KRYPTOS — Analysis of Sanborn K3 Worksheet [A clue to decode K4?]

“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)

April 24 2020 — Kryptos is a sculpture by the American artist Jim Sanborn located on the grounds of the Central Intelligence Agency (CIA) in Langley, Virginia. Of the four parts of the message, the first three — known as K1, K2 and K3 — have been solved. However K4, the last part of the message, remains one of the most famous unsolved code in the world.  Follow us on Twitter: @INTEL_TODAY

RELATED POST: KRYPTOS Week 2019 — Introduction : Sculpture Dedication Ceremony at the CIA (November 3 1990)

RELATED POST: KRYPTOS Week 2019 — How to Break a Vigenère Code

RELATED POST: KRYPTOS Week 2019 — The Solution of Section II

RELATED POST : KRYPTOS Week 2019 — History of the NSA Involvement

RELATED POST: KRYPTOS Week 2019 — SECTION I : A KEYED Vigenère Cipher [And why the CIA lies so much about it?]

RELATED POST: KRYPTOS Week 2019 — SECTION III : A Transposition Cipher

RELATED POST: KRYPTOS Week 2019 — SECTION IV : A Few Clues From Edward Scheidt

RELATED POST: KRYPTOS Week 2019 — New Clues from the CIA Website? [UPDATE : Get Ready for a New Clue!]

The series on KRYPTOS was the #2 top 2019 stories of Intel Today. To be honest, I am rather surprised — but certainly flattered — by this success.

This being said, we are not done yet, far from it!

RELATED POST: INTEL TODAY — Top 10 Stories of 2019

On November 3 2020, KRYPTOS will  30 years old. Since its dedication on November 3 1990, there has been much speculation about the meaning of the four encrypted messages it bears.

The sculpture continues to be of interest to crypto-analysts, both amateur and professional, who are attempting to decipher the fourth passage.

Will someone decode K4 before the 30th anniversary?

Over the years, Jim Sanborn has released several images of his Kryptos worksheets. Can we learn anything from these pics? You better believe it!

In my last post, I told what could be learned from Sanborn’s K1/K2 worksheet.

RELATED POST: KRYPTOS — Analysis of Jim Sanborn K1 Worksheet

RELATED POST: KRYPTOS — More Lies from the New York Times and C.I.A.

In this post, I will analyse his K3 worksheet.

First step

It is a grid 42 wide by 8 high (336 characters).

The plaintext is written out from normally (from left-to-right, top-to-bottom), starting at the top left. You see the first characters of the plaintext (“SLOWLY”) starting on the top left side.

The second line starts with the 43rd character of the plaintext (“EBRI” from “DEBRIS”).

Second step

It is a 14 by 24 grid (336 characters).

It is formed by starting with the lower-left character of Step 1 (“I”, covered by Sanborn’s hand in the picture above) and placing it in the top-left square of this grid.

Then, on the sheet from Step 1, you move up one space and on the Step 2 sheet you move right one space, taking the “L” from Step 1 and placing it in Step 2. You keep this up (moving right columns as needed in Step 1, and down rows in Step 2), until the grid is full.



In my previous posts,  I told you how the NSA people broke Section III.

RELATED POST : KRYPTOS Week 2019 — History of the NSA Involvement

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

I also presented a novel and rather unusual solution because I had great doubts that Sanborn — an artist — has used complex mathematical transformation to code Section III.

RELATED POST: The KRYPTOS Sculpture — SECTION III: A Transposition Cipher

Thus, I suggested that the plaintext was filling EXACTLY a (7*48) matrix. I did so because I felt that if KRYPTOS was a clue, the number 7 was relevant.

I was not entirely correct. Sanborn actually used a 14 by 24 grid, not a 7 by 48 grid.

But I was right about the number 7 and right about the use of a completely  filled matrix.

It is very clear that my method is equivalent to what Sanborn did. I will let a mathematician prove this.

It is also clear that the NSA people were wrong. (I guess they just got lucky?)


In a recent interview with NPR’s Mary Louise Kelly, Jim Sanborn volunteered the following (dis) information.

KELLY: When and if it is finally cracked, what is the puzzle? What is the mystery that will be revealed?

SANBORN: It’s a 97-character phrase. And that phrase is in itself a riddle. It’s mysterious. It’s going to lead to something else. It’s not going to be finished when it’s decoded.

In my post about this interview, I had a strong comment about this answer.

“Maybe, Sanborn has spent too much time with CIA people… Although I have no reason to suspect the last statement to be false, I believe that it may be seriously misleading those trying to crack the last part of KRYPTOS.”

I now believe that K4 is indeed 98 — NOT 97 — characters long! The question mark after the final Q of K3 is thus the first character of K4. [And that is already a nice clue!]


And this means that K4 can fill a 14 by 7 grid. With this information, I think that the K4 can be broken.



Original Decoding Charts for ‘Kryptos’ — New York Time (November 20 2010)




KRYPTOS — Analysis of Sanborn K3 Worksheet [A clue to decode K4?]

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