In a normal discussion/conclusions section you’d find all sorts of significant interpretations of the results.

We can summarize.

Basically I hypothesized that:

“Sanborn was to write a message of significant context (#1) with the total 97 letters.  Then, the separate 90 letters were enciphered with a digraphic substitution cipher (#2).  At this point the seven plaintext letters “kryptos” were added back into the message (#3) and a transposition cipher used to separate the digraphs (#4) which would make cryptanalysis difficult.”

I made some predictions.

I developed some methods.

I got some results.

It didn’t work.

That’s about as in-depth as it gets.  I didn’t find any significant lengths of plaintext fragments or a complete solution.  After about 10 iterations, can I be sure that I would have statistical probability on my side?  No.  It was a good enough effort in my mind.  So I was hoping for even a 9-10 letter fragment which isn’t much.  If you figure that there are 80 possible 10 letter fragments in a given orientation of K4 text and that I tried 10, I essentially was working with n=800.  I did three different cipher systems a grand total of 14 replicates for each n so my tested possible 10-letter fragment treatments were 11,200.  None of them worked.  Sure there are a ton of possible orientations for K4 and a lot of possible digraphic cipher-keyword combinations but I don’t really have access to the IBM Roadrunner at the Los Alamos National Laboratory.  What I did was good enough for me.

Is it impossible that K4 is still a digraphic subsitution/transposition cryptosystem.  Yes.  My recommendation to anyone else would be to not waste effort on it unless you’ve recovered the keywords needed for other sections and K4.  If we had solid hints/clues/directions then the process would be greatly simplified.  If wishes were horses everyone would ride.

Solution attempt complete, I plan on pursuing other deciphering strategies at this point in time.

Kryptosfan