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- From: choy@skorpio.usask.ca (I am a terminator.)
- Subject: Re: measuring complexity (was Re: Kolmogorov Complexity)
- Message-ID: <1992Aug12.173546.3298@access.usask.ca>
- Sender: choy@skorpio (I am a terminator.)
- Nntp-Posting-Host: skorpio.usask.ca
- Organization: University of Saskatchewan, Saskatoon, Canada
- References: <1992Aug10.104156.25017@julian.uwo.ca> <arms.713457373@spedden> <1992Aug10.163725.26986@julian.uwo.ca> <1992Aug10.115352.1@acad3.alaska.edu> <1992Aug11.124142@sees.bangor.ac.uk> <arms.713549466@spedden>
- Date: Wed, 12 Aug 1992 17:35:46 GMT
- Lines: 28
-
- In article <arms.713549466@spedden>, arms@cs.UAlberta.CA (Bill Armstrong) writes:
- |> mather@sees.bangor.ac.uk (Paul Mather) writes:
- |>
- |> ... lots of stuff I agree with ...
- |>
- |> >(*) PS (for the comp.theory people): Does an algorithm have to be finite?
- |> >Consider an infinite set of random instructions spat out by some other algorithm
- |> >running continuously on a Turing machine, say. Would the infinite stream of
- |> >executable instructions be classed as an algorithm?
- |>
- |> 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?
-
- |> That having been said, my gut feel is that mathematicians must have
- |> looked at various transfinite numbers of "instructions" to get
- |> functions on even larger spaces.
- |> --
- |> ***************************************************
- |> Prof. William W. Armstrong, Computing Science Dept.
- |> University of Alberta; Edmonton, Alberta, Canada T6G 2H1
- |> arms@cs.ualberta.ca Tel(403)492 2374 FAX 492 1071
-
- Henry Choy
- choy@cs.usask.ca
-