So if the 4th part of Kryptos is 98 characters after all, then what is the most likely candidate for a superenciphering additive?

Well, 98 factors into several sets but 7×14 is the most logical to me simply because KRYPTOS is 7-letters long.  What could be easier than adding (or subtracting) kryptos mod26 from the ciphertext of K4 to mask the linguistic characteristics?

I even included your very own spreadsheet with K4 and KRYPTOS as numbers if you wanted to do it yourself!-Kryptosfan

11 18 25 16 20 15 19 additive or subtractive
11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25 16 20 15
? 15 2 11 18 21 15 24 15 7 8 21 12 2 19 15 12 9 6 2
11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25 16 20 15
19 11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25 16 20
2 23 6 12 18 22 17 17 16 18 14 7 11 19 19 15 20 23 20 17
19 11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25 16 20
15 19 11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25 16
19 10 17 19 19 5 11 26 26 23 1 20 10 11 12 21 4 9 1 23
15 19 11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25 16
20 15 19 11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25
9 14 6 2 14 25 16 22 20 20 13 26 6 16 11 23 7 4 11 26
20 15 19 11 18 25 16 20 15 19 11 18 25 16 20 15 19 11 18 25
16 20 15 19 11 18 25 16 20 15 19 11 18 25 16 20 15 19
24 20 10 3 4 9 7 11 21 8 21 1 21 5 11 3 1 18
16 20 15 19 11 18 25 16 20 15 19 11 18 25 16 20 15 19
? O B K R U O X O G H U L B S O L I F B
B W F L R V Q Q P R N G K S S O T W T Q
S J Q S S E K Z Z W A T J K L U D I A W
I N F B N Y P V T T M Z F P K W G D K Z
X T J C D I G K U H U A U E K C A R  

 

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