Anyone know why Jim is still responding to emails and stuff? If the auction went through, it should all have been passed on by now. Did the buyer weasel out of the auction or something ? It sounded like there was to have been a new site for submissions and what not by now, but so far nothing. Just honestly curious what's gone on since the auction closed. Thoughts?
What insight do you think would be most helpful in solving K4?
I know my limitations in life and this far and exceeds it.
Enjoy the research and will keep doing it.
My question is out of curiosity for when I have and I feel that I have reached the end of my research.
If I was to share my theory with the community and told them what I have found if its a pattern or not. The community or some of the community agreed it was a pattern and was worth running with it to see how far it can go.
Guess what I am asking in around about way is would the community come as one and if it was the correct theory ( a very very very big if ) solve it as a community or would people go off on there own and solve it themselves.. if I offended anyone with this question then I am sorry about that,isnt my intention.
This post investigates whether K4 matches the speculative idea that HALF of K4 is transposed plaintext.
The proposed transposition would be induced by the K3 palimpsest idea; and the two halves would be the text written in the alternate rows that Sanborn indicated with P (plaintext) and C (ciphertext).
Let's start with typical frequencies for 48 English letters. I'm going to make one conceit, I'm going to swap "E"s and "K"s in the plaintext:
{'K': 6, 'T': 4, 'A': 4, 'O': 4, 'I': 4, 'N': 3, 'S': 3, 'R': 3, 'H': 2, 'L': 2, 'D': 2, 'C': 2, 'U': 1, 'M': 1, 'F': 1, 'G': 1, 'P': 1, 'W': 1, 'Y': 1, 'B': 1, 'V': 1, 'E': 0, 'J': 0, 'X': 0, 'Z': 0, 'Q': 0}
K4 frequencies after subtracting that:
{'U': 5, 'Z': 4, 'W': 4, 'Q': 4, 'B': 4, 'S': 3, 'J': 3, 'G': 3, 'F': 3, 'K': 2, 'E': 2, 'X': 2, 'T': 2, 'P': 2, 'L': 2, 'V': 1, 'R': 1, 'O': 1, 'D': 1, 'Y': 0, 'N': 0, 'M': 0, 'I': 0, 'H': 0, 'C': 0, 'A': 0}
Comparing these, it's plausible. Apart from Es (which would have to be changed to Ks) there are enough English-frequency letters in K4 for HALF of them to be transposed English and the other half to be transposed and substituted English.
So the complete story would be:
The puzzle then rests on understanding the order of overwriting nulls in step 2. For example, if writing vertically like the archimedes palimpsest, every other letter would be plaintext. If writing horizontally, there would be entire words of plaintext.
The pattern 011010 suggested by dYAhRo aligns uniquely with the nulls in 40 positions (5 per row). Ed or Jim suggested to Elonka it means "1-2-3". Are these the nulls we're supposed to fill? 3x40=120. when asked whether there are other "off by one" errors like XLAYERTWO, Ed said "well jim how are we going to answer that one?" and Jim replied "we don't". Ed said his secret is to "push into the square" (from the context, referencing this matrix) and "change the alphabet base" (the mask). Another comment was "sanborn seemed puzzled we hadn't recovered the original matrix and put it through all the shifts"
Speculation upon speculation, it's all we have. How could anyone possibly guess this method from the clues PALIMPSEST and LUCID MEMORY and dYAhRo and LAYERTWO ?
Okay, strap yourselves in. I'm going to show you something very palimpsesty. It's not a solution, just a demonstration of an idea.
S*LO*WL*YDhE*SP*AR*AThL*YS*LO*WLoY*TH*ER*EMoA*IN*SO*FPvA*SS*AG*EDo
E*BR*IS*THtA*TE*NC*UMtB*ER*ED*THpE*LO*WE*RPhA*RT*OF*THnE*DO*OR*WAt
YW*A*SR*EMdOV*E*DW*ITeHT*R*EM*BLtIN*G*HA*NDeSI*M*AD*EAiTI*N*YB*REs
AC*H*IN*THeEU*P*PE*RLlEF*T*HA*NDaCO*R*NE*RAlND*T*HE*NWyID*E*NI*NGa
TH*EH*O*LEsAL*IT*T*LEbII*NS*E*RThED*TH*E*CAbND*LE*A*NDlPE*ER*E*DIw
NT*HE*H*OTuAI*RE*S*CAiPI*NG*F*ROtMT*HE*C*HAiMB*ER*C*AUlSE*DT*H*EFt
LA*ME*TO*FyLI*CK*ER*BsUT*PR*ES*EsNT*LY*DE*TsAI*LS*OF*TaHE*RO*OM*Wi
IT*HI*NE*MeER*GE*DF*RsOM*TH*EM*IwST*XC*AN*YiOU*SE*EA*NtYT*HI*NG*Q?
