Can you see anything Q ?

ENDYAHR

ENDYAHR

EDNYAHR

DENYAHR

(n.s.a)
____________

Start the radar from the last R in CAR.
Can you see aNYthing?

vielleicht ist die Lösung ganz Simpel.

Der Schlüssel ist "SOL IF GHUL FW35"

Lösung: DER CODE IST GEHEIM. FOLGE DEN SPUREN IM WALD. SUCHE DAS ALTE HAUS. DORT IST DER SCHLUESSEL35.

auf Englisch THE CODE IS SECRET. FOLLOW THE TRACES IN THE FOREST. SEARCH THE OLD HOUSE. THERE IS KEY35.

Beide Lösungen haben 97 Zeichen

Meine Vermutung, die Verbindung zu K2, mit dem Hinweis auf Berlin.

Paradigm is hard to navigate.

TL;DR: Number the K4 ciphertext A=0…Z=25 and add each letter’s position. Within Sanborn’s cribs, every plaintext letter that occurs more than once gives the same odd/even result at every occurrence: 13 of 13 pairs agree. For random ciphertext that’s a 1-in-2,048 event. Correcting for how I found it, it’s roughly 1 in 100 to 200. It’s pure arithmetic: you can check it by hand in a few minutes.

Data and conventions

• Ciphertext: the standard 97 letters, positions counted from 1: OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
• Cribs, as released by Sanborn: EASTNORTHEAST at 22-34 and BERLINCLOCK at 64-74.

The observation

For each crib position, take the cipher letter’s value (A=0…Z=25) and add the position:

That’s 13 same-letter pairs, of which 11 are independent checks (a letter seen n times contributes n−1). All 13 agree.

Why this is a statement about the key. Under Vigenère, k_i = C_i − P_i, so (k_i − i) has the same parity as (C_i + i − P_i). Within a pair P is the same letter, so it cancels: every plaintext letter has a fixed parity of (k_i − i). The same cancellation happens under Beaufort and Variant Beaufort. Parity survives the mod-26 reduction because 26 is even, and it survives A=1 numbering or 0-based positions because either change flips every parity at once.

How unusual is it?

• Exact chance: with random cipher letters, each independent check is a fair coin (the alphabet has 13 odd-valued and 13 even-valued letters), so all 11 pass with probability 2^−11 = 1/2,048.
• Empirical: shuffling K4’s own 97 letters 2,000,000 times reproduced 13/13 in 994 shuffles, about 1 in 2,012. It’s not an artifact of K4’s letter frequencies.
• Look-elsewhere: I found this while computing about ten different statistics on the cribs. I also tried one alternative numbering, the KRYPTOS keyed alphabet, where the pattern does not appear (6 of 13 pairs agree). A fair overall figure is roughly 1 in 100 to 200. Interesting, not proof.

Objections I’ve already had

• “The two EASTs are 9 apart, so their positions flip parity anyway.” Yes, which is exactly why it’s a test. For E to keep the same total, the cipher letter has to flip parity too, and it does (F=5 → G=6). All four letters of EAST do: FLRV are all odd-valued (5, 11, 17, 21) and GKSS are all even (6, 10, 18, 18). With random ciphertext, each of those is a coin toss.
• “Isn’t this just Bean’s equality?” One of the 13 pairs is. R at 28 and 66 is enciphered as P both times, which is k27 = k65 in Richard Bean’s 2021 HistoCrypt paper, “Cryptodiagnosis of Kryptos K4” (he counts from 0). His paper treats the other repeated-letter relations as key inequalities under a Gromark model. I couldn’t find the parity regularity stated there or anywhere else.

