From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: *** X-Spam-Status: No, score=3.8 required=5.0 tests=BAYES_00,INVALID_MSGID, RATWARE_MS_HASH,RATWARE_OUTLOOK_NONAME autolearn=no autolearn_force=no version=3.4.4 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: fc89c,97188312486d4578 X-Google-Attributes: gidfc89c,public X-Google-Thread: 109fba,baaf5f793d03d420 X-Google-Attributes: gid109fba,public X-Google-Thread: 1014db,6154de2e240de72a X-Google-Attributes: gid1014db,public X-Google-Thread: 103376,97188312486d4578 X-Google-Attributes: gid103376,public From: "Tim Behrendsen" Subject: Re: What's the best language to start with? [was: Re: Should I learn C or Pascal?] Date: 1996/09/10 Message-ID: <01bb9f26$36c870e0$87ee6fce@timpent.a-sis.com>#1/1 X-Deja-AN: 179720863 references: <01bb8df1$2e19d420$87ee6fce@timpent.airshields.com> <4vcac4$gm6@zeus.orl.mmc.com> <01bb8f19$9a89d820$32ee6fce@timhome2> <841797763snz@genesis.demon.co.uk> <322f864d.42836625@news.demon.co.uk> <01bb9bf9$61e9e0e0$87ee6fce@timpent.airshields.com> <50sj6q$aci@mtinsc01-mgt.ops.worldnet.att.net> <01bb9d25$9cb3cb00$32ee6fcf@timhome2> <50v6k3$soo@mtinsc01-mgt.ops.worldnet.att.net> <01bb9ded$cd0fdf00$32ee6fcf@timhome2> <5136on$7qj@goanna.cs.rmit.edu.au> content-type: text/plain; charset=ISO-8859-1 organization: A-SIS mime-version: 1.0 newsgroups: comp.lang.c,comp.lang.c++,comp.unix.programmer,comp.lang.ada Date: 1996-09-10T00:00:00+00:00 List-Id: Richard A. O'Keefe wrote in article <5136on$7qj@goanna.cs.rmit.edu.au>... > "Tim Behrendsen" writes: > > >"If this process turned out to be non-procedural..." <--- this is > >impossible according to known laws of physics. > > I honestly don't see this. I don't personally know all the known laws > of physics; my Msc was in underswater acoustics which isn't exactly > TOE-of-the-month even if the refraction equation _is_ formally identical > to the one-dimensional Schroedinger equation. I have read a number of > papers on quantum computing, and think I understand the idea. Now > quantum computers cannot compute anything that a "procedural" computer > cannot compute, but they _can_ compute things asymptotically faster than > any possible 'procedural' computer (precisely because they are not discrete > one-step-at-a-time finite state beasts). > > So if Tim Behrendsen knows which presently known laws of physics make > quantum computers impossible, he should publish in Nature. I didn't say anything about "quantum computers", I said "non-procedural". See below. > >Going back to the SQL example, SQL is an expression of the algorithm, > >but it is not possible to "directly execute" SQL; it has to be > >translated into a procedural algorithm, and this is the same > >with all "non-procedural expression" languages. > > Tell me, do you regard optical computing as procedural? > If you don't (and since it is non-discrete, with no "time axis" that > is useful in understanding how it works), then SQL _can_ be translated > into a non-procedural but executable form. Unless I'm mistaken, you seem to regard "procedural" as meaning a discrete sequence of steps. My definition is a bit more broad; I see procedural as anything with a procedure; that is, an analog integrator is a procedural mechanism, even though there is no sequence of individual steps. It is similiar to the difference between summation and integration; one consists of individual sums, the other of an infinite number of sums. However, both are fundamentally adding. When I see "Optical Computing", I normally think of gates that use photons rather than electrons, but I'm guessing you are referring to what I know as a "data flow" computer, where the data is encoded in a stream and "flows" through various data transformation mechanisms. However, this still requires time. You make a claim above that "...and since it is non-discrete, with no 'time axis'...". Perhaps you could explain why non-discrete means it does not require a time axis, and give an example of *any* algorithm that does not require a time axis, i.e., does not require time. > If you _do_ regard optical computing as procedural, then you have > stretched the term to the point where you are no longer saying anything. Procedural means "has a procedure." Nothing in the real world is not procedural, but we can *express* algorithms non-procedurally. This is where the term is useful. The term is meaningless when it comes to implementation. -- Tim Behrendsen (tim@a-sis.com)