^ ^ ^ ^ ^ ^
Hopefully you can (almost) recognise this matrix as "the original matrix", K3 plaintext written as a 8 by 42 matrix. However, you will notice that a number of NULLS (asterisks) have been inserted. That means that this matrix is 8x66. If you look even closer, you will also notice that a second message (K2 plaintext) appears in lowercase, written upwards (the columns are indicated). This has been designed to resemble the Archimedes palimpsest. The second plaintext is sort of pushed in between the first. I chose columns that don't interfere with the K3 plaintext.
If we take THIS matrix, rotate it clockwise, and reshape as 24x22 then it looks like this:
ILNTAYESTATHCW********
BLHMHEHAROIEEH********
ISIWNTHONRSLEO********
OLTETYMFTEHMHDeyusedth
ELAAEOAERIILUV********
TSGCRIPEEPEKET********
PDNADESTEWCRFR********
CLRIUARBAELTMTssibleth
OUPIEHBLMTIIFT********
EYTPNNTRRSHRGS********
HEELEEFEAMDOMS********
RRNBTWIEOTDLHLwsthatpo
SNMECIEYTTTDON********
LTXLHTRGOHCYEH********
NHWEADCEEAERNE********
HCRNREYTAAADPMisibleho
OAMNNSAAUIBDDI********
RISLELTMTNESRE********
HAOSEOCAEDFOAF********
ANNETFNTUDWAHPtallyinv
YHSPITEATEEEDI********
DSHRDEENOSIOTR********
NYOANOHEIBRGGM********
EDNRWEQWFIGEAD?itwasto
Hopefully you also (almost) recognise this matrix as the "in between matrix", starting ILNTAYES... But here it has 8 added columns, and the K2 plaintext is clearly legible and - amazingly - it's cleanly separated from the K3 plaintext.
ENDYAHROHNLSRHEOCPTEOIBIDYSHNAI
ACHTNREYULDSLLSLLNOHSNOSMRWXMNE
TPRNGATIHNRARPESLNNELEBLPIIACAE
WMTWNDITEENRAHCTENEUDRETNHAEOET
FOLSEDTIWENHAEIOYTEYQHEENCTAYCR
EIFTBRSPAMHHEWENATAMATEGYEERLBT
EEFOASFIOTUETUAEOTOARMAEERTNRTI
BSEDDNIAAHTTMSTEWPIEROAGRIEWFEB
AECTDDHILCEIHSITEGOEAOSDDRYDLOR
ITRKLMLEHAGTDHARDPNEOHMGFMFEUHE
ECDMRIPFEIMEHNLSSTTRTVDOHW?***t
***i***w***s***e***i***a***s***
s***s***y***t***l***i***t***i**
*u***w***l***b***h***b***s***a*
**y***l***a***l***e***s***i***e
***t***e***d***t***n***h***p***
t***t***o***v***o***o***h***h**
Finally, this is what that matrix looks like when rotated clockwise. Everything up to the questionmark is exactly as seen on the copperplate (except the ragged edge).
After removing the nulls (asterisks), the "K4 ciphertext" at the end is just a transposed version of the K2 plaintext (read every 6th letter).
Now, there are only 48 letters of K4 here, including the question mark, so we're 50 letters short of demonstrating how K4 could have been encoded. Ed Scheidt said:
>"another math problem is transposition, transposition is again the name implies, you're transposing something. So instead of a direct correlation that you can visualize, transposition is a little harder to visualize. It's sort of like looking at a puzzle and you're defining the parameters of the puzzle in the sense of the square, and then now you have the square and you're going to transpose the characters or the letters that are in the square to something that's a secret. So my secret is push into this square. So that was another step. And then the last step which has been good for 30 years, which I didn't know at the time it would be, but I masked the framework, in other words if you can change the language base then it becomes in my favor and not your favor of trying to break it. It becomes more of a challenge now, when it was used as the mask it was current, 2020 secret."
