Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: olcott Newsgroups: comp.theory Subject: Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD) Date: Fri, 9 May 2025 23:44:48 -0500 Organization: A noiseless patient Spider Lines: 160 Message-ID: References: <87msbmeo3b.fsf@nosuchdomain.example.com> <87ecwyekg2.fsf@nosuchdomain.example.com> <87bjs2cyj6.fsf@nosuchdomain.example.com> <87r00xchn5.fsf@nosuchdomain.example.com> <23a27379d226b7b3b9f8c303a492f66edc9019ff.camel@gmail.com> <1020d30c2c5b5a7cce584777131d5ce414b480ea.camel@gmail.com> <0323d5ca6d757a1e35d7e4cf5eb4fc8f41bc866a.camel@gmail.com> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sat, 10 May 2025 06:44:49 +0200 (CEST) Injection-Info: dont-email.me; posting-host="7be348abb5bc2ec0a70724586a3ca680"; logging-data="3528092"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1+ZjWqg8fTEC71JDPcaklsh" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:hh3EDjUs+XM5X1E51NKIPPjKTvQ= X-Antivirus: Norton (VPS 250510-0, 5/9/2025), Outbound message In-Reply-To: <0323d5ca6d757a1e35d7e4cf5eb4fc8f41bc866a.camel@gmail.com> X-Antivirus-Status: Clean Content-Language: en-US Bytes: 8884 On 5/9/2025 11:32 PM, wij wrote: > On Fri, 2025-05-09 at 23:18 -0500, olcott wrote: >> On 5/9/2025 10:43 PM, wij wrote: >>> On Fri, 2025-05-09 at 22:24 -0500, olcott wrote: >>>> On 5/9/2025 10:13 PM, wij wrote: >>>>> On Fri, 2025-05-09 at 19:40 -0700, Keith Thompson wrote: >>>>>> olcott writes: >>>>>>> On 5/9/2025 4:40 PM, Richard Heathfield wrote: >>>>>>>> On 09/05/2025 21:15, olcott wrote: >>>>>>>>> On 5/9/2025 3:07 PM, Richard Heathfield wrote: >>>>>>>>>> On 09/05/2025 20:46, olcott wrote: >>>>>>>>>>> We have not begun to get into any of those points. >>>>>>>>>>> We are only asking can DDD correctly simulated >>>>>>>>>>> by any HHH that can exist ever reach its own >>>>>>>>>>> "return" instruction. >>>>>>>>>> >>>>>>>>>> DDD can't be correctly simulated by itself (which is effectively >>>>>>>>>> what you're trying to do when you fire up the simulation from >>>>>>>>>> inside DDD). >>>>>>>>> >>>>>>>>> How the Hell did you twist my words to say that? >>>>>>>> I haven't touched your words. What I have done is to observe that >>>>>>>> DDD's /only/ action is to call a simulator. Since DDD isn't itself a >>>>>>>> simulator, there is nothing to simulate except a call to a >>>>>>>> simulator. >>>>>>>> It's recursion without a base case - a rookie error. >>>>>>>> HHH cannot successfully complete its task, because it never regains >>>>>>>> control after the first recursion. To return, it must abort the >>>>>>>> simulation, which means the simulation fails. >>>>>>>> >>>>>>>>> void DDD() >>>>>>>>> { >>>>>>>>>      HHH(DDD); >>>>>>>>>      return; >>>>>>>>> } >>>>>>>>> >>>>>>>>> When 1 or more statements of DDD are correctly >>>>>>>>> simulated by HHH then this correctly simulated >>>>>>>>> DDD cannot possibly reach its own “return statement”. >>>>>>>> On what grounds can you persuade an extraordinarily sceptical >>>>>>>> readership that HHH 'correctly simulated' DDD? >>>>>>> >>>>>>> Any competent C programmer can see that >>>>>>> the call from DDD to HHH(DDD) (its own simulator) >>>>>>> is equivalent to infinite recursion. >>>>>>> >>>>>>> On 5/8/2025 8:30 PM, Keith Thompson wrote: >>>>>>>> Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it >>>>>>>> does nothing else, your code would be equivalent to this: >>>>>>>> >>>>>>>>        void DDD(void) { >>>>>>>>            DDD(); >>>>>>>>            return; >>>>>>>>        } >>>>>>>> >>>>>>>> Then the return statement (which is unnecessary anyway) will never be >>>>>>>> reached.  