So forget everything I’ve ever said and let’s return back to an observation that many have seen.

OBKR
UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO
TWTQSJQSSEKZZWATJKLUDIAWINFBNYP
VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

Reverse Row 2
Shift two letters to the right on lines 1 and 3 
Switch lines 2 & 3

KROB
YPTWTQSJQSSEKZZWATJKLUDIAWINFBN
OSSKGNRPQQVRLFWBBFILOSBLUHGOXOU
VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

The problem I have is that by assuming there is some tricksy substituting cipher of some kind at work, how do you explain the presence of “KRYPTOS” and what exactly does it mean to have it in there in the first place. 

Explaining the presence is just that.  The difficulty of manipulating the plaintext-cipher system-ciphertext-whatever other crap is going on with K4-and the final ciphertext output to allow KRYPTOS to be found is perhaps more than I’d expect.  Not impossible, just hard and curious.

For those who aren’t frothing at the mouth and shouting from the rooftops that K4 is strictly 97 letters long, take out KRYPTOS and you’ve got 90.  That’s grid-able my friends.

I’ve considered this little issue before but in light of my recent adoption of the fractionation/diffusion idea involving speculations on irregular polyalphabetic substitution… wouldn’t it be easier to believe that it’s just some kind of digraphic substitution like a Playfair/Polybius incarnation that uses 90 letters? 

Maybe the errant letters just help you arrange the text to reverse the encrypting transposition and have some interesting substituted ciphertext to hammer on?

Whatever the case, my previous proposal fails to take this detail into account and therefore the hypothesis must be rewritten to account for it.

No way it’s a coincidence you can find them.

No way, no how.

Kryptosfan

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