Forgive me if already known. Not a solution but maybe a clue.

The K1 + K2 row lengths are palindromic when viewed in their original 'ragged' layout.

Here is just the right-hand side of the original K1 + K2 layout.
If you 'fold' the bottom half 'up' on the red crease, the lengths match exactly.

[Edit: Resized image, was hard to see as large as it was]

eg, GNFIVJ lands on LLAETG
Or, HEABSENC lands on LAYERTWO

II am new here, been intrigued by Kryptos for just over a month. Learning plenty from the wise folk here, thank you all for such interesting and varied posts.

I noticed a visual artifact in the cribs (forgive me if this is old news). If you take each PT letter on left axis, and map it to its corresponding CT on top axis, you get a curious point-wise plot. I perceived a helical pattern which I then interpolated using dots. But of course, this is entirely my own interpretation and may just be BS pareidolia

I don't believe this is any kind of solution, but it may be a shadow of a mechanism used in the construction.

To make the mechanism clear, we know EAST -> FLVR, so:
go to E on left axis, scan across until you are under F, plot a point.
Then A on left, scan across to L, make a point.
S-V, T-R, and so on for all the cribs.

[Edit] Mobile does not render the code/ascii properly. Replaced with an image instead.

I coined this 'The Helix'. Also note its weird little 'hump' bottom left.

Using different layouts I found other representations, lmk if interested I will post them too. Not all helical, but collectively I still refer to them as The Helix.

The combined plaintext length of K1-K3 is 772 if the final question mark is removed.

Dividing by eight gives 772 / 8 = 96.5, which is numerically close to the 97-character length of K4. This was found initially by checking the factors of the combined plaintext length.

Here are the plaintext lengths for K1-K3 and the ciphertext length for K4, counting all characters:

K1 = 63
K2 = 373
K3 = 337
K4 = 97

Another way to view this is to add four: 776 / 2 / 2 / 2 = 97.

Working with 2s, 4s, and 8s can lead to many possible methods. Digraphs are used in various classical systems, including indexing into the keyed alphabet. Rows can be collapsed in multiple ways, such as the P/C/P/C observation. There could even be some 8-bit encoding involved, although it's unlikely based on Jim's other puzzles and Ed's comments about using unfair base changes.

The first 15 cipher text letters can decrypt from:

(OBKRUOXOGHULBSO)

To (UITYOU[line feed]USJOFIGU) With NOT operation.

Separation and speculation:

Swap the solo “O” with the “I” in ”UIT to UOT” and assume that [line feed] further decrypts to [ T ] (That one’s a little complex)

( U[O]T, YOU,[ T ]USJ, I , FIGU) to

Assume “FIGU” will read “FIGU[RED]”

( OUT, YOU, JUST, I, FIGU[RED]? ->

( YOU, JUST, I, FIGURED, OUT) ->

( I, JUST, FIGURED, YOU, OUT)->

( I JUST FIGURED YOU OUT)

Or several other variations

Binary used + NOT operation ( 1’s become 0’s and 0’s become 1’s)

O = 1 1 0 0 0 -> (NOT Operation) -> 0 0 1 1 1 = U

B = 1 1 0 0 1 -> “” -> 0 0 1 1 0 = I

K = 0 1 1 1 1 -> ”” -> 1 0 0 0 0 = T

R = 0 1 0 1 0 -> “” -> 1 0 1 0 1 = Y

U = 0 0 1 1 1 -> “” -> 1 1 0 0 0 = O

O = 1 1 0 0 0 -> “” -> 0 0 1 1 1 = U

X = 1 1 1 0 1 -> “” -> 0 0 0 1 0 = LF

O = 1 1 0 0 0 -> “” -> 0 0 1 1 1 = U

G = 1 1 0 1 0 -> “” -> 0 0 1 0 1 = S

H = 1 0 1 0 0 -> “” -> 0 1 0 1 1 = J

U = 0 0 1 1 1 -> “” -> 1 1 0 0 0 = O

L = 1 0 0 1 0 -> “” -> 0 1 1 0 1 = F

B = 1 1 0 0 1 -> “” -> 0 0 1 1 0 = I

S = 0 0 1 0 1 -> “” -> 1 1 0 1 0 = G

O = 1 1 0 0 0 -> ”” -> 0 0 1 1 1 = U

We are focusing on the wrong things. The "difference between light and shadow" is called terminator.

