However, it just so happened, that a member in the yahoo kryptos group had a conversation with Jim Sanborn due to a submitted solution. Sandborn's answer to the question contained again the last clue which surprisingly was EASTNORTHEAST at position 22-34. There is disagreement if Jim revealed this on purpose or he did it accidentially, but the new extended clue seem to be serious and valid.
Interestingly, EASTNORTHEAST is exactly the direction which is illustrated on the wind rose on one of the stones around kryptos, also created by Jim Sanborn. |
Below you can see that new hint at the position in the plaintext:
O B K R
U O X O G H U L B S O L I F B B W E A S T N O R T H E A S T 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 B E R
L I N
C L O C K W G D K Z X T J C D I G K
U H U A U E K C A R
Actually, i don't like the representation above, since it suggests that there is a 1-to-1 correspondence between the position of the letters in the ciphertext and plaintext. Although JS often said that this is the case, i actually think he did not really understood to question (See Post 2), due to his following statement he did lately:
Quotes from a March, 2019 lunch by Elonka: We spent quite a bit of time asking Jim about the correlation between the plaintext BERLIN and the ciphertext such as NYPVTT. Specifically, we were trying to find out if there was a 1:1 relationship from NYPVTT to BERLIN, or there was some other step, the masking technique. Jim was confused when we mentioned masking technique, evidently it's something that Ed said that Jim didn't understand. Jim said that yes the ciphertext and plaintext were connected, but when I tried to explain what exactly we were asking, like that in K1 EMUFPH is exactly BETWEE, but in K3 ENDYAH does not map exactly to SLOWLY, Jim backed off and said he would only commit to the fact that K4 is exactly 97 characters long, and that BERLIN is plaintext that starts at exactly the 64th character, but he wouldn't go further than that.
(kryptools.com/hints)
(kryptools.com/hints)
So transposition ciphers maybe back on the line. This gives me the chance to talk a little bit about the approach of Scott, who succeeded to create a reasonable method that makes BERLIN to appear at the correct position. His approach is as follows:
- Apply Quagmire3 with the two keywords KRYPTOS and EMUFPHZLRFA to the ciphertext of K4. Note that Quagmire3 was also used for K1 and K2. The second keyword are the first 11 characters of K1.
- In the plaintext of K2 the two coordinates 38° 57' 6.5'' N and 77° 8' 44'' W are mentioned. Scott used the linear congruence $y = 77 + 38x \pmod{97}$, that involves the two degree integers from the coordinates, whereof $x$ is the $x$-th character in the result of step 1 and $y$ is new position of $x$ in the output text.
The result is:
P G Z F
G T E Z P F C Y G F L J L Z G S O J A I L W D C C N Z F H S H
Q H I V Z K U C G O Z A A F J E V H O C Z V S N H Y E H B E R
L I N P E E K R R R H B U X J O P Y F F T X B M M Q M O F Q S
A closer look reveals that the characters that make up BERLIN actually only depend on the 6 characters MUFZLR of the second keyword: EMUFPHZLRFA. I played around with his approach, and i quickly realized, that all positions before applying the linear congruence of the characters BERLIN are $0 \pmod{7}$ and $4\pmod{7}$. The letter $K$ in the Quagmire3 method using KRYPTOS as the first keyword creates fixpoints. I also realized that every second Quagmire3 decryption of KRYPTOS and a second keyword that has length $7$ and has at the positions $4$ and $7$ the letter $K$, would also create BERLIN at the correct position (using the same linear congruence) after applying the two Quagmire3 decryptions. However, what actually surprised me is the following:
EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJ
YQTQUXQBQVYUVLLTREVJYQTMKYRDMFD
VFPJUDEEHZWETZYVGWHKKQETGFQJNCE
GGWHKK?DQMCPFQZDQMMIAGPFXHQRLG
TIMVMZJANQLVKQEDAGDVFRPJUNGEUNA
QZGZLECGYUXUEENJTBJLBQCRTBJDFHRR
YIZETKZEMVDUFKSJHKFWHKUWQLSZFTI
HHDDDUVH?DWKBFUFPWNTDFIYCUQZERE
EVLDKFEZMOQQJLTTUGSYQPFEUNLAVIDX
FLGGTEZ?FKZBSFDQVGOGIPUFXHHDRKF
FHQNTGPUAECNUVPDJMQCLQUMUNEDFQ
ELZZVRRGKFFVOEEXBDMVPNFQXEZLGRE
DNQFMPNZGLFLPMRJQYALMGNUVPDXVKP
DQUMEBEDMHDAFMJGZNUPLGESWJLLAETG
One of only two possible $7$ length strings that have two $K$s separated by $2$ starts in the middle of EMUFPHZLRFA going down vertically and forms a T. The keyword is HXDKZEK.
