But after the 2020 clues (another 13 letters), it became clear that it wasn't any single ACA cipher type, and it was probably something very difficult (because of K4's very low index of coincidence, i.e. 0.036 just below "random" 26-letter text at 1/26, plus the huge number of revealed plaintext letters "in place" i.e. letter-for-lettter correspondence).
That plausibly left a combination of two or more well-known cipher types, but if they were somewhat complex ciphers, the chance of solution would be rather remote.
Hence I always thought a "good" end to the puzzle would be like the book "Masquerade" by Kit Williams where the only guy in cahoots with the creator (Bamber Gascoigne) thought the initial puzzle was an unrealistic challenge, but Williams released clues which enabled two schoolteachers to solve it. So that part was satisfactory, even if hardly anybody remembers the solvers' names!
In contrast, the cribs for K4 haven't helped at all.
Edit: Unless the one time pad is a well known relative document, such as the Declaration of Independence.
Wikipedia has a good example: https://en.wikipedia.org/wiki/One-time_pad
In their example, "HELLO" is the plain text, "XMCKL" is the key, and the ciphertext is "EQNVZ". However, with a one time pad, an equally plausible plain text is "later" with the key "TQURI". Thus, without anymore data, it is simply impossible to know what the original message is.
Reasonable puzzles can be worked out (albeit maybe with a lot of work) with information provided by the puzzle or available somewhere in the environment.
Unreasonable puzzles (like some old Sierra games cough) are impossible without secret inside knowledge by the puzzle maker and/or brute force. And sometimes not even with brute force.
The hash/video example might just be an Easter egg hunt requiring looking across a wide set of videos (somewhat reasonable but boring), or completely unreasonable depending on circumstances.
Starting with the n-char plaintext, make it a loop. Now move the second letter two places to its right, the third three places, and so on ... until arriving at the original nth letter (painted red?) Or, starting with the digits of pi, move the second letter 3 to the right, the third 1, the fourth 4, und so weiter.
Doing a frequency on 97 weird letters wouldn't help much.
Good puzzles, even hard ones, should have some idea which way to approach them and should offer a method of attack other than brute force.
https://www.rrauction.com/jim-sanborn-kryptos-k4-solution-au...
The Wired piece has Sanborn saying the reserve should be "around $300,000."
It sounds like Sanborn really doesn't think it'll be solved before the auction date of 20 November (his 80th is on 14 November). If it does get solved due to this publicity bump, that's huge earnings foregone.
Perhaps he knows it is still an "unreasonable" challenge even with the 24 known letters.
You would think that one of the lessons of that is that someone could jump in right at the end and solve it after several clues were released. That hasn't worked with K4, which is increasing people's skepticism.
The article left me with a nagging question: Doesn’t the designer of the codes deserve a share of the proceeds of the auction? He’s still alive according to Wikipedia. It sounds like the unsolved code is what makes the art especially valuable. Was the cryptographer’s effort a “work for hire”, so he doesn’t get anything from the sale?
As Kryptos gots a huge amount of media attention in 1999, references to him changed from "chairman of A cryptographic center" to "chairman of [THE] CIA's cryptographic center" when it doesn't even seem that it has such a center.
And the featured story (around 52:00 of the video) has him apparently claiming credit for helping solve a Caesar cipher!
https://web.archive.org/web/19990501000000*/http://www.tecse...
He is an artist, not a mathematician. It’s a physical reveal for this layer of the copper onion.
Perhaps a 3D artist can model it and run some simulations with light.
So head north of it and use the shadows rim and 4/4/11/4 reveals and the Berlin clock sequence. Maybe on NOVEMBER NINE AT DAWN if you simulate it.
I don’t know if it ends.
Partial answer: FACE EAST THEN NORTHEAST. LET THE BERLIN CLOCK CHOOSE; READ EDGE TO EDGE TO FIND THE FINAL CODE.
K4 可能由类 Vigenère 层加柏林钟的基5转置构成；我以已知 crib 为锚点，运行搜索，并利用 IC、χ² 与 n-gram 收敛到英文分布。
NORTHWEST, EAST, NORTHEAST. BERLIN CLOCK TICKS EAST. READ EDGE TO EDGE; SEE TIME, USE THE NORTHEAST SHADOW. THE HIDDEN PLACE IS REVEALED.
Sorry for the bother
ATEQUINOXREADEDGETOEDEASTNORTHEASTFACETHENLETEDGEORDERBYTHERIMXBERLINCLOCKCHOOSEANDNAMETHEFINALCO
AT EQUINOX READ EDGE TO ED EAST, NORTHEAST FACE THEN LET EDGE ORDER BY THE RIM BERLIN CLOCK CHOOSE AND NAME THE FINAL CO
Edit- I think it might be K4. It is instructions for physical revealing k5.
Here is the intuition/clues (use your imagination) if you want to try too:
Edge to edge; Noon rim; Ignore flicker; Berlin clock beats
And remember the loadstone- that compass doesn’t point north
When you START searching for edges. Start with the YAR line. That’s what I did. But it goes on and on. Waypoint selectors…
The clues are in the K4 block of text, the actual answer is another layer using the rugged edges of all the panels. Oh and you need a timing program
If he responds I might write it up. But basically use compass order for edge stream in layer 1, use k0 (find the working sequence) for a digital interpretation (timing mask- diff from the E’s I used for clues) for layer 2, then start anchoring and rotating (use OOO for anchor after Ne block, one very small ordering move, and find your finisher cipher with clues.
THIRTY EIGHT DEGREES FIFTY SEVEN MINUTES SIX POINT FIVE SECONDS NORTH SEVENTY SEVEN DEGREES EIGHT MINUTES FORTY FOUR SECONDS WEST
Take care.
AT FIRST LIGHT FACE EAST. ALIGN WITH THE RIM SHADOW. READ EDGE TO EDGE TO FIND THE FINAL CODE.
You most certainly need to be there and k4 is instructions. I’m not sending the guy $50 to check my answer though
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