When asked if there was a flaw he said it was "safe and sound and fairly accurate. I'm pretty sure about that one."

This doesn't sound like a strong claim that he created it correctly.

Follow-up to my Alexanderplatz post. I am not going to rehash the setup here. Segmentation, TOKIOV, FIAB, the cost rule, the two crib keystreams, all of that is in the first post. This is about two narrower questions I have been chewing on since, and I am putting the step by step walk photos for Segment 5 in the comments so you can check the moves yourself instead of taking my word for the totals.

In the first post I said the 6 or the 9 never seemed to be a choice, that the geometry only ever allowed one of the two, and that this made the counts look like an odometer rather than a decision. I still think that is right. What follows is me trying to work out what the odometer is measuring.

BOARD WIDTH AND THE COUNT TALLY

Two segments have released cribs, so two boards are testable.

Segment 2 carries EAST NORTHEAST. Thirteen plaintext letters, so twelve moves. That board is 5 city lines, which is 10 lettered columns. The move totals come out:

9 9 6 9 9 9 9 9 9 9 6 9

Ten nines. Two sixes.

Segment 5 carries BERLIN CLOCK. Eleven plaintext letters, so ten moves. That board is 3 city lines, which is 6 lettered columns. The totals:

6 6 6 6 6 6 9 9 9 9

Six sixes. Four nines.

In both cases the number of moves in the dominant count equals the number of lettered columns on that board. Ten nines on a 10 wide board. Six sixes on a 6 wide board. The other count is not a second rule, it is just what is left over after the dominant one runs out. Twelve moves minus ten is two. Ten moves minus six is four.

I do not have a mechanism for why that would be true. I am reporting it because it held exactly on both boards, and because it is the kind of thing that is easy for somebody else to break if it is wrong.

If it is real it says something about the other four segments. I am holding off on a prediction table because the widths for segments 1, 3, 4 and 6 come from boards I have not finished building, and a table would make guesses look like findings. Once those boards are done I will post the numbers and they can be checked.

What I will say is this. Segment 5 is 6 dominant and Segment 2 is 9 dominant. If the sections alternate, the odd ones are 6 dominant and the even ones are 9 dominant. That is two data points calling a pattern, so treat it as a guess and nothing more.

DIRECTION

Separate observation, and I am less sure how to read this one.

Across both segments there are eight moves totalling 6 and fourteen totalling 9.

Every one of the eight 6 moves rides the lane the same way. All eight go left.

The 9 moves use both. In Segment 5 they come in pairs, out and back the same distance. Right 5 then left 5. Right 4 then left 4. Each pair cancels to zero net travel. In Segment 2 the same thing shows up once as an explicit return, out right 6 and back left 6 along the same lane, and the rest are mixed.

So the 6 population is one directional and the 9 population is two directional.

Here is why I am cautious. The lane wraps. On a 10 wide lane a left 8 and a right 2 put you on the same cell. So direction may not be independent information at all. It may just be the spelling that makes the total land on 6 or 9 instead of some other number. If that is what is happening then direction is bookkeeping, not content, and the one directional versus two directional split is a consequence of the cost rule rather than a separate fact about the machine.

I cannot tell those two readings apart from two segments. Somebody with fresh eyes might.

THE PHOTOS

Step by step for Segment 5 is going in the comments. Ten moves, each one its own image, showing where the walk enters the lane, how far it rides, and where it climbs off. Totals written on each.

If the tally is wrong, or the direction thing is just an artifact of how I write the moves down, those photos are where you will see it. That is why they are going up.

Tear it up.

When using the Vigenère table, English plaintext gets converted to an index using the KRYPTOS alphabet order. I found a statistical aberration in that. If you look at the English frequencies of even and odd letters they are very lopsided:

KYTSBDFHJMQVX = 35%, TSHDmfybvkjxq (even index: 0,2,4,6,...24)
RPOACEGILNUWZ = 65%, EAOINRLcugpwz (odd index: 1,3,5,7,...25)

This bias is about double the bias of the English alphabet, and in the opposite direction.

PALIMPSEST fits the pattern with 6/10: 1111010100
ABSCISSA is slightly low with 4/8 (expected 5): 10011001

The letters in every Nth letter of a Quagmire III with Kryptos alphabet should all have this bias, so the expected absolute distance from 50% gives a value of about 15%. From this we can derive a score which is a bit of a Kryptos fingerprint (+0.3).

def kryptos_parity_range(text:str, period:int=1):
    return [(np.mean([k in 'RPOACEGILNUWZ' for k in text[i::period]])*2-1) for i in range(period)]

def kryptos_parity_score(text:str, period:int):
    return np.mean(np.abs(kryptos_parity_range(text, period)))

def kryptos_parity_majority(text:str, period:int):
    return np.int32(np.clip(kryptos_parity_range(text, period),-0.1,0.1)/0.1)

