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From: joes <noreply@example.org>
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Subject: Re: Defining a correct simulating halt decider
Date: Tue, 3 Sep 2024 19:01:24 -0000 (UTC)
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Am Tue, 03 Sep 2024 13:40:08 -0500 schrieb 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.
But DDD halts, so it „specifies halting behaviour”.
HHH can’t simulate itself.

>>>>> 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.
How? It should work for all inputs.

>>>>> 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 counterexample to the false assumption that a decider exists.

>>>>> 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");
> }
I don’t understand what this is supposed to explain? The output is
clearly wrong, as evidenced by actually running HHH on it.

>>>> 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 !!!
Have you fixed the Root bug?

-- 
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.