Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.3 4.3bsd-beta 6/6/85; site harvard.UUCP Path: utzoo!watmath!clyde!burl!ulysses!bellcore!decvax!genrad!panda!talcott!harvard!draves From: draves@harvard.UUCP (Richard Draves) Newsgroups: net.ai,net.philosophy Subject: Re: A halting problem Message-ID: <633@harvard.UUCP> Date: Tue, 21-Jan-86 05:32:04 EST Article-I.D.: harvard.633 Posted: Tue Jan 21 05:32:04 1986 Date-Received: Thu, 23-Jan-86 20:54:31 EST References: <2175@aecom.UUCP> <14551@rochester.UUCP> <3978@kestrel.ARPA> <438@faron.UUCP> <4101@kestrel.ARPA> Reply-To: draves@harvard.UUCP (Richard draves) Distribution: net Organization: Aiken Comp Lab, Harvard Lines: 32 Xref: watmath net.ai:3202 net.philosophy:3892 Say that a Turing Machine is "in an infinite loop" if its computation repeats a configuration (and hence doesn't halt). If a TM M is guaranteed to either halt or go into an infinite loop on string w, then it is possible to decide which happens (i.e., we don't care what our decider does if the above condition isn't met). Simply have the deciding machine simulate machine M's computation on w, recording all configurations and checking for a repetition. Either M will halt, or a repetition will be found. Therefore the decider will be able to halt with a yes or no answer. On the other hand, given a TM M and a string w, it is not possible to decide yes if M goes into an infinite loop on w and no otherwise. If this were possible, we could decide the halting problem. We want to decide, given a machine M and string w, if M halts on w. Modify M to get M', a machine which computes as M does with the exception that where M halts, M' goes into a loop (10: goto 10). Now decide if M' has an infinite loop on w. If it doesn't, we know M doesn't halt on w so we say no. If it does, we know that M either halts or goes into an infinite loop some other way. However, from above we can decide which is the case. Therefore we can decide whether M halts on w, given the ability to decide if a machine goes into an infinite loop on a string. We have a contradiction, so the problem of deciding if a machine goes into an infinite loop must be undecidable. Rich -- "a picture in the head is a gory murder in an art gallery" -- Stephen Kosslyn