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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Who here understands that the last paragraph is Necessarily true?
Date: Sat, 20 Jul 2024 10:57:53 +0200
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Op 19.jul.2024 om 16:54 schreef olcott:
> On 7/19/2024 1:35 AM, Fred. Zwarts wrote:
>> Op 18.jul.2024 om 17:37 schreef olcott:
>>> On 7/18/2024 10:27 AM, joes wrote:
>>>> Am Thu, 18 Jul 2024 09:14:32 -0500 schrieb olcott:
>>>>> On 7/18/2024 3:25 AM, joes wrote:
>>>>>> Am Wed, 17 Jul 2024 15:36:24 -0500 schrieb olcott:
>>>>>>> On 7/17/2024 3:30 PM, joes wrote:
>>>>>>>> Am Wed, 17 Jul 2024 12:20:43 -0500 schrieb olcott:
>>>>>>>>> On 7/17/2024 12:16 PM, joes wrote:
>>>>>>>>>> Am Wed, 17 Jul 2024 08:27:08 -0500 schrieb olcott:
>>>>>>>>>>> On 7/17/2024 2:43 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-07-16 18:24:49 +0000, olcott said:
>>>>>>>>>>>>> On 7/16/2024 3:12 AM, Mikko wrote:
>>>>>>>>>>>>>> On 2024-07-15 02:33:28 +0000, olcott said:
>>>>>>>>>>>>>>> On 7/14/2024 9:04 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 7/14/24 9:27 PM, olcott wrote:
>>>>>>>>>>
>>>>>>>>>>>>>> You have already said that a decider is not allowed to answer
>>>>>>>>>>>>>> anything other than its input. Now you say that the the 
>>>>>>>>>>>>>> program
>>>>>>>>>>>>>> at 15c3 is not a part of the input. Therefore a decider is 
>>>>>>>>>>>>>> not
>>>>>>>>>>>>>> allowed consider it even to the extent to decide whether it
>>>>>>>>>>>>>> ever returns. But without that knowledge it is not 
>>>>>>>>>>>>>> possible to
>>>>>>>>>>>>>> determine whether DDD halts.
>>>>>>>>>>>>> It maps the finite string 558bec6863210000e853f4ffff83c4045dc3
>>>>>>>>>>>>> to non-halting behavior because this finite string calls
>>>>>>>>>>>>> HHH(DDD) in recursive simulation.
>>>>>>>>>> That string is meaningless outside of the execution 
>>>>>>>>>> environment of
>>>>>>>>>> HHH,
>>>>>>>>>> specifically the simulation of DDD it is doing. It does not 
>>>>>>>>>> encode
>>>>>>>>>> anything, DDD does not have access to that address. That string
>>>>>>>>>> doesn't call anything, the program in HHH's memory space does.
>>>>>>>>>> Ceterum censeo that HHH halts.
>>>>>>>>>>>> That mapping is not a part of the finite string and not a 
>>>>>>>>>>>> part of
>>>>>>>>>>>> the problem specification.
>>>>>>>>>>> decider/input pairs <are> a key element of the specification.
>>>>>>>>>>
>>>>>>>>>>>> The finite string does not reveal what is the effect of calling
>>>>>>>>>>>> whatever that address happens to contain.
>>>>>>>>>>> A simulating termination analyzer proves this.
>>>>>>>>>>>
>>>>>>>>>>>> The behaviour of HHH is specified outside of the input. 
>>>>>>>>>>>> Therefore
>>>>>>>>>>>> your "decider" decides about a non-input, which you said is not
>>>>>>>>>>>> allowed.
>>>>>>>>>>> HHH is not allowed to report on the behavior of it actual 
>>>>>>>>>>> self in
>>>>>>>>>>> its own directly executed process. HHH is allowed to report 
>>>>>>>>>>> on the
>>>>>>>>>>> effect of the behavior of the simulation of itself simulating 
>>>>>>>>>>> DDD.
>>>>>>>>>> HHH must report on itself if its input calls it.
>>>>>>>>>> HHH does not directly simulate itself, it just executes.
>>>>>>>>>> It reports on DDD by simulating it.
>>>>>>>>> Its input cannot call its actual self that exists in an entirely
>>>>>>>>> different process.
>>>>>>>> Of course it doesn't make sense to return to a higher stack frame.
>>>>>>>> And of course a function can recursively call itself.
>>>>>>> A separate process is like a different program on a different
>>>>>>> computer.
>>>>>> It makes no sense to call a running program. DDD creates a new 
>>>>>> instance
>>>>>> of the same code with its own memory and code pointer.
>>>>> It is not that it makes no sense it is that it is impossible.
>>>
>>>> I mean, why are you talking about that?
>>>
>>> All of the halting problem proofs are incorrectly anchored
>>> in the behavior of the direct execution of the input thus
>>> not the behavior that this input specifies to a decider that
>>> this input invokes.
>>
>> Exactly the same input is presented to the direct execution and the 
>> simulation, namely the x86 code of the program.
>> The semantics of the x86 language does not change in these two cases, 
>> so a correct simulator should interpret the x86 in the same way as the 
>> direct execution. 
> 
> When you are hungry you remain hungry until you eat.
>     Before HHH(DDD) aborts its emulation the directly
>     executed DDD() cannot possibly halt.
> 
> After you eat you are no longer hungry.
>     After HHH(DDD) aborts its emulation the directly
>     executed DDD() halts.
> 
> In other words you should remain hungry after you eat
> because you are the same person in both cases.
> 

So you think that you should keep eating, because you could dream of a 
situation where you were hungry.

You keep dreaming of an HHH that does not halt and therefore you keep 
thinking that an abort is necessary, even for a HHH that aborts and halts.

DDD is a misleading and unneeded complication. It is easy to eliminate DDD:

        int main() {
          return HHH(main);
        }

This has the same problem. This proves that the problem is not in DDD, 
but in HHH, which halts when it aborts the simulation, but it decides 
that the simulation of itself does not halt.

HHH is simply unable to decide about finite recursions.

void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}

It decides after N recursions that there is an infinite recursion, which 
is incorrect.

Olcott's HHH is programmed to abort the simulation after N cycles of 
recursive simulations. Therefore, it is incorrect to abort the 
simulation of HHH when the simulated HHH has performed only N-1 cycles, 
because that changes the behaviour of HHH.
Since the simulated HHH always runs one cycle behind the simulating HHH, 
it is clear that HHH can never simulate enough cycles for a correct 
simulation, as is required by the x86 language.
Therefore, the simulation is incorrect according to the criteria olcott 
stipulated.
The conclusion is simple:
HHH cannot possibly simulate itself correctly.

No matter how much olcott wants it to be correct, or how many times 
olcott repeats that it is correct, it does not change the fact that such 
a simulation is incorrect, because it is unable to reach the end.
Olcott's own claim that the simulated HHH does not reach its end 
confirms it. The trace he has shown also proves that HHH cannot reach 
the end of its own simulation. So, his own claims prove that it is true 
that HHH cannot possibly simulate itself up to the end, which makes the 
simulation incorrect.

Sipser would agree that this incorrect simulation cannot be used to 
detect a non-halting behaviour.

Olcott could not point to an error, but prefers to ignore it. So, I will 
repeat it, until either an error is found, or olcott admits that HHH 
cannot possibly simulate itself correctly.