Kryptos - The Cipher (Part 5)

ONE of the most surprising fact about Kryptos is, that nobody has yet discovered the anticipated way the keywords for K1 und K2 should be found. And maybe this missing fact is the key why we are stuck with K4. I don't think, that Sanborn wanted us to bruteforce the solutions (as we did), but somehow left a hint we didn't recognized so far.

Most people believe, that the morse code messages around Kryptos should somehow encode at least the first codeword for K1 (= Palimpsest). That's why the morse code message are often called K0.

The morse code messages are part of two pieces of granit with contained copper plates. I found two very interesting drawings of these two installations from Monet Friedrich on [1]:

The drawings are very insightful. You can see, that the messages can be read in the correct order only by standing on the rocks. Second, in  the lower left there is a lodestone. And the compass points exactly to the lodestone, as probably a real compass would do, while standing next to a powerful lodestone. So this means:

• The compass is off by around 247,5° (with some error)
  
• The orientation the compass shows is not EASTNORTHEAST but perhaps WESTSOUTHWEST.
  

Does this mean, that we have to rotate things by around 247,5°?

Further close up photos of each individual message can be found at [2]. 

Below you find messages with their most probable decoding.

1. e e V I R T U A L L Y e     e e e e e e I N V I S I B L E
2. D I G E T A L e e e     I N T E R P R E T A T I O N
3. e e S H A D O W e e      F O R C E S e e e e e
4. L U C I D e e e      M E M O R Y e
5. T I S Y O U R     P O S I T I O N
6. S O S
7. R Q
   

One note to 2. The copper plate actually one contains INTERPRETATI and a final dash. At this point the copper plate ends. But it is assumed that the dash is the beginning of the final 'ON'. And there is a mistake in DIGETAL, the fouth character should be a 'I'.

The first thing that jumps into the eye are the extra letters 'e'. There is no way, that Sanborn goofed up the dots and dashes while engraving them in a time consuming manner into the copper plates. There are many speculations about the 'e's like:

• The extra 'e's are forming some kind of background noise or heart beat from which the message emit
• The extra 'e's are a hint towards the fact, that the number 5 (since 'e' is the fifth letter) plays a crucial role for decoding the scultupre. At least the hint 'Berlin clock' gives another object, which uses base-5 to encode the time.
• The extra 'e's are just added for visual reasons
  

There are also some theories how to extract the code word 'ABSCISSA' (keyword for K2), from passages of the morse code messages, which however, i find very far-fetched.  

One could also argue that the messages are simply an indirect hint how to approach the ciphers K1 - K5 or hints for the related keywords [K5 is the final riddle which is revealed once K4 has been decrypted - according to Jim Sanborn]. So, let us see arguments for that:

The keywords for K1 and K2 are the following:

1. PALIMPSEST (K1)
   
2. ABSCISSA (K2)

Jim Sanborn is an artist and not expirienced in cryptography, he knows that in order to decrypt a message, you need a key. I don't think, that he wants us to bruteforce the keys (which is what we did) but to derive the key from some other informations Again, the morse code messages (without the extra 'e's) are:

1. LUCID MEMORY
2. T IS YOUR POSITION
3. SHADOW FORCES
4. DIGETAL INTERPRETATION
5. VIRTUALLY INVISIBLE
   

I am omitting the short ones: SOS and RQ. Take a look at the definition of PALIMPSEST:

Definition: Palimpsest (Oxford Languages)

    1) a manuscript or piece of writing material on which later writing has been superimposed on effaced earlier writing. 

    2) something reused or altered but still bearing visible traces of its earlier form.

The term "...bearing visible traces of its earlier form" maybe a hint to "a piece of writing having a lucid memory of its earlier writings". So perhaps Jim Sanborn tought, that if a crypto-guy thinks about LUCID MEMORY, he comes up with the word PALIMPSEST. I know this is also far-fetched, but perhaps this is Jim's way of thinking. So it connects LUCID MEMORY to K1.

Next, that the second keyword ABSCISSA:

Definition: Abscissa (Oxford Languages)

    1) the coordinate that gives the distance along the horizontal axis (= the one along the bottom)

Coordinates, at least in the plane, are often given as a pair $(x,y)$. Where of the $x$ is the point on the abscissa and the $y$ is on the ordinate. So the statement "T is your position" refers to a pair of integers, i.e., an integer on the abscissa and one on the ordinate. So we could argue that this connects ABSCISSA to K2.

Let's skip K3 for the moment. Of the three remaining words ('shadow forces', 'digital interpretation', 'virtually invisible') the term 'digital interpretation' could very well be a hint for K4. Note that, as already mentioned in the last post, a german guy scared Sanborn with his submitted solution, in which he used a 5-bit ASCII code. Eventually, it was wrong, but according to Sanborn, the approach scared him. And using ASCII is some kind of a digital interpretation. So we could connect DIGITAL INTERPRETATION to K4.

Now we are left with 'shadow forces' and 'virtually invisible'. If our reasoning is correct, one should be a hint for K3 and one for K5. 

 

For K3 the encryption method changed compared to K1 and K2. So similar to the term 'digital interpretation', the hint for K3 does probably not lead to a codeword, but to a hint for the decryption method. And we know the K3 is a transposition cipher. 

The only slightly meaningfull idea i came up with, is that the link could be that, 'shadow forces' (which Sanborn uses to refer to magnetic fields) can be used to change the orientation of objects. And changing the orientation can be seen as some kind of transposition. And maybe this connects SHADOW FORCES to K3.