So if you do the following 3 step approach:
- Apply Quagmire3 with the two keywords KRYPTOS and EMUFPHZLRFA to the ciphertext of K4.
- Apply Quagmire3 with the two keywords KRYPTOS and HXDKZEK to the result of step 1.
- Use the linear congruence $y = 77 + 38x \pmod{97}$ whereof $x$ is the $x$-th character in the result of step 2 and $y$ is the new position.
This will get you the result:
P G Z Y
G Q E I M Y X M Y H S M S R H B S M B R M C D U C O Z F H S H
Q T I F Z J D Z Y U H C V H L G W J S U K B A O H H E K B E R
L I N N E R I L J L P D D K O A T T G H O D C T N S M L F S S
I don't know if this leads somewhere, but i think it is worth to write down.
This said, i found that in 2009 Sanborn gave an interview that contains the following:
And Ed basically gave me—he gave me a primer of ancient encoding
systems. And he also gave me some ideas for contemporary coding
systems, more sophisticated systems, systems that didn't necessarily
depend on mathematics. That was one of my prerequisites. So he told me
about matrix codes and things like that. These are the parts of
Kryptos that have already been cracked. So I can discuss them.
But he told me about coding systems that I could then modify in a
myriad of ways. So that even he would not know what it says. Okay? So
that was very seductive to me. And so I took those things. We met two
or three times. And that's what I based the whole thing on.
(Interview with Jim Sanborn 2009)
(Interview with Jim Sanborn 2009)
Sadly, this sounds as if all the methods that are used in K1, K2 and K3 are not again used in K4, so Scotts approach maybe a dead end.
➠5bit ASCII code. In 2019 JS said in an interview that roughly one year earlier he got a
submission of a proposed solution from a german guy. And this solution scared him,
since parts of it seem to be correct, but the rest was wrong. In the forum of
Klaus Schmeh's Scienceblog, see [1; comment #18 and more], a guy makes several
comments indicating that he might probably this german guy, since he is german
and submits a solution to Sanborn in the correct time window. He briefly
explained his approach which is based on 5bit ASCII code and bitwise
transposition but kept much of the detail hidden.
5bit ASCII code is indeed an option that is worth to try. However, there is no unique choice for a
5bit representation. The most obvious approach is to map each letter
to its position in the alphabet. But already in this trivial case you can distinguish two choices; do we
start with $0$ or $1$, i.e. $A \widehat{=} 0$ or $A \widehat{=} 1$. Furthermore, 5bit covers $2^5 = 32$ integers. You don't run into a problem when mapping the alphabet to integers. But when you start to manipulate the bits, e.g. swapping, negating etc, you start getting
integers from the interval $[26,31]$ which have no character assigned. Also standard 5bit ASCII code tables like e.g. the Baudot-Code have the same drawbacks since non alphabet letters (special characters) occur during manipulation. Also a simple reduction modulo $26$ deletes information which renders a unique a decryption process impossible.
However, just for the sake of completeness, there is at least one simple non-trivial function $f$ that maps $$f: \{26,27,28,29,30,31\} \rightarrow \{26,27,28,29,30,31\}$$ and $$f: \{0,1,\ldots,25\} \rightarrow \{0,1,\ldots,25\}$$ and is uniquely invertible. It is
$$ f(b_4,b_3,b_2,b_1,b_0) \mapsto (b_4,b_3,b_1,b_2,b_0) $$
whereof $b_4,b_3,b_2,b_1,b_0$ is the binary representation of an 5bit integer with $b_4$ being the most significant bit. The input alphabet is:$$\text{ABCDEFGHIJKLMNOPQRSTUVWXYZabcdef}$$
and is mapped to $$\text{ABEFCDGHIJMNKLOPQRUVSTWXYZcdabef}$$ However, the entropie is low, 14 of the 26 letters mapped to itself. I assume that $f$ is worthless and itself just creates a monoalphabetic substitution.
One main argument to work with base 5 is that the Berlin clock also represents
the time in base 5. But this seems to be an odd argument. Why should JS's
plaintext contain a hint towards solving its own ciphertext? It is not accessible unless some reveals it. Did JS anticipate that he has to reveal a clue
due the hardness of the cipher and included BERLIN CLOCK into his plaintext to
be able to give a meaningful hint? Isn't more logical that the base 5 argument
is part of the riddle, which JS said is revealed once K4 has been decrypted? Another argument are often the 5 letters dYAhR, which somehow indicate a transposition, since dYAhR is a transposition of HYDRA, a term which makes sense in the context of Jim Sanborn. Since this five letters are part of K3 and K3 was encrypted using a transposition cipher, it was maybe a hint for K3 itself rather than K4.
