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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: HHH maps its input to the behavior specified by it --- never
 reaches its halt state
Date: Fri, 9 Aug 2024 10:04:36 -0500
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On 8/9/2024 1:46 AM, Fred. Zwarts wrote:
> Op 08.aug.2024 om 21:18 schreef olcott:
>> On 8/8/2024 2:04 PM, Fred. Zwarts wrote:
>>> Op 08.aug.2024 om 20:48 schreef olcott:
>>>> On 8/8/2024 1:44 PM, Fred. Zwarts wrote:
>>>>> Op 08.aug.2024 om 15:15 schreef olcott:
>>>>>> On 8/8/2024 3:24 AM, Fred. Zwarts wrote:
>>>>>>> Op 07.aug.2024 om 15:01 schreef olcott:
>>>>>>>> On 8/7/2024 3:16 AM, Fred. Zwarts wrote:
>>>>>>>>> Op 04.aug.2024 om 15:11 schreef olcott:
>>>>>>>>>> On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
>>>>>>>>>>> Op 03.aug.2024 om 17:20 schreef olcott:>>
>>>>>>>>>>>> When you try to show how DDD simulated by HHH does
>>>>>>>>>>>> reach its "return" instruction you must necessarily
>>>>>>>>>>>> must fail unless you cheat by disagreeing with the
>>>>>>>>>>>> semantics of C. That you fail to have a sufficient
>>>>>>>>>>>> understanding of the semantics of C is less than no
>>>>>>>>>>>> rebuttal what-so-ever.
>>>>>>>>>>>
>>>>>>>>>>> Fortunately that is not what I try, because I understand that 
>>>>>>>>>>> HHH cannot possibly simulate itself correctly.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> void DDD()
>>>>>>>>>> {
>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>    return;
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> In other words when HHH simulates itself simulating DDD it
>>>>>>>>>> is supposed to do something other than simulating itself
>>>>>>>>>> simulating DDD ???  Do you expect it to make a cup of coffee?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Is English too difficult for you. I said HHH cannot do it 
>>>>>>>>> correctly.
>>>>>>>>
>>>>>>>> *According to an incorrect criteria of correct*
>>>>>>>> You keep trying to get away with disagreeing with
>>>>>>>> the semantics of the x86 language. *That is not allowed*
>>>>>>>>
>>>>>>> Again accusations without evidence.
>>>>>>> We proved that HHH deviated from the semantics of the x86 
>>>>>>> language by skipping the last few instructions of a halting program.
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>>    return;
>>>>>> }
>>>>>>
>>>>>> Each HHH of every HHH that can possibly exist definitely
>>>>>> *emulates zero to infinity instructions correctly* In
>>>>>> none of these cases does the emulated DDD ever reach
>>>>>> its "return" instruction halt state.
>>>>>>
>>>>>> *There are no double-talk weasel words around this*
>>>>>> *There are no double-talk weasel words around this*
>>>>>> *There are no double-talk weasel words around this*
>>>>>
>>>>> Indeed. And this correctly proves that the simulation failed, not 
>>>>> because of an instruction simulated incorrectly, but because 
>>>>> instructions are skipped. 
>>>>
>>>> void Infinite_Recursion()
>>>> {
>>>>    Infinite_Recursion();
>>>>    return;
>>>> }
>>>
>>> Dreaming again of an infinite recursion?
>>>
>>>>
>>>> The return instruction in both cases is unreachable code.
>>>> DDD correctly emulated by HHH and Infinite_Recursion
>>>> correctly emulated by HHH cannot reach the "return"
>>>> instruction.
>>>
>>> It cannot reach it, because it was programmed to abort one cycle 
>>> before the program would end.
>>>
>>
>> *Maybe you have ADD like Richard has. I already said this above*
>> When zero to infinity steps of DDD are correctly emulated by
>> HHH no DDD ever reaches its own "return" instruction.
>>
>> Maybe the issue is that you don't know programming well enough
>> to understand that this is true.
> 
> Maybe you should try to learn English. I confirmed hat HHH cannot reach 
> the end of the simulation of itself.

Yes and cups of coffee are made from ground coffee beans.
Changing the subject is merely the dishonest dodge of the
strawman deception.

> Maybe you should learn to program. When the simulation of a halting 
> program is unable to reach the end, it proves that the simulation is 
> incorrect.


void Infinite_Loop()
{
   HERE: goto HERE;
}

The correct simulation of the above never halts.

> Everybody with sufficient programming knowledge understands that a 
> simulator cannot possibly simulate itself correctly up to the end, 
> because either it does not halt, or it misses the last cycle, the final 
> part of the simulation.

When a simulating termination analyzer is essentially called
in infinite recursion it is smart enough to abort.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer