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This is the second in a series of Unsolved Mysteries posts. These posts explore current unsolved real-world puzzles. Since these have been unsolved for years, it's uncertain whether anyone on on Puzzling.SE will be able to find a solution, but we do have a lot of brilliant minds here...

Related meta post: Posting famous unsolved puzzles


In 1988, as a new headquarters for the American Central Intelligence Agency (CIA) was being built in Langley, Virgina, sculptor Jim Sanborn was commissioned to create artwork for the courtyard of the new building. He designed a large copper monument, shaped somewhat like a flag, and engraved with hundreds of letters -- an encrypted message:

Kryptos sculpture at CIA headquarters

A soon as the sculpture was revealed, codebreakers both at the CIA and in the general public went to work on the puzzle. In 1999, a computer scientist named Jim Gillogly announced that he had solved most of the puzzle. There were four distinct parts to the code, and he had managed to solve the first three.

After his announcement, the CIA revealed that they had actually solved the first three parts internally, the year before. (Later, the U.S. National Security Administration (NSA) also claimed to have solved the first three parts in 1992.)

The fourth section of the code remained unsolved, and to date, no one has brought forth a credible solution for part 4.

The "tableau" on the sculpture has two sections. One side has the encrypted text, and the other is an alphabet table of sorts. (image credit: kryptosrevisited.com)

enter image description here

enter image description here

(a plaintext version of the tableaux is available here)

Other components of Kryptos

In addition to the copper sculpture pictured above, the artist created various other visual elements that are part of the same installation. There is a pool at the base of the sculpture, and a bench. There are also various large slabs of stone strewn about the area with various Morse Code messages. It is unclear whether these messages are related to the solving of the Kryptos cipher or whether they are simply adding to the artistic aspect of the sculpture. The Morse messages are as follows (some run right into the ground, so the text in brackets is a guess):

SOS
LUCID MEMORY
[WHA]T IS YOUR POSITION
SHADOW FORCES
VIRTUALLY INVISIBLE
DIGETAL INTERPRETAT[ION]
RQ

Solving the first 3 parts

Note: This section contains spoilers regarding the first three sections of the Kryptos code. If you want to try to solve these yourself, do not read this section. These are provided here to help set the context for the fourth (unsolved) section of the code.

Part 1

Though it was only discovered by solving Part 1, the ciphertext for Part 1 comprises the first two lines of the encrypted text:

EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJ
YQTQUXQBQVYUVLLTREVJYQTMKYRDMFD

It was encrypted using a Vigenere cipher, with a key of PALIMPSEST and the modified alphabet depicted on the second tableau.

(A palimpsest is a document (often vellum, but possibly parchment or paper), which has had its original text erased and has then been written over with new text.)

To decrypt the text, we reverse the process that was used to encrypt it. First, write out the keyword repeatedly underneath the ciphertext:

EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJYQTQUXQBQVYUVLLTREVJYQTMKYRDMFD PALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPAL

Then use the Kryptos alphabet tableau to look up the decryption for each letter (note that we ignore the top row and left column as depicted in the tableau above):

KRYPTOSABCDEFGHIJLMNQUVWXZKRYP
RYPTOSABCDEFGHIJLMNQUVWXZKRYPT
YPTOSABCDEFGHIJLMNQUVWXZKRYPTO
PTOSABCDEFGHIJLMNQUVWXZKRYPTOS
TOSABCDEFGHIJLMNQUVWXZKRYPTOSA
OSABCDEFGHIJLMNQUVWXZKRYPTOSAB
SABCDEFGHIJLMNQUVWXZKRYPTOSABC
ABCDEFGHIJLMNQUVWXZKRYPTOSABCD
BCDEFGHIJLMNQUVWXZKRYPTOSABCDE
CDEFGHIJLMNQUVWXZKRYPTOSABCDEF
DEFGHIJLMNQUVWXZKRYPTOSABCDEFG
EFGHIJLMNQUVWXZKRYPTOSABCDEFGH
FGHIJLMNQUVWXZKRYPTOSABCDEFGHI
GHIJLMNQUVWXZKRYPTOSABCDEFGHIJ
HIJLMNQUVWXZKRYPTOSABCDEFGHIJL
IJLMNQUVWXZKRYPTOSABCDEFGHIJLM
JLMNQUVWXZKRYPTOSABCDEFGHIJLMN
LMNQUVWXZKRYPTOSABCDEFGHIJLMNQ
MNQUVWXZKRYPTOSABCDEFGHIJLMNQU
NQUVWXZKRYPTOSABCDEFGHIJLMNQUV
QUVWXZKRYPTOSABCDEFGHIJLMNQUVW
UVWXZKRYPTOSABCDEFGHIJLMNQUVWX
VWXZKRYPTOSABCDEFGHIJLMNQUVWXZ
WXZKRYPTOSABCDEFGHIJLMNQUVWXZK
XZKRYPTOSABCDEFGHIJLMNQUVWXZKR
ZKRYPTOSABCDEFGHIJLMNQUVWXZKRY