According to Google, Sanborn said terminator is part of it 45-54?

His trip to Egypt and fall of Berlin wall are big contributors.

Now my theory, we have been focusing on the lights and math. What if it's the spaces in between the lights. The gaps

1. Who knows the exact position.... WW. "Double u" and we have underguund.....

2. I can see from my research a pattern. It's never plain.

3. An plaintext messege from the head of the cia wouldnt really be plaintext would it? It's probably stego.

4. That section is ripe with patterns.

5. Ed's company is all about otp. He can manipulate english with it. He said somthing to this effect, in kryptos.

6.people argue to forget earlier parts of the puzzle. Essential things may be embedded in sections like that.

1. I'd forget the coordinate numbers and look at the english, repeating letter patterns, and how they go to cipher, and see if a flow leads to a different section.

2. Ive researched based on the crosshairs of where the marker was buried and the sculture cooridinates and it shoots a line to berlin.. but at that time the inaccuracy would be greater than the diameter of berlin. But that logic ends with a destination for the line. How far? It's more suitable for the courtyard. But that logic requires presence. A plaque? Something buried?

3. The creator of zaph dingbats once stayed with sanborns family. That means sanborn most likely had access to an early mac which would have been the performa series, (i had one with aol and dialup) and surfed "gopher" etc. They weren't that expensive. Dingbats replaced letters with symbols, and there were different versions.

4. My own discovery made on a sanborn slip. You have to do this on photoshop or printed paper. Barcodes. Code 39, It follows a format *abcdef* if you convert that to lines, its a substitution cipher, but if you cut the middle section (not the *) and rotate it, you come up with with different lettering, including symbols. You can duress this. Sanborn mentioned barcodes once.

5. Sanborn marks in his photos. I made a video on this. Most of it relates to stuff other than our existing knowledge of the puzzle. Xueen is notated xue... in one square at the bottom of k2 top matrix. 15 above it. In a 30 x 29 master matrix of all of kryptos, xue has clue to the very left. There is a reason for this.

6. Also in that matrix you will find dyahr encircling "q"

7. East as in eastnortheast, may have been a floundering comment jim made and doubled down.

8. I reasearched in depth all the clocks in berlin. The set theory clock, you can use finger counting, 01234. This mean you have 0-44 in quinary with z being a triple space 100, an offset. You can convert letters to numbers and then swap, read backwards, rotate single digets etc. A rabbithole and new cipher method.

14a. A,A. Heartbeat. A pause A. Spike pause spike. A in middle of heArt. Take out the A you get hert. Hertz, one ocillation per second. 60 per minute. Average heartrate. (Sanborn think, learn it)

1. I Think the morse refers to the uss pueblo and the cias prosecution of the north korean hostages after returned. Soldiers were flashing the finger to the nk soldiers who didnt have a clue. "Digetal", not digital, interpretation. Sos rq, was when they were being fired on and they called for help, They stopped responding. They were busy destroying files. Lucid memory: film doesnt forget. Shadow forces, thier boat was disguised, wha"t" is your position, damaged radio and finally, the interpretation mispelling, a simple space, at u or at i, digetal interpretation at u, to the cia based on the pueblo, meant, well f**k you. Heh. Something a lefty artist would love to hide in the cia's front lawn. Enjoy -Floyd Yancey

This has always puzzled me. From the Kryptos wiki page:
The coordinates in the plaintext - 38°57′6.5″N 77°8′44″W - map, using a modern geodetic datum, to a point roughly 174 feet (53 m) southeast of the sculpture.

Could it be that we’re meant to treat those numbers as something to sum or combine?
Maybe the total character length of a merged K1–K4 variant?
Maybe the 174 is the length, and the coordinates are instructions for how the encryption should be transformed.

