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Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: olcott <polcott2@gmail.com> Newsgroups: comp.theory,sci.logic Subject: Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement-- Date: Thu, 7 Mar 2024 23:38:48 -0600 Organization: A noiseless patient Spider Lines: 210 Message-ID: <use899$1hhbj$1@dont-email.me> References: <us8shn$7g2d$1@dont-email.me> <us92f0$uvql$4@i2pn2.org> <us931e$8gmr$1@dont-email.me> <usa4rk$10ek4$3@i2pn2.org> <usa5to$gp0j$1@dont-email.me> <usa8lp$10ek5$5@i2pn2.org> <usa9o9$ho7b$1@dont-email.me> <usag21$118jg$1@i2pn2.org> <usanbu$klu7$1@dont-email.me> <usas0v$11q96$2@i2pn2.org> <usavq1$m7mn$1@dont-email.me> <usb01q$m897$1@dont-email.me> <usb0q0$m7mn$5@dont-email.me> <usb8d4$nksq$1@dont-email.me> <usb9e9$nkt8$4@dont-email.me> <usck1s$13k1e$2@dont-email.me> <uscs49$15f45$1@dont-email.me> <usdq1r$1be15$3@dont-email.me> <usdrjq$1bkg1$2@dont-email.me> <usdteu$15q44$1@i2pn2.org> <use0nb$1ga79$1@dont-email.me> <use249$15q44$6@i2pn2.org> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Fri, 8 Mar 2024 05:38:49 -0000 (UTC) Injection-Info: dont-email.me; posting-host="cbe692f823dc8310f00dd0aaf1f84978"; logging-data="1623411"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19RLcxqarM70Tf6bfyoxSRf" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:W/Vd5aJr5I8aPgdNP9q5MhuLo1Q= Content-Language: en-US In-Reply-To: <use249$15q44$6@i2pn2.org> Bytes: 9493 On 3/7/2024 9:53 PM, Richard Damon wrote: > On 3/7/24 7:29 PM, olcott wrote: >> On 3/7/2024 8:34 PM, Richard Damon wrote: >>> On 3/7/24 6:02 PM, olcott wrote: >>>> On 3/7/2024 7:35 PM, immibis wrote: >>>>> On 7/03/24 18:05, olcott wrote: >>>>>> On 3/7/2024 8:47 AM, immibis wrote: >>>>>>> On 7/03/24 03:40, olcott wrote: >>>>>>>> On 3/6/2024 8:22 PM, immibis wrote: >>>>>>>>> On 7/03/24 01:12, olcott wrote: >>>>>>>>>> On 3/6/2024 5:59 PM, immibis wrote: >>>>>>>>>>> On 7/03/24 00:55, olcott wrote: >>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn >>>>>>>>>>>> Correctly reports that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation. >>>>>>>>>>>> >>>>>>>>>>>> H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy >>>>>>>>>>>> Correctly reports that H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation. >>>>>>>>>>> >>>>>>>>>>> What are the exact steps which the exact same program with >>>>>>>>>>> the exact same input uses to get two different results? >>>>>>>>>>> I saw x86utm. In x86utm there is a mistake because Ĥ.H is not >>>>>>>>>>> defined to do exactly the same steps as H, which means you >>>>>>>>>>> failed to do the Linz procedure. >>>>>>>>>> >>>>>>>>>> Both H(D,D) and H1(D,D) answer the exact same question: >>>>>>>>>> Can I continue to simulate my input without ever aborting it? >>>>>>>>>> >>>>>>>>> >>>>>>>>> Both H(D,D) and H1(D,D) are computer programs (or Turing >>>>>>>>> machines). They execute instructions (or transitions) in >>>>>>>>> sequence, determined by their programming and their input. >>>>>>>> >>>>>>>> Yet because they both know their own machine address >>>>>>>> they can both correctly determine whether or not they >>>>>>>> themselves are called in recursive simulation. >>>>>>> >>>>>>> They cannot do anything except for exactly what they are >>>>>>> programmed to do. >>>>>> >>>>>> H1(D,D) and H(D,D) are programmed to do this. >>>>>> Because H1(D,D) simulates D(D) that calls H(D,D) that >>>>>> aborts its simulation of D(D). H1 can see that its >>>>>> own simulated D(D) returns from its call to H(D,D). >>>>>> >>>>>>>> >>>>>>>> An Olcott machine can perform an equivalent operation. >>>>>>>> >>>>>>>> Because Olcott machines are essentially nothing more than >>>>>>>> conventional UTM's combined with Conventional Turing machine >>>>>>>> descriptions their essence is already fully understood. >>>>>>>> >>>>>>>> The input to Olcott machines can simply be the conventional >>>>>>>> space delimited Turing Machine input followed by four spaces. >>>>>>>> >>>>>>>> This