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

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>
</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.