So my suggestion is: this is what Ed meant by "pushing into the square". He's talking about a palimpsest: writing both plaintexts in the same space.
The pattern of the "original matrix" at the top here, in terms of plaintext and nulls, is (10110110110). What I would wish it to be: (10010), because this is the pattern indicated by "dYAhR". Perhaps there's another way that I don't see to introduce this kind of "palimpsest". For example, adding 16 columns would work.
It was only recently that Sanborn pulled us away from the Mengenlehreuhr “set theory” clock.
Across many interviews he has kept pointing back to Berlin and the Berlin Wall as if that connection mattered more than people realised.
Stepping back and staring into the emptiness that is K4, I had a doorbell moment.
What if Sanborn is not pointing us to the Berlin Wall falling, but to a moment in time when the World Clock and the Wall shared a significant event.
Everyone assumes the fall of the Wall is the key, but is it.
Sanborn is an artist. He reacts to people, emotion, atmosphere and human courage.
The fall of the Wall was historic, but the events before it fell were the ones that would have struck him hardest.
This is where the World Clock and the Berlin Wall collided in time.
The human element was exposed.
Unity. Defiance.
Thousands on 7 October.
More than a million on 4 November.
All gathering in the square around the World Clock, standing together and demanding change.
That moment may be the real link to K4.
But the question is how we are meant to use it.
Do we need to be physically at Alexanderplatz, facing east north east, to understand the next step.
Do we need to be physically at the Kryptos site, standing the way those people stood, at the same date and time adjusted for time zones.
Or do we take the full date of the protest and use it as a key in something like Gronsfeld, Gromark or another transformation.
This idea ties the Berlin Clock and the Berlin Wall together at a moment that mattered more than the Wall coming down.
Hi all.
I not long ago found the kryptos puzzle and have been looking into it myself.( dipping my toes in as its way over my head.).
Been Looking at what people have been saying,there theories on how to solve k4.
I do have 3 questions I would like to ask the community..
I seen when I first looked at the page no AI solutions or you will be banned. Can I ask why AI wouldnt be able to solve it.
have noticed people talking about using supercomputers to solve it. I was wondering how that translates onto pen and paper as I assume a supercomputer can do multiple calculations at once.( think I read it could be done by pen and paper.)
As I am only dipping my toes into some research at the minute can I ask when does a coincidence become a pattern and when would a coincidence become your own bias. ( noticed people talking about coincidence now and then. Just curious about it.)
Would like to say thankyou in advance if anyone answers these questions.
I received one response and later a return of the fee
I want to raise something potentially troubling regarding the K4 submission and confirmation process. Sanborn's submission rules do not actually promise confirmation if a correct solution is submitted. Here's a cropped image of his rules with relevant text highlighted, and of course you can also go to his website to read the full sheet.
In its least optimistic interpretation, the last highlighted sentence could be read as a promise not to confirm a solution if one comes through. I also think the choice of "a solution" rather than "the solution" is a little strange.
I've had a number of submission interactions with him recently, and on the most recent one I tested out the "constructive dialogue" part and pushed him on the topic of confirmation. I'm sharing the email string here because I was able to get more than the "two back-and-forth emails", and he quickly addressed one question about the solution length (confirming what we already knew). But he ignored three times in a row the issue of confirming the correct solution. There was no response to my last message of May 8. The relevant portions are highlighted below. Disregard anything specifically related to my solution attempt, as that's not the point of this post:
I’ve seen ELY in K1 mentioned a few times as a possible K4-relevant artifact, but ELY is only one example of a broader pattern. I have not seen this mentioned elsewhere, so maybe it's useful information for someone.
[edit] I believe this is novel information, not a rehash of tired analysis, hence this post. I wrote it to help another user who was looking into ELY, I thought the following was well-known. Apologies if old news, please correct me. Also, see K4 relevance in later edit. This is NOT a solution to K4, but might help.
If you stack K1 as plaintext / key / ciphertext, several entire vertical columns repeat. In other words, it is not just that the ciphertext has repeated letters, but that the same PT+key+CT column triple blocks recur. Examples include:
That seems more specific than a normal text repeat. For a column triple to repeat, Sanborn’s plaintext wording, the repeating PALIMPSEST key alignment, and the resulting CT all have to line up.