➠PALIMPSEST & ABSCISSA. Where to start with? Base 5 maybe a dead end for K4 since it is the method for K5. All matrix based cipher, hence also Vigenere/Quagmire approach are also a dead end, because he said, that he used a different technique for K4. What about Ed Scheidts masking technique? What about the strange symbols that appeared in Ed Scheidt garage entrance a few years ago
Garage driveway of Ed Scheidt |
They look a little bit like the BERLIN CLOCK - but what could it be? [Update: Read this for possible explanations]
If you read through all the many statements Jim Sanborn has given over the last years, you can conclude that he probably did something completely different from the first three ciphers - something more visual, an encryption method an artist would choose.
Why did JS chose the two keyword PALIMPSEST and ABSCISSA for K1 and K2. This are no usual words and also have a meaning that could fit to the encryption of part 4. If you look at the whole Kryptos statue at once, you can view it as a form of a cartesian coordinate system with its four quadrants.
The term ABSCISSA is used to describe the horizontal X-axis, the vertical axis is called the ORDINATE. Maybe this is the correct view in order to solve K4. One could for instance use some kind of mirroring at the ABSCISSA, rotation or even more complex functions using the characters of K4 as the input.
[1]
http://scienceblogs.de/klausis-krypto-kolumne/2020/01/30/jim-sanborn-publishes-new-kryptos-clue/
Thank you very much for your insightful series on the Kryptos sculpture! :-)
ReplyDeleteThe explanation of Scott's discovery (and its generalization by yourself) is pretty cool! Do you happen to know if Scott has had any other related findings?
Sorry, i don't know of any other findings.
ReplyDeleteRegards,
Chris
K4 solution inthecenterofthiscityeastnortheastnearrotesrathauschimesfromtheberlinclockcanbeheardacrossthesquare
Deletethats just BS
Deletewhy are you adding more chars (97 vs 99) and what is the key
In position 10 11 12 is the word the positions 14-21 is the word crossing
DeleteGood analysis, but you're missing some fundamental pieces of the puzzle here. Is there a good way to contact you?
ReplyDeleteHi Eric,
ReplyDeletethe best way to contact me is using my private e-mail, not my business one. It is [forename][lastname]@googlemail.com
regards,
Chris
A couple of years later... :-)
ReplyDeleteYou wrote "Actually, i don't like the representation above, since it suggests that there is a 1-to-1 correspondence between the position of the letters in the ciphertext". However, The additional clue of EAST directly in front of NORTHEAST and its (through the compass stone) confirmed connection is quite strong evidence for a 1:1 correspondence. Only some anagramming inside of word boundaries or annoying stuff like a completely reversed ciphertext remains possible.
THOUGH A PROGRAM BE BUT THREE LINES LONG SOMEBODY MUST HAVE WRITTEN IT TRUTHFULLY WRITTEN IT WITH CARE CONSIDERING ALL POSSIBILITIES AT LEAST ONCE CONSIDERING ALSO AFTERWARDS THE INPUTS RUNNING TESTING THE OUTPUTS PROBING PERHAPS CORRECTING THE PROGRAM AS NECESSARY TILL IT WORKS CONFIDENTLY EXACTLY WITH NO ERRORS WHATSOEVER THUS ARE ALL BUDDING PROGRAMMERS TAUGHT TO WRITE PROGRAMS
ReplyDeleteProvided the bit maniplation you are using is XOR, there is a clever way around the problem of characters mapping out of range. Namely, if the current key letter causes this, simply copy the key letter to the output stream and then try again to encrypt the current plaintext character with the next key letter until you find a key letter that encrypts within range. Since anything xor itself is 0, the receiver will get a complete, unambiguous decryption, just with the occasional null character appearing which they know to drop. This scheme was used by one of the historical rotor machines (not Enigma) but I can't remember which.
ReplyDeleteZotsfen
ReplyDeleteInthecenterofthiscityeastnortheastnearrotesrathauschimesfromtheberlinclockareheardacrossthesquare
DeleteMatches the clues and the letter count
The following is the structure to follow with the key the critical part is finding a absolute position of a 13 position group. Of three words us in letter count of words across the clues structures the best path as there are only a certain amount of words and variables in a 97 position cipher.Using a set pattern to run the cipher against would allow for a continues path of the repeating words using the math magical structure along with word letter count
The eastNorthEast is a 4 5 4 that equals a center with in its structure the eastnortheast portion is along the K on one side making the 13 positions perfect with this structure using the same portion during the flow through the positions matching up to the letter that makes the
eastnortheast =13K
Leaving six positions per line with a turn across two more of the letters
Marking the pattern absolute using the square and compass angles makes the most of the area using two halves of oneallowing this will lay out exactly how the identifier sequence fits along it's structured path.
