Path: utzoo!utgpu!news-server.csri.toronto.edu!mailrus!cs.utexas.edu!usc!ucsd!rutgers!mcnc!rti!ntpdvp1!kenp From: kenp@ntpdvp1.UUCP (Ken Presting) Newsgroups: comp.ai Subject: Re: Searle and Radical Translation Summary: OK, I'll knock that chip off your shoulder Message-ID: <620@ntpdvp1.UUCP> Date: 9 Aug 90 00:51:59 GMT References: <612@ntpdvp1.UUCP> <5362@puggsly.cme.nist.gov> <614@ntpdvp1.UUCP> <5644@puggsly.cme.nist.gov> Organization: SNA Solutions Inc., Contract Programming Group Lines: 119 In article <5644@puggsly.cme.nist.gov>, kohout@cme.nist.gov (Robert Kohout) writes: > > To "formally" defend his assertion that "real" computers are "obviously" > more powerful that Turing Machines... Bob, you are now misquoting me. Let me remind you of my assertion: >. . . Each time a TM is given a particular input, the tape is erased >and the internal state is reset. So the TM computes a *function* - same >input => same output. That is obviously false for real computers, which >have updateable databases. . . . > >Theorems about TM's, which compute functions, do not apply to real machines, >for purely formal reasons. Real machines do not, in the technical sense, >compute functions. This assertion is by no means inconsistent with Church's Thesis, nor does it imply that real computers are more "powerful" than Turing Machines. It says simply that those real computers which make use of an updateable database are different from TM's. > } In article <614@ntpdvp1.UUCP> kenp@ntpdvp1.UUCP (Ken Presting) writes: > > } Since the TM is deterministic, the waiting state (if there is one) is a > } function of the initial state. There is no difference in computing power > } due to the new arrangement of the tape - the difference is only notational. > } The obvious point is that the TM does not, in this notation, compute 'o' > } as a function of 'i'. > } > Interesting, Ken, but what's the point? Do you intend to imply that this > relation represents something that computers "do" and Turing Machines don't? If you will recall the assertion that I set out to formalize, you will perhaps recognize that I have just made my point: when there is something on the tape besides "input", then the "output" is not a function of the "input". I hope that you can now agree that this is both obvious and trivial. What is interesting to me is the differences between the relations '|=' and '|-'. In the first place, there are non-denumerably many paths defined by |=, whereas there are only denumerably many defined by |-. Second, for any two states s, s' where s |- s', the information content (in the sense of Kolmogorov-Chaitin complexity) of s' is never greater than that of s. In other words, a TM can lose information but never gain any. This is not so under the relation |=, which holds not between TM states, but between whole *calculations*. Bob, these are formal claims, and I would be delighted to have you or anyone else correct me if they are mistaken. > Remember, non-determinism adds nothing to the computing power of a TM. If > a non-deterministic TM can compute a function, then there exists a deterministic > machine which can compute the same function. This fact is familiar to me also. The relation |= is defined over the calculations of a deterministic TM, so the D/ND issue is not directly relevant. But note that for any calculation , there are infinitely many other calculations such that |= . This is importantly different from |-, where only a finite number of configurations s' are accessible from any s. > > } 1. A computer is used repeatedly to compute different values. > } 2. The number of times a computer is used is unlimited in > } principle. > } 3. The device may modify its future operation in the course > } of performing the current task. (by updating the database) > } 4. The sequence of values input to the computer cannot be > } specified in advance. > } > None of these are beyond the capacity of a Turing Machine. It might help to avoid further confusion if you would take the trouble to explain this claim. > > } It may take a moment of thought to verify that while the set of paths > } for any TM is denumerable, the set of chains is not. > } > Do you mean to imply by this that a computer is more powerful than a TM? > Do you REALLY think you are refuting Church's Thesis? You must have been very excited to think that you found a "Pro-Searlite" (which I am not) who believes he can refute Church's Thesis (which I do not). Sorry to disappoint you, but you have only your reading skills to blame. > . . . You > certainly haven't shown anything that computers "do" that TM's don't. For > that matter, you haven't made a single assertion about this |= relation > insofar as it applies to computers and not Turing Machines. No. You have asserted without argument that my (1-4) are not "beyond the capacity of a Turing Machine." That is pure bluff. Or perhaps you are unfamiliar with the concept of "non-denumerable set". > > Why don't you just admit it Ken. What was "perfectly obvious" to you, is > just, plain wrong. Bob, it time for *you* to put up or shut up. I am anxious to discuss the formal side of the issues raised by the Chinese Room, especially in regard to radical interpretation, the representation of abstract concepts, and (most to the point) learning. I wish you would contribute to such a discussion. My impression is that you have no intention of offering us any mathematics. Your recent claim: > ... If it were possible to represent intelligence > by a purely formal definition, then of course we could program a purely > formal system to be 'intelligent' ... shows that your ability to apply the ideas of mathematics to the problems of your profession is severely restricted. Ken Presting ("Punchcards on the table")