is followed by the machine description of the machine >>>>>>>> that the UTM is simulating followed by four more spaces. >>>>>>> >>>>>>> To make the Linz proof work properly with Olcott machines, Ĥ >>>>>>> should search for 4 spaces, delete its own machine description, >>>>>>> and then insert the description of the original H. Then the Linz >>>>>>> proof works for Olcott machines. >>>>>> >>>>>> That someone can intentionally break an otherwise correct >>>>>> halt decider >>>>> >>>>> It always gives exactly the same answer as the working one, so how >>>>> is it possibly broken? >>>>> >>>> >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt >>>> >>>> When this is executed in an Olcott machine then >>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> is a different computation than H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> >>> >>> WHY? >>> >>> The Master UTM will not change the usage at H^.H, because that is >>> just an internal state of the Machine H^ >>> >> >> The ONLY thing that the master UTM does differently is append >> the TMD to the TMD's own tape. > > Right, so it gives H and H^ that input. > > H^ can erase that input and replace it. > >> >>> At that point we have the IDENTICAL set of transitions (with just an >>> equivalence mapping of state numbers) as H will have, and the EXACT >>> same input as H >> >> it is stipulated by the definition of Olcott machines >> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> // last element is <Ĥ> (not H) > > Nope. > > H^.H isn't a "Olcott Machine" it is a sub-machine of H^ You already know that there is no such thing as sub-machines of Turing machines. Turing machines have a set of states and a tape. They have no sub-machines. > > Your Master UTM can't touch the tape here or it totally breaks the system. > The master UTM is merely an ordinary UTM that always appends the slaves TMD to the slaves tape. That is all that there is to the master UTM. > It can only add the description at the beginning, as it doesn't know > what state in that machine are "sub-machines" and what are just parts of > the machine running. > An ordinary UTM writes the input to the slaves tape. The master UTM of Olcott machines simply adds one more step to this. The master UTM is merely an ordinary UTM that always appends the slaves TMD to the slaves tape. That is all that there is to the master UTM. > You don't seem to understand much about Turing machines. > > I seem to remember you bailed out after like the first part of the first > lesson. > You can't point out an errors because there are no errors. You keep saying things at least slightly incorrectly like sub-machines of Turing machines, there is no such thing. >> H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> // last element is <H> (not Ĥ) >> That the last element of the input to Ĥ.H and H <is> different. >> > > Until the code in H^ between q0 and H^.H changes it. I have no idea what you are saying yet the only category that it can be is that Ĥ tries to thwart its own internal termination analyzer. >>>> >>>> No matter how Ĥ ⟨Ĥ⟩ screws itself up this can have no >>>> effect on H ⟨Ĥ⟩ ⟨Ĥ⟩. >>>> >>>> The Olcott machine H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> can see that it does not >>>> call itself in recursive simulation. >>> >>> As will the machine at H^.H, since the description it thinks it is >>> doesn't match the machine that gets called. >> >> A simply string comparison of the finite strings >> ⟨Ĥ⟩ and <Ĥ> proves that they are the same. >> >>>> >>>> The Olcott machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> can see that it calls >>>> itself in recursive simulation unless it discards this >>>> ability. In either case it cannot fool H, it either halts >>>> or fails to halt. >>>> >>> >>> But that isn't what H^.H gets called with. >> >> Olcott machines always append the TMD to the end of the tape >> of this TMD. ========== REMAINDER OF ARTICLE TRUNCATED ==========