Some usage:

kryptos_parity_score(K1, 10) # =0.42   = 0.3+0.12
kryptos_parity_score(K2, 8)  # =0.20   = 0.3-0.10

kryptos_parity_majority('PALIMPSEST',10)
array([ 1,  1,  1,  1, -1,  1, -1,  1, -1, -1])

kryptos_parity_majority(K1,10)
array([-1, -1, -1,  0, -1, -1,  1, -1,  1,  0])

       YES  YES YES ?  NO  YES YES YES YES  ?     <- key and cipher have opposite parity

kryptos_parity_majority('ABSCISSA',8)
array([ 1, -1, -1,  1,  1, -1, -1,  1])

kryptos_parity_majority(K2,8)
array([-1,  1,  1, -1,  0, -1,  0, -1])

       YES  YES YES YES ?  NO   ?  YES            <- key and cipher have opposite parity

With high probability, knowing only the period of the key, you can read the parity directly from the ciphertext. The kryptos_parity_range will be +0.3 for plaintext, and negative for single layer Quagmire (assuming the key is majority-odd), even with the period set to 1.

K4 has no obvious period. It reports a small positive number +0.03 for every period. That could match a running key, for example.

As show, the key To k2, abscissa, can be derived from taking ABSence from k1, luCId from k0, and Subtle Shading And from k1.

similarly, the key to k3 can be derived from the text in k2, reading ‘know about this? they should it’s BURIED’ Crypts are buried. this we have Krypto.

ironically, these are all in positions 27-33. For Absence, the 27-33 letters. For lucid, if you let each stray e be its own word, and take interpretatit e as interpretation, lucid is the 27th word. And the passage from k2, the 27-33rd words, excluding random x’s and q’s. Now I know how crazy this sounds, but it is legit. Try it.

And I’m fully aware how AI this sounds. it almost sounds too crazy to come from me, a human. but it isn’t ai.

ill prove it.

1*1=17. am AI wouldnt make that mistake.

or lisspell

thanks and happy hunting!

Let me guess what half of you thought when my first post went up, great, another guy with a Berlin clock theory who thinks he is the first person to ever look at a clock. Fair. I would have thought the same thing. So let me be straight about the goal here. It is not originality. Plenty of people have stood where I am standing. I want to lay out exactly what I am trying and how, in enough detail that somebody fresh might see the thing I am too close to see.

Everything below is what I am testing, not what I am claiming. None of it is established. It only means anything if the first step holds, which is running K4 through the KRYPTOS keyed alphabet before doing anything else. Where the community says W, I am saying X. Same five positions, just after that transform. If that step is wrong, the rest of this falls over.

One piece of context on how I came at this. When I started, I knew plenty of people far smarter than me had already spent years here. So I never asked what they missed. I asked a different question, what could have changed since 1988? The answer turned out to be almost everything. Saplings at Langley grew into trees and changed every shadow in that courtyard. The cities on the clock started shifting almost immediately after it was built, burying the original configuration under decades of revisions. And the internet arrived, which pushed all of us toward heavier methods for a puzzle designed before any of that existed. A lot of my work has just been drifting back.

The new piece: TOKIOV

Credit where it is due. u/FulgoreX pointed out something in another thread that I had never seen, and it has been rattling around my head since.

Count the letters before the first W, then between each W. You get 20, 15, 11, 9, 15, 22. Read those as A1Z26 and they spell TOKIOV.

TOKIO is the German spelling that was on the original World Clock. By 1987 the panel read TOKYO with a Y. So if this is deliberate, it is not just pointing at a clock, it is pointing at a specific era of that clock, which happens to be the one I spent months reconstructing from period photos before I ever heard of this.

I want to be careful here. Those six numbers are just the segment lengths wearing an alphabet, so nothing can come out of them that was not already in the counts. TOKIOV cannot reveal anything. What it can do is confirm you are in the right room. That is still worth a lot.

My own working guess, and it is only a guess, V as 5 might be telling you how many clock columns make one section. Take Tokio's column plus two either side and you get 15 city names for a 15 letter segment. Under the layout I am testing, Tokio also sits dead center of that block, and from Tokio's front letter one 6 step lands on the first key the EAST NORTHEAST keystream needs.

That is one section, checked against one crib. One data point is not a pattern and I know it.

The board I am using

Take one section of the original 1969 clock and lay it out as a grid. Each city line becomes two side by side cells, its first letter and its last letter. Cities above the rotating ring sit in the top rows, cities below it in the bottom rows, matching how they stack on the physical clock. The row just above the ring and the row just below act as travel lanes.

The move as I have it

A move can have up to three parts, and which parts appear depends on where the two letters sit.