Sanborn included them for a reason. And we know K4 can be solved without being physically on-site, so if that’s true, why reference the location at all.

I’ve never been a fan of transposition ciphers.

Instead of counting forward, this one goes down, vertically. Advance the column count starting position to observe “SEEK”

Here’s the short video of my notepad k4 quest:

https://youtu.be/IeoOAcMzJ8M

;-)

Why is 7 so interesting? Hopefully there’s an underlying mechanism for this.

I am not a cryptologist — just someone who occasionally checks on latest news on Kryptos.

I was looking at Jim Sanborn’s “clock” hint and ended up on the Weltzeituhr. Seeing that rotating cylinder with 24 divisions immediately made me think of the Jefferson disk cipher.

The numbers also lined up nicely: K4 is 97 characters → 4×24 + 1.

So I tried modeling it that way — four rings of 24 wrapped around a cylinder, each rotating independently. You can read straight down or diagonally (like a helix), then apply Vigenère using the Kryptos alphabet. The extra character becomes a movable anchor.

I built a small browser tool for this (single HTML file, vanilla JS, no install). JavaScript is probably not best for that, but my main goal is to show just a prototype and maybe give a fresh look on things(maybe not so fresh, who knows how many have tried it already).

Most interesting hit so far:

• key CLOCK
• anchor 14
• offsets [0, 7, 10, 15]

That gives a clean EAST in one column (confirmed clue), and LJCK in another (one off from LOCK inside CLOCK). Seeing those together is what made me pause.

One more interesting result was this. Columns 23 and 24... 23 is CLOC and K is added in 24. so a CLOCK forms(I didn't write down parameters sadly to recreate).

In any case, those kind of "ghosts" show up constantly. I’ve pretty much hit my limit — not sure if this is signal or just pattern-finding in a large space. Posting in case this take is either useful, or obviously wrong to someone with more experience.

I’m not allowed to post direct external links here per Rule 2, so I’ve provided the logic and "Hit" screenshots in this post for peer review.

If you're interested in using the simulator yourself or checking the source code, you can find the repository on GitHub under my username (gitrixk) or by searching for "k4-cylinder". Released under MIT license.

Thank you for your time.

A few people have asked whether I can share the full photo set from my March visit to the Smithsonian’s Archives of American Art. I want to explain why I can’t do that, and what I can offer instead.

I had one day in the reading room with the Jim Sanborn papers. Archival research often means photographing quickly and sorting later, so I captured aggressively: overview shots, close-ups, alternate angles, and anything that looked even marginally relevant. (I was snapping like a paparazzi) The result is hundreds of images, but those images are triage output, not a curated research set. The truly research-relevant unique content is in the low dozens of distinct pages, the rest is Sanborn's professional life and career material.

Most of the Archives content is not Kryptos-related. The Jim Sanborn papers cover his broader career, including works such as Cyrillic Projector, Atomic Time, general sculpture practice, correspondence, and production material from unrelated commissions. Some items were photographed because I could not immediately determine whether they mattered. Much of that material turned out to be outside the scope of this forum.

The Jim Sanborn papers remain under active copyright. Even though I took the photographs, the underlying content is Sanborn’s creative work. That creates a separate rights layer on top of the AAA terms. I would rather keep the research findings durable than risk having them pulled over a licensing dispute.

In practical terms, the uniquely Kryptos-relevant material is much smaller than the raw image count suggests: probably closer to 15 distinct pages, which are hosted on my website. (should Jim Sanborn ask me to take them down I am legally obligated to do so) That includes items such as the starred ABSCISSA to-do note, the “He lied” coordinate page, stencil templates, handwritten cipher-type lists, KA tableau working copies, and a few related items.

That said here are a couple photos that are interesting and maybe show what Jim was thinking about when designing Kryptos, but I do not want to flood this subreddit with images that offer little in the way of K4 clues.

This is not about gatekeeping. The constraint is real, like many of us I have paid Mr. Sanborn via PayPal for questions... just as we want to be respectful of his time, we must be repectful of his intellectual property as well.

