end & just add an “x” at the end.
(like in playfair, if there’s an odd number of letters, add an “x” at the end).
good luck, I’m just some random person who’s also attempting to solve this in spare time. (just for fun.)
Does that mean you had some luck?
A little bit, I think.
If you used keyed alphabets, you could try adjusting the keys to capitalize on a break if you’ve found one.
Instead of trying to figure out how to deal with the “odd # of letters”, just do it & try it yourself. It’s easier that way.
Foursquare might work as the second layer but over and over again you hear how there is a masking method and I can’t help but think that it will be “over” the more traditional cipher method. I’ll probably hold off unless I find something that makes me think I’ve got the masking method sorted out.
In regard to that “masking” thing,
would it happen that the plaintext was “DEcrypted” instead of “ENcrypted”?
Like “Decoding” the plaintext, and then encoding that cipher text to make the present ciphertext.
I think that would be a good idea to take into consideration.
Hope this helps & good luck with it!
It’s a good idea and when possible I’ve tried running it both ways but I haven’t always been good about that. If you put the ciphertext from K4 into a frequency program or even just do your own tick marks, you can see that there was some funny business. The text isn’t completely flat which would indicate a high amount of randomness but it’s also not like plaintext (transposition) or plaintext-ish (monoalphabetic substitution). It’s more like it’s muted. There are some things that stick out like the fact that every letter from A-Z is in K4 at least once which would be unlikely but not impossible in such a short text. This is the observable effect of Ed Scheidt’s masking technique which was intentional to make it harder to finish the last section (ref: interviews, anecdotal gossip).
Say, after I thought about it for a while, I was thinking that maybe Kryptos might NOT be a Transposition Cipher at all, because of when JS gave that “Berlin hint”. Although I’m not sure if this is correct because there might a other ways to do for the Berlin ciphertext to be in the same position after the transitions. For example:
(This is one way that I think might work to make the ciphertext letter positioning corresponding to the plaintext letter positioning in the same positions)
Plaintext –> Transposition –> Encryption –> then finally another Transposition back to plaintext order (corresponding plaintext-ciphertext letters are placed back into the same positioning as the original plaintext order)
A bit vague but you get the idea right?
Basically, if a Transposition was used, I think it should be done twice to make the cipher-plaintext letter positioning the same as the original plaintext letter positioning.
(Phew! Now that was a bit of typing there. – I think… Maybe? – I can’t really tell from where I’m typing it.)
Note: Also, please know that I have not tested it yet.
Actually, now that I noticed, that make things a bit “redundant”.
Also, we’re replying a bit too much to the same comment!
I’d start a new comment for the reply.
That’s quitter talk, I’m sure there’s squeezable room until things are just
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Hehe, okay fine then. So, any new leads yet? Did you get any contact info yet? If not, I can “try” to help thought I’m not sure how successful I’ll be at that. Still, Good Luck either way!
I’m waiting on Sanborn at this point but have had no luck with Ed Scheidt, David Stein and the NSA guys. At least the NSA said they’d forward my email even if they couldn’t promise an answer.
Say, I want to know something. Has anybody tried a “2-Square”? (also called a double playfair)
Just wondering.
Also, that “layer two” thing, (after hearing about a “double encryption”), got me thinking, so I’ll probably have to rework my some things out.
Good Luck on your side!
I think yes but it’s usually wrapped into some really unnecessarily complicated many-step process that still yields nothing. I would say it’s worth considering to be sure, the problem I’ve always run into is picking keywords for the keyed alphabets. Everything I do is unfortunately by hand but if you’re a programmer you could probably automate something to check a lot of them.
Definitely good luck to you too!
I also do everything by hand as well. (& I’m quite used to it too)
(Also, I mainly do it on Notepad, so I don’t have a bunch of scattered papers)
I’m not a Programmer (yet), but I’m (currently) learning how to be one.
Say, by the way, could I ask if you know (and could list/send me a list) all the cipher methods used/tried to decrypt Kryptos?
Just wondering, just so I’d know what all has been tried.
Thanks!
Here’s my email address if you’re going to send it to me:
unknown.anonymous.random.user0@gmail.com
(best to copy & paste that – kind of long, right?)