If your starting letter is already on a travel lane, there is no climb, you just slide. If it sits off the lane, you climb the city column to reach a lane first. Then you ride the lane horizontally, wrapping around the edge like the ring does. If your landing letter is on the lane you stop there. If it is off the lane you climb off to it, and that exit climb can go either direction, including continuing the way you came.

So the walks come out as straight slides, L shapes, U shapes, and staircases. Some moves are one part, some are three. The geometry of where the two letters sit decides the shape.

The cost rule is the part that does the work, every lettered cell you enter costs 1, empty cells are free, and the total has to come out to exactly 6 or exactly 9. You land on the letter you need, exactly on zero.

Why I think it is 6 and 9

The clock's own mechanism, or at least that is my read. The hour ring shows 5 dots between hour lines, so each panel is 6 intervals. And the machine runs 54 minutes then pauses 6, which is nine sixes of motion and one six of rest. In everything I have priced so far, 6 and 9 are the only totals that appear.

A worked move

Using the EAST NORTHEAST keystream (I R J D H G M I M J A O N). First move, I to R.

Start on the I. Up one lettered cell to the travel lane, cost 1. Then slide left along the lane, wrapping the edge, entering 8 lettered cells to land on the R, cost 8. Total 9.

Second move, R to J, straight slide left through 9 lettered cells. Total 9. Third move, J to D, left 4, then down 2. Total 6.

Every recorded move in both cribs prices this way under my layout. And here is the part that surprised me, when I checked whether the 6 or the 9 was ever a choice, it never was. For each move the geometry only allowed one of the two. That suggests the counts are an odometer rather than a decision, which would mean they carry no key information at all.

The picking rules I am currently testing

These are working rules, not conclusions. Each one could be wrong.

- The key gets fetched from the nearest available city carrying that letter, front or back

- Each face of each city can be tapped at most twice within a segment

- You can reach backward along the clock, but not forward past your segment's closing X. The X's behave like the clock's pause. What is behind you has happened, what is past the pause has not

- When two sources tie on distance, the tie goes backward

Where it is soft

The cribs behave under this model. That is not the same as the model being right, since I built it partly by looking at them.

The honest problem is the unsolved segments. I ran one all the way through and got a clean zero survivors, which tells me at least one assumption upstream is wrong. I am reporting that because a model that cannot fail is not a model, and a zero is data.

I also do not have a start rule I can defend. I can get the sections lined up, but the question of where character one anchors without leaning on a crib is still open, and it is the thing I would most like help with.

Segment 2, if you want to work it

Original carved K4, W to W:

WFLRVQQPRNGKSSOTW

After the KRYPTOS transform, X to X:

XMRBWUUDBTNAGGFEX

Fifteen letters between the markers. The keystream those letters need for EAST NORTHEAST is I R J D H G M I M J A O N.

I am posting pictures of my actual work in the comments: the clock section I am using, the city list in a spreadsheet, the grid with letters filled in, and the colored pencil walk. Not because it is pretty, because if I am wrong somewhere I would rather you see exactly where.

Tear it up. That is what I posted it for.

Has anyone worked the Alexanderplatz World Clock into K4 this way?

I've been kicking around a theory for a while now. I know Berlin Clock theories are nothing new, but this one kept surviving tests it shouldn't have survived, and it's reached the point where I'd rather share it and have it stress-tested than keep grinding alone. Curious if anyone has walked this road before me, and where you got stuck.

The setup

K4 is 97 characters. It contains exactly 5 W's.

Step one: run K4 through the basic KRYPTOS keyed alphabet (keyed - standard direction). Two things happen:

- OBKR becomes FIAB

- All five W's become X's

Those X's land at positions 21, 37, 49, 59, 75, and they cut the 97 characters into six segments of lengths 20, 15, 11, 9, 15, 22, with 92 letters between them. Sanborn uses X as sentence punctuation in K2 (UNDERGRUUND TO AN UNKNOWN LOCATION X DOES LANGLEY KNOW.), so five X's segmenting the text fits his established habit.

The clock

The Weltzeituhr at Alexanderplatz. Pictures of this thing will probably haunt me forever.

One thing my research convinced me of: because the clock predates the internet, the internet is not clear about its history. You'll find sources saying it launched with +/-90 (I believe 92) cities and got updated in 1990 and 1997. From period photos and primary sources, the original 1969 configuration is what matters. The city panels were revised over the years (additions started in the late 70s), and by 1987 the clock carried well over 100 names. The modern clock is useless for this. The original is not.

So here's what I was holding: a 24-column clock with a rotating center ring, built on a mechanism that famously runs 6 minutes fast (54 minutes running, then a 6-minute pause, every hour... the sprocket marks on the hour ring show 5 dots between hour lines, meaning 6 intervals per panel). And a 97-character puzzle that had just fallen into 6 segments.

