> C=A^B is also pure noise
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.
> Is C really "pure noise" if you can get A back out of it?
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.
No, C is really noise, fundamentally.
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.
I'm not sure I agree. In the case of new source D, C is being used as the key, not the encoded data.
> B^C gives you back A
If both B and C are pure noise, where did the information for A come from?
XOR is commutative so for a one-time pad key and ciphertext can be swapped.
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.
Thanks for engaging in this discussion and explaining your thoughts. I'm still trying to understand.
> 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?