The first letter is E, and the first letter of the key (PALIMPSEST) is P. So we find P in the leftmost column, move right until we hit E, and then move up to the top, ending at B. This makes the first letter of our plaintext a B.

            |
            V
    KRYPTOSABCDEFG...
    RYPTOSABCDEFGH...
    YPTOSABCDEFGHI...
--> PTOSABCDEFGHIJ...
    TOSABCDEFGHIJL...
    OSABCDEFGHIJLM...
    ...

For the next letter, we find A in the leftmost column, and follow it to the right until we hit M, then up, ending at E.

Continuing on in this way, we end up with the following decryption:

EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJYQTQUXQBQVYUVLLTREVJYQTMKYRDMFD
PALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPALIMPSESTPAL
BETWEENSUBTLESHADINGANDTHEABSENCEOFLIGHTLIESTHENUANCEOFIQLUSION

With spaces and proper case:

Between subtle shading and the absence of light lies the nuance of iqlusion

Part 2

The ciphertext for Part 2 comprises the remaining text in the top half of the tableau (until the break halfway down, i.e. lines 3 to 12):

VFPJUDEEHZWETZYVGWHKKQETGFQJNCE
GGWHKK?DQMCPFQZDQMMIAGPFXHQRLG
TIMVMZJANQLVKQEDAGDVFRPJUNGEUNA
QZGZLECGYUXUEENJTBJLBQCRTBJDFHRR
YIZETKZEMVDUFKSJHKFWHKUWQLSZFTI
HHDDDUVH?DWKBFUFPWNTDFIYCUQZERE
EVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDX
FLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKF
FHQNTGPUAECNUVPDJMQCLQUMUNEDFQ
ELZZVRRGKFFVOEEXBDMVPNFQXEZLGRE
DNQFMPNZGLFLPMRJQYALMGNUVPDXVKP
DQUMEBEDMHDAFMJGZNUPLGEWJLLAETG

It is solved in exactly the same way as the first part, except that it uses the keyword ABSCISSA instead of PALIMPSEST

VFPJUDEEHZWETZYVGWHKKQETGFQJNCEGGWHKK?DQMCPFQZDQMMIAGPFXHQRLGTIMVMZJANQLVKQEDAGDVFRPJUNGEUNAQZGZLECGYUXUEENJTBJLBQCRTBJDFHRRYIZETKZEMVDUFKSJHKFWHKUWQLSZFTIHHDDDUVH?DWKBFUFPWNTDFIYCUQZEREEVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDXFLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKFFHQNTGPUAECNUVPDJMQCLQUMUNEDFQELZZVRRGKFFVOEEXBDMVPNFQXEZLGREDNQFMPNZGLFLPMRJQYALMGNUVPDXVKPDQUMEBEDMHDAFMJGZNUPLGEWJLLAETG
ABSCISSAABSCISSAABSCISSAABSCISSAABSCI?SSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAAB?SCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISS?AABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAABSCISSAA
ITWASTOTALLYINVISIBLEHOWSTHATPOSSIBLE?THEYUSEDTHEEARTHSMAGNETICFIELDXTHEINFORMATIONWASGATHEREDANDTRANSMITTEDUNDERGRUUNDTOANUNKNOWNLOCATIONXDOESLANGLEYKNOWABOUTTHIS?THEYSHOULDITSBURIEDOUTTHERESOMEWHEREXWHOKNOWSTHEEXACTLOCATION?ONLYWWTHISWASHISLASTMESSAGEXTHIRTYEIGHTDEGREESFIFTYSEVENMINUTESSIXPOINTFIVESECONDSNORTHSEVENTYSEVENDEGREESEIGHTMINUTESFORTYFOURSECONDSWESTIDBYROWS

