Re: Functions computed by Turing Machines MUST apply finite string transformations to inputs --- MT

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Sujet : Re: Functions computed by Turing Machines MUST apply finite string transformations to inputs --- MT
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 05. May 2025, 02:23:05
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vv93tq$383jd$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
User-Agent : Mozilla Thunderbird
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.

I know you'll not understand what I've just said, because it is all too
abstract and you don't understand the concepts involved, and consequently
you probably don't agree with my Sipser interpretation, and even if you did
I doubt you would be able to work out its consequences.  So I don't expect
to be posting any further.
 Not you then!  I sympathise, though my reason for not talking to him is
his unacceptable insults.
 
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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