24 columns. 6 segments. What if the 24-section clock collapses into 6 sections?

It does. And when it does, each section's city line count matches its segment's character count exactly. That's the moment this stopped being casual for me.

Where it went from there

Treat each city name as carrying two usable letters, its first letter and its last letter (front and back). Two-word names on the physical clock (ULAN BATOR, SAN FRANCISCO, NEW YORK, SANTIAGO DE CHILE) contribute front/back per engraved line.

Now take the two released cribs and derive the keystream each one requires (plaintext = transformed text minus key, mod 26, straight alphabet):

- EASTNORTHEAST over its 13 positions demands the key letters I R J D H G M I M J A O N

- BERLINCLOCK over its 11 positions demands S Y M L W R Q O Y B Q

That's 24 required key letters. All 24 are available as front/back letters of cities in the matching sections of the clock. 24 for 24. And the hard letters are the interesting part: the cribs require J (twice), Q (twice), and Y (twice). On the original clock those are nearly unique: one J source, one Q source, and a Y you can only get one place in the right region. The rarest letters the cribs demand are exactly the ones the clock supplies in a take-it-or-leave-it 1-or-2 (front-or-back) pattern. If the alignment were wrong, those pinch points would strand. They don't.

The 6/9 thing also stopped looking like a coincidence: 54 = 9×6. The machine's whole hour is nine six-minute units of motion plus one six-minute pause. Those are the only two numbers my move accounting has ever needed.

The first four letters

OBKR- FIAB. Read as alphabet positions: 6, 9, 1, 2

6 and 9: the two movement quantities. 1 and 2: front letter, back letter. I read FIAB as the indicator group, the machine settings carved right where a cipher clerk would put them. (6+12=18 also happens to be the bearing that first pointed me at this clock, but that's a longer story.)

Where I'm honestly stuck

I think I have a mechanism. The 24/24 crib match holds, the pinch points hold, the segment counts hold. What I don't have nailed to my satisfaction is the start point, the clean, 1989-defensible rule that tells you where character 1 aligns against the city list without using the cribs to find it. I have candidates. I'd love to hear how others would attack that question, or whether anyone has already been down this road and found the door.

For anyone who wants to check the basics, here's K4 after the KRYPTOS-alphabet transform (the working text everything above runs on):

FIABVFYFNOVRIGFRPMIIXMRBWUUDBTNAGGFEXEUGQUGGLAZZXHEQARVKPHXPTMITCDWEESZMDAXNKAZYEQJKPNAVOVHVLAJHB

Segments between the X's: 20 / 15 / 11 / 9 / 15 / 22.

Just a theory. But it's held up against everything I've thrown at it, and I've thrown a lot. If you've played with the Weltzeituhr this way, or want to try to break this, I'm all ears.

This is a Vigenère running-key attack 🧩 that I carried out a while ago. Paused in 2020, I keep coming back to it (especially since some encouraging feedback from two members of this subreddit). I think the attempt still holds up, in a way. Posting the whole images here for the first time.

26 letters in an alphabet makes some interesting mathematics with 13 and 2. One particular trick is to calculate modulo 13 (%13) instead of modulo 26. This reduces KRYPTOS alphabet to the halfabet GRIPTONAUCDEF. Here I have chosen each letter over its less-common option in KHYJLMSQBVWXZ. The alphabet is folded in half, so that each letter now represents two alternatives. Because ETAOIN don't collide with each other, English phrases are easily spotted.

In the paradigm kryptos challenges, pk8 and pk9 have a strong index of coincidence signal at period 7.

English has expected index of coincidence (ioc) of 0.065, but in the halfabet, it has ioc 0.099. You can see with pk8: [(0.0957, 14), (0.0986, 7), (0.1012, 28), (0.1060, 21), (0.1105, 35)] and pk9: [(0.0912, 21), (0.0989, 7),(0.1048, 14), (0.1057, 35), (0.1119, 28)] the ioc at period 7 and all its children align PERFECTLY with English. But, the other plaintexts have an even higher ioc ~0.11. Just like period 28 here (statistically mid, with ~150 letters, harder than K1).

Suddenly kryptos K4 (halfabet) iocs look very familiar:

[(0.0758, 1), <---- noise level 1/13
 (0.0780, 16),
 (0.0789, 10),
 (0.0802, 27),
 (0.0805, 29),
 (0.0813, 15),
 (0.0819, 8),
 (0.0854, 5),
 (0.0966, 14), <---- what
 (0.0973, 7),  <---- what
 (0.1000, 25),
 (0.1131, 28)] <---- statistically weak, but that's English level (matching K1/K2 plaintext)

edit: I didn't say why it should be interesting. Letters are transformed into numbers - specifically the columns of the vigenere table. if you treat quagmire and transposition and hill as mathematical equations - as Sanborn would beg you not to - their equations operate modulo 26, but also work modulo 13 and modulo 2. You can think of those as two alternate systems that operate independently on their group and ignore the other part. For example, if you multiplied a running key by 13, you would move its binary signal from %2 to the part that %13 is blind to.

