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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Functions computed by Turing Machines MUST apply finite string
 transformations to inputs --- MT
Date: Tue, 6 May 2025 12:49:13 -0500
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On 5/6/2025 6:23 AM, Richard Damon wrote:
> On 5/5/25 10:36 PM, olcott wrote:
>> On 5/5/2025 8:23 PM, Richard Damon wrote:
>>> On 5/4/25 9:23 PM, olcott wrote:
>>>> On 5/4/2025 8:04 PM, Ben Bacarisse wrote:
>>>>> Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:
>>>>> ...
>>>>>> As explained above, UTM(⟨Ĥ⟩ ⟨Ĥ⟩) simulates Ĥ run with input Ĥ 
>>>>>> (having the
>>>>>> same halting behaviour) and Ĥ run with input Ĥ HALTS.  So 
>>>>>> embedded_H does
>>>>>> not "gather enough information to deduce that UTM(⟨Ĥ⟩ ⟨Ĥ⟩) would 
>>>>>> never
>>>>>> halt".  THAT IS JUST A FANTASY THAT YOU HAVE.
>>>>>>
>>>>>> UTM(⟨Ĥ⟩ ⟨Ĥ⟩) DOES halt, so embedded_H can't possibly gather 
>>>>>> information
>>>>>> that genuinely implies it DOESN'T halt.  The explanation is obvious:
>>>>>> embedded_H gathers information that *YOU* believe implies that 
>>>>>> UTM(⟨Ĥ⟩ ⟨Ĥ⟩)
>>>>>> would never halt, but *YOU ARE SIMPLY WRONG*.
>>>>>
>>>>> He used to claim that false ("does not halt") was the correct answer,
>>>>> /even though/ the computation in question halts!  Those were simpler
>>>>> days.  Of course cranks will never admit to having been wrong about
>>>>> anything other than a detail or two, so anyone who could be bothered
>>>>> could try to get him to retract that old claim.
>>>>>
>>>>
>>>> <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>
>>>>
>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> In other words embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct to
>>>> reject its input if
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> Would not halt.
>>>
>>> Nope, because that isn't the input that it was given. 
>>
>> *Wrong*
>> <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
>>
>> *simulated D would never stop running unless aborted*
>> simulated D (the actual input)
>> never stop running unless aborted (hypothetical H/D pair)
>>
> 
> No, that is changing the input.
> 

*would never stop running unless aborted*
means the hypothetical same HHH that DD calls
except that this HHH does not abort.

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