In practice, the program will likely crash due to a stack >>>>>>>> overflow, unless the compiler implements tail-call optimization, in >>>>>>>> which case the program might just run forever -- which also means the >>>>>>>> unnecessary return statement will never be reached. >>>>>> >>>>>> I had not intended to post again, but I feel the need to make >>>>>> a clarification. >>>>>> >>>>>> I acknowledged that the return statement would never be reached >>>>>> *given the assumption* that HHH correctly simulates DDD.  Given >>>>>> that assumption, a call to DDD() should be equivalent to a call >>>>>> to HHH(DDD). >>>>>> >>>>>> I did not address whether the assumption is valid.  I merely >>>>>> temporarily accepted it for the sake of discussion, just as I would >>>>>> accept that if I were ten feet tall I would bump my head against >>>>>> the ceiling in my house. >>>>>> >>>>>> The discussion I had with olcott did not reach the point of >>>>>> discussing *how* HHH could correctly simulate DDD, or whether it >>>>>> would even be logically possible for it to do so.  I also did not >>>>>> address any issues of partial simulation, where olcott claims that >>>>>> HHH can "accurately simulate" only a few x86 instructions rather >>>>>> than simulating its entire execution.  I did not participate in >>>>>> any discussion that would require knowledge of x86 machine or >>>>>> assembly code.  (I have no doubt that I could learn x86 machine >>>>>> and assembly code reasonably well if motivated to do so, but I am >>>>>> not so motivated.) >>>>>> >>>>>> What I acknowledged was barely more than "if HHH correctly simulates >>>>>> DDD, then HHH correctly simulates DDD".  (My understanding from >>>>>> posts by others, whom I presume to be sufficiently knowledgeable, >>>>>> is that HHH logically cannot accurately simulate DDD.)  I would >>>>>> prefer that olcott refrain from using my words to support any of >>>>>> his arguments beyond the scope of what he and I directly discussed. >>>>> >>>>> Don't know why you people stick on the 'simulation' stuff so long. >>>>> The HP simply asks for such an H (in function form. POOH does not >>>>> resemble TM): >>>>>    H(D)=1 if D() halt. >>>>>    H(D)=0 if D() not halt. >>>> >>>> My invention of a simulating termination >>>> analyzer shows exactly how to compute the >>>> mapping that the input that HHH(DD) specifies >>>> into a correct answer for the halting problem's >>>> otherwise impossible input. >>>> >>>> All rebuttals are based on failing to compute >>>> this mapping correctly. >>>> >>> >>> What is the correct mapping? >>> >> >> _DDD() >> [00002172] 55         push ebp      ; housekeeping >> [00002173] 8bec       mov ebp,esp   ; housekeeping >> [00002175] 6872210000 push 00002172 ; push DDD >> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD) >> [0000217f] 83c404     add esp,+04 >> [00002182] 5d         pop ebp >> [00002183] c3         ret >> Size in bytes:(0018) [00002183] >> >> Computing the mapping of DDD emulated by HHH >> according the the rules of the x86 language >> to its behavior by HHH actually emulating DDD. > > The above says you have no idea what the mapping is. > >>> If POOH are not talking about the mapping: >>> >>>    H(D)=1 if D() halt. >>>    H(D)=0 if D() not halt. >>> >> >> The way that simulating termination analyzers process >> their input by showing all of the steps of how the mapping >> must be computed refutes the above simplistic view. > > No (real) problem with that. But the HP asks: > > H(D)=1 if D() halt. > H(D)=0 if D() not halt. > > You still evade the question: Is POO H anything to do with the HP? > I have recently proven that the above requirements are incorrect. That people consistently ignore this proof with pure bluster is not actually any rebuttal at all. >>> POOH is likely nothing to do with HP >>> >> >> > > -- Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius hits a target no one else can see." Arthur Schopenhauer