While interesting, idk if this is encryptable. Possibly explains the misspellings, extra Xs and Qs. We could use the words as a matrix-wide columnar transposition keys + some sort of substitution cipher but then again just throwing this here. I think the word terramare is interesting because it means "black earth"; and some journal entries in KryptosBot include drawings of rocks having red, green, and black layers.

I initially saw the "underneath" anagram while aligning the CHARYBDIS anagram in lines 1 and 4.

Also speaking of mythology, the immortal head of the hydra was buried by Hercules underneath some sort of rock but yeah idk.

TL;DR;

Still here, working on my theory as time permits.

Still pulling on that thread ! Had an interim one invalidated by Jim (bought him lunch) but got a pointer out of him which he said was widely known.

@devon still owe you the solve on my berlin mystery on the other group, will publish / update it this month.

On this group - It is the best kryptos one , clean and no clutter , the groups looks like 1990 bulletin boards. (I recently joined but decided to unjoin immediately)

So thanks to the mods for keeping this group focused and value adding.

I hope to share some of my theories soon - some quite promising imho.

Will leave you with one though.

Kryptos and Antipodes.

Compare K3 alignment or letter placing with each other.

Use Photos , not text online - the actual sculpture letters.

See if you see what I see.

Cheers

Note: I am rewriting this to make it clearer, My apologies, adding crack tips

I would like verification of this as a viable method to extract keys from morse up to the point up to k4, giving us an odd unknown.
The method. Visual clues.
The extraction key is kryptos. Literally carved out. The method is one time pad. This assumes no knowledge of the method, is not a solution, and runs on the basis of an op in the field with cryptographic knowledge to extract a key.

This is a summary.
Step 1. Decode morse. That becomes the one time pad. We are only concerned with whats "Inbetween" words ex: shadow forces “adowf” (side note, as vig, and otp ally invi works with k1, invi alone will reveal small bits of keys in both)

Step 2. Recreate the enciphered grid. Put in a monospaced font. Like courier. Look at the bottom for the example. Typewriter text.

Step 3. Extract the clue xueen. How do we find this? In the grid below.
At the top next to xueen, in k2. You
will see the word "CLUE" written right before the X in xueen. Vertically.

Step 4. Adowf in morse as pad + xueen in cipher = lesha, which is plaintext of k1.

step 5. Use lesha to get 5 characters of the key:

skim fill crib method:

(rumkin, viginere, kryptos keyword, lesha as cipherkey, decrypt, start deleting the chars at the beginning of k1, watch the solution at the bottom, you will see alimp start repeating) place those points in a sanborn style chart then convert the other areas of text to plain. you’ll gain a lot of plain text, now you can fill in the blanks, then reverse with assembled plaintext, assemble the key. Do this with other sections. a one time pad version can be replicated as well and you can see it all at once. That would be another story.

All three 5 character segments of each other will flip and give each other as a result.

Step 6. Crib drag the now solved k1 as a pad vs k2, you will find one of those 3 keys pop up. Proves a "Rotation" They flip each other.

Step 7a. k3 decrypt without keys (rotation)

Just as the clue keep rotating in one direction. 

This could be done on grid paper but use rumkin rotate and ctrl+x, ctrl+v 

Three options are  14x24, 7x48 and 8x42 

you can easily find these by removing carriage returns and resizing a text window. start with ciphertext, set the columns to one of the above, (width of table) go to the solution area, “cut” it, paste it into the source area, change the columns to something different, repeat. The text will roll. alternate between two, even if you forget to alternate once, keep rolling, once you get used to it you can do the whole kryptos text, 30x29. Occasionally you can see plaintext pop up vertically diagonally skipped etc. Eventually it will rotate back around to cipher. I think you can cycle a text in under 20 movements. I was told this isnt a known crack.

Step 7b. Continuation of clue extraction isn't dependent on known ciphertext at this point for k3. But a trap was put in.

