I dunno, maybe.
I think Vorlath will figure it out before me but it’s worth considering.
Straddling checkerboards fall under a substitution cipher because they convert plaintext into numbers. These numbers can also be converted back into letters and also modified by keywords.
You build a 4×11 grid with numbers across the top row, the 8 most common letters under those, label the last two rows with the numbers that didn’t get letters and then fill in the rest of the grid:
Each letter can now be replaced by its coordinates. This diffuses the message across a monome-dinome cipher and also suppresses letter frequencies that might otherwise help us.
It’s a good bid for K4.
I can’t attempt it because I don’t know how to set up the encryption/decryption grid for Kryptos, I don’t know if a keyword was applied to the letters and what it would be, and I don’t know whether this is the only ciphering mechanism used on K4 or if it comes before or after the masking technique. I could probably spend a lot of time on it but it would be more of a matter of hoping that I got lucky and hoping someone found a solid method for identifying keywords/clues.
There are some folks who are trying to sort this one out but I don’t have enough solid evidence to make my own attempt. If it’s doable, let’s hope they have some luck.
Again, this would be a very logical method for Sanborn to choose to use so it should be considered.
This gets my vote
VIC Cipher.
A VIC cipher would explain why no one has broken it but it’s pretty much a ridiculously hard cipher to solve with only 97 letters on pencil and paper. One can only hope that Sanborn actually wanted us to be able to solve this.
Agreed, unless he gave you all/most of the keys. Just glad to see you replied, I’ve used your site countless times throughout my research even though you were out shortly before I started.
A theory is set up and it’s housed within the Yahoo Group if you’re interested in taking a peak. Although i understand if you’re still retired.
Take care!
Here’s a thought I had. JS would have had to leave us some clue how the checkerboard was organized. So..
K1 -> Palimpsest -> JATFB (based on my observation from his notes)
K2 -> Abscissa -> 77844 (the longitude, from the K2 plaintext)
K3 -> RAY (the upshifted characters, in plaintext order of occurrence)
JATFB + 77844 = RAY?
JATFB – 77844 = RAY?
Am trying various checkerboards based on this rule but haven’t found it yet. I assume the alphabet is keyed but cannot determine the keyword.
Checkerboard ciphers can be solved without knowing the board. For a board like the one shown in this post there are only 45 possible permutations for every ciphertext. You can quickly find the correct permutation using IC calculations.