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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Wed, 4 Sep 2024 11:34:37 +0200
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Op 03.sep.2024 om 20:40 schreef olcott:
> On 9/3/2024 9:42 AM, joes wrote:
>> Am Mon, 02 Sep 2024 16:06:24 -0500 schrieb olcott:
>>> On 9/2/2024 12:52 PM, Fred. Zwarts wrote:
>>>> Op 02.sep.2024 om 18:38 schreef olcott:
>>>>> A halt decider is a Turing machine that computes the mapping from its
>>>>> finite string input to the behavior that this finite string specifies.
>>>>> If the finite string machine string machine description specifies that
>>>>> it cannot possibly reach its own final halt state then this machine
>>>>> description specifies non-halting behavior.
>> Which DDD does not.
> 
> DDD emulated by HHH cannot possibly reach
> its final halt state no matter what HHH does.
> 
>>>>> A halt decider never ever computes the mapping for the computation
>>>>> that itself is contained within.
>> Then it is not total.
> 
> Yes it is you are wrong.
> 
>>>>> Unless there is a pathological relationship between the halt decider H
>>>>> and its input D the direct execution of this input D will always have
>>>>> identical behavior to D correctly simulated by simulating halt decider
>>>>> H.
>> Which makes this pathological input a counterexample.
> 
> Which makes the pathological input a counter-example
> to the false assumption that the direct execution of
> a machine always has the same behavior as the machine
> simulated by its pathological simulator.
> 
>>>>> A correct emulation of DDD by HHH only requires that HHH emulate the
>>>>> instructions of DDD** including when DDD calls HHH in recursive
>>>>> emulation such that HHH emulates itself emulating DDD.
>>>> Indeed, it should simulate *itself* and not a hypothetical other HHH
>>>> with different behaviour.
>>> It is emulating the exact same freaking machine code that the x86utm
>>> operating system is emulating.
> 
>> It is not simulating the abort because of a static variable. Why?
>>
> 
> void DDD()
> {
>    HHH(DDD);
>    OutputString("This code is unreachable by DDD emulated by HHH");
> }
> 
>>>> If HHH includes code to see a 'special condition' and aborts and halts,
>>>> then it should also simulate the HHH that includes this same code and
>>> DDD has itself and the emulated HHH stuck in recursive emulation.
> 
>> Your HHH incorrectly changes behaviour.
>>
> 
> No you are wrong !!!
> 

Yes it does. HHH simulates only a few recursions, then it sees a 
'special condition', stops the simulation, returns to DDD and DDD halts.

There is no unreachable code, because there are only a few recursions, 
just as in:

void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
   printf ("Olcott thinks this is never printed.\n");
}

This is proved by the direct execution and by the correct simulation by 
a world class simulator and even by HHH1.
But HHH fails to reach this aborting code in the simulation.


-- 
Paradoxes in the relation between Creator and creature.
<http://www.wirholt.nl/English>.