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Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: How Partial Simulations correctly determine non-halting ---Ben's
 10/2022 analysis
Date: Sat, 8 Jun 2024 09:11:17 -0500
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On 6/8/2024 8:56 AM, Fred. Zwarts wrote:
> Op 08.jun.2024 om 15:32 schreef olcott:
>> On 6/8/2024 8:20 AM, Fred. Zwarts wrote:
>>> Op 08.jun.2024 om 15:04 schreef olcott:
>>>> On 6/8/2024 1:42 AM, Mikko wrote:
>>>>> On 2024-06-07 22:26:05 +0000, olcott said:
>>>>>
>>>>>> On 6/7/2024 4:00 PM, joes wrote:
>>>>>>> Am Fri, 07 Jun 2024 09:47:35 -0500 schrieb olcott:
>>>>>>>> On 6/7/2024 1:30 AM, Mikko wrote:
>>>>>>>>> On 2024-06-06 15:31:36 +0000, olcott said:
>>>>>>>>>> On 6/6/2024 10:14 AM, Mikko wrote:
>>>>>>>>>>> On 2024-06-06 13:53:58 +0000, olcott said:
>>>>>>>>>>>> On 6/6/2024 5:15 AM, Mikko wrote:
>>>>>>>>>>>>> On 2024-06-05 13:29:28 +0000, olcott said:
>>>>>>>>>>>>>> On 6/5/2024 2:37 AM, Mikko wrote:
>>>>>>>>>>>>>>> On 2024-06-04 18:02:03 +0000, olcott said:
>>>>>>>
>>>>>>>> A simulating halt decider cannot report on what the behavior of a
>>>>>>>> non-terminating input actually is because this would take forever.
>>>>>>> Exactly. Didn't you say it is allowed to abort?
>>>>>>>
>>>>>>>> H is not allowed to report on any computation containing its 
>>>>>>>> actual self
>>>>>>>> because Turing machines can only take finite string inputs thus 
>>>>>>>> cannot
>>>>>>>> take Turing machines as inputs.
>>>>>>> Bullshit. It can take other machines just fine. It doesn't know 
>>>>>>> about
>>>>>>> itself.
>>>>>>>
>>>>>>
>>>>>> No actual Turing machine can be the input to any other actual
>>>>>> Turing machine. Turing machines only take finite string inputs.
>>>>>
>>>>> Any finite string can be an input to some Turing machine.
>>>>> Can you prove that a Turing machine is not a finite string?
>>>>>
>>>>
>>>> By definition Turing Machines are not finite strings in the
>>>> conventional model. In my x86utm model of computation x86
>>>> machine language <is> the input to another function written
>>>> in the x86 language.
>>>>
>>>>> Your own attempts of a conter-proof are not about Turing machines
>>>>> but C programs. C programs are finite strings, so a C program
>>>>> is a valid input to a C program (and a Turing machine, too).
>>>>>
>>>>
>>>> They are about Turing Machines yet cannot be sufficiently understood
>>>> with less than the 100% compete precision of the x86 language. They
>>>> x86utm model is required to prove that false assumptions about the
>>>> nature of correct simulation are false assumptions.
>>>>
>>>> void DDD(int (*x)())
>>>> {
>>>>    HH(x, x);
>>>> }
>>>>
>>>> DDD correctly simulated by any HH cannot possibly stop running
>>>> without being aborted. When this is understood and accepted then
>>>>
>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by any embedded_H is understood
>>>> to not possibly stop running until aborted.
>>>>
>>>> I have all of the details of the machine code and C code for HH and 
>>>> DDD.
>>>> I can't to the same thing for embedded_H and ⟨Ĥ⟩ ⟨Ĥ⟩ so we have to 
>>>> learn
>>>> by analogy.
>>>>
>>>>>>>> This means that H is not allowed to report on the behavior of the
>>>>>>>> directly executed P(P).
>>>>>>> So on which program does it report then?
>>>>>
>>>>
>>>> DD(DD) is just like infinite recursion that gets terminated at
>>>> its second recursive call. DD(DD) halts only because HH(DD,DD)
>>>> correctly determines that its input DOES NOT HALT >
>>>> If HH(DD,DD) did not correctly determine that its input
>>>> DOES NOT HALT then DD(DD) would never halt.
>>>>
>>>>
>>>
>>> Indeed, *only if*. if HH does not halt, then DD does not halt. But 
>>> your claim is that HH halts, because of that DD(DD) *DOES* halt. You 
>>> cannot base a conclusion on something that does not happen.
>>
>> You are twisting the meaning of my words.
> 
> No, your are twisting your own words.
> 
>> *Verified facts* (some may be beyond your technical competence)
> 
> Up to now, your have not been able to tell how they were verified. It 
> seems beyond your competence to see the difference between verified 
> facts and wishes. That is twisting the meaning of words.
> 
>> (1) The input to HH DOES NOT HALT.
> 
> Only if HH (as part of the input) does not halt.
> You even proved that this conclusion is a false negative on 05.jun.2024 
> at 15:59 (CET)
> 
>>
>> (2) HH correctly recognizes that its input DOES NOT HALT.
> 
> Only if HH would not abort, but it does abort. 

<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 words10/13/2022>

I stop at the first big mistake so that we can focus on correcting
this big mistake.

When
   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.



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