> However, I think that a lossless model can
reasonably derive this information by observing that p(x, x') is approximately
equal to p(x) or p(x'). In other words, knowing both x and x' does not
tell you any more than x or x' alone, or CDM(x, x') ~ 0.5. I think this is
a reasonable way to model lossy behavior in humans.
How does a lossless model observe that "Jim is
extremely fat" and "James continues to be morbidly obese" are approximately
equal? I would assume that it would have to be via the same world model
that a lossy model would -- which is waaaay above the bitstream
level.
Also, I think that going at this via a probability
model is not the way to go.
> knowing both x and x' does not tell you any
more than x or x' alone
Can't you rephrase this with the following
approximately equal phrases:
- You need to discard either x or x' to reach a canonical form, or
- Discarding either x or x' is not a lossy operation?
Mark
----- Original Message -----
From: "Matt Mahoney" <[EMAIL PROTECTED]>
To: <agi@v2.listbox.com>
Sent: Sunday, August 27, 2006 10:32 PM
Subject: Re: [agi] Lossy *&* lossless
compressi
>
> But I see your point. I argued that a lossless model knows everything that a lossy model does, plus more, because the lossless model knows p(x) and p(x'), while a lossy model only knows p(x) + p(x'). However I missed that the lossy model knows that x and x' are equivalent, while the lossless model does not.
>
> However, I think that a lossless model can reasonably derive this information by observing that p(x, x') is approximately equal to p(x) or p(x'). In other words, knowing both x and x' does not tell you any more than x or x' alone, or CDM(x, x') ~ 0.5. I think this is a reasonable way to model lossy behavior in humans.
>
> -- Matt Mahoney, [EMAIL PROTECTED]
>
> ----- Original Message ----
> From: Philip Goetz <[EMAIL PROTECTED]>
> To: agi@v2.listbox.com
> Sent: Sunday, August 27, 2006 9:23:25 PM
> Subject: Re: [agi] Lossy *&* lossless compressi
>
> On 8/25/06, Matt Mahoney <[EMAIL PROTECTED]> wrote:
>> As I stated earlier, the fact that there is normal variation in human language models makes it easier for a machine to pass the Turing test. However, a machine with a lossless model will still outperform one with a lossy model because the lossless model has more knowledge.
>
> That would be true only if there were one correct language model, AND
> you knew what it was.
> Besides which, every human has a lossy model. It seems to me that by
> your argument, a machine with a lossless model would "out-perform" a
> human, and thus /fail/ the Turing test.
>
> - Phil
>
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