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From: bruck@actcom.co.il (Uri Bruck)
Subject: Re: Is Turing Wrong about the Limits of Computation?
Organization: ACTCOM - Internet Services in Israel
Date: Wed, 22 Feb 1995 22:47:49 GMT
Message-ID: <D4FBBq.5yL@actcom.co.il>
References: <3h53g7$e18@highway.LeidenUniv.nl> <Pine.SUN.3.91.950207154346.8460A-100000@sun1> <3hcpei$85k@highway.LeidenUniv.nl> <D41xMB.496@gremlin.nrtc.northrop.com>
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Martin Cohen (mcohen@charming.nrtc.northrop.com) wrote:
: In article <3hcpei$85k@highway.LeidenUniv.nl> vosse@ruls41.LeidenUniv.nl (Theo Vosse) writes:
: >Eugen Leitl (ui22204@sunmail.lrz-muenchen.de) wrote:
: ............
: >If the
: >TM has got Q states, you get a total of at least QM(M+1)/2 possible
: >configurations that the second TM has to remember in order to
: >detect the cycle. But then that TM would have to have a tape
: >larger than that of the original TM (i.e., it has got more memory),
: >so you actually have a TM being controlled by one that is more
: >powerful! This is not very acceptable (or would you get yourself
: >a Cray to check the parity on your old XT's memory banks?).
: >...................

: There is an ingenious method of checking to see if a system is
: in a loop (not original by me by any means):

: Run two copies of the system, with the second running twice
: as fast as the first. At each step of the first step,
: check to see if the states of the two copies are the same.
: If so, the machines are in a loop, and the period of the
: loop divides the number of steps taken by the slower machine.

: Proof: After n steps, machine 1 is in state M(n) and machine
: 2 is in state M(2n). If M(n)=M(2n) then we will have
: M(2n)=M(3n)=...=M(kn) for all integral k>=1. If p is the
: actual smallest period of M, then, if n=ap+b, 0<=b<p,
: M(n)=M(n+ap)=M(2n)=M(n+ap+b), a contradiction unless b=0,
: so p divides n evenly.

: We thus trade time (have to run up to twice as long, but
: usually not too much longer) and resources (have to run
: two copies) for lots of space (to store all the states)
: and lots of time (have to compare the current state
: with all the previous states).

: In general, this technique is a big win (IMHO).


: --
: Marty Cohen (mcohen@nrtc.northrop.com) - Not the guy in Philly
:   This is my opinion and is probably not Northrop Grumman's!
:           Use this material of your own free will
The technizue maybe ingenious, it has however a slight flaw, it doesn't work.
using the simple represntation * 1
				** 2
				*** 3
				**** 4  etc.
try the basic Turing adding machine, add ****** and ********
TAM-1 and TAM-2 will be in the same state many times, still this is not a 
a loop, they just happen to repeat the same state until the input changes.

	Uri Bruck 
