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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Fri, 9 May 2025 09:48:49 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 08 May 2025 22:34:35 -0500 schrieb olcott:
> On 5/8/2025 10:14 PM, Mike Terry wrote:
>> On 09/05/2025 03:13, olcott wrote:
>>> On 5/8/2025 8:30 PM, Keith Thompson wrote:
>>>> olcott <polcott333@gmail.com> writes:
>>>>> On 5/8/2025 6:49 PM, Keith Thompson wrote:
>>>>>> olcott <polcott333@gmail.com> writes:

>> His simulation is in fact a single-stepped x86 instruction simulation,
>> where the stepping of each x86 instruction is under the HHH's control.
>> HHH can continue stepping the simulation until its target returns, in
>> which case the situation is logically just like direct call, as you
>> have described.  Or HHH could step just 3 x86 instructions (say) and
>> then decide to return (aka "abort" its simulation).  Let's call that /
>> partial/ simulation in contrast with /full/ simulation which you've
>> been supposing.
> A full simulation of infinite recursion?
> I am only doing one tiny idea at a time here.
Yeah, so not a full simulation.

>>>> 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.
>>> Yes you totally have this correctly.
>>> None of the dozens of comp.theory people could ever achieve that level
>>> of understanding even after three years. That is why I needed to post
>>> on comp.lang.c.
>> Everybody on comp.theory understands this much.
> No one here ever agreed that when 1 or more instructions of DDD are
> correctly simulated by HHH that DDD cannot possibly reach its own
> "return" instruction.
That's wrong as written. HHH cannot simulate DDD returning in a
finite number of instructions, it takes infinitely many.

>>>> This conclusion relies on my understanding of what you've said about
>>>> your code, which I consider to be unreliable.
>> 
>> Hmm, did PO make it clear that when he says
>>     "..DDD correctly simulated by HHH cannot
>>      possibly REACH its own "return" instruction."
>> he is not talking about whether "DDD halts"?  [I.e. halts when run
>> directly from main() outside of a simulator.]  No, what he is talking
>> about is whether the /step-by-step partial simuation/ of DDD performed
>> by HHH proceeds as far as DDD returning.
> 
> When 1 or more steps of DDD are correctly simulated by HHH the simulated
> DDD cannot possibly reach its "return" instruction (final halt state).
> No one here has agreed to that. Not in several years of coaxing and
> elaboration.
It's true for a finite number. Aborting is not correct simulation, even
if HHH did return that DDD halts.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.