With spaces and proper case (punctuation in grey added by me):

It was totally invisible. How's that possible? They used the Earth's magnetic field. X
The information was gathered and transmitted undergruund to an unknown location. X
Does Langley know about this? They should; it's buried out there somewhere. X
Who knows the exact location? Only WW. This was his last message: X
Thirty eight degrees fifty seven minutes six point five seconds north,
Seventy seven degrees eight minutes forty four seconds west.
ID by rows.

Some years after this was originally solved, the sculptor Sanborn made it known that there was an error in the ciphertext, where a single character had been omitted (right after the word "west" on the second-last line). He assumed that it would be obvious to a solver, because it would cause the remaining text to decipher into gibberish, but by what seems to be coincidence, the text actually deciphered into the phrase "ID by rows."

It has since been determined (and confirmed) that the correct decryption ends with XLAYERTWO instead of IDBYROWS.

Part 3

The ciphertext for Part 3 comprises the majority of the bottom half of the tableau, up to (but not including) the ? character on the fourth-last line:

ENDYAHROHNLSRHEOCPTEOIBIDYSHNAIA
CHTNREYULDSLLSLLNOHSNOSMRWXMNE
TPRNGATIHNRARPESLNNELEBLPIIACAE
WMTWNDITEENRAHCTENEUDRETNHAEOE
TFOLSEDTIWENHAEIOYTEYQHEENCTAYCR
EIFTBRSPAMHHEWENATAMATEGYEERLB
TEEFOASFIOTUETUAEOTOARMAEERTNRTI
BSEDDNIAAHTTMSTEWPIEROAGRIEWFEB
AECTDDHILCEIHSITEGOEAOSDDRYDLORIT
RKLMLEHAGTDHARDPNEOHMGFMFEUHE
ECDMRIPFEIMEHNLSSTTRTVDOHW

This turns out to be a very different kind of encryption. It uses a transposition cipher which doesn't actually change any of the letters, but simply rearranges them. The task is to find the method used for the rearrangement.

There have been many different attempts at explaining the transposition. For this kind of transposition cipher, there are many different but cryptographically equivalent ways of describing and performing the cipher that all produce the same text, and there is no way to know which one the cryptographer had used without looking over his shoulder.

Luckily, photographs of the worksheets Sanborn used while enciphering K3 (pictured here) make it clear that the original transposition method is a double route transposition using two completely-filled rectangles. The rectangles are of different sizes, but the same route paths (in by rows and out by upwards columns) are used each time. This cipher does not use a literal key; the route paths and grid sizes are what make up the keys for the two route transpositions.

To encipher, the plaintext is first written into a 42-letter wide and 8-letter high grid, from left to right starting on the top row (that is, the usual reading order). The intermediate ciphertext is read out from that grid in upwards columns, reading each column upwards from the bottom row, starting with the leftmost column.

The intermediate text is then written into a different grid, 14 letters wide and 24 letters high. As before, it is written in by rows in the usual reading order, and read out by upwards columns to produce the text seen on the sculpture.

Deciphering is the inverse of enciphering. First, write the K3 ciphertext into a 14x24 grid by upwards columns. Note the beginning of the ciphertext (ENDYAHRO...) reads upwards from the bottom of the first column.