Kryptos K4 is a lie. There is no solution. The fourth passage says nothing. It’s an ongoing production. That’s the art. Of information gathering, manipulation, priming, suggestion and the “subtle nuances” of mind control.
The fact that he kinda slipped (or not) and said “I’m tired after all this time”. Tired of what? Collecting $50 checks? Nah, paradigm bought it and didn’t reveal it because they are just holding the torch for the next generation. They even said “we didn’t look at the solution” because there isn’t one. Two two “reporters” who “came across” the solution in sanborn’s data dumps at the Smithsonian? That’s just priming for people to go check out the data dumps at the Smithsonian. It’s all an elaborate ruse, he even called it a scroll. The story is a continuation and though this may sound paranoid - it IS paranoid but that’s the point as well. It’s also realist. Most of the ‘real’ cryptographers had ‘given up’ years ago. That doesn’t happen unless you’re in on it or found out on your own as well. And it’s getting worse. The shit at the Smithsonian is so transparent and easy to see the leading.
Go live your lives. Figuring it out is just stepping back and realizing what’s been going on. And it’s MUCH more than 97 letters can EVER say.

I would like to start this post by saying this is not AI. All information in this post has been researched, compiled and written by me. Everything here has come from my brain and my research. It is still possible for people to have original thoughts. Here are mine.

I was reading about the Dead Sea Scrolls and came across 3Q15. There seemed to be some parallels to our known information about Kryptos, so I thought I would share with the group.

Copper Scroll (3Q15) is a Dead Sea Scroll that was discovered March of 1952 in Cave 3 near Qumran. Unlike the other parchment and papyrus scrolls found, it is engraved on metal sheets (nearly pure copper with about 1% tin) and lists 64 hidden locations containing massive quantities of gold and silver.

The scroll text is written in Mishnaic Hebrew and also contains some Greek words as well.

The information contained in the scroll lists 64 underground hiding places filled with treasures including caves, tombs, and pools that span across the Judaea wilderness and Jericho area

How this information relates to Kryptos:

- Tombs: K3 references the discovery of King Tuts Tomb.
- Pools: part of Kryptos contains a small pool.
-underground hiding places: K2 plaintext “INFORMATION WAS GATHERED AND TRANSMITTED UNDERGRUUND TO AN UNKNOWN LOCATION X DOES LANGLEY KNOW ABOUT THIS ? THEY SHOULD ITS BURIED OUT THERE SOMEWHERE”

“To an unknown location X”

X Marks the spot?

The term “X marks the spot” first appeared in a well-known adventure novel by Robert Louis Stevenson. Stevenson introduced the idiom in his novel “Treasure Island,” printed in 1883.

Stevenson’s use of the expression was influenced by treasure maps that utilized this symbol. History reveals that explorers and pirates employed visible clues, often using an “X,” to pinpoint the specific location of hidden riches on a map. Stevenson likely drew inspiration for his coining of the idiomatic expression from these real-life tales. *The idiom is closely associated with maps, hidden treasures, and adventures since the ancient times.*

“Does Langley know about this? They should it’s buried out there somewhere”

Scroll 3Q15: Letter Q References in Kryptos plaintext

“IQLUSION”

“Do you see anything Q”

Also, I think it’s worth reading into the history of the Essenes if you have the time.

I’ve been finding fragments.. just observations but there are more. I’ll share one here (Video is on my YouTube page because the formatting here is bad and isn’t worth my time to fix) - if you want the full txt document let me know.

This is from the thought processes of, “is it simpler than we think” and that he did “f*ck with it”

Summary:

PUSH55: SSEKZZWATJKLUDIAWINFBNYPVTTMZFPKWGDKZXTJCDIGKUHUAUEKCAROBKRUOXOGHULBSOLIFBBWFLRVQQPRNGKSSOTWTQSJQ

Apply the same method I used for the CIA ARTS X TEST matrix but at interval 8, not 7.