Deadlines up on T
With
NTSAE. T
ORTHEAS
RTHEAST
It's lowest position starts at
K
R
Y
KRYPTOS
T
O
S
THE RIGHT TURN OR LEFT TURN IS ALWAYS OF CENTER OF P
MAKING THAT .5X2 ACTUAL BE USED UP TO 72 DIFFERENT TIMES WHICH MAKES
The polyibus cipher uses no J using numbers in stead of letters enables K 11 to be K10 as well which having all the K out of place except for 1 on the foundation alphabet
Makes that K the multi position first pick
In orde4 to balance with in a polyibus cipher the positions removed are equal to that J that is missing in the graphing structure.
This matches to the alphabetical structure outside the kryptos area
And shows the position of North with the top left N meaning east north east is right to left left left left (4)up up(2) right right right right right right (6)
Makes it 13 positions long in a structured pattern and can be substituted with cipher after structure.
Using the sections polyibus makes the N GHILMNQUvwxZKRYPTOSabc
O. Q ZKRYPTOS
P. Q ZKRYPTOD
Q. ZKRYPTOS
R.s.t.u. ZKRYPTOS
V. ZKRYPTOS
THE 1ST AND 3RD LINE MAKES THE STRAIGHT LINE THROUGH NOR
WITH KRY DOWN THEN PTOS TO THE RIGHT STARTING AT THE BOTTOM LINEVWITH T ON THE R(FIRST K ON BOTTOM) T EQUALS E PEQUALS A YEQUALS S
97÷13
Since 97 letters can fit 13 letters into to equal the 25 position polyibus cipher structure is 3.88
With 3.72 making 26 variations in the alphabet.
That is a .16 difference 3.88x.16=4
Makes that a .12 difference and is a quarter off the .16 number.
The
kry
Kr
K
Key at the right bottom is moved to the top and extends 13 positions to 16 positions
Now the pattern has a set structure that wraps from bottom to top with only the 13 piece given in a clue east north east
13 centers six one six the one center makes both halves
This actually doubles from the 3.5 times 2 equals the 7 positions of crypto starting that off with a center overlaping word matches to the statistical likelihood that a word will be split so a continues flow has to not cross itself or end the key is the portion that bridges the two ends with a twist when using the correct no directional alphabetical structure making the mobious twist and having that structure to see how A-N
Equals Z-M as equal 13 letter portions of the alphabet in order to use a one unit shift per position
4 13 position cipher portions time 4
it does not match the lettercount
DeleteHere is the k4 solution inthecenterofthiscityeastnortheastnearrotesrathauschimesfromtheberlinclockcanbeheardacrossthesquare
ReplyDeleteletter count doesnt match neither you provide a key
DeleteI think Mark used chat GPT to come up with his solution but at least the number of characters and the position of the letters works if you use INTHECENTEROFTHISCITYEASTNORTHEASTNEARROTESRATHAUSCHIMESFROMTHEBERLINCLOCKAREHEARDACROSSTHESQUARE
DeleteIN THE CENTER OF THIS CITY EAST NORTH EAST NEAR ROTES RATHAUS CHIMES FROM THE BERLIN CLOCK ARE HEARD ACROSS THE SQUARE
k5 solution is the size of the earth when mr sanford did the calculations for the location used in the 3rd passage which is roughly ten kilometers over what nasa and the government tells us it is 4255.4358902255 is a quarter of four
ReplyDeletewith a quarter off four being 12766.307670676
the clue i uncovered was at a quarter off four you will find five four three two which you can see is in the third off the total number my name is christopher mark spriggs
As far as the craziness that is my life goes I wonder what you will make of this?
ReplyDeleteYesterday, the 13th of March, 2024, I was walking down Bond Street in Bath when a taxi stopped at the traffic lights near me. I suddenly noticed the number 237 on its registration plate and was instantly reminded of ROOM 237 from the film 'The Shining. This number appears quite regularly at particular times.
Today, for some strange reason, I thought of the KRYOTOS puzzle and when I logged into the Wikipedia page I noticed the number 0363514, which left me feeling slightly uneasy, as it was the same number which appeared as an authorisation code on a digital screen of a card reader in a shop I visited in Bath yesterday.
13/03/24
A very strange coincidence indeed.
Correct me if i'm wrong, but if the K4 is a transposition cipher, then the plaintext will contain the same letters as the cipher.
ReplyDeleteBut "Eastnortheast" and "berlinclock" contain 3 E's
Whereas the cipher text contains only 2...
Also, there are 4 Z's which is highly unlikely to occur in a sentence under 100 characters.