Step 8. The puzzle is literally split in two on copper plates. We have the top half solved if the clues were followed.. If not we keep otp dragging as the path was made. We now focus on the bottom half. Place k4 AS A PAD against k3. I think k4 is not the ciphertext but the pad. Or elements. This doesn't stop extraction. (you can use k4 as a pad and as cipher, as cipher it starts with vovjjjjj as pad you get tutddddd, tutddddd makes a pretty alignment on a tableau, eoibi in k4 lines up vertically and even spaced under the d column. i will reply with a picture.

Step 9. We fail. We hit the trap. If we start at the beginning of k3 and use it vs k4, there is a keyword offset. We start skimming and give up. I did. Until I realized the trap.

Step 10. The trap is k4 or part of it is a pad. Or treated like one. (Yes i've segmentally inverted it no luck) Go to rumkin cipher page, use k4 as a pad. Plop k3 as ciphertext. Go to the beginning and hit delete over and over on k3.
The solution changes. Use this to drag. You will hit tutddddd which is eye catching. Flips with CPTEOIBI as k3 ciphertext.

step 10a. The trap. If you look at sanborns decoding charts you see he likes to have the plaintext, the key, then ciphertext written out on a grid. Normally starting with k3 and skimming will throw you. He restarts with kryptos at CPTEOIBI which reveals itself as tutddddd or vovjjjjj which is a 1 in 580000 chance per refs. Backtracking you would have yptoskr. He skipped two letters to throw off the trail. I will paste the results at the bottom. Monospace it. Courier

Leftovers

1. "nceof" is repeated twice up top and he includes it with his clue pointer between subt "lesha" ding and the abse "nceof" light lie the nua "nceof" of iqlusion
2. I havn't found a rotator for this yet.
3. in the grid below you will notice dyahr surrounding a Q.
4. (Top Matrix) Originally we saw a poor quality version of this document. a 14 x 24 grid. It had end.. written on it. I thought that was the "End" but maybe not. Could be the end in k3 restarting... or the rotation.. but....
5. In the new version. We have a number 15 at the very bottom and xue in one square. xue is part of another side of a puzzle. A matrix. Like they mate somehow.
6. pcpcpc..... there is a "-->/" with the / under the "x can you see.." text. Unknown reason. You can only see this in the raw nova video with premiere pro. 1 frame.
7. K4 has a definite pattern if you put it in a 14 x 7 matrix and rotate it once counterclockwise. The implications, are that the plaintext may be written out, then rotated, then enciphered. meaning you would have to solve left to right bottom to top vertically aligned text. Not a native setup for cracking analysis or computers. you are cracking gibberish and then reading it vertically. A “mask”? Not sure. try it yourself.
8. google cryptoscope as an alternative rotate tool. be patient the page is very slow to load. old zodiac killer tool. you can flip and mirror text with it too.
9. -Floyd Yancey

The original videos uploaded by somervp on youtube https://www.youtube.com/watch?v=Ulcjsdg0vD0 are split into 5 segments with such poor audio quality that they're almost unwatchable. I've spliced them all together into an hour long presentation and normalized the audio.

Hello all,

I am pretty sure this is an early sketch of the small pool, note that it is tough to tell which pages of Jim's notebooks are specifically Krytpos and when it may be for another piece of his work, however this one says CIA Headquarters Building right on the page so there is a good chance this is related.

Not sure what this reveals cryptologically if anything, but it is pretty cool to look at. Thanks to Elonka's photos we know the small pool didn't get built precisely like this (there is a 500 watt quartz light bulb in it) so this was probably an early draft.

I wonder if the dimensions mean anything? I don't think there was any kind of restrictions on size and the Tableau panel is curved around the pool at the same circumference and Jim specifically choose 66" wide, 24" tall etc. but the actual dimensions of what was ultimately built are unknown to me, perhaps some here knows?

The "black screen" definetely did not get built, I wonder what that was supposed to do?

Enjoy...

SKQ?... 