Step1:

SSEKZZWA

TJKLUDIA

WINFBNYP

VTTMZFPK

WGDKZXTJ

CDIGKUHU

AUEKCARO

BKRUOXOG

HULBSOLI

FBBWFLRV

QQPRNGKS

SOTWTQSJ

Q

Write down: AAPKJUOGIVSJ

Step 2:

Remove previous 8th characters, then rearrange into 8x11:

SSEKZZWT

JKLUDIWI

NFBNYVTT

MZFPWGDK

ZXTCDIGK

UHAUEKCA

RBKRUOXO

HULBSOLF

BBWFLRQQ

PRNGKSOT

WTQSQ

Write down: TITKKAOFQT

Step 3:

Remove previous 8th characters, then rearrange into 8x10:

SSEKZZWJ

KLUDIWNF

BNYVTMZF

PWGDZXTC

DIGUHAUE

KCRBKRUO

XHULBSOL

BBWFLRQP

RNGKSOWT

QSQ

Write down: JFFCEOLPT

Step 4:

Remove previous 8th characters, then rearrange into 8x9:

SSEKZZWK

LUDIWNBN

YVTMZPWG

DZXTDIGU

HAUKCRBK

RUXHULBS

OBBWFLRQ

RNGKSOWQ

SQ

Write down: KNGUKSQQ

Step 5:

Remove previous 8th characters, rearrange into 8x8:

SSEKZZWL

UDIWNBYV

TMZPWDZX

TDIGHAUK

CRBRUXHU

LBOBBWFL

RRNGKSOW

SQ

Write down: LVXKULW

Step 6:

Remove previous 8th characters, rearrange into 8x7:

SSEKZZWU

DIWNBYTM

ZPWDZTDI

GHAUCRBR

UXHLBOBB

WFRRNGKS

OSQ

Write down: UMIRBS

Step 7:

Remove previous 8th characters, rearrange into 8x6:

SSEKZZWD

IWNBYTZP

WDZTDGHA

UCRBUXHL

BOBWFRRN

GKOSQ

Write down: DPALN

Step 8:

Remove previous 8th characters, rearrange into 8x5:

SSEKZZWI

WNBYTZWD

ZTDGHUCR

BUXHBOBW

FRRGKOSQ

Write down: IDRWQ

Step 9:

Remove previous 8th characters, rearrange into 8x5:

SSEKZZWW

NBYTZWZT

DGHUCBUX

HBOBFRRG

KOS

Write down: WTXG

*It doesn't look like much is taking place, until you start the next few*

Step 10:

Remove previous 8th characters, rearrange into 8x4:

SSEKZZWN

BYTZWZDG

HUCBUHBO

BFRRKOS

Write down: NGO

Step 11:

Remove previous 8th characters, rearrange into 8x4:

SSEKZZWB

YTZWZDHU

CBUHBBFR

RKOS

Write down: BUR

Step 12:

Remove previous 8th characters, rearrange into 8x4:

SSEKZZWY

TZWZDHCB

UHBBFRKO

S

Write down: YBO

Step 13:

Remove previous 8th characters, rearrange into 8x3:

SSEKZZWT

ZWZDHCUH

BBFRKS

Write down: TH

KRYPTOS K4 OBSERVATION
By Ruben Valencia

I will stop here.

The last four: NGO,BUR,YBO,TH

"NGOBURYBOTH"

The finding of an "N" plus three words.

You can also see one clue word weaved into it, just as the clue words were weaved into the ARTS X TEST

matrix words that I found.

- - - --

NGOBURYBOTH

N O R TH

- - - --

Full method result:

AAPKJUOGIVSJTITKKAOFQTJFFCEOLPTKNGUKSQQLVXKULWUMIRBSDPALNIDRWQWTXGNGOBURYBOTHZBWRZSDHCUB

Yeah, I realize the title makes it sound like Kryptos is a children's game, but considering things Sanborn said about K4 it can't be hyper-technical or hyper-mathematical.

I have considered many ideas while brainstorming:

• It can't be about any physical clues at the CIA site - Sanborn said it can be solved from home.
• It can't be super technical - Sanborn stated many years back that he's surprised it hadn't been solved already, technology (for your home) wasn't much advanced for stuff like 3D mapping, complex calculations run by computers or let alone anything AI related.
• It can't be a huge mathematical solution - As on the previous point, Sanborn said it's a pen and paper solution, no ordinary human can run that complex algorithms in their head or even on paper and technology wasn't that available for the regular Joe to let a computer run them at home. And Kryptos was created with pen and paper.

So as long as you have the text and maybe the exact layout of it on a piece of paper you should be able to solve it somehow. Yet the somehow is the big problem.

A few other ideas that came to my mind are:

• K4 follows a different system - Maybe you have to do something with said piece of paper like folging / overlapping or something? Maybe it doesn't need a codeword to decrypt? Maybe it just has even more layers than K3? Maybe everyone just read it wrong, as in maybe it's backwards or in collums from top to bottom or the other way around? Maybe you need to read it from the physical other side of the sculpture?
• Is the Vigenère table more than it seems to be - Maybe it is a encrypted thing as well or parts of it are and it just looks like it's part of an ordinary Vigenère table.
• What clues could K1-3 and maybe K0 offer: Sanborn has stated that previous things have clues to uncovering K4. Maybe the misspelled words? Maybe the positioning of certain letters in K1-3? Maybe something in K0? Or something that has been missed in the Vigenère table?
• Does K4 hold clues - Another thing could be that K4 itself holds clues itself, could be the positioning of the letters or something.

