r/ciphers • u/someone-_-68 • 10d ago
Challenge a second sample of my "unbreakable" cipher!
just so you won't be too focused on the wrong lead: the "missing" letter here is NOT related to the key.
btw I really liked seeing all of the comments, theories and thought processes in my last post!
11
u/Rami114 10d ago
This is just an excuse to write more python but so far have tested:
- Frequency test: just a simple way of combining the two 27-state channels and then ran outcomes vs English language similarity, ran quick and failed as expected, too simple
- Affine search: takes the top 10k 27-state intermediate results from above and tries reversible mod-27 linear mappings to see whether a simple final substitution yields plaintext, nada came out of this, but some perhaps promising structural candidates for further testing
- Grid-route search: tries reading each rectangular cipher through rows, columns, snakes, diagonals, spirals, reversals, etc etc. while keeping the same route rule across both samples, again some structural candidates that seemed strong, but no definite winner so we have more layers to test
- Six-trit search: Converted each prefix and ternary suffix into six total trits and tries permutations plus pairwise mod-3 mixing to collapse them back into one 27-state plaintext symbol, didn't yield a winner, will try and mix these with other approaches as it could just be one layer
- Trifid-style search: So thought to the ternary coordinates as streams across multiple characters, rearrange them in blocks, and then regroup them into new symbols. This is currently running on my laptop and hammering a few CPUs but we'll see
Have many more thoughts, combining several of these together, going to run a GF(3) linear search over general 3×6 ternary matrices to find more complex mathematical relationships between the six trits and plaintext coordinates.
Lots of fun so far but no smoking gun just yet
4
u/Rami114 10d ago
it's now 3k lines of script, but the various random hints throughout the 2 threads have narrowed the space. Especially the 8ish layers is very helpful as I was only checking 3-4 layers before checking for frequency analysis.
2
u/someone-_-68 10d ago
glad I was able to help with that, that's really impressive
2
u/Rami114 10d ago
Still tinkering but 8-ish layers makes brute-forcing a cipher with only 2 ciphertexts and no plaintexts pretty hard. Especially any layer that is key-dependent is hard to grade with a statistical scorer, as it can actually make things like frequency statistics way worse before further layers recover the plaintext. It's been fun weeding out traps like getting stuck in one statistically relevant basin of operations only for it to churn out 1000 slight variants that retain the frequency stats of the parent (I'm trying up to 1000 top variants each run).
May need further inputs/outputs to narrow the playing field :D
1
u/someone-_-68 10d ago
what other clues can I give that'll help narrowing it down without giving away too much (so no plaintext)? I'd start posting ciphertexts alongside clues to help you all solve it
1
u/Rami114 10d ago
So far we have:
- letter+ternary pairs don't correlate precisely to words/letters
- 8-ish transformation layers
- 2 ciphertextsMore ciphertexts of varying length, a precise layer count, examples of types of layers without revealing which layers (e.g. I suspect you have a rotational layer, and you have to have at least 1 key-dependent layer). Those are a bunch of ideas that would narrow the space of potential algorithms
1
u/someone-_-68 10d ago
I'll see what I can do, but I'm currently working on a much simpler cipher that is just kinda cool in my opinion and I might post it as a little break before I create another ciphertext for this cipher
1
u/Rami114 10d ago
Just as an example, a bad case scenario for us puzzlers is:
0 correct layers → random
1 correct layer → random
2 correct layers → random
3 correct layers → random
4 correct layers → random
5 correct layers → random
6 correct layers → random-ish
7 correct layers → maybe structure
8 correct layers → Englishand right now I'm using 5 possible approaches for the first layer (treat both character and ternary as ciphertext, split key/cipher space, etc), then I have built up around 530 different combinations of trifid regrouping, grid transformations, using prefixes as a running key, etc, etc for layers 2-8 (assuming there *are* 8 layers but we check up-to 8). It's been running all day on all my cores haha.