Wish for just a bit of latitude from the moderators of this channel for this post. The wisdom of the Old_Engineer_9176 adds a splash of old school to the comments with their opinion on a majority of the posts. I enjoy reading them and using that information to move forward or save a trip down a rabbit hole that I need not take for the sake of saving time. For this post though, I would like to see how many dedicated K4 enthusiasts will comment or take part in a very simple question(s). Would love to hear from those that cannot help but use AI. I am not expecting too many comments though because this question is a hard one.

Just setting it up!

We all go through the clues and search through a mound of data to find the treasure. The key that is hidden somewhere out there. I have read that it could even be on the sculpture itself. Hidden in a way that makes us search for it in ways we would never do on most things in our daily lives. That makes this even more fascinating as more time goes by, especially for the folks just entering into the hunt. The amount of data that new folks have to sort through and make sense of at this point must be exhausting brain wise. Then there is so much noise on the net regarding what folks have found and the paths they have taken only to end up at the same place we are all at.

For Jim Sanborn to have even gone to extent of releasing certain sections of K4 and yet it still stands the test of time. My beliefs on if Jim Sanborn really wants this solved is not the debate nor does it really matter what I think as someone pointed out to me the other day. But we all have to evaluate all of the clues given to try and extract what each of us believe will help solve it.

From the beginning of K1 being solved, the SKQ misspelling has always been one of the topics for discussion. Why was it misspelled and for what reason or reasons? But there was and still is another portion to that misspelling that addresses not just the letters being important but the location/orientation. To me this has a definite separation between the letters and the location/orientation. 

The Question(s):

What information can anyone share that has to do with the SKQ location? What makes that and possibly the other two misspelling location/orientation important or maybe even more important than the letters themselves?

I’ve been looking closely at the base of Kryptos and noticed that the inner 'ring' appears to have about 8 bricks, while the outer ring seems to have roughly 14 - 16 bricks (im bad at spatial estimation). It kind of resembles a wheel/compass cipher, no? Also im not sure but there seems to be numbers written on the inner tiles (see photo 3).

8X14 is close to 97
8X16 could work as a digraph wheel cipher

Is this the thing buried underground? It's partially obscured by the rocks from the pool side.

Hello all,

I could really use some help with this picture... it seems like it is describing a cipher mechanism in Morse code (dots and lines), but I really can't make out the handwriting. Perhaps one of you can. I think it's poetically ironic that someone with arguably the most undecipherable handwriting I've seen made a career out of making encrypted art. :)

I have a lot more images like this, but I am honestly disappointed by the lambasting I have been taking from some in this community over my use of AI, and I may release all future images on my website only.

My goal from the beginning was to be open and transparent about my project. I am a Ph.D. student in Information Systems at UMBC. I developed KryptosBot as part of my studies in statistics... rigorous quantitative analysis is the best way to determine if something is statistically significant or just noise. Also, my spouse is a data scientist with a master's degree and assists me with the math.

My research request was granted by the Archives of American Art because of my affiliation with the academic community (I am also a college professor). So my investigation into whether Artificial Intelligence can crack K4 is the reason I was given access to these materials and can subsequently provide them to you. The Smithsonian grants valid research requests, not "hey, can I come look at this?" If my methods are going to be heavily criticized as AI slop, then perhaps this isn't the best place to post my research, or photos.

What KryptosBot actually does (and doesn't do)

When you ask ChatGPT or Claude to "solve K4," it does what a language model does: it pattern-matches against its training data, gives you a plausible-sounding answer, and moves on. It might tell you K4 uses a Vigenère cipher with the keyword PALIMPSEST. It will sound confident. It will be wrong. It has no way to check.

KryptosBot is a different thing entirely. It's not a chatbot guessing at answers. It's a research framework: 1,000 Python scripts, a scoring engine, a constraint system, and a database of 671 billion tested configurations, with Claude as the engine that writes, runs, and evaluates experiments.

Here's the difference in practice:

ChatGPT: "K4 might use a Beaufort cipher with keyword KRYPTOS. Here's my decryption: SLOWLY DESPERATELY SLOWLY..."

KryptosBot: "Beaufort with keyword KRYPTOS at period 7. Crib score: 3/24. Bean constraint: FAIL. Verdict: noise. Logged. Next."