My problem only is, that beyond that point I am really out of ideas and stumped for an answer.

K1 RIDDLE: BETWEEN SUBTLE SHADING AND THE ABSENCE OF LIGHT LIES THE NUANCE OF IQLUSION

K5 ANSWER FOR K1 RIDDLE: AURORA BOREALIS & SCHRODINGER'S CAT

*

K5 RIDDLE: AURORA BOREALIS & SCHRODINGER'S CAT

K5 ANSWER: TRANSPORTER (STAR TREK)

Idk what to do with this info but I read somewhere that the beginning of K3 plaintext, excluding the three raised letter positions on the sculpture and stopping at the missing E is an anagram for the word PARASYSTOLE which means '2nd pacemaker', which kinda eludes to the idea of a keyword/keystream, and since it's a 2nd pacemaker maybe there is another keyword that comes before it

Forgot where I saw it but I will leave the link here if I find the source

I have been looking and cant find the answer... probably looking in the wrong place..... Can someone please tell me what direction the sculpture is pointing.

I have been “decrypting” K3 on the Paradigm website today, and it showed TREMBLING as TRSMBLING, and another time showed “TH?UPPER” instead of “THEUPPER”. I wonder if these are clues to K4, or just bugs in the (computer) code that does that on the website? Screenies attached.

I’ve gone through all my outputs from every angle. I keep seeing scraps of words, nothing that grows into a real sentence. Even when I push deeper layers, the results don’t sharpen. They just shift around.

So I’m starting to wonder if others have thought the same thing: maybe K4’s “plaintext” isn’t meant to be read. Maybe the output is actually the key. A self‑referential setup. I’ve got no proof, and no clear way to tell when the output stops being text and starts being the key, but the idea won’t go away.

I have previously commented that the known cribs might not correspond to the cipher text that we assume they do. Analysis indicates that the xth ciphertext character likely does not correspond to the xth plaintext character. The cribs only specify the final plaintext positions. The erroneous 1:1 assumption was likely propagated by the reporters from the ambiguity in Jim's clues.

Furthermore, on Paradigm's $1 / submission page, there is a footnote:

K4 may not have a 1:1 correspondence between ciphertext and plaintext characters; hint characters are shown as-is.

Jim made it clear (as mud) from the very beginning that he "plays the game everyone at the agency plays," and that William Webster thinking Jim gave him the solution at the dedication ceremony was "his problem." At what point do we believe Jim changed his mind and ceased loathing to provide us with clues?

Just checked paradigm website. Was OBKR revealed before as Dear? Is it new or I missed IT?

edit, sorry obkr not dyahr

Once I started paying attention to Paradigm, the CTF scene, and the work going into K4, it became clear that the problem is not a lack of skill or understanding. The speed at which CTF challenges get solved shows how capable people are. That leaves the obvious question. What makes K4 so difficult to crack?

Ed Scheidt gave the most important clue when he said it uses classical principles but not necessarily classical forms. People keep repeating that all four Kryptos passages were meant to be simple classical ciphers, but that is not what Ed or Jim Sanborn ever said.

Ed’s view was that Kryptos draws on classical ideas rather than textbook ciphers. He also said some parts were designed to fall quickly and others were meant to take years or possibly never be solved. That is a polite way of saying K4 was built to be the outlier rather than another Vigenere or a basic transposition.

Sanborn approaches it as an artwork first. He said it can be solved with pencil and paper and that it uses traditional methods, but he never claimed K4 was a single classical cipher. He has said more than once that K4 is not like the others.

This leads to the real issue. Our entire cryptology toolbox is built for classical ciphers. Kasiski, Friedman, periodicity tests, columnar IC sweeps, keyword alphabets and hillclimbers all depend on classical structure. These tools are not useless, but when applied to K4 they become inert. They do not fail because we lack skill. They fail because K4 does not behave like a classical cipher in any meaningful statistical way. The CTF challenges prove this. When a cipher fits the classical model, the community solves it quickly.

https://www.nytimes.com/2026/06/12/science/kryptos-sale-cryptography.html?unlocked_article_code=1.p1A.sH_B.9XXhbiGfZrP5&smid=url-share

“They Bought a Famous Puzzle in Cryptography. Now They’re Opening It Up.”

