November 8 2020 — In a previous post, we looked at the inside story of the NSA people who took the challenge to decrypt — part of — the KRYPTOS code. I will tell you later how the NSA people broke Section III but today I will present a novel and rather unusual solution.
Why? Because I have great doubts that Sanborn — an artist — has used complex mathematical transformation to code Section III. Who cares? Does it matter anyway? One would think that as long as the solution is correct, the method used to get it is irrelevant. One could be wrong. The devil IS in the detail. Follow us on Twitter: @INTEL_TODAY
RELATED POST: KRYPTOS 30-Year Anniversary — Introduction : Sculpture Dedication Ceremony at the CIA (November 3 1990)
RELATED POST: KRYPTOS 30 Years Anniversary — How to Break a Vigenère Code
RELATED POST: KRYPTOS 30 Years Anniversary — The Solution of Section II
RELATED POST: KRYPTOS 30 Years Anniversary — History of the NSA Involvement
RELATED POST: KRYPTOS 30 Years Anniversary — SECTION I : A KEYED Vigenère Cipher [And why does the CIA lie so much about it?]
The ciphertext on the left-hand side of the sculpture (as seen from the courtyard) of the main sculpture contains 869 characters in total (865 letters and 4 question marks).
The right-hand side of the sculpture comprises a keyed Vigenère encryption tableau, consisting of 867 letters.
In our last posts about KRYPTOS, we learned how to break a Vigenère code and we apply this knowledge to decode the entire section I and II. The section III involves a different kind of encryption method.
Encrypted Text of KRYPTOS Section III
Slowly, desparatly slowly, the remains of passage debris that encumbered the lower part of the doorway was removed. With trembling hands i made a tiny breach in the upper lefthand corner and then widening the hole a little I inserted the candle and peered in. The hot air escaping from the chamber caused the flame to flicker but presently details of the room within emerged from the mist X Can you see anything Q?
This is a paraphrased quotation from Howard Carter‘s account of the opening of the tomb of Tutankhamun on November 26, 1922, as described in his 1923 book The Tomb of Tutankhamun.
The question with which it ends is asked by Lord Carnarvon, to which Carter (in the book) famously replied “wonderful things”.
In the November 26, 1922 field notes, however, his reply was, “Yes, it is wonderful.” [Wikipedia]
Comment: The word “DESPARATLY” is obviously misspelled and stands for “desperately”. I do not know if Jim Sanborn has made any comment regarding this mistake.
I suspect that if Sanborn had spelled this word correctly, the symbol “Q” at the end of the plaintext would not be there!
Letters Frequency Analysis: Remember what the IC is!
The Index of Coincidence [IC] measures the probability that any two randomly chosen source-language letters are the same.
This probability — also known as the index — is about 0.067 for monocase English while the probability of a coincidence for a uniform random selection from the alphabet is 1/26 = 0.0385.
where c is the size of the alphabet (26 for English), N is the length of the text, and through are the observed ciphertext letter frequencies, as integers. [Tutorial]
The IC for the entire Section III is 0.0662. That is exactly what we would expect for an English text. You can check this result with an online frequency analysis tool.
Obviously, the letters have been scrambled around? The question is how? As I said in the introduction, I will tell you how the NSA solved this transposition cipher in a following post, but today I wish to describe an “artistic” rather than “mathematical” solution! Hang on! You will soon understand why…
The text is 336 characters long. Could KRYPTOS be a clue again, as in Section I and II? Transposition ciphers are about numbers.
So, if KRYPTOS is a clue, perhaps it is the lucky number 7?
Well, it turns out that 336 is equal to 7 X 48. Let us therefore write the cipher in 48 rows of 7 characters. Fill the first column (ENDY…), then the second and so on until the end (…DOHW).
Stairway to Heaven
This is still a meaningless message but notice what happens if you start with the “S” in the middle of the last row and count every fourth letter moving to the right and up.
Here is the result:
The sequence is:
S L O W L Y D E S P A R ….
And so on! Clearly, you can read the entire plaintext.
The last symbol
Perhaps, you have noticed that the plaintext ends with the symbol “Q” before the question mark.
Can you see anything Q ?
Is the symbol “Q” really part of section III? Or is it the first symbol of section IV?
According to the NSA solution, the text was encoded by keyed columnar transposition technique with an incompletely filled matrix size 4*86.
I strongly believe that this is not true. Stay tuned!
KRYPTOS in 3D — VIDEO
Kryptos — Wikipedia
Stein, David D. (1999). “The Puzzle at CIA Headquarters: Cracking the Courtyard Crypto” (pdf). Studies in Intelligence. 43 (1).
The puzzle at CIA headquarters. Cracking the courtyard crypto — CIA Website
Vigenère cipher — Wikipedia
KRYPTOS Week 2019 — SECTION III : A Transposition Cipher
KRYPTOS 30 Years Anniversary — SECTION III : A Transposition Cipher