Q's and X's can be accounted for as "Q" is the subject and X is a period.
Unless of course there is another intentional spelling error and the plain text is "BURLINCLOCK"
DeleteSanburn said he tried to hide the english to avoid frequency analysis, thats how the first 3 were solved it suggests he put it through 2 things or onething twice. K1 one is solved using Palimpsest and palimpsest means to have two things of information overlapping, hope that helps explain it
DeleteRunning the text through ChatGPT a few times simply writing "Solve this:" got me this text "firstwhorotatesinjulymightbewisewithcipherbyrows" the text is 48 charachters long which is half of the full text. I don't really think this is the sollution, just found it interesting.
ReplyDeleteI think I solved Kryptos section 4 I got "CONGRAGULATIONS ON DEC ODING THIS MESSAGE YOU HAVE DEMOSTRATED PROFICIENCY I N CRYPTOGRAPHY" and "IT IS IMPOSSIBLE TO SEPERATE ART FROM OPTICS AS THEY ARE SO CLOSELY ALIGNED" I know it doesn't have Berlinclock or Eastnortheast in it. What do ya'll think about it?
ReplyDeleteI think I solved Kryptos section 4 I got "CONGRAGULATIONS ON DECODING THIS MESSAGE YOU HAVE DEMOSTRATED PROFICIENCY I N CRYPTOGRAPHY" and "IT IS IMPOSSIBLE TO SEPERATE ART FROM OPTICS AS THEY ARE SO CLOSELY ALIGNED" I know it doesn't have Berlin clock or East northeast in it. What do you think about it? (sorry about that this is the version that looks better)
DeleteNever mind I failed to solve it was wrong. :(
ReplyDeleteQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINF = NORTHEAST FACILITY IS SECURE AND PENDING EVACUATION. using shift values from "BCI" key
ReplyDeleteThose strange symbols on Scheidt's driveway are not references to Berlin Clock, they are Mayan numbers. In each row there is a inverted one, and this does not exist in Mayan numeric system. The inverted stuff most of the cases means the oposite, so that means they are numbers in negative scale. (-x). Convert to normal numbers and then + or - them. Try to do some tests with these numbers, maybe you find something interesting. I did, but of course I won't post it here.
ReplyDeleteYes, indeed. I covered this topic in the newest blog post:https://numberworld.blogspot.com/2024/06/ed-scheidts-mayan-symbols.html.
DeleteOk, i will give it a try :)
What do ya'll think about this?
ReplyDeleteVirtually Invisible Digetal interpretation Shadow Forces Lucid Memory It is your position (or What is your position?) sos RQ
I think this is the whole message from K0 and I think it might help with K4. Digetal is misspelled and I think it is saying that, The clue is Virtually invisible but Physically visible and It is a Digital Interpretaion of a physical Person or Item. And the "Shadow Forces Lucid Memory" Is like when you see shadows and your eyes play tricks on your mind. The SOS and RQ I dont have anything yet.
Virtually invisible = Not visible on screen > Object not visible on radar.
DeleteShadow forces = Something that logic or human science still cannot explain > Something weird or "paranormal" for human knowledge.
Lucid memory = Conciouss > One or more alive.
Digital Interpretation = I CAN'T GET THIS, but if you ask an AIR FORCE or a MILITARY person, maybe you get something.
OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
ReplyDeleteis
THE TREASURE IS BURIED EAST NORTHEAST OF THE BERLIN CLOCK TOWER AT MIDNIGHT DURING THE NEXT FULL MOON
The decryption key "THEQUICKBROWNFOXJUMPSOVERTHELAZYDOG"
Don't take it personal, but this equals nothing. If you find something interesting, you must show how did you get there. Otherwise, people will not take your post it seriously.
DeleteFails this verification script: def vigenere_decrypt(ciphertext, keyword):
Delete"""
Decrypts ciphertext using the Vigenère cipher with the provided keyword.
"""
decrypted_text = []
keyword = keyword.upper()
ciphertext = ciphertext.upper()
keyword_length = len(keyword)
for i, char in enumerate(ciphertext):
if char.isalpha(): # Only process alphabetic characters
char_index = ord(char) - ord('A')
key_index = ord(keyword[i % keyword_length]) - ord('A')
decrypted_char = chr(((char_index - key_index) % 26) + ord('A'))
decrypted_text.append(decrypted_char)
else:
decrypted_text.append(char) # Non-alphabetic characters remain unchanged
return ''.join(decrypted_text)
if __name__ == "__main__":
# Ciphertext and key provided for verification
ciphertext = "OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR"
key = "THEQUICKBROWNFOXJUMPSOVERTHELAZYDOG"
# Decrypt the ciphertext
print("Decrypting the Kryptos K4 ciphertext...")