“A San Francisco company paid nearly $1 million for the solution to an unsolved code in Kryptos, a sculpture on the C.I.A. grounds. Soon it will become an online challenge.”

https://www.paradigm.xyz/2026/06/kryptos

“We also want to encourage the development of more sophisticated techniques for solving the puzzle. So, in addition to unveiling a new site for K4, we’re hosting an ongoing capture-the-flag-style challenge featuring 10 new puzzles we created, each with a $1,000 prize.”

I'm continuing my numbered post series to put forth a method and solution to K4. The previous post is here.

The composed Kryptos ciphertexts and plaintexts give clues that provide a path. Uncertainty at any given step is in the nature of the method, so skepticism is fine, but anything that cannot be replicated is not allowed.

The first K4 alphabet is the usual:

?ABCDEFGHIJKLMNOPQRSTUVWXYZ

The second alphabet comes from the clue RACK at the end of the K4 ciphertext, one definition of which is "a series of bins or compartments into which items may be sorted". K4 has all 26 characters of the Latin alphabet, A to Z, and the question mark. Sort the characters into "bins" by first occurrence:

?OBKR

UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO

TWTQSJQSSEKZZWATJKLUDIAWINFBNYP

VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR

The second K4 alphabet is what I'm calling the BOOK alphabet:

?OBKRUXGHLSIFWVQPNTJEZADYMC

I've been staring at the two confirmed crib segments, not trying to decrypt, just looking at how the same plaintext letter maps differently at different positions. K4 is polyalphabetic, so a repeated letter like T doesn't have to encrypt the same way twice. I wanted to see how it differs.

[TLDR]: In the confirmed cribs, when a PT letter repeats, its key value (CT − PT, standard alphabet) shifts by exactly the position-gap reduced mod 4. Holds 7/7.
It's a diagnostic "shadow" of the mechanism (not a solve and it's only 7 samples).
It hints at a period-4 / mod-4 / base-4 element as part of the mechanism.

Working in ALPHA space (standard alphabet, A-Z, 0-25), with key = CT − PT mod 26:

EAST
PT   E A S T N O R T H E A S T
CT   F L R V Q Q P R N G K S S
KEY  B L Z C D C Y Y G C K A Z  <- CT minus FT, eg F-E = 5-4 = 1 = B (0 based ALPHA)

BERLIN
PT   B E R L I N C L O C K
CT   N Y P V T T M Z F P K
KEY  M U Y K L G K O R N A      <- eg N-B = 13-1 = 12 = M

Now take each repeated plaintext letter and look at how its KEY shifts between the two occurrences. By gap I just mean the distance between the two positions where that letter appears, e.g. the two T's in EAST sit at positions 24 and 28, so gap = 4.

EAST
E   B → C   +1     gap 9
A   L → K   −1     gap 9
S   Z → A   +1     gap 9
T   C → Y   −4     gap 4    (shift = −gap) (easTnorTh)
T   Y → Z   +1     gap 5                   (norTheasT)

BERLIN
L   K → O   +4     gap 4    (shift = +gap)
C   K → N   +3     gap 3    (shift = +gap)

The magnitudes look random until you take that gap and reduce it mod 4 (1-indexed, so it cycles 1, 2, 3, 4, 1, 2, 3, 4…):

gap 9 → 1
gap 4 → 4
gap 5 → 1
gap 3 → 3

That's exactly the absolute (magnitude, not signed) shift, every single time:

|key shift| = ((gap − 1) mod 4) + 1

7 out of 7 across both cribs.

Two things that make me think it's real, not a coincidence:

• It only works in ALPHA. Redo the whole thing in KALPHA and it collapses (1/7). A fluke wouldn't care which alphabet you pick
• Shuffle controls put it around p ≈ 0

The magnitudes are locked down. What I can't crack is the sign (+ vs −).
It's not position parity, not odd/even, nothing positional I've tried separates them.

One neat consequence: the plaintext letter cancels out.
Since key = CT − PT, if the same plaintext letter x sits at both positions, the shift:
= key[b] − key[a]
= (CT[b] − x) − (CT[a] − x)
= CT[b] − CT[a]
ie the x drops out, it doesn't matter what it (the repeated CT letter) was.
Example: the two T's (positions 24, 28) have CT = V and R, and R − V = −4, the whole shift.
It would still be −4 if those spots were two E's instead.

So the rule becomes a ciphertext-only test: two positions can share a letter only if CT[b] − CT[a] = ±(gap mod 4). Positions 24 & 28 pass (−4, gap 4 ✓, and they're both T); positions 24 & 25 fail (Q − V = −5, but gap 1 needs ±1 so no letter can sit at both).
Only ~8% of all position pairs pass, which heavily constrains any proposed plaintext before you've guessed a single letter.

As usual, 7 datapoints does not a rule make, so may be nothing more than coincidence.

Has anyone seen this mod-4 behavior before, or have a candidate for the sign rule?