Re: Bruno's argument

From: 1Z <peterdjones.domain.name.hidden>
Date: Tue, 01 Aug 2006 04:27:06 -0700

Brent Meeker wrote:
> 1Z wrote:
> >
> > Brent Meeker wrote:
> >
> >>1Z wrote:
> >>
> >>>Brent Meeker wrote:
> >>>
> >>>
> >>>
> >>>>I'm considering rejecting the idea that a computation can be
> >>>>distinguished from noise by some internal characteristic of the
> >>>>computation. I don't think you can make the idea of "information hidden
> >>>>in noise" well defined. By Shannon's measure noise is information.
> >>>
> >>>
> >>>You can easily distinguish computation from noise using counterfactuals
> >>
> >>Can you make that more concrete - an example perhaps?
> >
> >
> > Counterfactuals come from the undertlying physics of the computation.
> > Cups of coffee don't have any woth speaking about-- you can't force
> > them into the same state twice.
>
> Sorry, but I still don't understand the counterfactual aspect.

You have to be able to say what *would* have happened if
the computation had gone down the other fork of an if-then. That
requires some causal stability.

http://en.wikipedia.org/wiki/Causality#Counterfactual_theories_of_causation

> > Whether they are part of the "internal characteristitcs of a
> > computation"
> > depends, question-beggingly , ont what you mean by "computation".
>
> I think I agree with that. I'm trying to come up with a non-question
> begging definition of computation and I think the idea that a rock
> implements all computations implies that computation can't be defined in
> terms of some chracteristic of its sequence of internal states.

I think the idea that a rock implements all computations is the wrong
place to start.

> > If you think a computation is nothing but a string of 1's and 0's,
> > counterfactuals
> > will be very difficulty to find.
>
> So you're agreeing with me that it's impossible to distinguish noise and
> computation based their sequence of internal states (e.g. 1's and 0's)?

No: I'm saying you do need to find counterfactuals,
and since they aren't in bit-strings ("movies" or recordings),
bit-strings aren't computations. Therefore, rocks don't compute
merely by going through a succession of internal states.

> Brent Meeker


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Received on Tue Aug 01 2006 - 07:28:07 PDT

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