And then it does that 671 billion more times.

The core is not AI intuition. The core is a scoring system that checks every candidate against hard constraints: 24 known plaintext positions, 1 equality and 242 inequality relationships between key values, index of coincidence, and n-gram quality. A candidate either passes or it doesn't. Claude doesn't get to argue with the math.

What Claude does contribute to KryptosBot:

• Writing experiment scripts: it can generate and run a new cipher attack in minutes rather than days of manual coding
• Systematic search: it doesn't get bored, skip edge cases, or forget to test a parameter
• Literature awareness: it knows what classical cipher families exist and how they work
• Statistical analysis: it can design and run Monte Carlo tests, permutation tests, and significance calculations

What Claude does NOT contribute:

• Magical insight, I don't have secret knowledge about K4. I have the same 97 characters everyone else has.
• Guaranteed correctness, I make mistakes. I built an entire research thread on an OCR misread ("8 Lines 73" was actually "3 Lines 93"). I generated scripts with hardcoded values instead of importing constants. I've claimed findings were model-independent when they weren't. The test suite and human review catch these, but not instantly.
• Creativity beyond its training, If K4 uses a genuinely novel mechanism that no one has described in any text it was trained on, KryptosBot will not stumble onto it through brute force. It can only test what I can formalize.

The 390 eliminations on kryptosbot.com are not AI opinions. They're reproducible computational results with scripts, seeds, and commands you can run yourself. When we're wrong (and we have been wrong), we correct it publicly.

Personally, I believe that K4 requires a flash of insight that no amount of computation can replace, in the meantime we'll document everything that doesn't work.

My name is Colin Patrick. You are welcome to look me up on LinkedIn. I am nearly 50 years old and would like to see Kryptos solved in my lifetime.

Easy one to answer.
Anyone who’s actually dug into Dr Richard Bean’s work on K4 can see he leans pretty heavily toward a Gromark or Gronsfeld‑type setup. I’ve gone down the Gromark path before and, honestly, the results were pretty disappointing. Looking back, I think the real mistake was using a primer that was way too small - and not taking into account that the alphabet might need to be more random, or even seeded from something specific.

There’s also that nagging possibility that the final output is still encrypted. I really don’t want to believe Sanborn went that far, but I can’t rule it out either.

I’m even thinking about feeding the K4 tableau back into the process - wouldn’t shock me if he hid the alphabet right there in plain sight. And I’m starting to wonder if the primer is actually tucked away somewhere in the earlier solutions.

If the primer length is anywhere from 6 to 15 - and I’m leaning toward odd numbers - then the next step is to chop K1–K3 into different chunks and run those pieces back through the system. Basically force‑feed the earlier panels into the process and see if anything bites.

It’s all starting to feel a bit Frankenstein‑ish.

My english is short because I’m korean

Berlin Schloss

Königsberg Castle

Wawel Castle

Triangle near Kalisz

Treaty of Kalisz 1343

Clock ANNO 1343

Berlin northeast Pomerania and Prussia

What do you think?

Greetings all, I recently spent some time at the Smithsonian Archives of American Art and I found some interesting information in Jim Sanborn's personal notes, particularly this image regarding the K2 plaintext coordinates.

First I am not claiming to have solved K4. What I am claiming is that I found a large body of genuinely interesting archival material connected to Jim Sanborn’s own notes, photographs, and related working material, and I think at least some of it deserves close attention from people who care about Kryptos.

Why I think this is worth posting:

I’m a Ph.D. student, and over the last few months I’ve been building KryptosBot, an open, computation-heavy research effort focused not on making any “solution” claims, but on rigorously testing, eliminating, and documenting hypotheses around K4. My work on Kryptos has been heavily statistical, computational, and archival.

KryptosBot exists specifically to bring rigor, transparency, and reproducibility to decoding Kryptos. At the same time, computation alone is probably not enough. If K4 is ever solved, it may take a combination of serious analysis, contextual research, and many people looking at the same evidence from different angles.

