There are several assumptions we can make (have to for this to matter):
- The keywords are retrievable from the Kryptos materials
- The likeliest location is the Morse Code
- It is very unlikely that a substitution cipher was used to hide them
- They are hidden because no one knows how to find them
- It’s not an easy, cut and dried transposition (that we know of)
- It’s possible to brute force palimpsest, abscissa, vigenere and two rotations out of the Morse Code letters by anagramming
- We’re not really sure which orientation to use the Morse code in or whether to include the E’s
- We need a retrieval method to obtain a complete keyword message so we can be sure what the clue for K4 is
- There’s a hint for solving each part of Kryptos
- We were expected to figure this part out first
- It’s possible the words/phrases aren’t perfect plaintext (i.e. xkeyxxwoxrdxxxx)
- The 25 extra E’s probably change the location of certain letters around, perhaps in a grid of some kind.
- No one has figured out the keyword retrieval yet
- It’s very important
- I want to name it something funny
So, figuring the retrieval is not an irregular method (1st letter, 45th letter, 48th letter, 56th letter) but more of a regular one (every 5th letter), we can attempt to detect a transposition at work within the Morse Code letters. We need some method of detecting a regular distribution of the keyword across some version of the Morse Code.
P(i)MP Test Introduction
The Morse Code has only 2 P’s and 2 M’s. Palimpsest uses both P’s and one M. We can use this to develop a method for not only determining the keyword message distribution but also the orientation/version of the Morse Code to use when extracting the message.
We know Palimpsest is a keyword and we know it can be found in the Morse Code. This will be the initial basis for the Test.
Ignoring the more common letters, we can simplify palimpsest into P***MP****.
With P***MP**** we can make some important observations. The first is that the two P’s will be 5 units apart. The second is that the M will only be 1 unit from a P. So P-to-P = 4 and M-to-P = 1. I’m defining a unit here as the block of ciphertext that lies between each letter of the keyword. Now you know why I’m assuming a regular transposition or a simple transposition.
Example: If a unit ends up equaling 5 letters, then there will be 25 letters between the P’s and only 5 letters between an M and a P. Remember that we are counting the space between each letter. P-5 letters-A-5 letters-L-5 letters-I-5 letters-M-5 letters-P.
So how can we simplify this further to make a fast diagnostic test?
P(i)MP Test Diagnosis
So, take any string of the Morse Code in any direction in any arrangement, anything, anything, anything. Count the number of letters between the two P’s, count in both directions.
Subtract 4 (ALIM) and divide that number by 5, the number of unit gaps (*) between p*a*l*i*m*p. If it’s a whole number then this is the encryption/decryption key you want to use.
A fast test of the key is to see if M is one key-length from a P. To allow for the possibility of muddled plaintext to throw us off, if the M is more than one key-length away for it to be considered possible, it must be some (multiple of the key-length+1, 2, 3) in order to do a fast confirmation of the retrieval.
Use your new-found key and try to retrieve the rest of Palimpsest.
Use your new-found confidence and successful method to retrieve the keyword message and share with the world.
If it failed but you’re really, really, really sure it should have worked. Go back to the math bit (2nd step) and consider if there may be some intentional spelling mistake. See if there’s at least an A, L, I and M between the P’s and reconfigure what the unit spacing could be.
Give me an example!
Distance between P’s? 24
24 – 4 = 20. 20/5 = 4. 4 = keylength.
Check distance of M? = to one key length
Peel letters off by every fourth to find palimpsest
Retrieve the rest of the message
I exaggerated how the message would read slightly to emphasize the point. In reality it would give us parts of the plaintext until we applied our method to the Morse Code string as a whole.
So why not just grid it out and get the same result?
My way is faster. It doesn’t require checking every grid size and length. It allows for distortion of the plaintext by intentional misspellings. It gives the key length. It also allows for irregular grid sizes. Without the e’s, we can make a nice 9×9 grid, or a 3×27 grid. With the e’s, there’s no easy gridding. It allows fast checking of various types of perspectives on the Morse code as well. It does however assume there is a regular algorithm at work that encrypted the message. It also assumes there is a message to recover.
- Pick your arrangement of Morse Code.
- Count the distance between the P’s in both directions.
- Subtract 4.
- Divide by 5.
- If a whole number, check the distance of M
- Try and recover palimpsest
- Use key to retrieve the rest of the Kryptos Keyword message from your Morse Code String
- Announce it to the world so we can make some progress solving K4