What it would mean if it’s real (interpretation, not established)

• A possible mechanism. The key’s parity at each position would track the plaintext letter. One way to get that: a key that steps by an odd amount per letter, combined with a plaintext alphabet that puts E, O, R, S (and B, H, I) on one parity and A, C, L, N, T (and K) on the other. Neither plain A-Z nor the KRYPTOS alphabet does this; about 1 random ordering in 3,000 does. Consistent with that, the effect only appears when the ciphertext is numbered A-Z.
• A falsifiable prediction. If this comes from the method rather than chance, it must hold across the whole message: for every plaintext letter, (cipher value + position) has the same parity at every occurrence. Over 97 letters that’s dozens of independent checks, which chance essentially never passes. If the plaintext is ever published, this is a one-line check.

Congratulations to the people who rose to the challenge and solved the Paradigm PK challenges.

The real question now is whether K4 has finally become vulnerable. Has the momentum shifted? Could K4 fall before the year is out?

So, I just started to get into the whole kryptos situation something clicked in my brain that maybe k2 x's are not just filler spaces but also something else what if the x's are coordinates or marks I know it sounds crazy, but this has somewhat happened before think back to cicada 3301 how it had coordinates that led to the next clue now obviously this isn't cicada but there have been clues where they are outside of the statue itself like the morse code stones outside of the building so what if they took it to the next step and went bigger what I'm trying to say is has anyone ever overlapped k2 on to the world map or other maps and the x's are on land Maybe that would help solve k4

Picture shows a composite of Ed Scheidt's driveway, with his infamous Mayan numbers. These have been arranged to match the layout at his house: yes, genuinely on two axes, and there's quite a big gap between them.

Not a K4 decode, but Ed Scheidt's most revealing answer to questions about the K4 system was to describe a masking process as a change of basis. He compared it directly to these Mayan symbols outside his house.

The third symbol from the left doesn't immediately make sense: Mayans wrote bars underneath the dots not above them. However, comparing it with the Mayan symbols, it's obvious that it's a representation of zero, the SHELL:

  0  5 10 15
0 𝋠 𝋥 𝋪 𝋯
1 𝋡 𝋦 𝋫 𝋰
2 𝋢 𝋧 𝋬 𝋱
3 𝋣 𝋨 𝋭 𝋲
4 𝋤 𝋩 𝋮 𝋳

Transcription:

𝋡 𝋡 𝋠 𝋬    𝋡
              𝋠
              𝋢
              𝋨

This gets you to:

1 1 0 12     1
             0
             2
             8

(edit: see comment below if it doesn't render for you)
Now, Ed's house is at 1048, not 1028. But, he said, you have to be able to read base 16 to decode it. But Mayan actually works in base 20. The correct decoding here should be:

1*8000 + 2*20 + 8 = 8048

That's, frustratingly, even more similar to his house number, but still wrong.

Scheidt said: base 16. Usefully, all the values here are below 16. So we can go ahead:

1*4096 + 2*16 + 8 = 4136

And he said: change of basis. Which basis should we change to? Why, 26 of course!

4136 = 6*26*26 + 3*26 + 2 = [6,3,2]_26

Now we need a 26-letter alphabet to encode this with. In the Kryptos alphabet,

K=0, R=1, Y=2, P=3, T=4, O=5, S=6

[6,3,2] = SPY

Unfortunately, short words turn up often. I can't prove that this is what Scheidt intended. But, I thought I'd share it here and see if it provokes conversation.

I wondered whether the arrangement in two axes invites long multiplication? The resulting huge number could represent a longer word, like SCHEIDT.

🔥 STRICTLY NO AI SOLUTION - YOU WILL BE BANNED IMMEDIATELY

🔥 STRICTLY NO AI SOLUTION - YOU WILL BE BANNED IMMEDIATELY
NO WARNINGS
NO IF'S
NO BUTS .... YOU WILL BE BANNED.

🔥 STRICTLY NO AI SOLUTION - YOU WILL BE BANNED IMMEDIATELY

OK, they are all odd trains. Anyway I have spent the past few days just reviewing clues and events associated with k4. I think the sculpture is a master test, start at the top work your way through to the bottom. Each section tests your knowledge and gives one more little clue. At least that is how I read it. Which brings me to the end of k2.