ILNTAYESTATHCW
BLHMHEHAROIEEH
ISIWNTHONRSLEO
OLTETYMFTEHMHD
ELAAEOAERIILUV
TSGCRIPEEPEKET
PDNADESTEWCRFR
CLRIUARBAELTMT
OUPIEHBLMTIIFT
EYTPNNTRRSHRGS
HEELEEFEAMDOMS
RRNBTWIEOTDLHL
SNMECIEYTTTDON
LTXLHTRGOHCYEH
NHWEADCEEAERNE
HCRNREYTAAADPM
OAMNNSAAUIBDDI
RISLELTMTNESRE
HAOSEOCAEDFOAF
ANNETFNTUDWAHP
YHSPITEATEEEDI
DSHRDEENOSIOTR
NYOANOHEIBRGGM
EDNRWEQWFIGEAD

Read out the intermediate text (ILNTAYES...) by rows. Then, write the intermediate text into a 42x8 grid by upwards columns. The plaintext will be evident on the rows. Note the first letters of the intermediate text (ILNTAYES...) read upwards from the bottom of the first column.

SLOWLYDESPARATLYSLOWLYTHEREMAINSOFPASSAGED
EBRISTHATENCUMBEREDTHELOWERPARTOFTHEDOORWA
YWASREMOVEDWITHTREMBLINGHANDSIMADEATINYBRE
ACHINTHEUPPERLEFTHANDCORNERANDTHENWIDENING
THEHOLEALITTLEIINSERTEDTHECANDLEANDPEEREDI
NTHEHOTAIRESCAPINGFROMTHECHAMBERCAUSEDTHEF
LAMETOFLICKERBUTPRESENTLYDETAILSOFTHEROOMW
ITHINEMERGEDFROMTHEMISTXCANYOUSEEANYTHINGQ

Sanborn's worksheets also make it clear that this text is exactly 336 letters long, equal to the areas of both rectangles. The question mark is not part of this ciphertext, and it is unclear whether it is the first "letter" of the following Part 4 ciphertext or whether it is simply a delimiter or separator.

With spaces and proper case (punctuation in grey added by me):

Slowly, desparat[e]ly slowly, the remains of the passage debris that encumbered the lower part of the doorway was removed. With trembling hands, I made a tiny breach in the upper-left-hand 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. "Can you see anything?" Q

Part 4

The ciphertext for Part 4 is the remaining 97 (or 98 if the question mark is included) characters on the tableau:

                          ?OBKR
UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO
TWTQSJQSSEKZZWATJKLUDIAWINFBNYP
VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

Most Kryptos fans assume that decrypting it will involve some combination of the methods used for the first three sections. To date, no one has been able to solve it.

Will you be the first?


There is a plethora of information about Kryptos on the web. A great place to start if you want more information is Elonka Dunin's Kryptos Page (especially the FAQ).


Like all of my Unsolved Mysteries puzzles, I will give a 500 rep bounty to anyone who manages to find a credible solution. Alconja and question_asker have both agreed to match this bounty, so the total bounty is 1500 reputation.

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  • $\begingroup$ I'll match your bounty offer with an additional 500 rep for this and any other Unsolved Mysteries solutions. I've edited the open bounties meta post, but feel free to edit the offer into your questions as well. $\endgroup$
    – Alconja
    Commented Jan 30, 2016 at 5:12
  • 1
    $\begingroup$ For those that didn't directly go to Wikipedia, here is some clues given by the artist on part 4: When commenting in 2006 about his error in section 2, Sanborn said that the answers to the first three sections contain clues to the fourth section. In November 2010, Sanborn released a clue, publicly stating that "NYPVTT", the 64th-69th letters in part four, become "BERLIN" after decryption.[18][19] Sanborn gave The New York Times another clue in November 2014: the letters "MZFPK", the 70th-74th letters in part four, become "CLOCK" after decryption. (1/2) $\endgroup$
    – nine9
    Commented Feb 1, 2016 at 8:48
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    $\begingroup$ This may be a direct reference to the Berlin Clock. Sanborn further stated that in order to solve section 4, "You'd better delve into that particular clock," but added, "There are several really interesting clocks in Berlin." (2/2) $\endgroup$
    – nine9
    Commented Feb 1, 2016 at 8:48
  • 1
    $\begingroup$ @nine9 I've added a chat room to discuss this puzzle. I took the liberty of copying your comments into it. $\endgroup$ Commented Feb 2, 2016 at 21:15
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    $\begingroup$ Probably convoluted, but there are 97 characters, which is prime. The coordinates, if placed on a clock instead of GPS, give you 1:13, which 113 is also prime. Meanwhile, 113 is the ASCII code for q, which showed up strangely in two locations within the previous panels. If you build a square-to-square number pyramid the primes create two prime-rich rings resembling a pendulum-clock pointing to 7:50. And aren't prime numbers used a lot in cryptography? $\endgroup$
    – mkinson
    Commented Jul 16, 2019 at 17:07