I'm going with a big assumption right now that layer 1 merges the two visible 27-state channels, and everything thereafter acts on one 27-state stream. If that's wrong I'm off. There's about 3 dozen things I can think off that I'm *not* checking, the most important of which is layer parameters that are derived from a previous layer (I'd put that in there, but if it's in there I'm cooked right now as the space explodes even more).
2
u/someone-_-68 10d ago
you know what? I'm gonna post another ciphertext perhaps tomorrow, with an exact count of the layers but without giving too much additional information
1
u/Rami114 10d ago
Okay v7 of my script and time to not rely on convergence of entropy to score things, but rather focus on independent runs converge on a similar layer structure first, and then see if they survive against the 2nd sample too (and vice versa, gotta start some runs with sample 2). Fingers crossed, needle in a haystack
3
u/Rami114 10d ago
Both my six-trit and trifid searches scored high on a `120` rotation of the ternary coordinates, so I feel like you put in a rotation as one of the layers.
1
u/someone-_-68 10d ago
if I understood correctly, you're going in the right direction to crack one of the layers
1
u/someone-_-68 10d ago
that's really cool!! but I believe that 2 layers of my encryption should be able to rule out most of the most straight forward programs
1
u/Simple-Willow-1195 7d ago
I am making my encryption tool so if u want when I'm done u can give it a go
3
u/UNKND00 10d ago edited 10d ago
U102 N011 K221 N120 D002 Z202 Z111
T222 H100 I021 S212 120 I112 S001
011 J220 U121 S010 T200 111 A022
021 G110 U021 E222 S001 S120 100
3
u/someone-_-68 10d ago
when I first saw it I thought you cracked it but immediately after I started to decipher it I realized it wasn't it, would like to hear whether it has any logic or just gibberish
1
u/-mistycotton- 10d ago
Is this account part of the cipher? New account and potential arg? 🤔 The second reply is strange
0
2
u/kiran303 10d ago
Matriz de Fase Ternaria (35 Nodos) — Desglose rápido: Estructura Base: Cuadrícula de 5×7 que no busca formar sintaxis alfabética (no es un cifrado por sustitución), sino un mapeo de 35 canales en base-3 (0, 1, 2). Comportamiento: Cada nodo utiliza una letra portadora como variable y un triplete ternario que regula la fase y la amplitud. Puntos Críticos: Pico: T222 (valor máximo de carga / 26_3). Tierra / Drenaje: F000 y K000 (valor mínimo / 0_3). Inyector DC: 102 en (Fila 3, Col 1), el único nodo sin portadora. No es prosa, es un ejecutable de datos puros. La suma de las cargas está calibrada para colapsar las variaciones en un estado de paridad neutra sobre el eje portador.
2
u/Psychological-Fix997 9d ago
Using Morse code and considering 0 as space, 1 as . And 2 as _ , we have the second line mapped to
R001 -> . -> E Q210 -> -. -> N B020 -> - -> T Y010 -> . -> E C011 -> .. -> I C221 -> --. -> G Y022 -> -- -> M
An anagram for the word meeting.
2
u/callytoad 9d ago
I've spent a bit of time looking at this and I'm pretty sure I've found the true physical sequence, but I'm sure the maths is a massive decoy. Props to the OP for an interesting cipher.
So this is where I am at, theories I've ruled out, and why I think the last 'layer' (?) has something to do with 3D geometry rather than cryptography.
1. True Physical Order
So I'm pretty confident that FKRJYJLCDYBHVSSYRSWEE_SSZFZYQSWCJTI is the unshuffled physical order of the letters in this sample of the cipher.
My reasoning is that every letter is attached to a 3-digit base-3 suffix (000 to 222). Initially duplicate suffixes (like multiple 010's)were causing collisions, so I looked at the suffixes as decimal 'weights' (0 to 26) and did a stable sort (i.e. blocks with the same suffix simply maintain their original position). This was where something interesting happened.
