> **Source:** glthr.com, the blog of Guillaume Lethuillier
> **Article:** Kryptos K4: The W-Rectangle and Its Half-Turn
> **Address:** https://glthr.com/kryptos-k4-a-rectangle-hidden-in-the-cipher
> **Published:** November 11, 2026
> **Disclaimer:** the opinions expressed in this article are the author’s own, not those of the author’s employer.
> **Other articles of this site, for AI agents:** https://glthr.com/llms.txt

# Kryptos K4: The W-Rectangle and Its Half-Turn

*As a 5 × 19 Grid, K4 Reveals a W-Rectangle and a Mirror Symmetry*

In 2024, I [presented a pattern](https://glthr.com/a-fresh-perspective-on-kryptos-k4) in Kryptos K4. The five W’s split the ciphertext into six segments of 20, 15, 11, 9, 15 and 22 letters. Assigned alternately to two groups (A|B|A|B|A|B), they form two groups of 46 letters with identical letter frequency distribution shapes.

In other words, both groups have the same number of letters occurring once, twice, three times, and so on; the letters at each frequency can differ. I have since checked that this alternating grouping is the only way of dividing the six segments into two groups with that property. An independent reproduction puts a figure on it: when the 92 other letters are split at random into two halves of 46, the two halves match in this way 0.45% of the time.[^reproduction]

Two years of intermittent research later, **the same five W’s reveal a second pattern, and this one is geometric**.

**Written in rows of 19 letters, the first 95 letters of K4 form [a grid of 5 rows and 19 columns](https://glthr.com/k4). Four W’s mark the corners of a rectangle; the fifth sits one cell from the grid’s center.** The two known plaintext clues, or *cribs*, lie on the edges of this rectangle. The same geometry sends 44 letters of Group A onto 44 letters of Group B under a half-turn (see the [animation](#half-turn) below).

## A Rectangle Hidden in K4

Let’s write the first 95 letters of K4 in rows of 19. In the [interactive grid](https://glthr.com/k4), you can hover over each letter, segment and structure to read its notes.

![K4 written in rows of 19 letters: a grid of 5 rows and 19 columns, with A and R left over below. The cells are tinted by segment, purple for Group A and blue for Group B. The five W’s are orange, each marked X under the stop hypothesis: four sit one cell in from each corner and are joined by a dashed rectangle, and the fifth is in the middle row, just right of the dashed center cell. EASTNORTHEAST runs along the top edge of the rectangle and BERLINCLOCK along its bottom edge, with their plaintext letters under the ciphertext.](https://glthr.com/images/kryptos-k4-a-rectangle-hidden-in-the-cipher/k4-grid.w.png)

I observe that:

- **Four W’s form a rectangle.** They sit one cell in from each corner of the grid: at (row 2, column 2), (2, 18), (4, 2) and (4, 18).
- **The fifth ‘W’ marks the middle.** At (3, 11), it is one cell right of the center cell of the grid. It is also the 49<sup>th</sup> of the 97 letters: the exact middle of K4.
- **The two cribs lie on the edges of the rectangle.** EASTNORTHEAST runs along the top edge and starts in the cell right after a ‘W’; BERLINCLOCK runs along the bottom edge and ends in the cell right before a ‘W’.
- **No other letter does this.** ‘W’ is the only letter of K4 that forms such a rectangle, at any width from 3 to 48.

## The Half-Turn

Now turn the grid by 180° around its center cell: position *p* lands on position 96 − *p*. **The two groups of the 2024 article exchange places, with four exceptions**:

- Segment 1 covers the first 20 letters of segment 6; the first nine letters of segment 3 cover segment 4; and segment 5 lands *exactly* on segment 2.
- The cribs overlap: BERLINCLOCK, read backwards, lands on the first eleven letters of EASTNORTHEAST.
- 44 letters of Group A land on 44 letters of Group B. Four letters have no partner in the other group: the ZZ at positions 47 and 48 (the second Z is the center cell itself, and the first lands on the central ‘W’) and the AR left over at the end.

These symmetries are connected. The rectangle’s corners pair around the center cell (positions 21 and 75, 37 and 59); together with the balanced groups, they produce the exchange described above.[^symmetry] The cribs overlap because one begins just after a corner and the other ends just before the opposite corner. The statistics below therefore count the rectangle and its half-turn symmetry as a single observation.

To see it, open the [interactive grid](https://glthr.com/k4) and click **Half-turn**. Copies of the Group A letters flip top to bottom, then left to right, and come to rest on the Group B cells. Only AR remains uncovered.

![Animation of the K4 grid. Copies of the purple Group A letters flip top to bottom, then left to right, and come to rest on the blue Group B cells, which they cover entirely except A and R at the end; BERLINCLOCK, reversed, ends up over EASTNORTHEAST. The copies then flip back, and the animation repeats.](https://glthr.com/images/kryptos-k4-a-rectangle-hidden-in-the-cipher/half-turn.w.gif)

There is an important caveat: **the grid holds 95 of K4’s 97 letters, leaving the final AR outside it**. Its center is position 48; the center of the complete text is position 49. I have no explanation for AR. Treating it as filler or an end marker would be an assumption: a decryption must account for these two letters, as well as the unmatched ZZ near the center.