1 Answer 1

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This is not a solution, only a potential step along the way recently proposed on the Yahoo group by void solid and extended by dankuda. I'm reposting it here in hopes that someone can take it another step forward.


Take the text of K4 and split it on W's:

OBKRUOXOGHULBSOLIFBB
FLRVQQPRNGKSSOT
TQSJQSSEKZZ
ATJKLUDIA
INFBNYPVTTMZFPK
GDKZXTJCDIGKUHUAUEKCAR

This gives us six letter strings. Group them into two alternating strings (i.e. 1, 3, 5 and 2, 4, 6):

OBKRUOXOGHULBSOLIFBB TQSJQSSEKZZ INFBNYPVTTMZFPK
FLRVQQPRNGKSSOT ATJKLUDIA GDKZXTJCDIGKUHUAUEKCAR

This gives us two strings of exactly the same length. Now, lets check the letter frequencies in each string:

   evens            odds
        K  5 each  B
       AU  4 each  OS
     RGTD  3 each  KFTZ
   LQSJIC  2 each  ULIQNP
FVPNOZXHE  1 each  RXGHJEYVM

There are exactly the same number of letters of each frequency in each string! This looks to me like more than mere chance, but I'm not enough of a cryptographer to know where to go from here.


Of note, it's been officially stated that NYPVTTMZFPK (the last 11 characters of the 5th string) will decode to BERLINCLOCK.


There's an additional argument for this being the right first step. One of the stone strata that make up the extended Krpytos installation has the following Morse code on it (mostly in pairs):

VIRTUALLY INVISIBLE
SHADOW FORCES
LUCID MEMORY
SOS
RQ

If you apply the same even/odd logic to this, you end up with:

VIRTUALLY SHADOW LUCID SOS
INVISIBLE FORCES MEMORY RQ

The letter frequencies clearly doesn't match here, but since they produce same-length strings, it makes sense as a clue to apply the same logic to K4 (as I did above).

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  • 2
    $\begingroup$ For what it's worth, there have been a lot of theories for decrypting K4 that haven't gone anywhere. I particularly like this one and think it's on the right track because it A) can be derived from elsewhere in the sculpture, B) works out so symmetrically (remember, this was created by an artist), and C) doesn't rely on the BERLINCLOCK crib to figure out what to do. $\endgroup$
    – Bobson
    Commented Apr 10, 2016 at 16:51
  • $\begingroup$ It may or may not mean anything, but the two doubled letters in the "evens" line both show up only those two times (even), and the three doubled letters in the "odd" line show up 3, 3, and 5 times each. $\endgroup$
    – Bobson
    Commented Apr 12, 2016 at 2:52
  • $\begingroup$ But I don't think it's the correct answer as the original OP (not GRP, the sculptor) gave hints that there should BERLINCLOCK in the deciphered text. $\endgroup$
    – user27395
    Commented Sep 28, 2016 at 13:04
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    $\begingroup$ The frequencies of the stone strata do match if you replace the last “R” by a “C” (as suggested in a comment on Lemmino’s video), i.e. if you add a single bar in front of the Morse letter (i.e. “.–.” becomes “–.–.”). Maybe that bar is missing because it should have been (partially) covered? CQ is used to make a general call in Morse Code. Also note that another clue has been given, FLRVQQPRNGKSSO should translate to EASTNORTHEAST. $\endgroup$
    – Didier L
    Commented May 20 at 22:36
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    $\begingroup$ @DrXorile it seems I made a copy-paste error, the last “O” shouldn’t be included, just FLRVQQPRNGKSS $\endgroup$
    – Didier L
    Commented Jun 5 at 21:02

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