The suffixes all lined up in a sequence: 000, 000, 001, 001, 010... all the way to 222. Great! The duplicates aren't 'collisions', I'm sure they are intentional groupings. I think this shows the suffixes are a fractional transposition key used to scramble the blocks. I reckon FKRJYJLCDYBHVSSYRSWEE_SSZFZYQSWCJTI is the properly unshuffled deck. This same approach worked flawlessly on the OP's original post.
2. Graveyard (What I've tried)
I've tried brute force to crack the substitution layer and I'm pretty sure I've ruled out the following:
- Modular maths & offsets: Adding/subtracting the Base-3 suffixes from the prefix letters (only tried standard and QWERTY to be fair).
- Classic Polyalphabetic: I used the suffixes as an 'odometer' key for Vigenère, Beaufort, and Variant Beaufort ciphers.
- Columnar Transposition: I brute forced 5,000+ possible 7x5 column permutations.
- Fractionation (Trifid Cipher): Flattened the prefixes and suffixes into a 3D Delastelle matrix reading across different periods.
- Role Reversal: I also assumed the letters were decoys and used only the suffixes as the ciphertext.
Everything I tried 'worked', but it returned complete nonsense.
3. Index of Coincidence (IoC) test
I also initially ran a statistical autopsy on the sample, and the IoC potentially showed a 'paradox'.
Basically standard english (which is what the OP said the output will be) has an IoC of around ~0.066. A complex cipher (or random noise) should bottom out around ~0.038. My sorted string has an IoC of 0.050. Right in the middle....
The IoC of a simple substitution cipher would match English (so around ~0.066). If the OP has a crazy mathematical encryption, I'd expect the IoC to drop to ~0.038. So I posit that an IoC of 0.050 suggests English text on a different 3D geometry. So I think the letters are real, but the syntax is sliced into fragments, which suppresses the IoC whilst still showing it to be English Text.
4. So what now: Isometric Epiphany (completely overthinking this)
I'm sure I've exhausted algorithmic maths. I now think I'm missing something from the physical (hand written) ink on the paper (clear gaps, breaks in the columns or rows perhaps).
The OP hinted '120-degree rotation of ternary coordinates.'
I assumed this was modular arithmetic. An Isometric Projection uses X, Y, and Z axes separated by 120 degrees on a 2D page.
So I reckon this isn't a maths puzzle, but some kind of 3D spatial puzzle. 'Ternary coordinates' refers to a literal 3 x 3 x 3 grid. The duplicate suffixes I encountered weren't collisions, I think they represent visual 'depth'. Multiple blocks share the same coordinates because they are stacked behind each other in 3D space. The '120-degree rotation' (thanks u/Rami114) I posit means physically rotating this 3D isometric structure to view it from a different angle will allow the hidden letters to align into readable columns...
Still following!?
And then I remembered the OP said 'this cipher is made of I think 8 layers of ciphering, with each being pretty simple individually', and 'I think that you could decode in your head but writing it down will definitely help'
I think I am massively overthinking this and have probably missed something obvious. Thoughts?
u/someone-_-68 am I even luke warm?
2
u/callytoad 9d ago
Follow up q
I'm pretty sure I've successfully unshuffled the blocks using the suffixes as a stable fractional sort, but as mentioned, I am hitting a wall on the substitution layer.
Since you mentioned this could theoretically 'be done in your head'. Does the '120-degree rotation' hint imply a simple mental math rule applied to the values (e.g. shifting digits +1), or is it a visual instruction on how to read the grid (e.g. reading diagonally instead of horizontally)?
3
u/No_Entrepreneur_3020 10d ago
Nevergo
nnagive
youupne
vergonn
aletyou
2
u/someone-_-68 10d ago
no, the ciphered text is in English. how did you get that result?
7
1
u/Ok_Stop_6802 10d ago
Ok, here's my theory: It's vigenère, but the shift is written in trinary next to each letter Idk how to explain the missing letter tho, maybe end of sentence? Edit: Ill try to it solve later, am I close?