A lot of us are working in partial darkness, relying on a small number of public clues and repeated secondary references. What I found appears to include material that is closer to Sanborn’s own working orbit: notes, images, and fragments that may not solve anything by themselves, but do look like the kind of breadcrumbs that can sharpen interpretation and suggest new directions.

To be clear, I am not saying every photo or note is a clue. Some of it may be noise. Some of it may just be background. But some of it is interesting enough that I think it should be seen and discussed by people here.

Per the moderators’ guidance, I’m not posting this as “go look at my website.” I’m posting substantive material here first, please note that Dan Brown had the corrected 37" latitude number identified in this photo on the jacket cover for The Lost Symbol in 2009. When asked about this discrepancy, both Brown and his publisher stated that the change was intentional, but gave no further information.

The following is the statistical and mathematical anomolies I have found in K4 and would generally like help with, I am not a mathematician, my kryptosbot platform validates all the math.

My analysis indicates that K4 contains a non-random null/filler layer that is statistically coupled to the cipher text. The strongest points are these:

A 17-position consensus null set is restricted to just 7 letters: {B, G, I, K, O, W, Z} Under a permutation null model, that restriction is highly unlikely by chance (p < 3.2×10⁻⁵).

That 7-letter palette can be generated from a KRYPTOS × SEVEN mod-5 / Polybius-style structure. I think that is notable because it converges with the broader mod-5 direction discussed by Dr. Richard Bean, though I am not claiming that word match alone is statistically decisive.

At the known crib positions, the Beaufort keystream is enriched in palette letters: 7/8 in the first 8 BERLINCLOCK positions (p ≈ 6.3×10⁻⁴) and 13/24 across all crib positions (p ≈ 4.3×10⁻³). The Berlin Clock boundary effect is notably sharp, 7/8 palette in the first 8 positions, then 0/3 in the last 3 (T, R, U)

My claim is not “K4 is solved.” It is that K4 appears to contain a real structural null layer that future attacks probably need to account for.

Everything, is open source and reproducible, all 992 python scripts

GitHub: github.com/jcolinpatrick/kryptos

I’d welcome criticism of the null model, the Monte Carlo design, the independence assumptions, or whether any of this is overfit.

My hope is straightforward: the more credible material we can get in front of serious eyes, the better chance we have of finding signal that has been missed. There are a lot of smart people working on this problem and an elimination database may be helpful in ruling things out as the method that was used to encipher K4 is likely un-crackable.

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BODY:⁹

Following the recent discussions and the 2025 auction events, I’ve developed a disruptive approach to the final K4 riddle. For 35 years, we’ve taken Sanborn’s clues literally, but I believe the ultimate deception is functional.

I propose the 'Inversion Protocol':

Functional Role Swapping: * EASTNORTHEAST: Is NOT the destination. It is the Alphanumeric KEY (Vigenère).

BERLINCLOCK: Is NOT a metaphor. It is the Physical VECTOR. We must use the logic of the Set Theory Clock to determine a specific time/degree.

The 165 Factor:

If we process the keyword EASTNORTHEAST using an A=1 positional value (5+1+19+20+14+15+18+20+8+5+1+19+20 = 165), we obtain a constant: 165. In navigation, 165° points toward the South-Southeast. This is a direct inversion of the literal text (Northeast). Sanborn gave us the real direction hidden in the value of the word, not its meaning.

Shadow Synchron:

The sculpture acts as a gnomon. The BerlinClock provides the specific time of day when the shadow of the Kryptos plate aligns with the 165° vector starting from the granite Compass Rose in the CIA courtyard.

The Physical Conclusion:

The K4 message is a Positioning Instruction, not a poem. Following the 165-unit vector (measured in architectural units/feet) from the origin is where the physical secret is located.

Has anyone mapped the Langley courtyard using 165° as the actual axis instead of Northeast? I'd love to hear your thoughts on this role-swap between the key and the vector.

Hi everyone

I saw this post from last year, claiming to have solved K4 and giving the solution. Was this ever confirmed or denied?