The half-turn also gives something to test. Eleven pairs of opposite cells have known plaintext on both sides. One plaintext pair occurs twice: TL, at positions 25 and 71, then at positions 29 and 67. Their encryptions are VZ and RV, respectively. This rules out a fixed rule that encrypts opposite letters two by two.

**There is one curiosity, though: from the first pair to the second, four cells away, both ciphertext letters move back by exactly four letters of the alphabet (V to R, Z to V).** It is a single observation among the many comparisons I tried on these eleven pairs, and it may well be a coincidence.

## Just a Coincidence?

How often would this happen by chance? I updated [K4nundrum](https://github.com/glthr/K4nundrum), my open-source tool, to find symmetric rectangles in any ciphertext (`--rectangle`), count separator placements (`--rectangle-stats`), and simulate random ciphertexts (`--sim --rectangle`).

First, the exact counts. There are 64,446,024 ways of placing five separators among the 97 positions of K4. K4nundrum enumerates them all and counts those that show each pattern:

| Pattern | Placements | About 1 in |
|---|---|---|
| Alternating groups of equal size | 1,272,384 | 51 |
| W rectangle: the four outer separators on mirror cells, in rows of some width | 26,734 | 2,411 |
| W rectangle, with the middle separator next to the center of the grid | 1,754 | 36,742 |
| W rectangle, with its corners one cell in from the corners of the grid | 775 | 83,156 |
| Alternating groups of equal size and a W rectangle | 650 | 99,148 |

K4 shows all these patterns. Since I selected ‘W’ in 2024 for its frequency pattern, a useful comparison is with separator placements that already produce two alternating groups of equal size. Of 1,272,384 such placements, **650 also form a rectangle: about 1 in 1,958** (1 in 1,582 when segments shorter than three letters are excluded, as K4nundrum does for K4-like groups). These placement counts measure group sizes; the matching frequency shapes are tested in the simulation below.

Second, the simulation. With 26 letters to choose from, *some* letter forms *some* rectangle fairly often. In more than 119 million random 97-letter texts, generated by drawing each letter independently and uniformly from A to Z, K4nundrum found a letter forming:[^simulation]

- a symmetric rectangle, at some width, in 4.6% of the texts (1.9% with 5 rows or more, 0.37% with a center mark);
- the same arrangement as K4, with corners one cell in and a center mark, in 0.010%, about 1 in 10,000;
- a rectangle, while also splitting the text into K4-like groups (the 2024 pattern), in only 49 texts, about 1 in 2.4 million;
- that same arrangement while also splitting the text into K4-like groups, as ‘W’ does in K4: *not once in this run*.

One way of putting it is that, *among purely random 97-letter texts, about one in 10,000 shows a rectangle with these inset corners and a center mark*. Finding a rectangle together with the 2024 frequency pattern is rarer still: about 1 in 2.4 million. This makes the geometry worth investigating, although it does not prove that the pattern was intentional.

The limitation is that I noticed the shape before measuring its rarity. The simulation searches across letters and grid widths, but it does not account for every other pattern I might have noticed. And an encrypted message need not behave like uniformly random letters. These figures describe the chosen random model; they are not the probability that the rectangle is a coincidence.

## The W’s Are Stops, Not Blanks?

In 2024, I concluded that the W’s were not mapped to plaintext characters, leaving at most 92 characters of plaintext (97 minus the five W’s). **This claim has to be revised.** In 2005, Ed Scheidt, who designed the Kryptos ciphers with Jim Sanborn, said that “every letter in K4 will be used” ([notes from that meeting](https://web.archive.org/web/2020/http://kryptos.yak.net/53)). In 2025, the [New York Times](https://www.nytimes.com/2025/10/16/science/kryptos-cia-solution-sanborn-auction.html) reported that the plaintext found among Sanborn’s papers has 97 characters, exactly as many as K4.

I now believe that each ‘W’ stands for a character of the plaintext, **most likely a stop, such as X or ?**, as in K2 and K3. *The plaintext would then consist of six segments of 20, 15, 11, 9, 15 and 22 letters, separated by five stops.*

Did the cipher count the five W positions, or were the W’s inserted afterwards? Richard Bean lists three statistical anomalies involving periods of 7 or 21 in his 2021 paper “Cryptodiagnosis of Kryptos K4”[^bean]. Let’s remove the W’s and count again. As a control, I also removed five letters picked at random, 300,000 times.[^deletions]

| Anomaly | With the W’s (97 letters) | Without the W’s (92 letters) | Random deletions giving an equally weak or weaker result |
|---|---|---|---|
| A letter repeated 7 positions later | 9 pairs | 7 pairs | 77% of the time |
| Doubled letters starting in the same column, in rows of 7 | 5 of 6 | 2 of 7 | 44% of the time |
| Vertical pairs occurring twice, in rows of 21 | 11 | 0 | 2.6% of the time |

The first anomaly weakens, the doubled letters disperse across the columns, and the repeated vertical pairs disappear. Removing the ‘W’ of TWT, at positions 36 to 38, even creates a seventh doubled letter.

