Kryptos - The Cipher (Part 4)

EASTNORTHEAST - This is not exactly the hint Jim Sanborn (JS) gave for K4 on the 29th of January this year. He only gave NORTHEAST - which refers to the positions 26-34 of K4's plaintext.  Beside BERLIN and CLOCK it is the third revealed plaintext word of K4. However, also this hint does not seem to help much. 

However, it just so happened, that a member in the yahoo kryptos group had a conversation with Jim Sanborn due to a submitted solution. Sandborn's answer to the question contained again the last clue which surprisingly was EASTNORTHEAST at position 22-34. There is disagreement if Jim revealed this on purpose or he did it accidentially, but the new extended clue seem to be serious and valid. 

Interestingly, EASTNORTHEAST is exactly the direction which is illustrated on the
wind rose on one of the stones around kryptos, also created by Jim Sanborn.

Below you can see that new hint at the position in the plaintext:

                                                       O B K R 

 U O X O G H U L B S O L I F B B W E A S T N O R T H E A S T O 

T W T Q S J Q S S E K Z Z W A T J K L U D I A W I N F B B E R

L I N C L O C K W G D K Z X T J C D I G K U H U A U E K C A R

 
Actually, i don't like the representation above, since it suggests that there is a 1-to-1 correspondence between the position of the letters in the ciphertext and plaintext. Although JS often said that this is the case, i actually think he did not really understood to question (See Post 2), due to his following statement he did lately:

Quotes from a March, 2019 lunch by Elonka: We spent quite a bit of time asking Jim about the correlation between the plaintext BERLIN and the ciphertext such as NYPVTT. Specifically, we were trying to find out if there was a 1:1 relationship from NYPVTT to BERLIN, or there was some other step, the masking technique. Jim was confused when we mentioned masking technique, evidently it's something that Ed said that Jim didn't understand. Jim said that yes the ciphertext and plaintext were connected, but when I tried to explain what exactly we were asking, like that in K1 EMUFPH is exactly BETWEE, but in K3 ENDYAH does not map exactly to SLOWLY, Jim backed off and said he would only commit to the fact that K4 is exactly 97 characters long, and that BERLIN is plaintext that starts at exactly the 64th character, but he wouldn't go further than that.

(kryptools.com/hints)

So transposition ciphers maybe back on the line. This gives me the chance to talk a little bit about the approach of Scott, who succeeded to create a reasonable method that makes BERLIN to appear at the correct position. His approach is as follows:

1. Apply Quagmire3 with the two keywords KRYPTOS and EMUFPHZLRFA to the ciphertext of K4. Note that Quagmire3 was also used for K1 and K2. The second keyword are the first 11 characters of K1.
2. In the plaintext of K2 the two coordinates 38° 57' 6.5'' N and 77° 8' 44'' W are mentioned. Scott used the linear congruence $y = 77 + 38x \pmod{97}$, that involves the two degree integers from the coordinates, whereof $x$ is the $x$-th character in the result of step 1 and $y$ is new position of $x$ in the output text.

The result is:

                                                      P G Z F

G T E Z P F C Y G F L J L Z G S O J A I L W D C C N Z F H S H

Q H I V Z K U C G O Z A A F J E V H O C Z V S N H Y E H B E R

L I N P E E K R R R H B U X J O P Y F F T X B M M Q M O F Q S

A closer look reveals that the characters that make up BERLIN actually only depend on the 6  characters MUFZLR of the second keyword: EMUFPHZLRFA. I played around with his approach, and i quickly realized, that all positions before applying the linear congruence of the characters BERLIN are $0 \pmod{7}$ and $4\pmod{7}$. The letter $K$ in the Quagmire3 method using KRYPTOS as the first keyword creates fixpoints. I also realized that every second Quagmire3 decryption of KRYPTOS and a second keyword that has length $7$ and has at the positions $4$ and $7$ the letter $K$, would also create BERLIN at the correct position (using the same linear congruence) after applying the two Quagmire3 decryptions. However, what actually surprised me is the following:

      EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJ

      YQTQUXQBQVYUVLLTREVJYQTMKYRDMFD

      VFPJUDEEHZWETZYVGWHKKQETGFQJNCE

      GGWHKK?DQMCPFQZDQMMIAGPFXHQRLG

      TIMVMZJANQLVKQEDAGDVFRPJUNGEUNA

      QZGZLECGYUXUEENJTBJLBQCRTBJDFHRR

      YIZETKZEMVDUFKSJHKFWHKUWQLSZFTI

      HHDDDUVH?DWKBFUFPWNTDFIYCUQZERE

      EVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDX

      FLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKF

      FHQNTGPUAECNUVPDJMQCLQUMUNEDFQ

      ELZZVRRGKFFVOEEXBDMVPNFQXEZLGRE

      DNQFMPNZGLFLPMRJQYALMGNUVPDXVKP

      DQUMEBEDMHDAFMJGZNUPLGESWJLLAETG

One of only two possible $7$ length strings that have two $K$s separated by $2$ starts in the middle of EMUFPHZLRFA going down vertically and forms a T. The keyword is HXDKZEK.

So if you do the following 3 step approach:

1. Apply Quagmire3 with the two keywords KRYPTOS and EMUFPHZLRFA to the ciphertext of K4. 
2. Apply Quagmire3 with the two keywords KRYPTOS and HXDKZEK to the result of step 1. 
3. Use the linear congruence $y = 77 + 38x \pmod{97}$ whereof $x$ is the $x$-th character in the result of step 2 and $y$ is the new position.