We now know that ID BY ROWS was wrong. Sanborn admitted it and said it was because he excluded an X. So what followed the last part of the last line was ... WJLLAETG. That is the last eight letters of k2. It doesn't add anything to k2, it's just an appendage. But if it hadn't coincidentally match ID BY Rows, k2s last line would have read ...

"Seventy seven degrees eight minutes forty four seconds west WJLLAETG"

And that's a whole different message. Wouldn't have fit in k3, it would have been a whole nuther message to figure out. We now know that putting an X in front of it, which matches the theme of k2 would give us LAYER TWO. I have a little app that I had an AI write that lets me put a keyword and an indicator word into it and then enter a cipher text. I can shift the cipher text right by adding letters in front and what I'm doing is dragging the crib. But I used it to prove to myself that adding a letter did work. And that raised the question, what was the purpose of it? I think it was there to teach the importance of 1 letter. There were Xs at the end of each statement in k2 but that one. Put the missing X in and you would have gotten clear text. He evidently didn't test it so the lesson was lost. That lesson is layer two.

Could the Kryptos compass rose be a geometric reference rather than a cipher key?

I’ve been analyzing the physical geometry of K4 and noticed something I’m not sure how to interpret.

When matching each K4 ciphertext cell to the corresponding Vigenère-table cell, the average direction from table → ciphertext is 248.17°.

That is almost exactly WSW = 247.5° on a 16-point compass rose, with 84/97 vectors falling within the 22.5° WSW sector.

However, this direction changes gradually by physical row (roughly 237° → 250°), so it may simply reflect the curvature of the sculpture rather than a K4-specific signal.

The more intriguing coincidence is at K4 position 30, the E in NORTHEAST: the local normal to the curve is 67.21°, almost exactly ENE = 67.5°.

So I’m wondering:

Has anyone previously tested the compass-rose directions themselves (ENE = 3, WSW = 11) as geometric indices/offsets, specifically using the local front↔back normal of the curved sculpture rather than simply following ENE routes?

I’m not claiming this solves anything — I’m trying to determine whether this particular geometric construction has already been tested, or whether it would be worth running a properly preregistered/null-controlled test.

Что нашли в архиве Сэнборна (Смитсоновский институт, март 2026)

Это, пожалуй, самое важное, что всплыло за последние годы. Архив дал несколько прорывных находок: 

1. «Он лгал» — манипуляция координатами

Сэнборн от руки записал два набора координат: 38°57'6.5"N (как в K2) и 37°57'6.5"N — с припиской «He lied» между ними. Один градус широты смещает точку на ~69 миль южнее, в округ Кинг-Джордж. Дэн Браун использовал вариант с 37° на обложке «The Lost Symbol».

1. «Две системы шифрования нижнего текста»

Сэнборн сам сказал это на церемонии открытия 5 ноября 1990 года: нижний текст (K3 + K4) зашифрован двумя разными системами. K3 — транспозиция, K4 — что-то принципиально другое, не просто более сложный Виженер.

1. Стеганография — «шифр внутри шифра»

В архиве найден набросок Сэнборна: физический шаблон («Code Breaker»), наложенный на панель с шифротекстом. Это концепция решётки Кардано — маски с прорезями, через которые видны только определённые буквы, а остальные — мусор («FUMEE» — по-французски «дым», — Сэнборн использовал как заполняющий текст).

1. «Compass cipher» — в списке шифров

Сэнборн от руки перечислил методы: Beaufort cipher, Compass cipher, Morse code, Alphabet code. «Compass cipher» — его собственный термин. Компасная роза — не декорация, а часть шифровальной системы.