But to be honest, it is not very convincing. So, maybe you have a better idea for K3?

 

The last word 'virtually invisible' may be a hint that there is a fifth riddle at all, which was not known, before Sanborn confirmed it. So K5 was 'virtually invisible' to us. So maybe the fact that Sanborn engraved a fifth morse code message should be a hint that there is also a fifth riddle. So we have a connection from VIRTUALLY INVISIBLE to K5.

So we covered all morse code message and linked them to a cipher K1-K5. Is this correct? I doubt it, but worth to mention.

▉ The approach of 'german guesser' ▉

 

The  following is a citation of the comment from [1]:

The following happened during the CNN meeting in spring 2019: “When asked about the solutions that have been sent to him (Jim Sanborn), and whether any of them were close, Jim said he got one from Germany about a year ago that was “kind of scary”, that the first part had started to look right, but then the rest didn’t.”

So what did he do? 

 

He used a method from WW2 where the encrypted message itself contained the decryption key in an again encrypted form. The first 21 characters of K4 are (with two extra '-'): 

O B K R - U O X O - G H U L B S O L I F B B W...

 He assumed that OBKR is the decryption key (DK) for the message key (MK). The encrypted message key is UOXO. He then used DK to decrypt MK (with some unknown method, that uses 5-bit ASCII encoding with transposition). The result, i.e., the decrypted MK, is either CIAX or CIAW. Using MK he then was able to recover 10 plaintext characters afterwards, but it is not know if he used the same method as for decrypting MK itself. But then the method fails. And it is not known what the plaintext was. Were there clear english words or some kind of interpretable string?

 

To test his approach, one must first find the method he used to decrypt MK using DK to get CIAX or CIAW using some kind of 5-bit ASCII encoding. 

 

Note: The reason why he thinks that 5-bit is correct are the extra letters 'e' in the morse code.

 

I thought about this quite a while, and i don't think that 5-bit ASCII is the correct approach. As already explained previously, using ASCII you have to agree onto a decoding table first. We have 5bit, so we have 32 possible bit strings. 

 

Obviously, you could use something like $$\text{ABCDEFGHIJKLMNOPQRSTUVWXYABCDEF}$$ with double letters A-F, where e.g. A is represented by $A \widehat{=} 00000$ and $A \widehat{=} 01011$. And this ambiguity migh cause problems. If you have a ciphertext and you are going backwards, you have to decide which of the two 5-bit strings to use for $A$. If you choose the wrong one, you might end with a wrong decoding. 

 

For a proper encryption system, you have to guarantee, that the decoding is  unique. For example, assume that we are using the simple encrypting system that uses XOR with the secret bitstring $N \widehat{=} 01101$. When we encrypt the two letters $$N, G \widehat{=} 01101, 00110$$, we get $$A, A \widehat{=} 00000, 01011$$ However, the receiver, which wants to decrypt the message only gets $AA$ as the ciphertext (as we get alphabetic letters for K4 and not its binary representation). Then he has four options:

 

$\begin{align}
A, A & \widehat{=} 00000, 00000 \rightarrow \text{XOR with}\; 01101 \rightarrow 01101, 01101 \widehat{=} N, N\\
A, A & \widehat{=} 00000, 01011 \rightarrow \text{XOR with}\; 01101 \rightarrow 01101, 00110 \widehat{=} N, G\\
A, A & \widehat{=} 01011, 00000 \rightarrow \text{XOR with}\; 01101 \rightarrow 00110, 01101 \widehat{=} G, N\\
A, A & \widehat{=} 01011, 01011 \rightarrow \text{XOR with}\; 01101 \rightarrow 00110, 00110 \widehat{=} G, G
\end{align}$

But only the decoding in the second row is correct. And this ambiguity holds for all 6 letters ABCDEF.

Maybe what Sanborn had scared was not the 5-bit ASCII approach, but that he identified the first 4 characters of K4 (OBKR) as the decryption key. There is an argument, why this can be true.

If you take a look at all the row length of the left side of Kryptos, you get

$\begin{align*}
32,31,31,30,31,32,31,31,32,31,30,31,31,32\\
32,30,31,30,32,30,32,31,33,29,31,31,31,31
\end{align*}$ 

So, there is no more than a successiv repetition of two amoung the first 24 lines, but the last four lines, which contain K4, are all equal in length.This form makes 'O B K R' stand out, and maybe is another visiual hint for "take me, i am the key".

* * * * * * * * * * * * * * * * * * * * * * * * * * * 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

 

And maybe we should not use 5-bit ASCII, but using base-5 encoding, like the Berlin clock does. Base-5 is easier to handle. In base-5 a letter is a tuple with integers from $\{0,1,2,3,4\}$. E.g. $A = (0,0)$,  $N=(2,3)$ and $Z=(4,4)$. You can create 25 tuples, so one less than 26, but this is not a problem. Other, famous encryption methods (e.g. Polybius square) only use a 25-alphabet. They set either 'i' = 'j' or 'u' = 'v'. The pairs are made up of letters that are very similar (visually) and one can read a text, where a 'j' is replaced by 'i' without any problems. 

One might rise the argument that K4 contains all letters from A to Z and not only 25. Yes, but one could easily change some randomly 'j's back to 'i's (or 'v's to 'u's) after encryption, since the decoder will revert this changes anyway. 

 

And also base-5 is a digital interpretation, since the term only means that information is described by digits.

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

[2] https://kryptosfan.wordpress.com/morse-code/