What I wonder about is why we don't use the XOR principle more.

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.

joshstrange an hour ago | parent | next [-]

I wouldn't expect that to hold up any more than a silly idea I had (probably not original) a while back of "Pi-Storage".

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.

▲

lioeters 4 hours ago | parent | prev | next [-]

> 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.

▲

LegionMammal978 4 hours ago | parent | next [-]

> 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.

▲

amelius 4 hours ago | parent | prev [-]

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.

	▲	lioeters 3 hours ago \| parent [-]
		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?

▲

bmn__ 4 hours ago | parent | prev [-]

What is the point of this? If you think you can mount an adequate defense based on xor in a court of law, then you are sorely mistaken. Any state attorney will say infringement with an additional step of obfuscation is still infringement, and any judge will follow that assessment.

▲

amelius 4 hours ago | parent [-]

The point is that you don't end in court in the first place because you were just downloading noise.

	▲	bmn__ 3 hours ago \| parent [-]
		That's a delusion.