r/KryptosK4

Doubts About The Solution Confirmation Process: K4 X/

Doubts About The Solution Confirmation Process: K4 X/

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.

https://preview.redd.it/mx2wsspw9q0h1.png?width=1604&format=png&auto=webp&s=2195e54657c4e6eb4eaa4ff614bcbb672fa82729

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:

https://preview.redd.it/i9m1iaw81r0h1.png?width=1700&format=png&auto=webp&s=268fe4ef489f62406973b9cedea4d9e82339801d

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u/nideht — 1 day ago

Repeated column trigrams in K1

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:

  • ELY appears 4 times (position is almost regular - 13, 23, 33, 43, 53)
  • ELY SIU appears twice
  • NPV CAJ ELY OIQ FMT appears twice
  • other single column triples like EPH, HMS, etc. repeat too

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
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u/Decent_Note_1964 — 2 days ago

​[Theory] K4 Temporal Analysis: The Berlin Clock "Rise" Model

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.

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u/PossibleAnt9461 — 4 days ago

K4 CT to PT lookup experiment

This is not a solution, but it was an interesting result.

To reproduce:

- Take K1-K3 PT, align it to be 31x25 writing column-wise

- Use Kryptos alphabet as the headers

- Look up each digraph in K4 ciphertext to find corresponding PT clusters

I attempted this first with the known plaintext BERLINCLOCK, and CT NYPVTTMZFPKWGD, which can be seen in this example.

Here's what I found at each:

# 20, 3 = BER
# 4, 24 (obo) = LI
# 19, 26 = NC
# 4, 12 (obo) = LOC
# 14, 11 = TED / WHO 
# 1, 26 = HE / HAC
# 5, 25 = HOLE / HE / CAR / CITE / YEC
# 17, 10 = IN / NAI / RIE / URI / IED _ FY / FLE / FOUR
# 11, 16 = KO/KR / OF / DOOM / DO / OURS / FOUR
# 14, 1 = DIVUL / DE / DIA / SHI / SEL / SHA
# 22, 15 = TT / TTI / TTE / TRON / TH / THS

NY = RBE
PV = LI
TT = 
MZ = NC
FP = LOC
KW = 
GD = HO K

There are multiple issues with this procedure, which show up in most failed K4 attempts:

  1. Off-by-one issues
  2. No deterministic system, although the directions seem like they could have a pattern
  3. Most of the characters are very common in the plaintext (other than K, there are only 4), using a random string instead of K4, this may produce similar results. I haven't calculated the probability or tested this yet.
  4. Some look-ups required row/column others column/row, and some seemed like they were nulls or related to unrelated characters to the known plaintext. Also the first result requires permutation of RBE into BER

I tried expanding this beyond BERLINCLOCK to see possible plaintext with this method. There were a few solutions, but this gets into anagramming and without a clear idea of a system within a sea of Plaintext, there's a high chance of false positives.

Some examples of plaintext (but probably incorrect):

BERLIN CLOCK HOUR OF DETH .. 
BERLIN CLOCK WHO HE CARRIED ..
BERLIN CLOCK THE HOLE IN OUR ..

"HOUR OF DETH" was interesting, because it could refer to Die Todesstunde (The Hour of Death) by Alfred Kubin, which is an eerie drawing of a clock. Alfred Kubin had multiple exhibitions in Berlin.

But again, this doesn't seem close enough, or it would require discovering the system in place, if any exists. If this is close to the correct path forward, it seems pretty wacky unless alignment is off or some missing information. For now, I'm filing this in the creative anagram folder.

u/petrified-wax — 6 days ago

K4 -Aristocrat decryption brute‑force run. Merit: none. But it did surface some strong English fragments - urgent, sun, watt, handbook, dpr, is, such, cosmo, ss, fill. Anyone seen this any where before?

#0 FVTRUEHWKMPZJLQNYGSBOCIXDA utideuxurgentsunwatthandbookdprissuchcosmossfillhzcmineywzhwpatpqkbccjlakihryilxcmvywriegezefivzd 948

#1 FVTRUEHWKYPZJLQNMGSBOCIXDA utideuxurgentsunwatthandbookdprissuchcosmossfillhzcmineywzhwpatpjkbccqlakihryilxcmvywriegezefivzd 948

#2 FVTRUEHWKAPZJLQNYGSBOCIXDM utideuxurgentsunwatthandbookdprissuchcosmossfillhjcmineywjhwpatpqkbcczlakihryilxcmvywriegejefivjd 947