This will get you the result:

                                                       P G Z Y 

G Q E I M Y X M Y H S M S R H B S M B R M C D U C O Z F H S H

Q T I F Z J D Z Y U H C V H L G W J S U K B A O H H E K B E R

L I N N E R I L J L P D D K O A T T G H O D C T N S M L F S S

I don't know if this leads somewhere, but i think it is worth to write down.

This said, i found that in 2009 Sanborn gave an interview that contains the following:

And Ed basically gave me—he gave me a primer of ancient encoding systems. And he also gave me some ideas for contemporary coding systems, more sophisticated systems, systems that didn't necessarily depend on mathematics. That was one of my prerequisites. So he told me about matrix codes and things like that. These are the parts of Kryptos that have already been cracked. So I can discuss them. But he told me about coding systems that I could then modify in a myriad of ways. So that even he would not know what it says. Okay? So that was very seductive to me. And so I took those things. We met two or three times. And that's what I based the whole thing on.

(Interview with Jim Sanborn 2009)

Sadly, this sounds as if all the methods that are used in K1, K2 and K3 are not again used in K4, so Scotts approach maybe a dead end.

➠5bit ASCII code. In 2019 JS said in an interview that roughly one year earlier he got a submission of a proposed solution from a german guy. And this solution scared him, since parts of it seem to be correct, but the rest was wrong. In the forum of Klaus Schmeh's Scienceblog, see [1; comment #18 and more], a guy makes several comments indicating that he might probably this german guy, since he is german and submits a solution to Sanborn in the correct time window. He briefly explained his approach which is based on 5bit ASCII code and bitwise transposition but kept much of the detail hidden. 

5bit ASCII code is indeed an option that is worth to try. However, there is no unique choice for a 5bit representation. The most obvious approach is to map each letter to its position in the alphabet. But already in this trivial case you can distinguish two choices; do we start with $0$ or $1$, i.e. $A \widehat{=} 0$ or $A \widehat{=} 1$. Furthermore, 5bit covers $2^5 = 32$ integers. You don't run into a problem when mapping the alphabet to integers. But when you start to manipulate the bits, e.g. swapping, negating etc, you start getting integers from the interval $[26,31]$ which have no character assigned. Also standard 5bit ASCII code tables like e.g. the Baudot-Code have the same drawbacks since non alphabet letters (special characters) occur during manipulation. Also a simple reduction modulo $26$ deletes information which renders a unique a decryption process impossible.

However, just for the sake of completeness, there is at least one simple non-trivial function $f$ that maps $$f: \{26,27,28,29,30,31\} \rightarrow \{26,27,28,29,30,31\}$$ and $$f: \{0,1,\ldots,25\} \rightarrow  \{0,1,\ldots,25\}$$ and is uniquely invertible. It is 

$$ f(b_4,b_3,b_2,b_1,b_0) \mapsto (b_4,b_3,b_1,b_2,b_0) $$

whereof $b_4,b_3,b_2,b_1,b_0$ is the binary representation of an 5bit integer with $b_4$ being the most significant bit. The input alphabet is:$$\text{ABCDEFGHIJKLMNOPQRSTUVWXYZabcdef}$$

and is mapped to $$\text{ABEFCDGHIJMNKLOPQRUVSTWXYZcdabef}$$ However, the entropie is low, 14 of the 26 letters mapped to itself. I assume that $f$ is worthless and itself just creates a monoalphabetic substitution.

One main argument to work with base 5 is that the Berlin clock also represents the time in base 5. But this seems to be an odd argument. Why should JS's plaintext contain a hint towards solving its own ciphertext? It is not accessible unless some reveals it. Did JS anticipate that he has to reveal a clue due the hardness of the cipher and included BERLIN CLOCK into his plaintext to be able to give a meaningful hint? Isn't more logical that the base 5 argument is part of the riddle, which JS said is revealed once K4 has been decrypted? Another argument are often the 5 letters dYAhR, which somehow indicate a transposition, since dYAhR is a transposition of HYDRA, a term which makes sense in the context of Jim Sanborn. Since this five letters are part of K3 and K3 was encrypted using a transposition cipher, it was maybe a hint for K3 itself rather than K4.

➠PALIMPSEST & ABSCISSA. Where to start with? Base 5 maybe a dead end for K4 since it is the method for K5. All matrix based cipher, hence also Vigenere/Quagmire approach are also a dead end, because he said, that he used a different technique for K4. What about Ed Scheidts masking technique? What about the strange symbols that appeared in Ed Scheidt garage entrance a few years ago

Garage driveway of Ed Scheidt

They look a little bit like the BERLIN CLOCK - but what could it be?

If you read through all the many statements Jim Sanborn has given over the last years, you can conclude that he probably did something completely different from the first three ciphers - something more visual, an encryption method an artist would choose. 

Why did JS chose the two keyword PALIMPSEST and ABSCISSA for K1 and K2. This are no usual words and also have a meaning that could fit to the encryption of part 4. If you look at the whole Kryptos statue at once, you can view it as a form of a cartesian coordinate system with its four quadrants.

Kryptos - Seen as a coordinate system

 

The term ABSCISSA is used to describe the horizontal X-axis, the vertical axis is called the ORDINATE. Maybe this is the correct view in order to solve K4. One could for instance use some kind of mirroring at the ABSCISSA, rotation or even more complex functions using the characters of K4 as the input.

[1] http://scienceblogs.de/klausis-krypto-kolumne/2020/01/30/jim-sanborn-publishes-new-kryptos-clue/