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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: A state transition diagram proves ... GOOD PROGRESS -- I only
 wanted to cross post this key break through once.
Date: Fri, 18 Oct 2024 11:41:11 -0500
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On 10/18/2024 11:39 AM, olcott wrote:
> On 10/18/2024 9:41 AM, joes wrote:
>> Am Fri, 18 Oct 2024 09:10:04 -0500 schrieb olcott:
>>> On 10/18/2024 6:17 AM, Richard Damon wrote:
>>>> On 10/17/24 11:47 PM, olcott wrote:
>>>>> On 10/17/2024 10:27 PM, Richard Damon wrote:
>>>>>> On 10/17/24 9:47 PM, olcott wrote:
>>>>>>> On 10/17/2024 8:13 PM, Richard Damon wrote:
>>>>>>>> On 10/17/24 7:31 PM, olcott wrote:
>>
>>>>>>>>> When DDD is correctly emulated by HHH according to the semantics
>>>>>>>>> of the x86 language DDD cannot possibly reach its own machine
>>>>>>>>> address [00002183] no matter what HHH does.
>>>>>>>>> +-->[00002172]-->[00002173]-->[00002175]-->[0000217a]--+
>>
>>>>>>>> Except that 0000217a doesn't go to 00002172, but to 000015d2
>>
>>>> The Emulating HHH sees those addresses at its begining and then never
>>>> again.
>>>> Then the HHH that it is emulating will see those addresses, but not the
>>>> outer one that is doing that emulation of HHH.
>>>> And so on.
>>>> Which HHH do you think EVER gets back to 00002172?
>>>> What instruction do you think that it emulates that would tell it to do
>>>> so?
>>
>>>> At best the trace is:
>>>> 00002172 00002173 00002175 0000217a conditional emulation of 00002172
>>>> conditional emulation of 00002173 conditional emulation of 00002175
>>>> conditional emulation of 0000217a CE of CE of 00002172 ...
>>> OK great this is finally good progress.
>> The more interesting part is HHH simulating itself, specifically the
>> if(Root) check on line 502.
>>
> 
> That has nothing to do with any aspect of the emulation
> until HHH has correctly emulated itself emulating DDD.
> 
>>>> and if HHH decides to abort its emulation, it also should know that
>>>> every level of condition emulation it say will also do the same thing,
>>> If I understand his words correctly Mike has already disagreed with
>>> this.
>> He hasn't.
>>
>>> Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
>>> On 3/1/2024 12:41 PM, Mike Terry wrote:
>>>   > Obviously a simulator has access to the internal state (tape 
>>> contents
>>>   > etc.) of the simulated machine. No problem there.
>>> This seems to indicate that the Turing machine UTM version of HHH can
>>> somehow see each of the state transitions of the DDD resulting from
>>> emulating its own Turing machine description emulating DDD.
> 
>> Of course. It needs to, in order to simulate it. Strictly speaking
>> it has no idea of its simulation of a simulation two levels down,
>> only of the immediate simulation; the rest is just part of whatever
>> program the simulated simulator is simulating, which happens to be
>> itself.
>>
> 
>  From the concrete execution trace of DDD emulated by HHH
> according to the semantics of the x86 language people with
> sufficient technical competence can see that the halt status
> criteria that professor Sipser agreed to has been met.
> 
> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>      If simulating halt decider H correctly simulates its input D
>      until H correctly determines that its simulated D would never
>      stop running unless aborted then
> 
>      H can abort its simulation of D and correctly report that D
>      specifies a non-halting sequence of configurations.
> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
> 
> I will paraphrase this to use clearer language that directly applies
> to HHH and DDD.
> 
>      If emulating termination analyzer HHH emulates its input DDD
>      according to the semantics of the x86 language (including HHH
>      emulating itself emulating DDD) until HHH correctly determines
>      that its emulated DDD would never stop running unless aborted
>      then ...
> 
>      HHH can abort its emulation of DDD and correctly report that DDD
>      specifies a non-terminating sequence of x86 instructions.
> 
>>> *Joes can't seem to understand this*
>>> Only the outer-most HHH meets its abort criteria first, thus unless it
>>> aborts as soon as it meets this criteria none of them will ever abort.
> 
>> This is very simple to understand. Almost as simple as: even if only
>> the outermost HHH didn't abort, it would still halt, 
> 
> Yet that is based on the factually incorrect assumption
> that every instance of HHH does not use the exact same
> machine code.
> 
> Since you should know that this assumption is factually
> incorrect I could it as flat out dishonestly on your part.
> 
>> since it is
>> simulating a halting program: the nested version will abort.
>>
>>>> and thus the call HHH at 0000217a will be returned from, > and HHH has
>>>> no idea what will happen after that, so it KNOWS it is ignorant of the
>>>> answer.
> 
> 


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