#3 FVTRUEHWKAPZJLQNMGSBOCIXDY utideuxurgentsunwatthandbookdprissuchcosmossfillhjcmineywjhwpatpzkbccqlakihryilxcmvywriegejefivjd 945

#4 FVTRUEHWKCPZJLQNYGSBOMIXDA utideuxurgentsunwatthandbookdprissuchcosmossfillhzcmineywzhwpatpqkbccvlakihryilxcmjywriegezefijzd 943

#5 FVTRUEHWKCPZJLQNYGSBOXIADM utideuvurgentsunwatthandbookdprissuchcosmossfillhxcmineywxhwpatpqkbcczlakihryilvcmjywriegexefijxd 943

#6 FVTRUEHWKCPZJLQNMGSBOYIXDA utideuxurgentsunwatthandbookdprissuchcosmossfillhzcmineywzhwpatpvkbccqlakihryilxcmjywriegezefijzd 942

#7 FVTRUEHWKXPZJLQNYGSBOCIADM utideujurgentsunwatthandbookdprissuchcosmossfillhxcmineywxhwpatpqkbcczlakihryiljcmvywriegexefivxd 942

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u/Old_Engineer_9176 — 7 days ago

Do the Shadows Force an Ambigram

I was reading this post (thanks u/Sorry_Adeptness1021) which has the below Vig image from the Smithsonian archives. I realized I could read it (almost) legibly even though it was upside down/rearward.

I believe this is known as an ambigram (an approach where letters are designed to be readable from multiple orientations - eg SOS reads the same upside-down, as does SWIMS, and WOW reads MOM).

I interpret the Vig letters as:
ELCHI?KT <-- idea is freshly-baked, no idea what J would represent
DELCHI?K
GDELCHI?

The serifs help make this interpretation possible (grateful he didn't use comic-sans)

Then I remembered he once said something about (paraphrase) 'turn it around, shine a light on it, ...' and I thought - if a light was placed behind the scroll, it would project the text reversed. Furthermore, if you consider a camera lens (what is a photo but a LUCID MEMORY) it would project the text as inverted (note that camera OBSCURA letters can be found in same K4 block as KRYPTOSY)

This is not far-fetched, he demonstrates exactly this with another artwork, Cyrillic Projector, see 2nd picture. Note how MEDUSA is projected with correct orientation, ie he must have engraved it in reverse - ASUDEM

In other words, is there a potential encoding when read this way? Obviously it's the K4 text I am referring to; the Vig image just makes the idea clear.

Just an idea, maybe it helps someone - I have not had time to investigate it myself yet.

Per my usual warning, I could also be hallucinating, the toast has so many (clock) faces.

Smithsonian archives, annotated by OP

[Edit]: Here is a photo from Elonka's site of the projection. Apologies for the multiple edits - I could not get all 3 images to render. Hopefully working now.

Projection of MEDUSA (Elonka, annotated by OP)

Reversed engraving ASUDEM (Elonka, annotated by OP)

Ugh - images may refuse to render - See images in my reply below

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u/Decent_Note_1964 — 9 days ago

LAYERTWO and 'L' row are both extended on row-14

A curious alignment of clues. I haven't seen this mentioned, apologies if it's old news. There are two 'extended' rows on Kryptos, one on the left & one on the right. And they're both row 14.

  • The 'mistake' in K2 for IDBYROWS / LAYERTWO is on row 14.
  • The row on the Vig with the extra L is on row N == 14.

So we had to 'fix'/add an extra letter to the left-row, extending it by 1.
Already had an extra letter on same right-row, extending it by 1.

                                      + extra
N(14) GHIJLMNQUVWXZkRYPTOSABCDEF GHIJ L  Vig row N

14    DQUMEBEDMHDAFMJGZNUPLGEWJL LAETG   C (idbyrows)
      TESFORTYFOURSECONDSWESTIDB YROWS   P

14    DQUMEBEDMHDAFMJGZNUPLGEsWJ LLAETG  C (layertwo)
      TESFORTYFOURSECONDSWESTxLA YERTWO  P
                             + extra

Aligns with my belief that it was not a mistake, it was a clue.

Remember the P/C thing on his photo - maybe this reinforces it - tells us how to anchor P/C/Vig? Not sure. But maybe it helps someone.

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u/Decent_Note_1964 — 9 days ago

Using the phonetic alphabet in the mask?

> Ed Scheidt: So substitution was one I had to use, 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.

I've already talked a bit about how the square is the same thing as Sanborn's "original matrix"; and I think "pushing in" is a reading order directive (aka "washing machine"). That's the K4 puzzle: how to extract this running key from K3. I think there are clues.