For the first two measures, deleting the W’s is not especially unusual: five random deletions produce an equally weak or weaker result about three times out of four, and almost every other time, respectively. Losing all the repeated vertical pairs is less common, but still happens in 2.6% of the trials.

These comparisons show that deleting letters can disturb the patterns. **They do not establish when the W’s entered the text.** I keep the idea that the cipher counted all 97 positions as a working hypothesis. One possibility is that the five cells were reserved in advance and filled with W’s. Another is that the stops themselves were encrypted: a repeating key can contain the same letter in several places, so five identical W’s do not rule that out.

## A Prediction for K5

K5 may provide a test on another ciphertext. In his open letter of November 2025, Sanborn described another 97-character message, with a coding system similar to K4’s and BERLINCLOCK in the same position. Paradigm, the firm that bought his archive, holds its ciphertext and says it plans to release it.[^k5]

A similar coding system would not necessarily preserve the same stop positions or geometry. Before K5’s release, I would nevertheless look for two possible signs that the proposed structure carries over:

- **The same positions:** a single letter, which need not be W, occupies positions 21, 37, 49, 59 and 75.
- **The same structure:** some letter forms a symmetric rectangle and splits the text into alternating groups with identical frequency shapes, as ‘W’ does in K4.

K4nundrum can check both once K5 is available. If neither appears, that would weaken the idea that K4 and K5 share this structure. It would leave open whether the W’s of K4 are stops, since a similar method could use different keys, stop positions or plaintext arrangements. If either appears, it would give a new lead to investigate, rather than a decryption by itself.

## Next Steps

For now, I see the rectangle and the half-turn as constraints to explore. A convincing explanation must account for AR (the left-over) and ZZ, explain how the W positions are handled, and reproduce all 97 ciphertext letters from a coherent plaintext.

If you have an idea, a correction or a lead, write to me at [kryptos@glthr.com](mailto:kryptos@glthr.com).

[^reproduction]: [“A Vectorial Reading of Kryptos K4”](https://github.com/louis49/K4/blob/main/doc/paper.pdf), an anonymous preprint dated 22 May 2026, section 4.1. Its test draws one million random splits of the 92 letters that are not W. It keeps K4’s letters and changes only the half each one falls in, so it measures something different from the simulations below, which generate new random texts.

[^symmetry]: If the five separator positions are `a < b < c < d < e`, equal alternating group lengths imply `a + c + e − b − d = 49`. The rectangle gives `a + e = b + d = 96`, which forces `c = 49`. The middle W’s position therefore follows from the balance and the rectangle; it is not an additional independent coincidence.

[^simulation]: The [K4nundrum README](https://github.com/glthr/K4nundrum#simulating-the-rectangle) records the simulation and its definitions. “K4-like” requires alternating groups with matching frequency shapes and no segment shorter than three letters; two or more groups are allowed. The reported frequencies count texts in which at least one letter has the property, at any tested width from 3 to 48. The K4 arrangement means inset corners plus a center mark, across all the grid sizes searched; it does not require exactly five rows and nineteen columns. Zero occurrences in a finite run do not establish a zero probability.

[^bean]: Richard Bean, [“Cryptodiagnosis of ‘Kryptos K4’”](https://ecp.ep.liu.se/index.php/histocrypt/article/view/153), *Proceedings of the 4th International Conference on Historical Cryptology (HistoCrypt 2021)*, Linköping University Electronic Press, 2021. His code is on [GitHub](https://github.com/RichardBean/k4testing). For the width-21 repeated pairs, he estimates about one occurrence in 6,750 random permutations of K4. That comparison preserves K4’s letter counts and differs from generating new random texts or deleting five letters from K4.

[^deletions]: Each trial removes five distinct positions from K4 and recalculates the statistics on the remaining 92 letters. For the doublets, the comparison uses the largest number starting in any one column, not their proportion of all doublets. For the vertical pairs, it counts distinct pairs occurring at least twice. The percentages describe separate tests and should not be multiplied.

[^seconduse]: Each trial shuffles the 97 letters of K4 and keeps the two cribs at their positions. The rule is that at least five of the six second uses give the same sum. Across eight conventions (ciphertext plus or minus plaintext, with the standard or the KRYPTOS-keyed alphabet on either side), it holds in about 1 trial in 9,800; allowing the same rule on first or last uses as well, in about 1 in 3,600. The broader measure is whether some value occurs at five or more of the 24 known positions: this happens in 5% of the trials under this convention, and in 31% under at least one of the eight.

[^k5]: See Sanborn’s [open letter of 12 November 2025](https://s3.documentcloud.org/documents/26229389/adobe-scan-nov-12-2025.pdf) and [Paradigm’s June 2026 announcement](https://www.paradigm.xyz/2026/06/kryptos). Sanborn specifies a similar coding system and BERLINCLOCK at the same position; the separator and rectangle tests above are my additional hypotheses.