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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: The actual truth is that ... Turing computability issues have
 been addressed --- marathon winner
Date: Wed, 16 Oct 2024 14:50:38 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Wed, 16 Oct 2024 08:43:42 -0500 schrieb olcott:
> On 10/16/2024 8:10 AM, joes wrote:
>> Am Wed, 16 Oct 2024 07:52:00 -0500 schrieb olcott:
>>> On 10/16/2024 1:32 AM, joes wrote:
>>>> Am Tue, 15 Oct 2024 22:52:00 -0500 schrieb olcott:
>>>>> On 10/15/2024 9:11 PM, Richard Damon wrote:
>>>>>> On 10/15/24 8:39 AM, olcott wrote:
>>>>>>> On 10/15/2024 4:58 AM, joes wrote:
>>>>>>>> Am Mon, 14 Oct 2024 20:12:37 -0500 schrieb olcott:
>>>>>>>>> On 10/14/2024 6:50 PM, Richard Damon wrote:
>>>>>>>>>> On 10/14/24 12:05 PM, olcott wrote:
>>>>>>>>>>> On 10/14/2024 6:21 AM, Richard Damon wrote:
>>>>>>>>>>>> On 10/14/24 5:53 AM, olcott wrote:
>>>>>>>>>>>>> On 10/14/2024 3:21 AM, Mikko wrote:
>>>>>>>>>>>>>> On 2024-10-13 12:49:01 +0000, Richard Damon said:
>>>>>>>>>>>>>>> On 10/12/24 8:11 PM, olcott wrote:
>>>>>>>>
>>>>>>>>>>> Trying to change to a different analytical framework than the
>>>>>>>>>>> one that I am stipulating is the strawman deception.
>>>>>>>>>>> *Essentially an intentional fallacy of equivocation error*
>>>>>>>>>> But, you claim to be working on that Halting Problem,
>>>>>>>>> I quit claiming this many messages ago and you didn't bother to
>>>>>>>>> notice.
>>>>>>>> Can you please give the date and time? Did you also explicitly
>>>>>>>> disclaim it or just silently leave it out?
>>>>>>> Even people of low intelligence that are not trying to be as
>>>>>>> disagreeable as possible would be able to notice that a specified
>>>>>>> C function is not a Turing machine.
>>>>>> But it needs to be computationally equivalent to one to ask about
>>>>>> Termination.
>>>>> Not at all. A termination analyzer need not be a Turing computable
>>>>> function.
>>>> It definitely does. An uncomputable analyser is useless.
>>> It is true that a termination analyzer is not required to work
>>> correctly for all inputs.
>> But it should work for DDD.

>>> That there is one way that HHH can consistently catch the
>>> non-terminating pattern of its input proves that this can be done.
>> DDD does terminate. Otherwise it would contradict the HP.
>> 
>>> Mike suggested some different ways that would seem to be Turing
>>> computable yet too convoluted to be time consuming for me to implement
>>> in practice.
>> What are those ways?

>>> The basic approach involves the idea that every state change of the
>>> emulations of emulations is data that belongs to the outermost
>>> directly executed HHH.
>> Nothing special. They aren't even running on the hardware proper.
>> 
>>> It is too convoluted for me to provide a way for HHH to look inside
>>> all of the emulations of emulations and pull out the data that it
>>> needs, so knowing that this is possible is enough to know that it is
>>> Turing computable.
>> It cannot dig into an infinite simulation chain.
> It need not dig into an infinite emulation chain. It merely needs to see
> that DDD calls HHH(DDD) twice in sequence having no termination
> condition within DDD.
If it sees no termination condition in itself (it's a recursion), being
the same program, it also doesn't have one, so cannot abort.

>>> When HHH is an x86 emulation based termination analyzer then each DDD
>>> *correctly_emulated_by* any HHH that it calls never returns.
>>> Each of the directly executed HHH emulator/analyzers that returns 0
>>> correctly reports the above *non_terminating _behavior* of its input.
>> If HHH doesn't return, DDD doesn't either, not even the directly
>> executed one.
> The emulated HHH is merely data to the executed termination analyzer.

>>>>> When HHH is an x86 emulation based termination analyzer then each
>>>>> DDD *correctly_emulated_by* any HHH that it calls never returns.
>>>> Only because the nested HHH doesn't abort.
>>> Every nested HHH has seen one less execution trace than the next outer
>>> one. The outermost one aborts its emulation as soon as it has seen
>>> enough. Thus each inner HHH cannot possibly abort its own emulation.
>> Each inner HHH must abort if the outer does,
> In the same way that each person in a marathon that are running at the
> same speed and ten feet behind the person in front of them must win the
> race.
> The first person does not win the marathon they merely tie with everyone
> else.
> The fact that they start out ahead and remain ahead in the whole
> marathon has no effect on whether they reach the finish line first.
> 
>> since they are the same program. Of course, the outer doesn't simulate
>> the inner abort, because it has already aborted. Therefore the outer
>> HHH doesn't need to abort, because the inner HHHs already halt by
>> themselves. Reverse the if(Root) check on line 500 or what in Halt7.c.
>> As long as at least one of the infinitely many HHHs aborts, the whole
>> chain terminates (but the nested HHHs don't get simulated completely).
>> But they are all the same program,
>> so all of them must have the abort.
-- 
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.