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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 09:46:05 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 28 Mar 2025 13:56:31 -0500 schrieb olcott:
> On 3/27/2025 5:01 PM, joes wrote:
>> Am Thu, 27 Mar 2025 12:50:12 -0500 schrieb olcott:
>>> On 3/27/2025 2:18 AM, Fred. Zwarts wrote:
>>>> Op 27.mrt.2025 om 04:09 schreef olcott:
>>>>> On 3/26/2025 8:22 PM, Richard Damon wrote:
>> 
>>>>>> Non-Halting is that the machine won't reach its final staste even
>>>>>> if an unbounded number of steps are emulated. Since HHH doesn't do
>>>>>> that, it isn't showing non-halting.
>>>>> DDD emulated by any HHH will never reach its final state in an
>>>>> unbounded number of steps.
>>>>> DDD emulated by HHH1 reaches its final state in a finite number of
>>>>> steps.
>>>> It is not very interesting to know whether a simulator reports that
>>>> it is unable to reach the end of the simulation of a program that
>>>> halts in direct execution.
>>> That IS NOT what HHH is reporting.
>> That is exactly what it does, and you have said so before(tm).
> You are saying that HHH is reporting that HHH is screwing up THAT IS
> FALSE. HHH IS REPORTING THAT DDD IS SCREWING UP.
How so? It mostly calls HHH.

>>> HHH correctly rejects DDD because DDD correctly emulated by HHH cannot
>>> possibly reach its own final halt state.
>> DDD doesn't *do* anything, it is being simulated. HHH can't reach DDD's
>> existing halt state.
> DDD specifies a recursive emulation relationship with HHH
No, only when HHH simulates it.

>>>> It is interesting to know:
>>>> 'Is there an algorithm that can determine for all possible inputs
>>>> whether the input specifies a program that [...]
>>>> halts when directly executed?'
>>>> This question seems undecidable for Olcott.
>>> It is the halts while directly executed that is impossible for all
>>> inputs. No TM can ever report on the behavior of the direct execution
>>> of any other TM.
>> The direct execution of a TM is obviously computable from its
>> description.
It seems you disagreed.

>>> A TM can only report on the behavior that the machine code of another
>>> TM specifies. When it specifies a pathological relationship then the
>>> behavior caused by the pathological relationship MUST BE REPORTED.
>> No, the machine code doesn't "specify a pathological relationship",
>> that is purely a feature of trying to simulate it with the included
>> simulator.
> The classic HP counter-example input HAS ALWAYS SPECIFIED A PATHOLOGICAL
> RELATIONSHIP TO ITS DECIDER.
Only when simulated.

> The question has always been what Boolean value can H correctly return
> when D is able to do the opposite of whatever value that H returns?
None.

> When we prove that it is impossible for D to do the opposite of whatever
> value that H returns the original question becomes moot.
How is it impossible? H by definition returns a boolean, and D halts or
enters an infinite loop based on that. Nothing magical.

-- 
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.