I know, I know, supposedly the letter frequency will not do us any good (according to Ed Scheidt).

Continuing the assumption that there are a portion of the letters in K4 that are meaningless, is it possible to estimate the size of the hidden message?  At first one would say no but consider again.  If it’s just a transposition with null letters blocked in, the E will still have the highest frequency.  In K4, the number of E’s is two so unless it’s a really short message, it will need to be substituted.  Therefore, following the same reasoning, the most frequent letter will be E which corresponds to K in the fourth part of Kryptos.

Speculations as to the letter identities after this point becomes muddled as we don’t have many clues from the K4 letter frequencies.  One thing that we can do is estimate the size of a plaintext message that would have 8 K’s (CT) or, 8 E’s (PT).

How?

Message length proof:

Assumption: K=E
Observation: in K4 there are 8 K’s, therefore there are 8 E’s
Observation: the plaintext frequency of E is 12.7% or 0.127
Equation:  PT frequency = # of E’s/# of letters

PT frequency = # of E’s/# of letters
PT freq.(# of letters) = # of E’s
# of letters = # of E’s/PT freq.
# of letters = 8/0.127
# of letters = 63

Conclusion: Assuming K=3 in Kryptos’ K4 section, the message length is 63 hidden by 34 null characters.

 
It is possible to determine the number of PT letters by multiplying their frequency by 63 but this is no guarantee you will be able to guess them correctly.  The only value they would have is a slightly more focused brute force attack and I have neither the time nor the energy to attempt them.

Besides, every indication is that you can’t just brute force it.  I would imagine the CIA, NSA and Elonka will have tried that by now.

I think I’ve reached the end of the road with my ideas about null ciphers.  Even though they would prove quite useful in many situations, enough of the ciphertext would remain exposed or continuos that you would have small recoveries of plaintext while you were trying everything and anything you could.  Even if you had 7 letters that did nothing more than change letter frequencies, you’d still be able to see the message underneath.  Can I promise I’ll never try a null cipher again?  No, but I think I need to at least consider other options for a while. 

BTW, my internet intuition says that a lot of the interest in Kryptos has died down after the 2005 WIRED interviews and after Dan Brown’s book causing a resurgence of interest.

Folks just don’t know what to do with it and a lot of folks just don’t really care anymore.

Kryptos Fan

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