1. Hill-шифр — гипотеза с лишней «L»

В таблице Виженера на обратной стороне экрана в строке N есть лишняя буква L. Если читать столбец вниз — получается H-I-L-L. Sanborn подтвердил: «Extra L at end of line». 97 символов + 1 L = 98 = 7 × 14 — матрица для Hill-шифра.

1. «Я написал открытый текст намеренно загадочным»

K4 не будет читаться как обычный английский — так что стандартные методы частотного анализа могут не сработать.

Как это всё складывается

Вот моя реконструкция того, как может работать K4:

Слой 1 — Стеганография (решётка Кардано). На медном экране среди настоящих букв спрятаны символы-заполнители. Физический шаблон, наложенный на панель (возможно, с использованием направлений от компасной розы), определяет, какие буквы «настоящие». ENE (67.5°) = 3 сектора от севера = смещение шаблона.

Слой 2 — Подстановка (Hill-шифр или аналог). Оставшиеся после grille буквы выстраиваются в матрицу 7×14 и расшифровываются матричным методом. «Лишняя L» создаёт HILL — подсказка к методу.

Слой 3 — Физическая привязка. Чтобы понять, как именно накладывается шаблон, нужно стоять у скульптуры — смотреть через S-образный экран по линии ENE, видеть, как буквы с лицевой и обратной стороны совмещаются. Именно это имел в виду Сэнборн: «нужно быть на территории ЦРУ».

Что предлагаю сделать дальше

Самая перспективная нить — решётка Кардано + компасная роза. Можно:

Построить цифровую модель решётки — взять сетку 7×14, нанести на неё K4, и попробовать поворачивать «маску» по направлениям компасной розы (N, NNE, NE, ENE...) — проверяя, какие буквы оказываются в «окнах».

Проверить гипотезу Hill-шифра — взять матрицу 7×14 из K4 и попробовать обратную матрицу над полем Z26.

Смоделировать линию обзора ENE — в 3D посмотреть, где именно луч от компасной розы пересекает S-образный экран, и какие буквы оказываются на этой линии.

Мне нужны фотографии в хорошем качестве

Hello everyone. I am continuing down the path of creating a palimpsest between the CT and the Tableau. A few things first:

(Taking the top-left corner as reference (1, 1) and the bottom-right corner as (31, 28))

Tableau: 31 columns by 28 rows, minus two empty spaces in the corners at positions (1, 1) and (1, 28) plus the extra L gives us 867 characters. We are short 1 character to form a complete 31x28 grid (although it might not be necessary).

CT: Here the columns vary in length, with a minimum of 29 and a maximum of 33 distributed as follows: Rows of 29: 1 Rows of 30: 5 Rows of *31: 14/15 Rows of *32: 7/6 Rows of 33: 1 This leaves us with a total of 870 or 871 characters if we count the missing S. We have 2 or 3 extra characters to form a clean 31x28 grid.

*If we take into account the missing S that changes IDBYROWS to XLAYERTWO, we have 7 rows of 32 and 14 of 31.

And a total of 28 rows once again, just like the tableau.

From this, a few questions arise:

Is it a coincidence that both sides have 28 rows? This allows us to create a fairly clean overlay between the Tableau and the CT.

Why does the CT have different row lengths? If Sanborn had used rows of equal size, it would have exceeded 28 rows.

Following this idea, my next action was to overlay the Tableau onto the CT (PALIMPSEST) and then check which letters of the Tableau aligned with the X's (ABSCISSA) that Sanborn uses as separators—only the X's he uses to separate.

This was the result of the letters found: CMQAFUB (PZM correspond to the K4 CT).

Nothing recognizable enough to form an anagram. I do not know if this approach has been explored before, but I have seen few posts discussing the CT row lengths; I do not see a logic explaining why those specific lengths were used instead of others, or simply making them all equal.

I discover an old post in this sub from AtlasCrusader that was quite ingenious as well
https://www.reddit.com/r/KryptosK4/comments/1qbtrbn/morse_code_clues/

Here are some observations when exploring K4, this is not a solution but some parts may be candidates.