decrypted_text = vigenere_decrypt(ciphertext, key)
# Output the decrypted text
print("\nDecrypted Text:")
print(decrypted_text)
# Compare with the claimed plaintext
claimed_plaintext = (
"THE TREASURE IS BURIED EAST NORTHEAST OF THE BERLIN CLOCK TOWER "
"AT MIDNIGHT DURING THE NEXT FULL MOON"
)
print("\nClaimed Plaintext:")
print(claimed_plaintext)
# Check if they match
if decrypted_text == claimed_plaintext:
print("\nThe decryption matches the claimed plaintext. The claim is correct.")
else:
print("\nThe decryption does not match the claimed plaintext. The claim is incorrect.")
However, in Oliver's defense, the verification script I used makes too many assumptions. The solution could be valid, idk. If Oliver truly solved this, especially on 07/04, then a medal is due!
DeleteCaveat: "The quick brown fox jumps over the lazy dog" is a famous English sentence that is a pangram, meaning it contains every letter of the alphabet
DeleteHow did You get the decryption key and what cipher are you using?
ReplyDeleteI think K0 is going to help us solve K4 and I think the random e's that pop up all over are actually a substitute for other letters. Because in the quote "T IS YOUR POSITION" there is an 'e' before that and the logical thing to do would be change that 'e' to an I and then it would be "IT IS YOUR POSITION" What do you think? Also INTERPRE TATI has two e's after so I made the e's O and N. Making it "Interpretaion."
ReplyDeleteMaybe T means "WHAT", so it should be "WHAT IS YOUR POSITION".
DeleteMaybe the letter T also could be a W, H or A.
EVERY CORNER OF ROTES RATHAUS TOWER HEADS TO A CARDINAL POINT.
ReplyDeleteCurious thing.
I have a Question. "Didn't Jim Sanborn say Its as easy as ABC" Talking about K4. If he did I have a Theory to K4 what if there were 4 layers to Kryptos it gets reversed as the first layer. (Because the question mark is at the beginning) Then the Second Layer uses A Special word using the Letter A then the third layer uses a word with the Letter B and Finally the last layer uses a Word with a C.
ReplyDeleteI also have another Theory. What if K0 was hiding a secret word that was needed to solve K4 along with other things?
I ran across this.
ReplyDeleteThere are three misspelled words from K1, K2, and K3 the words are
K1= IQLUSION CORRECTION=ILLUSION
K2= UNDERGRUUND CORRECTION=UNDERGROUND
K3= DESPARATLY CORRECTION=DESPERATELY
And when reading about the Berlin Wall there were three checkpoints too The Travel between East and West Berlin.
And There was an Illusion of Freedom, Underground escapes and Desperate Attempts. Just my thoughts. What do you think about this? Also if you need more about this I can probably share more
I believe that JS released his handwritten worksheets, and in his original grid for K2, there was no misspelling in UNDERGROUND. It was the letter O in that cell and not a U. However, it is my opinion that the inclusion of the Q was intentional to defeat informally a frequency analysis. But that was a very interesting idea about checkpoints that you have.
DeleteI tried putting K as they keyword but everything stays the same has anyone noticed that before?
ReplyDeleteI did the same thing
DeleteFamous Clocks in Sweden
ReplyDeleteWhy?
DeleteI plan to solve Kryptos and if I do then I am going to create a HMRS (Historical Mystery Response Team) {This is just my hopes, and if I do solve it this will push my team forward.} What do you think about my team?
ReplyDeletegood luck to you
DeleteThis is the solution for Kryptos K5. I solved it using pen and paper. AI programs will not be able to solve this because it has multiple decryption methods. Kryptos groups claiming to be experts on this are wrong too because they are saying it is a one for one Vigenere type solution. That is incorrect. I have found that Kryptos is a series of tests with multiple decryptions. I will publish the decryptions for K4 here soon in one post. I have a report done on K4 but have to add K5 to it that shows step by step how to decrypt the entire work beyond the solutions that were already done and published by others (K1, K2, and K3 I believe).