In the last step, the mask, Scheidt changed the language base. This post is a suggestion for what that could mean.

Give the string BERLIN, an everyday way you could change the language base is by replacing each letter with its phonetic alphabet equivalent:

BRAVO ECHO ROMEO LIMA INDIA NOVEMBER

The problem is, this makes the string much, much longer. And, the frequency of the letters will still match English.

I can also stack it like this:

BERLINCLOCK
RCOINOHISHI
AHMMDVAMCAL
VOEAIERAARO
O O AML RL
     BI  I
     EE  E
     R

Still the same codes, but now the phonetic alphabet appears in columns. We generated a bunch of extra codewords like RCOINOHISHI AHMMDVAMCAL, VOEAIERAARO and so on. If you note, those are alphabet substitutions of the original plaintext (the nth letter of the phonetic alphabet keyword).

One last step. I'm going to fill all the spaces with K (which is reminiscent of the Morse code) and I'm going to shift each successive row to the right in a staircase pattern.

BERLINCLOCK
KRCOINOHISH
KKAHMMDVAMC
KKKVOEAIERA
KKKKOKOKAML
KKKKKKKKKKB

Now I have used the phonetic alphabet to create 8 keystreams, which I can combine together using the vigenere table (addition in columns modulo 26). The information of each letter is spread across the following 4-8 symbols of ciphertext.

I've changed the language base in a way that has masked the original plaintext. I guarantee you will have 26 distinct letters in your ciphertext, and very random distribution. Even repeated short words will be corrupted.

But, now the problem is: how to reverse this process? For me this is the best part. There is only one way to reverse this: slowly, very desperately slowly. The weakness is that the first letter B is uncorrupted, and if you subtract the keyword BRAVO from the first five letters, then the second letter is revealed to be E, with keyword ECHO starting at the second position. And so on to the end.

Pencil and paper. Changed the letter base. Very hard, minimal effort. Is this what Scheidt has been talking about?

I'm not sure. This doesn't generate kryptossy letters. But it does generate random ciphertext distributions. If you used an alphabet order like ETAOIN as your plaintext alphabet? In that case, you would pad with E not K.

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u/colski — 8 days ago

The K Walk on Vig

A walk on the Vig tableau exactly transforms EAST → FLRV. It works partially for other cribs too, but breaks down quickly. Probably not the answer, but maybe a useful seed for someone.

I'm not claiming this solves K4. It almost certainly doesn't. But the EAST/FLRVmapping is exact and reproducible, so I'm putting it out there.

I found this while trying to see what IDBYROWS could mean geometrically.

TLDR; navigate the Vig surface in a specific way using EAST and it emits FLRV

EAST / FLRV:

1 - First ct is F. Go to F row on Vig, find corresponding pt, E
2 - scan across until you find K. Then +1 since you moved right (left=0)
3 - scan down until you find the next pt, A. You are on row L
4 - scan left to K. Scan down to find S. Row is R.
5 - scan left to K. Scan down to find T. Row is V.

A KRYPTOSABCDEFGHIJLMNQUVWXZKRYP
   
F       E       →      KR  - start on F row, letter E
        ^               ↓  - scan right to K (then +1 since right)
                        ↓  - scan down to find A -> is on L row
L                K   ←  A  - scan left to K (+0 since left) 
                 ↓      ^
                 ↓         - scan down to find S -> is on R row
R          K  ←  S         - K on left (+0)
           ↓     ^
           ↓               - scan down to find T -> is on V row
V          T               - K on left (+0)
           ^

Three consecutive hops, all clean, no fudging. Input EAST, output FLRV

Where it breaks:

  • BERLINCLOCK / NYPVTTMZFPK: only the opening B/E step lands on a K cleanly; the rest need much larger post-K offsets that don't fit the "+1" rule.
  • CLOCK / MZFPK: middle two contacts (L/O on Z→F and O/C on F→P) work, but the rest don't. However it DOES work if you use TXXT instead of K
  • NORTHEAST / QQPRNGKSS: no full path under the strict rule.

What's interesting about it:

  • It's purely geometric — no key, just the literal letter K on the tableau.
  • The K-letter overrepresentation in K4 CT (8 K's, ~12× English) has a natural explanation if K is structurally an anchor.

What it's NOT:

  • A full decryption method.
  • Tested for letters past the four official cribs.
  • Robust beyond EAST. Though maybe a tweak will 'fix it'

Analysis says this result is ~10x random expectation, so I believe there's a hidden nugget here. But could still be coincidence.

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u/Decent_Note_1964 — 6 days ago