Yup, I just got done talking about this.

So the argument is tenuous, plausible but fragile and easily disputed…

The vigenere side of Kryptos is 866 letters plus a rogue L which I will refer to as “Ricky”. Now if we take those 866 letters and “Ricky”, we can check the ciphertext side of the Copperplate. Here’s where we need some background on Kryptos. There is an X in K2 that was missing and cryptanalysts added the Q at the end of K3. With these two taken out, we have 866 letters of ciphertext/plaintext.

So, just to emphasize the uncertainty of my argument: we leave out the X that Sanborn himself omitted, we leave out the Q that the original solvers added, and we ignore “Ricky” and we are left with the exact same number of characters on each side of the Kryptos Copperplate, he exact same number. Did I mention that they’re the exact same number?

Why does this matter?

Well, the ciphertext isn’t gridded out evenly in rows so what’s the point of keeping the exact number of letters the same? It would matter if the plaintext letters were the important letters because we can grid them out ourselves into the right rows and columns.

If the numbers were off by even 5 or 10 letters I would argue that it was just aesthetic motives for keeping the numbers similar in order to have a symmetric sculpture.

But if they’re the same, or for arguments sake within 2-3 letters of being the same, then it is probably (although not definitely) significant.

So, character numbers are important…
If a ciphering method was used that allows fractionation of any kind, then the number count is thrown off.

That’s the argument against fractionation.
Modest, isn’t it?

So by extenuating circumstances, I am arguing that if Sanborn went through all this trouble to make it so both halves are exactly the same then it is doubtful he would change that in the fourth part. Especially when that fourth part is in rows of 31 already.

Yeah, I could be wrong.

Makes you wonder though, doesn’t it?

(Additions after some emails)

  • So it doesn’t necessarily mean you’d lose any characters if you chose your rows and plaintext carefully. By adjusting which letters get a single digit/letter and which get the two, it’s possible to modulate the ciphertext character count.
  • Also, the extra characters would need to make an appearance in K4 unless you carefully planned the text so they weren’t used.
  • All 26 letters of the alphabet are used in K4 but no symbols…
  • Is it possible to use a fractionation method with extra letters, like two A’s, B’s and the regular bits?  (I’m sure but how would you know to do this?)

These soften the odds that fractionation is at play but introduce quite a bit of work on Sanborn’s part