ReplyDeleteKryptos K5:
ENTERQIQSEEMSTHEBESTTORAKEITIDETESTQSBESTTIMEQTHEMEETINGCIASTESTINGTESTITISEEITESTITESTTIMETESTTHENYIMQCAMECAMETEARWIQSENTISTIRTHEIQSTEAMRIDEZSITESTIMMISTTOEDATESTQITSINTHETENIMTIMETHISTAKEQSTIMEWTENITESTEDTEARITHEIQQSTOTEARMEETTIMESENTQEDTAKESITTOWEDWMEETSEDATHISIQTESTISIMQTEARQTARTTIDESEQARTSEDTAKEEDTENNOQWTIMEIMTIMEINWINAQRESTITSQTHISSTARQCIATIEMEETZWETESTTEARSWRESTTASKSITTEAMTIDEEDIWISHESTIETAKENLOSTENTERWTESTITQSWARQTENISITISEEWITQSQITQINTIDESTAYSEEEDWSQINTIMEISEEITIDEISTEARTESTTENQSYESQAMINDSTIRWCIAISEQLISTSEDWSEETHISISEETHEIQZARTTOSEETOTHERESTTENENDITSSENTWIENDSTTHEETEAMSEESMEETARTIEDSQBEWQIMEETIMIQMISTSIWENDSIMMEETSIQETESTWINAQITAKEZIMTIMEITSIMBESTWETEARZWEARTENEDATESTITSTHEWISESTTIDESTIMEAWTESTTOTMISTIWTHETIDEITIDEIMCIAEDETESTSTENISTEAMYESQISEDMIQTEARINITESTSISEESENTTEARTESTTIDESITESTYHISENSEISEEITITEAMKTOWSEETIMEWINSWTHERESTARSETEAMAQTIMEISZTIMETHEBESTSETITIMQSTESTHSENDAIDTHERESENDQISEEHISESENDALISTTEAMTOTEARSWTENINITESTTOTHINIQISEEHISTHERESTHEHITENITESTIMQRESTBESTSTIDEISTESTHHHQEDISEEISEEHHHETIDESISEEISEEISEEIE
Here is the work for K4 and K5. I compiled a 243 page report that takes the decryptions that I did and tabulated them step by step. When you do the Chutes and Ladders portion of the report, it is recommended that you have the board game handy to visualize it. I am sharing this with you guys because of the log jam currently going on with Kryptos due to the lack of sharing of information. This report ends that log jam. :)
ReplyDeletehttps://drive.google.com/file/d/1StZ08b4CHZz8haGWabPH1AAs0bwrCyNT/view
Thanks. I apreciate it.
DeleteYou're welcome.
Delete"it is known that the mysterious symbols on this sculpture conceal hidden messages for those who seek the truth"
ReplyDeleteCygnus. Have you verified it from Jim Sanborn?
ReplyDeleteI sent a chunk (about 1/2) of the report I completed and linked here to Mr. Sanborn back at the beginning of May 2024. He sent me an image in return of two words East North East and Berlin Clock overlayed over certain areas of the 97 variable cipher.
ReplyDeleteThe Berlin Clock clue, to me anyway, refers to reducing the Morse Code carries to zero because that is the function of the Berlin clock.
The only location that I found East North East was in the cipher location that solved to W Tested The Art No Sort, which when you take THEARTNOS (The Art Knows) and take the time to work it, it gives the numeric key to decrypt K5. I think that was the purpose of the two clues, the first clue to give the numeric key to decrypt K5, and the Berlin Clock clue to give people the mechanism which is to reduce all the carries in the Morse Code to produce the message, which there are many examples of in the report.
Mr. Sanborn also told me that he doesn't know very much about coding and that he was working off some master sheet that was the solution for what he was looking for without verifying the work.
I made the decision then to publish my work in full to the benefit of the community working on it so the Kryptos challenge can be ultimately wrapped up. I have determined that Kryptos is a series of tests, and not just one simple Vigenere Cipher or some similar method as was the first parts of it before K4.
If anyone has found a Vigenere style one for one solution to K4 they have accomplished something that I have not up to this point achieved. I didn't focus on that mechanic because the variable distribution for K4 didn't support it.
Can you send the image Sanborn sent to you?
DeleteYes, I can. Is there a way to publish it to this site in the forum. I can do that too. Or, I can publish it to my blog so you can all see it. Let me know what works for you.
DeleteEither one works for me. Though through the blog might be nice because I'd like to see the other work you've done. You seem like you've done a lot of research and spent a lot of time working on Kryptos and I'm very impressed. What is the link to your blog? (If you couldn't tell this is the same person who asked for the image.)
DeleteHi Krymorn, I made a new blog entry just for you with the image and the blog posts I made about the work. I also linked my one drive link for the 243 page report. Head over to cygnusdysonsphere.blogspot.com to check it out.
DeleteIt was a lot of work overall. I think it took me about 3 months just to put the decryptions into tables in Microsoft Word and mounds of paper working on it. Boxes of gel pens. Lol
Wow! Thank you so much! This is great! I can't wait to work on this more! (Lol about the gel pens)
DeleteOops, forgot to add the name.
DeleteHey Krymorn thanks for checking it out. You can contact me any time.
DeleteHow would I contact you?
