The basic idea being, can you using the digits of Pi to encode data, or rather, can you find ranges of Pi that map to data you have and use it for "compression".
A very simple example, let's take this portion of Pi:
> 3.14159265358979323846264338327950288419716939937
Then let's say we have a piece of data that, when encoded and just numbers, results in: 15926535897997169626433832
Can we encode that as: 4–15, 39–43, 21–25, 26–29 and save space? The "compression" step would take a long time (at some point you have to stop searching for overlap as Pi goes on for forever).
Anyways, a silly thought experiment that your idea reminded me of.
Is C really "pure noise" if you can get A back out of it?
It's like an encoding format or primitive encryption, where A is merely transformed into unrecognizable data, meaningful noise, which still retains the entirety of the information.
If you throw out B, then there's no possible way to get A out of C (short of blindly guessing what A is): that's one of the properties of a one-time pad.
But distributing both B and C is no different than distributing A in two parts, and I'd have a hard time imagining it would be treated any differently on a legal level.
Imagine another copyrighted work D.
E=C^D, therefore C=D^E
As you see, the same noise can be used to recover a completely different work.
Since you can do this with any D, C is really noise and not related to any D or A.
> B^C gives you back A
If both B and C are pure noise, where did the information for A come from?
In this example given ciphertext C = A ^ B you can decrypt plaintext A using key B or you can decrypt plaintext D using key E = C ^ D:
( A ^ B ) ^ B = A
( A ^ B ) ^ ( C ^ D ) = ( A ^ B ) ^ ( A ^ B ) ^ D = D
B and C are pure noise individually but they are not uncorrelated just like E and C are not uncorrellated.
> given ciphertext C = A ^ B you can decrypt plaintext A using key B
Right, C ^ B = A, to get back the original.
Already here, it seems to me that C is not pure random noise if A can be recovered from it. C contains the encoded information of A, so it's not random. Is that wrong?
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> or you can decrypt plaintext D using key E = C ^ D
In this example, isn't E the cyphertext and C is the key? E (cyphertext) = D (original) ^ C (key).
Then E ^ C = D, to get back the original.
Here, it seems to me that E contains the encoded information of D, so it's not random. And C plays the role of a key similar to B, so it's not being decoded as a ciphertext in this case, and nothing is implied whether it's random or not.
> XOR is commutative so for a one-time pad key and ciphertext can be swapped.
Maybe this is what I'm missing. In the second example, I'll swap the ciphertext and key:
C ^ E = D
Hmm, so both C and E are required to recover D the original information. Does that mean that the information was somehow distributed into C and E, so that they are both meaningful data, not random?
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But what about B in the first example, a key that was supposed to be pure random? If I swap the ciphertext and key:
B ^ C = A
The information in A is recovered from the interaction of both B and C. So the entirety of the information is not in C, the ciphertext (or key in this case).
Does that mean, B is no longer pure noise? When it was used as a key, it became meaningful in relation to A. That's information, isn't it?
If A is a copyrighted work, and B is pure noise, then C=A^B is also pure noise.
Distribute B and C. Both of these files have nothing to do with A, because they are both pure noise.
However, B^C gives you back A.