One Year Ago — The KRYPTOS Sculpture — SECTION III : A Transposition Cipher

With trembling hands, I made a tiny breach in the upper left hand corner… widening the hole a little, I inserted the candle and peered in… at first I could see nothing, the hot air escaping from the chamber causing the candle to flicker. Presently, details of the room emerged slowly from the mist, strange animals, statues and gold – everywhere the glint of gold. For the moment – an eternity it must have seemed to the others standing by – I was struck dumb with amazement, and when Lord Carnarvon, unable to stand in suspense any longer, inquired anxiously “Can you see anything?”, it was all I could do to get out the words “Yes, wonderful things”.

Howard Carter — The Tomb of Tutankhamen

Diary — November 26 1922

September 3 2017 — 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 you 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 details.

PS: This post will soon be updated to explain carefully several technical points. Follow us on Twitter: @INTEL_TODAY

RELATED POST: The KRYPTOS Sculpture — An Introduction

RELATED POST: The KRYPTOS Code — How to Break a Vigenère Code

RELATED POST: The KRYPTOS Code — The Solution of Section II

RELATED POST : The KRYPTOS Sculpture — History of the NSA Involvement

RELATED POST: The KRYPTOS Sculpture — SECTION I : A KEYED Vigenère Cipher [And why the CIA lies so much about it?] — UPDATE

Background

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

ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIA
CHTNREYULDSLLSLLNOHSNOSMRWXMNE
TPRNGATIHNRARPESLNNELEBLPIIACAE
WMTWNDITEENRAHCTENEUDRETNHAEOE
TFOLSEDTIWENHAEIOYTEYQHEENCTAYCR
EIFTBRSPAMHHEWENATAMATEGYEERLB
TEEFOASFIOTUETUAEOTOARMAEERTNRTI
BSEDDNIAAHTTMSTEWPIEROAGRIEWFEB
AECTDDHILCEIHSITEGOEAOSDDRYDLORIT
RKLMLEHAGTDHARDPNEOHMGFMFEUHE
ECDMRIPFEIMEHNLSSTTRTVDOHW?

Decrypted Text

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 misspelt and stands for “desperately”. I do not know if Jim Sanborn has made any comment regarding this mistake. I suspect that if Sanborn had spelt 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 \kappa 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.

\displaystyle \kappa =\frac{\sum_{i=1}^{c}n_i(n_i -1)}{N(N-1)}

where c is the size of the alphabet (26 for English), N is the length of the text, and n_1 through n_c 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…

336 Characters

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).

E N W Q F G A
N O N H I R G
D H D E O I T
Y S I E T E D
A N T N U W H
H O E C E F A
R S E T T E R
O M N A U B D
H R R Y A A P
N W A C E E N
L X H R O C E
S M C E T T O
R N T I O D H
H E E F A D M
E T N T R H G
O P E B M I F
C R U R A L M
P N D S E C F
T G R P E E E
E A E A R I U
O T T M T H H
I I N H N S E
B H H H R I E
I N A E T T C
D R E W I E D
Y A O E B G M
S R E N S O R
H P T A E E I
N E F T D A P
A S O A D O F
I L L M N S E
A N S A I D I
C N E T A D M
H E D E A R E
T L T G H Y H
N E I Y T D N
R B W E T L L
E L E E M O S
Y P N R S R S
U I H L T I T
L I A B E T T
D A E T W R R
S C I E P K T
L A O E I L V
L E Y F E M D
S W T O R L O
L M E A O E H
L T Y S A H W

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 Solution

The last symbol

Perhaps, you have noticed that the plaintext ends with the symbol “Q” before the question mark.

Can you see anything Q ?

I do not believe that the symbol Q carries any meaning. In my opinion, it is just a “filling symbol” added to the plaintext to reach the number 336 which is necessary to fill the (7*48) matrix. Thus, its presence in the plaintext can be explained.

On the other hand, the NSA solution argues that the text was encoded by keyed columnar transposition technique with an incompletely  filled matrix size 4*86.

The problem with the NSA solution is that it does NOT explain the presence of the symbol Q, if indeed it is a filling symbol as I believe. Time will tell…

KRYPTOS in 3D — VIDEO

 

REFERENCES

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

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One Year Ago — The KRYPTOS Sculpture — SECTION III : A Transposition Cipher

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