Sorry for not being brief… as a reward for tolerating this post, there is a method to convert BERLIN into the first few letters of K4 in here. Its just another coincidental occurrence, but I find it intriguing.
As I recall, Ed said:
(1) the masking technique masks the English language
(2) the method may or may not be known
(3) the technique(s) for K4 does not reveal the identity of the sender or receiver.
I suggest some matching hypotheses:
(1) means the method masks the letter frequency – but is not useful as an encryption method.
(2) suggest that it uses a simple method but one that is not utilized in the real world of spies – probably because it is too easy for computers to decode.
(3) May mean that there is no keyword. The tableau that occupies half of Kryptos is probably OK to use because it is there for everyone to see. Possibly, it may be a running key sequence based on text within K3.
Also, there were some conversations with Sanborn where he made a point of not saying if K4 was 97 or 98 letters. And what is the “?” for anyway?
So I tried the following idea: shift the plain text by one character and use that as a running key sequence for encoding. This will mask the letter frequency but it is too easy to decode to be useful for spies. It also messes up the beginning/end; maybe that is the origin of the question mark.
Consider:
P) BERLIN
K) ERLIN*
Find the letter in P in the ABC column at the far left of the tableau. Move across until the letter in K is found in the KRYPTOSABC row (below the ABC row at the top). Get the letter from the tableau. [Sorry I don’t know the Quagmire X or whatever nomenclature for this] This yields:
FOBKR*
To get even more imaginative, suppose Sanborn first obfuscated the plain-text by moving the first letter of a word to the end. Use the above method to get:
P) *ERLINB
K) ERLINB*
C) ?OBKRU*
That’s the first few characters of K4!
Unfortunately, as these things always turn out… its just a random tease. Continuing the method yields garbage after that point. And trying to encode CLOCK yields letter sequences that do not appear in K4. Not to mention that its in the wrong place. So this appears to be just a coincidence. But its awfully intriguing.
Maybe this will stimulate someone to think of a method that works for the whole thing!