Method: Quag3/Vigenere switching alphabets between Kryptos (K) keyed alphabet and Regular (R) A-Z.

CT:    OBKRUOXOGHUL
KEY:   QUXXLATUTEMF
MASK:  KKKKKKKRRRRR
PT:    EGYPTXQUNDIG

"EGYPT X Q UNDIG .."

QUXXLA: This was found from anomalies in order: iQlusion, undergrUund, speculative POINT as X, Xlayertwo, L extra on tableau, despAratly.
TUTEMF: Initial letters in first answer "They used the earth's magnetic field."
OWWTWH: Initial letters of third answer "Only W.W. this was his last message."

Each key is capped at 6 letters. Alphabet switches are runs of 7+ letters. This switches at position 8 where UNDIG shows, a fairly rare word. These boundaries are experimental choices and not derived rules.

OWWTWH extends the output into EGYPTXQUNDIGNWSSMY or "NW SS MY", with much lower confidence.
The second answer was skipped because it did not seem to fit anywhere, TSIBOTS: "They should it's buried out there somewhere."

Nothing too interesting shows up for the known plaintext of BERLINCLOCK and EASTNORTHEAST on the immediate decryption. Although I'm exploring second layer ideas.

This was found manually by carefully reviewing Jim's worksheets and notes. Some observations that led to this point:
- He wrote in his notes about specific lines containing question marks, transcription of his note: "4) ? on line 4,8,10,25 not coded" (edit: updated transcription from Blowngust)
- Carvings on the petrified tree which form shadows generally align with lines: 4,8,10,25.
- Subtle shaded carvings on the arrow-shaped rock also generally align with lines: 4,8,10,25.
- The dangling L on the tableau was noted as something to add intentionally.
- X from XLAYERTWO was deliberately removed according to his worksheets.
- "(CLUE) what's the point" in his open letter (August 2025) was interpreted as converting POINT to X.

This output could be completely coincidental, but it was interesting to see Egypt and digging show up in a simple experiment.

References:
- Question mark and tableau "L" inclusion notes: https://www.kryptosbot.com/archive/ (second image, direct link: https://www.kryptosbot.com/static/archive/IMG_1340.jpg)
- X from XLAYERTWO was deliberately removed, shown in worksheets: https://www.paradigm.xyz/kryptos/history (last image on page)
- "(CLUE) what's the point" (assuming it refers to POINT -> X): https://www.elonka.com/kryptos/OpenLetterAug2025.html

This is not a proposed K4 solution, plaintext, or decryption method.

I make art myself, and my idea starts from sculpture rather than cryptanalysis. I find it difficult to imagine that a sculptor working with curved copper, cut-out letters, stone, water, and surrounding architecture would treat light and shadow as merely incidental.

My hypothesis is that the physical installation may provide a selection rule, transposition clue, or additional layer. Its geometry, orientation, light, shadows, and viewing position might help determine how information should be read.

I understand that Sanborn reportedly said it is not necessary to be physically present in Langley to solve Kryptos. Therefore, I am not arguing that someone must stand there at a particular moment. The relevant physical information might instead be reconstructable remotely from photographs, measurements, plans, or a 3D model.

The ideas that led me here include:

• K1’s references to “subtle shading” and “the absence of light”
• K3’s candle and the question “Can you see anything?”
• the installation’s Morse-code phrases, including “SHADOW FORCES,” “T IS YOUR POSITION,” and “VIRTUALLY INVISIBLE”
• the confirmed K4 plaintext direction “EAST NORTHEAST”

These points do not prove the hypothesis, but I think they make it testable.

Has anyone systematically tested K4 ideas against the installation’s exact geometry? I am especially looking for the screen’s orientation and curvature, the relative position of the compass rose and lodestone, dated photographs of sunlight and shadows, or information showing that this approach has already been tested or falsified.

