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Path: ...!weretis.net!feeder9.news.weretis.net!news.nk.ca!rocksolid2!i2pn2.org!.POSTED!not-for-mail From: Richard Damon <richard@damon-family.org> Newsgroups: comp.theory Subject: Re: Philosophy of Computation: Three seem to agree how emulating termination analyzers are supposed to work Date: Sun, 10 Nov 2024 16:02:06 -0500 Organization: i2pn2 (i2pn.org) Message-ID: <114d7d0cb5266295ec2c9e9097158d78e5f51dea@i2pn2.org> References: <vgr1gs$hc36$1@dont-email.me> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 7bit Injection-Date: Sun, 10 Nov 2024 21:02:06 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="1871941"; mail-complaints-to="usenet@i2pn2.org"; posting-account="diqKR1lalukngNWEqoq9/uFtbkm5U+w3w6FQ0yesrXg"; User-Agent: Mozilla Thunderbird Content-Language: en-US In-Reply-To: <vgr1gs$hc36$1@dont-email.me> X-Spam-Checker-Version: SpamAssassin 4.0.0 Bytes: 4282 Lines: 90 On 11/10/24 2:28 PM, olcott wrote: > *The best selling author of theory of computation textbooks* > <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022> > If simulating halt decider H correctly simulates its input D > until H correctly determines that its simulated D would never > stop running unless aborted then Right, if the correct (and thus complete) emulation of this precise input would not halt. > > H can abort its simulation of D and correctly report that D > specifies a non-halting sequence of configurations. > </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022> Which your H doesn't do. > > Correct simulation is defined as D is emulated by H according to > the semantics of the x86 language thus includes H emulating itself > emulating D. And also means that it can not be aborted, as "stopping" in the middle is not to the semantics of the x86 language. An thus, your H fails to determine that the CORRECT emulation by H will not terminate, since it doesn't do one. > > I made D simpler so that the key essence of recursive simulation > could be analyzed separately. ChatGPT totally understood this. Nope, your broke the rules of the field, and thus invalidates your proof. Either by passing the address of DDD to HHH implies passing the FULL MEMORY that DDD is in (or at least every part accessed in the emulation of DDD) and thus changed in your > > void DDD() > { > HHH(DDD); > return; > } > > ChatGPT > Simplified Analogy: > Think of HHH as a "watchdog" that steps in during real execution > to stop DDD() from running forever. But when HHH simulates DDD(), > it's analyzing an "idealized" version of DDD() where nothing stops the > recursion. In the simulation, DDD() is seen as endlessly recursive, so > HHH concludes that it would not halt without external intervention. But DDD doesn't call an "ideaized" verision of HHH, it calls the exact function defined as HHH, s0 your arguemet is based on false premises, and thus is just a :OE/ > > https://chatgpt.com/share/67158ec6-3398-8011-98d1-41198baa29f2 > This link is live so you can try to convince ChatGPT that its wrong. > > On 11/3/2024 12:20 PM, Richard Damon wrote: > > On 11/3/24 9:39 AM, olcott wrote: > >> > >> The finite string input to HHH specifies that HHH > >> MUST EMULATE ITSELF emulating DDD. > > > > Right, and it must CORRECTLY determine what an unbounded > > emulation of that input would do, even if its own programming > > only lets it emulate a part of that. > > > > *Breaking that down into its key element* > > [This bounded HHH] must CORRECTLY determine what > > an unbounded emulation of that input would do... > > When that input is unbounded that means it is never > aborted at any level, otherwise it is bounded at some > level thus not unbounded. > No, because there aren't "levels" of emulation under consideration here. Only does the emulation that the top level HHH is doing, since everything else is just fixed by the problem. I guess you are just doubling down on your committement to prove that you are nothing but an pathetic pathological lying idiot that doesn't understand a thing that he is talking about.