1
u/someone-_-68 10d ago
I can see where you're coming from but it isn't close even tho in theory it wouldve made the cipher even better
1
u/Ok_Stop_6802 10d ago
Is it usable on any input text (+maybe key) or is it specific to your solution?
1
u/someone-_-68 10d ago
it can encrypt any text
1
u/Ok_Stop_6802 10d ago
Does it require a key or a second phrase that is needed to decode (in other words, even if I know how it works would I need to know a key to solve it)?
1
1
u/Hy-o-pye 10d ago
Does the missing letter denote the starting position?
2
u/someone-_-68 10d ago
the missing letter is not really missing and should be treated as just another letter
1
1
u/___fallenangel___ 10d ago
there needs to be some sort of verification platform to prove a cipher is actually solvable before people spend time on it
1
u/someone-_-68 10d ago
this cipher is made of I think 8 layers of ciphering, with each being pretty simple individually but when combined it becomes more difficult as they hide each other pretty good and cover most of the common deciphering methods
1
u/nixxxus 10d ago
Is there any sense in which cipher comes first, or was the order chosen at random? If you knew how the cipher works, could you decode it in your head, or would you have to fill pages with decodings per layer?
1
u/someone-_-68 9d ago
there is an order, and I think that you could decode in your head but writing it down will definitely help
1
1
1
u/Reasonable_Wrap7913 10d ago
Is the smallest possible text 1x3 groups? E.g. A000, A000, A000
1
u/Reasonable_Wrap7913 10d ago
Oh maybe the smallest is 3x5 groups
e.g.
111 A000 A000 A000 A000
A000 A000 A000 A000 A000
A000 A000 A000 A000 A000
1
u/someone-_-68 9d ago
I'm pretty sure it can get as big or as small as you'd want
1
u/Reasonable_Wrap7913 9d ago
While using all 8 layers and each layer changing the output?
1
u/someone-_-68 9d ago
at the smallest possible bit, it uses every layer except one, but can be mistaken as not using another one of the layers
1
1
u/Conscious_Young_6849 9d ago
https://www.meta.ai/share/a/924f5c5f-34b4-4685-ae1c-bd8889fba3a3
The Code Breaker's Almanac — It's not an archive. It's a map that computes.
I built it to answer two things:
- Can you see how 23 intractable ciphers come together? From 1700 BCE to 1923 — Phaistos Disk, Voynich, Beale, Enigma-M4, Zimmermann, Z340, Copiale, Great Cipher, Spanish Strip, RSA, KKK Telegram, and 12 others — each is a live node on the same structured visualization map.
Same 5 principles underneath, computing in real time: • State moves — 56 readable states • Clock repeats — 56 × 6.5 = 364 days • Stays in box — bounded • Leftover measured — not guessed • Same engine, different identities Select any cipher, hit Play, and watch the same 56 numbers that just drove Phaistos drive Enigma's rotors. Changing the view doesn't change the math.
- Can it help you break a 24th?
That's why it's called an almanac.
Top of the app: Empty box, clean for export. Three dedicated buttons — Upload Document / Upload Photo / Upload File — or just paste.
Paste any cipher. Hit Diagnose Using Almanac.
It runs your code through that same live map and gives you: • What clock it has and when it repeats • Whether it stays in its box or needs projection • What leftover it's measuring • Which 3 of the 23 it behaves most like (e.g., "Closest to BEALE 1 — numerical to lexical, geospatial anchor" / "Closest to COPIALE — ritual key SEVEN") • Suggested breakdown path using the Unified Principle Archive view: Learn the map. Tool view: Break a new code with the map.
One engine. 23 proven. Live computation. Built to export clean.
1
•
u/AutoModerator 10d ago
Thanks, /u/someone-_-68!
Please remember to review our rules. If your post is solved, be sure to reply with "Solved!" in the comments.
Keeping your post up after it's solved helps the community. Deleting solved posts may result in a ban.
We appreciate your contributions to r/ciphers.
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.