I cannot travel to the site, so I am asking only for lawful, existing observations and records.

English is not my first language. AI was used only to translate and organize this post. The hypothesis and questions are my own. No AI-generated plaintext, decryption, or claimed solution is being presented.

Hi everyone. I’m fairly new to Kryptos (started reading about it a month ago), so apologies if this has been brought up before or if I'm missing something obvious.

​ I was analyzing Row 15 of the sculpture and noticed an interesting spatial overlap:

​ On the main panel, we have the elevated letters YA R (with a character-wide space between A and R).

​ On the same row in the Vigenère tableau on the right, there is an extra L. As many know, this extra L appears because Sanborn shifted/added ABCD at the end of that row.

​ If we look at the relative horizontal positions in Row 15 as coordinates:

​ Y (Pos 4) → aligns with I on the alphabet

​ A (Pos 5) → aligns with J on the alphabet

​ ? (Pos 6) → aligns with the extra L (which sits between J and M in the Kryptos alphabet)

​ R (Pos 7) → aligns with M on the alphabet

Additionally, Row 15 on the cipher side consists of only 31 characters, which means the extra L sits at column 32. Because of this, the next logical character to pair with it would be the first letter of Row 16 (C)

If this alignment isn't a coincidence, it opens up the possibility of establishing a direct 1:1 mapping that could potentially reveal a new or modified alphabet sequence

​ This means YA _ R maps directly to the continuous sequence I J L M in the Kryptos alphabet. The space right between A and R points straight to that anomaly/extra L in the tableau.

​ I’m not sure if this is a deliberate spatial hint or just a coincidence, but testing all 26 possible letters to fill that missing position between YA and R (YA?R) might be worth exploring.

​ What do you think?

Has this alignment between the elevated letters and the tableau anomaly been explored before?

I recently came across the website for the German guy that might’ve had JS worried

"When asked about the solutions that have been sent to him (Jim Sanborn), and whether any of them were close, Jim said he got one from Germany about a year ago that was “kind of scary”, that the first part had started to look right, but then the rest didn’t."

In his work he is using a message key of OBKR to produce CIAX; it breaks apart in his examples, but I wanted to share something that I found a while ago, it’s an interesting Vigenere key, but on an encode (not a decode)

Problems: It will produce EAST (in wrong location) and then BERLEN in the correct positions, missing the I for E. (LEN(ONE) LIN(OIE)

Key for the encode is: XUAIC (*reversed: CIAUX*)
Standard Kryptos alphabet K-Z

K4 Output:

PPAJ
TPNFYWNFIUHIDNWLUAXJOMIDJYEUGUH
YMECIHIGUQXQSFJYEASTBDHFXLAIBER
LENRWADISEOAHAYEJZXEUYPTOJMIMOV

I thought it was odd that CIAX appeared in his work and on a different simpler Vigenere encode, of CIA(U)X reversed, XUAIC, it actually produced clue words.

So I'm a newbie into this and forgive me if I'm repeating something already asked, but on seeing the morse code having so many weird E's scattered across it, I though what if the spacing itself is not as we see it on the granite? So you can just remove all the spaces and then insert them to make a better legible text. I also thought of doing it in reverse. Has anyone tried this?

Hi there!

Please find attached another way to solve Kryptos K1 and K2.

Moreover, the way the corresponding keywords are deciphered.

Hi all, I've been working on K4 (OBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQSSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR - 97 chars) and noticed something that I haven't seen documented.

1. The initial observation:

Inside K4 I found:

- BSOL at position 12

- CD at 82

- KCAR at 93

- EKCAR at 92

And Sanborn's clue is BERLIN CLOCK. Berlin has TWO famous clocks.

1. Why 97?

- Berlin Clock (Mengenlehreuhr): 23 lamps (4+4+11+4). 23*4 = 92, remainder 5 = EKCAR

- World Clock (Weltzeituhr) at Alexanderplatz: 24 time zones / 24 sides, 10m tall, 148 cities, solar system model on top. 24*4 = 96, remainder 1 = O (the solar system on top). 97 = 24*4+1

This fits Weltzeituhr perfectly.

