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Article 4724 of comp.ai.philosophy:
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>From: orourke@unix1.cs.umass.edu (Joseph O'Rourke)
Newsgroups: sci.philosophy.tech,comp.ai.philosophy
Subject: Re: A rock implements every FSA
Message-ID: <45426@dime.cs.umass.edu>
Date: 25 Mar 92 16:58:38 GMT
References: <1992Mar24.051654.18747@bronze.ucs.indiana.edu> <1992Mar24.112548.10215@husc3.harvard.edu> <45390@dime.cs.umass.edu> <1992Mar24.231518.10230@husc3.harvard.edu>
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Reply-To: orourke@sophia.smith.edu (Joseph O'Rourke)
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In article <1992Mar24.231518.10230@husc3.harvard.edu> 
	zeleny@zariski.harvard.edu (Mikhail Zeleny) writes:

>In article <45390@dime.cs.umass.edu> 
>orourke@sophia.smith.edu (Joseph O'Rourke) writes:
>
>>In article <1992Mar24.112548.10215@husc3.harvard.edu> 
>>zeleny@zariski.harvard.edu (Mikhail Zeleny) writes:

>MZ:
>> >...as I've said earlier, all
>> >that remains to be done is to interpret the states of Putnam's automaton as
>> >ordered pairs <state, input> of a FSA (cf. the relevant comments on p.124);
>> >follow this by running through enough input/state combinations to exhaust
>> >the finite combinatorial possibilities afforded by the machine's table.
>> >Finally, you do the mapping.  [...]  Just string all possible traces
>> >together in a sequential order.  

>JO'R:
>>Let's face it:  your rock doesn't realize or
>>implement the FSA in an interesting sense.  It just sits there, whereas
>>the FSA is active: it responds to the input.

>MZ:
>I don't believe you are saying this seriously, but I'll bite.  Please give
>me a precise definition of the essential difference between "just sitting
>there", and "actively responding to the input".

It may be that I misinterpreted your proposal.  David Chalmer's rebuttal
seems to interpret it differently.  So perhaps you could clarify
with the following simple example.  I'll lay out what I thought you had
in mind, and you correct where necessary.
	Here is a two-state FSA, call it FSA1, with two possible input 
symbols:

        FSA state   Input   ==>   FSA state
	---------   -----         ---------
	    A         0               A
            A         1               B
	    B         0               B
            B         1               A

You make an new FSA, call it FSA2, that has no inputs, and whose states
are <A,0>, <A,1>, <B,0>, <B,1>.  The transition table for this automaton
is not clear to me.  What did you intend?  Let me plow on nevertheless.

Now you fix a limit to the length of possible inputs; let's use n=2,
so the possible input streams are 00,01,10,11.  For each possible
input stream, you make a trace of the states of FSA2 (let's assume
A is the start state):

        00:  <A,0>, <A,0>, <A,x>
        01:  <A,0>, <A,1>, <B,x>
        10:  <A,1>, <B,0>, <B,x>
        11:  <A,1>, <B,1>, <A,x>

I don't know what to put for the last state, so I just used x.
You list all these traces one after the other:

        <A,0>, <A,0>, <A,x>; <A,0>, <A,1>, <B,x>; 
        <A,1>, <B,0>, <B,x>; <A,1>, <B,1>, <A,x>

Now we take a rock, and identify 12 time consecutive intervals,
and call these rock states S1, S2, ..., S12.  And we map the states
of the traces to these physical states:

	<A,0> -> S1
	<A,0> -> S2
	...
	<A,x> -> S12

Now the rock "realizes" FSA1.