Since Jim only had a budget of 63 characters (K1 length) it would be unusual to force 29 columns (46%) to occur in a repeated triple unless there was good reason.
For example, imagine he needed ALICE to repeat in 2 specific spots, including exact corresponding keyword fragment (say ALIMP) and CT. He'd need to plot ALICE on said positions (strongly constrained by where (p)ALIMP(sest) happens to land, it is predetermined) then build up a coherent sentence around them. Here is an example. Now imagine having to fill in the space surrounding ALICE/ALIMP with lucid prose.
what here
? |A L I C E| ? ? ? ? ? |A L I C E| ? ? ? ? ... CT
p |A L I M P| s e s t p |A L I M P| s e s t ... keyword
Given that K1’s phrasing was apparently “carefully worded,” I don’t think this should be dismissed as just "ELY appearing by chance". I don’t know what, if anything, it implies for K4, but the phenomenon is better described as repeated PT/key/CT column-triple blocks, not merely repeated CT fragments.
[edit, per mod]: Consider that these repeated sections are useful as alignment edges or strides (see the 13/23/33/43/53 pattern). Or that their content is elements of a transposition or (say) a Polybius square. Or perhaps concatenated as a keystream. Or maybe under some shift they give us the full RDUM strings. Or the indexes and lengths frame text extents in K4. Not sure, but perhaps they connect some dots someone else has been looking for.
Nomenclature: Repeated sections delineated using |x|, eg at position 13, you can see ELY SIU, which you can see again at 43. The more interesting/longer one is at 31 & 51
|13 | <-- See repeat of this 2 column block at 43
B E T W E|E|N S U B T L|e s|H A D
P A L I M|P|S E S T P A|l i|M P S
E M U F P|H|Z L R F A X|y u|S D J
|6|
|31 35| <-- See repeat of this 5 column block at 51
I N G A N D T H|E|A B S E|n c e o f|
E S T P A L I M|P|S E S T|p a l i m|
K Z L D K R N S|H|G N F I|v j y q t|
^ 26 ^
|43 | |51 55| *
L I G H T L I|e s| T H|E N|U A|n c e o f|I q L U S I O N
P S E S T P A|l i| M P|S E|S T|p a l i m|P c E S T P A L
Q U X Q B Q V|y u| V L|L T|R E|v j y q t|M k Y R D M F D
|47 |
Inventory (29 columns):
EPH x2 positions 6, 26
NSZ x2 positions 7, 19
USR x3 positions 9, 49, 59
ELY x4 positions 13, 33, 43, 53
SIU x2 positions 14, 44
HMS x2 positions 15, 25
APD x2 positions 16, 21
NPV x2 positions 31, 51
CAJ x2 positions 32, 52
OIQ x2 positions 34, 54
FMT x2 positions 35, 55
LPQ x2 positions 36, 41
IPM x2 positions 56, 61
Researcher: Mohamed Osman Abdelbagi
Discipline: Civil Engineering / Structural Analysis
Environment: Python 3.11 (Termux Linux)
The Thesis:
The unsolved K4 segment of the Kryptos monument is not a standard letter-substitution cipher. Based on my engineering analysis, it functions as a Temporal Chronometer linked to the physical and historical context of the Mengenlehreuhr (The Berlin Clock). I propose the "Berlin Rise" Model, which suggests that the solution is a projection of the 14-year delta between the clock's inception (1975) and the fall of the Berlin Wall (1989).
Technical Observations & Methodology:
The Pivot Point (6): In the ciphertext fragment VVTREOTREHBT6FDALVWPWASWQE, the number 6 is not a placeholder but a geospatial and temporal fulcrum. In structural engineering, this acts as the "Origin Point" for the rotational shift.
Temporal Offset (The Rise): Applying a 14-point shift (representing the 14 years from 1975 to 1989) as a rotational key to the characters following the pivot.
Linguistic Emergence: Under this specific temporal rotation, the ciphertext begins to yield patterns that align with phonetic structures like "FLCN" (Falcon) and concepts of "TIME".
Geospatial Calibration: The alignment occurs when factoring in Berlin's coordinates (52°N, 13°E). The theory suggests K4 describes the "East Region" relative to the clock's shadow at a specific moment in time.