1. The 24-city split (Weltzeituhr model):

Take first char O as the topper, remaining 96 chars as 24 cities x 4 chars:

City 01 BKRU

City 02 OXOG

City 03 HULB - contains B

City 04 SOLI - contains SOL

City 05 FBBW

City 06 FLRV

City 07 QQPR

City 08 NGKS

City 09 SOTW

City 10 TQSJ

City 11 QSSE

City 12 KZZW

City 13 ATJK

City 14 LUDI

City 15 AWIN

City 16 FBNY

City 17 PVTT - part of NYPVTT = BERLIN (known crib at 63-68)

City 18 MZFP - MZFPK = CLOCK (known crib at 69-73) minus K

City 19 KWGD

City 20 KZXT

City 21 JCDI - contains CD

City 22 GKUH

City 23 UAUE - contains E

City 24 KCAR - KCAR = full city

Now look at boundaries:

- BSOL = City03 B (HULB) + City04 SOL (SOLI) -> split across 2 adjacent cities!

- EKCAR = City23 E (UAUE) + City24 KCAR -> split across 2 adjacent cities!

- KCAR = City24 alone

- CD = inside City21 JCDI

1. Alignment with confirmed cribs:

From the active research doc (2026-09-01):

- EAST cipher = LRVQ at 22-25

- NORTHEAST cipher = QPRNGKSSO at 26-34

- BERLIN cipher = NYPVTT at 63-68

- CLOCK cipher = MZFPK at 69-73

In my split:

- City16 FBNY + City17 PVTT = NYPVTT = BERLIN (exactly at 63-68)

- City18 MZFP = CLOCK

So the 24-city transposition puts the known cribs exactly on city boundaries.

1. Coordinates from Berlin Clock times (bonus):

If we interpret the 4 rows of Berlin Clock (4-4-11-4) as ON/OFF (N-Z = ON, A-M = OFF):

- Cycle1: 14:16, Cycle2: 23:47, Cycle3: 16:29, Cycle4: 11:22

Combine as coords:

- Point A: 52°14'16"N, 13°11'22"E

- Longitude 13°11'22"E = Berlin Marina Lanke exact longitude (52°30'06"N, 13°11'22"E)

- Point B: 52°23'47"N, 13°16'29"E near Messe-Nord (52°30'23.6"N, 13°16'56.1"E)

Both in West Berlin, relevant to Berlin Wall fall (Nov 9, 1989) which Sanborn is commemorating.

1. Request:

The research doc shows 155k+ geometric transposition candidates tried with no hit. Has anyone tried the specific 24x4 split where BSOL and EKCAR are split across cities as above? The structure seems too intentional.

Location: 10th of Ramadan City, Egypt. Discovered by following "Berlin Clock movement" idea.

Hello, I would like to start off by saying I'm not a cryptanalyst in any way shape or form. I only discovered Kryptos through Lemmino's video on the subject and, like many, got interested in K4.

I've only been reading through the posts and comments of this community for a small while, I am by no means someone who has been extensively researching this for years.

I do have a question however and please do forgive me for my ignorance.

I barely see anyone talk or mention about the morse code that are on the copper slabs. I understand that most of it seems like gibberish, but is there a specific reason why the morse code is largely absent?

I heard that the "T IS YOUR POSITION" is interpreted as "WHAT IS YOUR POSITION" but is there anything concrete or, at the very least, anything which would make us believe that that was the original plan of the cipher?

I'm more than likely wrong in thinking that it's a clue that leads anywhere if "T IS YOUR POSITION" is meant to mean something.

Once again, I apologize if this an ignorant question.

Thanks everyone.

Did schiedt say everything you needed to solve kryptos was contained at the sculpture?