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- From: crawford@church.mitre.org (Randy Crawford)
- Subject: Re: measuring complexity (was Re: Kolmogorov Complexity)
- Message-ID: <1992Aug13.033835.4492@linus.mitre.org>
- Sender: news@linus.mitre.org (News Service)
- Nntp-Posting-Host: church.mitre.org
- Organization: The MITRE Corporation, McLean, VA
- References: <arms.713549466@spedden> <1992Aug12.173546.3298@access.usask.ca>
- Date: Thu, 13 Aug 1992 03:38:35 GMT
- Lines: 40
-
- In article <1992Aug12.173546.3298@access.usask.ca> choy@skorpio.usask.ca (I am a terminator.) writes:
- >In article <arms.713549466@spedden>, arms@cs.UAlberta.CA (Bill Armstrong) writes:
- >|> The trouble is that all you need to define any function is a
- >|> (complete) table of input-output pairs. If you allow this table as
- >|> (part of) the "instruction set", all of computability theory becomes
- >|> uninteresting. So yes, normally algorithms have to be finite.
- >
- > Why would computability theory become uninteresting?
-
- Because any machine may be made arbitrarily powerful (that is, of less or equal
- power to the original machine) by passing it more or less complex input (instr-
- uctions). In evaluating the machine as an interpreter of the instructions,
- you are measuring the complexity of the input which the modified machine will
- recognize, and not the complexity of the machine itself. The arguments will
- direct the interpreting machine to behave like a different machine. Therefore
- stack machines may become finite-state machines, and turing machines may become
- _any_ kind of machine.
-
- This is not so different from the undecidability problem. If you pass arguments
- into a machine which direct the machine to act like a different machine, not
- only can you not decide whether the original machine terminates for all input,
- but the new machine no longer behaves with the same level of complexity or
- power that it did before the instructions were introduced.
-
- That begs a question. Are all universal turing machines of indeterminate
- power? Or must they be considered turing machines because they _may_ be
- configured as turing machines, even though they also may not be so configured?
- If they _are_ still turing machines, then once again, why does computibility
- theory become uninteresting once a turing machine is passed instructions?
-
- (Do I detect a bit of circular reasoning here...)
-
-
- Randy
- crawford@mitre.org
- --
-
- | Randy Crawford crawford@mitre.org The MITRE Corporation
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