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

On 5/6/25 3:15 PM, olcott wrote:

On 5/6/2025 2:01 PM, Fred. Zwarts wrote:

Op 06.mei.2025 om 19:49 schreef olcott:

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.
>

>
Which makes it a fundamentally different HHH.

 None-the-less it is the words that the best selling
author of theory of computation textbooks agreed to:
*would never stop running unless aborted*
 

And it DOES stop running without being aborted.

is the hypothetical HHH/DD pair where the same HHH
that DD calls does not abort the simulation of its input.
 

Which is isn't except in your stupid mind. The truth is you don't get to change the input, which to be a valid input, contains the code for the original HHH which does abort, and thus will halt when correctly emulated.
You just don't understand what you are talking about, and seem to be too stupid to learn it.