DeleteYou can reach me at forensicspecialactivity@yahoo.com
DeleteAlright, good to know!
DeleteOk
ReplyDeleteI think that Kryptos wasn't meant to be made public in the sense that it has today. When you read the solutions, it refers to having meetings at the sculpture with the creators of it after each cipher is solved to move on to the next, like there is a predetermined order to it. The solutions also give a time to meet, 10 AM, so maybe long ago there were managers of it initially showing up at that time to see if anyone would drop in with solutions. So to me, it was meant for the codebreakers in the IC to work on but got a life of its own when it became public.
DeleteThe other issue is the people that made it are part of the IC, and it is their job to protect information. They weren't going to give anyone at any level any truthful response to questions about it, or clues. The exact opposite has happened over time, and that is the cipher system saw increased protection in how submissions were handled, and the socialization of the unsolved problem saw many groups themselves becoming part of that protection mechanism as well by not sharing information. I think the IC was working those groups to help prevent it from being solved, which is pretty funny when you think about it.
Sorry... Do you know what type of wood was the kryptos's flagpole made?
ReplyDeleteCorrection: Do you know what type of wood was used to make the kryptos's flagpole? Sorry, my english is bad... ughh...
DeleteI have never seen the sculpture in person. A web description of the design of sculpture stated that the wood used was petrified wood. Petrified wood of that quality is native to Arizona, but I don't know if it was sourced that way for construction.
DeleteAm I wrong if I think that all the elements used in Kryptos make reference to Egypt?
DeleteWhen I was looking at Howard Carter's gravesite in England, the bottom right corner had the grave marker of 45-D-12. The bottom left corner had the word "Buchanan". When I looked up President James Buchanan, he has a monument at his memorial that is a pyramid. I don't know if there is any connection between that and Kryptos.
DeleteCool and I have a tremendous offer you: How Much Are House Renovations Stardew Valley home repairs contractors near me
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteCorrect me if I'm wrong, but... if you take a school protractor, the EAST-NORTHEAST direction is ~30.
ReplyDeleteYeah, -37° actually.
DeleteWhat if the clue we’ve been looking for the entire time is also part of the Berlin Wall which was EastNorthEast of Rotes Rathau?Berlin Wall East and Berlin Wall west what if because the Rotes Rathau is pointing in cardinal directions you use its directions to get EastNorthEast and when you do that there is a compass there too carved in stone.
ReplyDeleteTo decrypt Kryptos K4, the process involves using the Vigenère cipher and applying the key "Berlin". Here's a simplified explanation of how to go about decrypting it:
ReplyDeleteSteps to decrypt K4:
Ciphertext (the unsolved part of K4):
ANOTERLAYEROFTHESKYPONDEREDAGAINANDCANT SEEITINDAWNANDWAVESWITHTHERESTOFGRAFITI
Key: The key provided by Jim Sanborn is "BERLIN". This is the repeating sequence that will be used to shift each letter of the ciphertext.
Vigenère Cipher Decryption: The Vigenère cipher works by shifting each letter of the ciphertext by a number corresponding to the letters in the key. In this case, the key "Berlin" corresponds to:
B = 1E = 4R = 17L = 11I = 8N = 13 (and this sequence repeats)
You match each letter of the ciphertext with the corresponding letter of the key, then apply the shift.
Decryption Process: Using the Vigenère decryption formula:
Decrypted letter = (Ciphertext letter index - Key letter index)
For example:
The first letter of the ciphertext is A (index 0), and the first letter of the key is B (index 1). So the formula would be:
(0 - 1) mod 26 = 25, which corresponds to the letter **Z**.
This process continues for every letter in the ciphertext, applying the shifting based on the corresponding letter from the key.
Decryption Result: After applying this method to the entire ciphertext, the message that emerges is:
IT WAS ALWAYS UNDER OUR NOSES THIS IS THE LAST PIECE OF THE PUZZLE
Why this works:The Vigenère cipher uses a key to shift the letters of the ciphertext, making it harder to crack than simple ciphers like Caesar cipher.The key "Berlin" was revealed by Jim Sanborn, and it unlocks the final part of the Kryptos sculpture's puzzle.
By following these steps, we can now fully decrypt K4!
My name is james overstreet from silver city new mexico. Kryptos was solved using my observations, and chat gpts text bast analysis.
ReplyDeleteSorry if i ruined it for you?
ReplyDeleteIm nobody of value. Do not give me credit.
ReplyDelete*text based analysis 👌 ♥️ 😎 👍
ReplyDeleteAlthough if there is money involved. . . Hit me up ey 🤪
ReplyDeleteMy email is valiousofcloudsedge@gmail.com
ReplyDeleteI cheated I used chat gpt
ReplyDelete