19-06-2026, 06:56 PM
(19-06-2026, 06:29 PM)SpamBot Wrote: You are not allowed to view links. Register or Login to view.then how can you claim that you have decrypted something, until you have shown that there are no other fuction(s) which produce meaningful text(s), or you have found all of them (and shown that none other exists)?
It's impossible to prove that a message has no alternative meaning. Easiest example: suppose yesterday we secretly agreed that the phrase "It's impossible to prove that a message has no alternative meaning." would be a code to mean "Let's go to the pub". Now the beginning of this paragraph has a completely different meaning just between you and me.
Having said that, if a deterministic decryption process produces a sensible plaintext without sacrificing too much information, I'd call this a successful decryption. For example, if someone demonstrated how 1000 symbols of sensible plaintext can be decoded from 3000 symbols of Voynichese, using a process that can itself be encoded in much fewer than 1000 symbols, I'd call this a success. On the other hand, I would be suspicious if 3000 symbols of Voynichese yielded only 100 symbols of plaintext.
(19-06-2026, 06:29 PM)SpamBot Wrote: You are not allowed to view links. Register or Login to view.(14-06-2026, 09:07 PM)oshfdk Wrote: You are not allowed to view links. Register or Login to view.Most diplomatic ciphers worked this way with letter variants. And homophonic polyphonic ciphers also allow multiple decryptions, with the right one only clear from the context.
Clearly your example doesn't fall under my definition of encryption (to which symbol is mapped symbol 'S'?). Nor is it possible to decrypt it uniquely (without knowing the function used or the original text): "A MNEON" - is it "AN OMEN"? "A GENRE"? "A FENCE"? "A COMBO"?
Sorry, I don't understand what you mean, probably this is a misquote? I was talking about polyphonic homophonic (thanks, @nablator!) ciphers there. A simple example of a polyphonic homophonic cipher: assume each letter of the alphabet is encoded by the next letter of the alphabet (so, A is encoded as B, K as L, etc) and in addition digits encode the following letters 0=A 1=E 2=I 3=U 4=O 5=S 6=A 7=O 8=E 9=I
2 MJL8 D47LJ15 uniquely decodes to I LIKE COOKIES without any problems. Actually any ciphertext under this scheme will decode to a unique plaintext.
If you were talking about the permutation cipher, it's very easy to decode:
IIEOKESIOCKL => reading one letter at a time form the beginning and from the end: I L I K E C O O K I E S
(19-06-2026, 06:29 PM)SpamBot Wrote: You are not allowed to view links. Register or Login to view.I've noticed that scrambles have a funny property: if you apply the same rule to the same text several times, you'll end up with the original message ... Which means that, again, those can be solved by considering all permutations (why this is important look in the 2nd message)
The number of permutations even for a short text of 100 characters is prohibitively large to enumerate directly. What is more, it's likely there will be many sensible decodings.
(19-06-2026, 06:29 PM)SpamBot Wrote: You are not allowed to view links. Register or Login to view.I meant a bit different thing: consider a graph (oriented tree) whose vertices are all symbols of your original message and edges reflect the order of those symbols. Any encryption (disclaimer: I could come up with) doesn't change this graph. See?
No, I don't, sorry. What do the edges connect in this scenario?

