Kind of. Okay look, ‘K4 is finished’ sounds much better than ‘we’re decades in and finally have a crib uncovered’
What was uncovered:
> How K1 password was derived using the morse code
> How K2 password was derived using K1 [which reinforced what we already know about K2 and K3 working together]
> K3 was solved using information from K2 and a clue from the morse code [we knew this already it just confirmed that all puzzles are self referential]
> K4 involves layering everything we’ve seen before: section, quagIII, movement,
> Jim was wrong, this is solvable with AI, and it can be brute forced but neither happen without context
How I figured it out:
[You only need to tweak the solution to K1]
YECRNGLYDKXVFGWFBTEHOYGPXFOJICJBLOFNICKXFKOVLVBBWOIGCZTJYDNDLGALCCHHZNOJRDKSVETCIASKMFNTBTQYLPTTPDOKMGXOVDYGHMVWXPMJDBGGKFPMHZSJUPLBLTLQILCDFVBQRDDEXEXGUKYDCIAWFQZLIHUHDLLABYWDYGHFSKCRET.
Github for verification:
https://github.com/ElodineOfficial/KrackersAndCheese <Slider proves intended solutions behind K2 with an animation
Click here for more information about K1+2 Password Recovery.
https://github.com/ElodineOfficial/KryptosK4 < Solve for K4, up until step5 [anything past that is just theories for the final steps]
Technical overview:
This is a technical text dump of how these scripts work and solve the first steps of K4.
Raw K4: OBKR UOXOGHULBSOLIFBBWFLRVQQPRNGKSSO TWTQSJQSSEKZZWATJKLUDIAWINFBNYP VTTMZFPKWGDKZXTJCDIGKUHUAUEKCAR
Here’s a step‑by‑step recipe you can execute entirely by hand. Wherever you see “letter→number,” use A = 0, B = 1, …, Z = 25, and whenever you “mod 26,” if you get a negative result add 26, or if ≥ 26 subtract 26.
- Clean and split the raw K4 text
Strip out spaces so you have one 97‑letter string.
Remove the 49th character (the extra “W” in the exact center). You now have 96 letters. >Reminder for later maths<: This became our crosshair and axis, it did not necessarily get ‘removed’
Cut into four blocks of 24 letters, in reading order:
NW = letters 1–24
NE = letters 25–48
SW = letters 49–72
SE = letters 73–96 - First‐layer decryption of each quadrant
For each block (NW, NE, SW, SE), do:
Write it in a 4 rows × 6 cols grid, filling each row left→right, top→bottom.
Re‑label the columns by applying the quadrant’s permutation. If perm = [p₀,p₁,…,p₅], that means “new column 0 is old column p₀,” “new column 1 is old p₁,” etc.
Quadrant
Permutation
Caesar shift
Vigenère key
NW
[1, 3, 4, 0, 2, 5]
+ 1
KRYPTOS
NE
[2, 0, 1, 5, 3, 4]
− 7 (≡ + 19)
ABSCISSA
SW
[4, 3, 5, 1, 2, 0]
+ 15
PALIMPSEST
SE
[3, 1, 2, 4, 0, 5]
+ 8
PALIMPSEST - Flatten the permuted grid back into one 24‑letter string (row by row).
Apply the Caesar shift to each letter: convert letter→number, add the shift, mod 26, convert back.
Vigenère‑decrypt with the given key:
Repeat the key to length 24.
For each position i, compute (cipherᵢ – keyᵢ) mod 26 → plaintext letter.
Record the resulting 24‑letter “quadrant plaintext.”
After this you’ll have:
NW_pt = SBXASBXYEASTFUZLRRJSUSGU
NE_pt = JNREASTZDLTFBXUJLBFJDALS
SW_pt = LZYAMAFHKEASTFYBKQCVPIQA
SE_pt = WEDKGSSXZSQQEKQNWESTDIRR
Where the we’ve been going wrong [myself included]:
> Focused so hard on LATIN that we didn’t realize there were english passcodes
> Focused so hard on K4 that the methodology behind K1 and K2 solutions were glossed by, afterall, they’d been solved
> Not enough focus on the in puzzle clues, [with respect I’ve seen some people try and tie in the bible or books outside of the quoted Carter Howard passage and that’s just not going to go anywhere, this is a contextually heavy puzzle but it is all internal to itself and general cultural knowledge]
> Not enough focus on the access of information during the time, as Sanborn wasn’t making this with internet popularity in mind. He was thinking about local libraries, internal clues, and perhaps even some pop culture osmosis. Pen and paper were always the focus of this solve.
> We assume corrections are corrections, instead of clues that’ve been messed up so badly that the puzzle may be unsolved without correction [XIDBYROWS vs XLAYERTWO]. Sanborn has said this mixup was a highly unfortunate probability event. I can’t stress enough how every little thing here is intentional even when it seems insignificant. But artist intention and perfection in execution are wholly different, we can’t treat this as an infallible puzzle. This ended up being a hint to how the palimpsest in k4 functioned. So it was critical to correct, I do appreciate Sanborn doing that.
On that note. Allow me to be the bad guy for a moment: the Kryptos community in general is composed of good faith actors, and that isn’t always how decryption, hacking, or puzzle breaking happens. This is a social engineering problem, where Sanborn himself has to be considered as he has done so many self referential clues throughout his entire career [not uncommon, I do this too] and we see them begin to bleed through here as well. By having a better understanding of the personal knowledge base, we get a greater idea of what would’ve made sense to use at the time as a first, second, or even obscure third option. The better you understand the man [ethical considerations aside] the easier this puzzle is to undo. K4 is being solved by being the bad guy. By being the hacker, the social engineer, and the goblin. So if you’re going to solve the other half, grab your grey hat!
Where we’re at:
K4 is cracked. But there’s still a lot of good work to do. Science is about running tests, validation through more trials, and getting rid of junk data. This page has been edited multiple times in order to remove junk data and lucky numbers while preserving everything we can validate.