Path: ...!weretis.net!feeder9.news.weretis.net!news.nk.ca!rocksolid2!i2pn2.org!.POSTED!not-for-mail From: Richard Damon Newsgroups: comp.theory Subject: Re: The philosophy of computation reformulates existing ideas on a new basis Date: Wed, 6 Nov 2024 19:45:06 -0500 Organization: i2pn2 (i2pn.org) Message-ID: References: <086fc32f14bcc004466d3128b0fe585b27377399@i2pn2.org> <11408789ed30027f4bc9a743f353dfa9b4712109@i2pn2.org> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Thu, 7 Nov 2024 00:45:08 -0000 (UTC) Injection-Info: i2pn2.org; logging-data="1293004"; mail-complaints-to="usenet@i2pn2.org"; posting-account="diqKR1lalukngNWEqoq9/uFtbkm5U+w3w6FQ0yesrXg"; User-Agent: Mozilla Thunderbird Content-Language: en-US X-Spam-Checker-Version: SpamAssassin 4.0.0 In-Reply-To: Bytes: 7039 Lines: 138 On 11/6/24 10:26 AM, olcott wrote: > On 11/6/2024 8:39 AM, Mikko wrote: >> On 2024-11-05 13:18:43 +0000, olcott said: >> >>> On 11/5/2024 3:01 AM, Mikko wrote: >>>> On 2024-11-03 15:13:56 +0000, olcott said: >>>> >>>>> On 11/3/2024 7:04 AM, Mikko wrote: >>>>>> On 2024-11-02 12:24:29 +0000, olcott said: >>>>>> >>>>>>> HHH does compute the mapping from its input DDD >>>>>>> to the actual behavior that DDD specifies and this >>>>>>> DOES INCLUDE HHH emulating itself emulating DDD. >>>>>> >>>>>> Yes but not the particular mapping required by the halting problem. >>>>> >>>>> Yes it is the particular mapping required by the halting problem. >>>>> The exact same process occurs in the Linz proof. >>>> >>>> The halting probelm requires that every halt decider terminates. >>>> If HHH(DDD) terminates so does DDD. The halting problmen requires >>>> that if DDD terminates then HHH(DDD) accepts as halting. >>> >>> void Infinite_Loop() >>> { >>>    HERE: goto HERE; >>>    return; >>> } >>> >>> No that is false. >>> The measure is whether a C function can possibly >>> reach its "return" instruction final state. >> >> Not in the original problem but the question whether a particular >> strictly >> C function will ever reach its return instruction is equally hard. About > > It has always been about whether or not a finite string input > specifies a computation that reaches its final state. Which means the input needs to actually specify a computation, which yours doesn't, and when we add that code of HHH, we see that the program so described WILL halt as long as the HHH is like the one you present that returns an answer. > > _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] > > DDD emulated by HHH cannot possibly reach its own "return" instruction. WHich is just your equivlocation again. DDD does reach the return instruction. The emulation by HHH (that answers) doesn't yield a result that qualfies to show "can't possibly reach", as partal emulations don't give that sort of fact. > > When Ĥ is applied to ⟨Ĥ⟩ > Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ > Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn > > (a) Ĥ copies its input ⟨Ĥ⟩ > (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ > (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩ > (d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩ > (e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ > (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩ > (g) goto (d) with one more level of simulation until the embedded_H that started in (c) decides to abort its emulation, and then the whole thing stops. If it doesn't, then H can't abort its either, as they are supposed to be the same Turing Machine. > > ⟨Ĥ⟩ simulated by embedded_H cannot possibly reach either > of its states ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩ thus cannot possibly reach > its final state of ⟨Ĥ.qn⟩. Which is irrelevent, as only the emulation by a UTM matters, and embedded_H can't be a UTM if H aborts its emulation to go to qn, as embedded_H is an exact copy of that H. > > Thus when the measure of correct is the behavior that > the finite input string specifies and the measure of > halting is reaching the final state of this finite > string input then HHH and embedded_H are correct to > reject their input as non-halting. But the "behavior" that the finite input string specifies is the behavior of a UTM emulation that input, not that of a PARTIAL emulation. > > If these measures are believed to be incorrect then sound > reasoning must be applied to show this. No, the DEFINITION must be applied, something you have rejected, but still imply you follow, as you haven't put yourself outside the system, since you still talk about refuting things in the system, which can only be done in the system. So, you are just shown to be a stupid liar. > > That is just not what most everyone else believes IS NOT > sound reasoning. Nope, just shows that you don't understand the definitions, and are too stupid to understand when people correct you. > >> a C function that is not strictly conforming the question may have a >> third answer that it may or may not reach its "return" instruction >> dpending >> on the C implementation. >> >> A C function can terminate without reaching its return statement. >> The C standard specifies several other possibilities. >> >>> Your measure determines that Infinite_Loop() halts. >> >> No, it does not. You should not present claims without justification. >> To look like liar is not the best defence against being called a liar. >> > >