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

On 5/6/25 4:38 PM, olcott wrote:

On 5/6/2025 3:25 PM, joes wrote:

Am Tue, 06 May 2025 12:49:13 -0500 schrieb 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>
</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.

Yes, that is not the same HHH.
>

 Yet is the HHH that Professor Sipser agreed would be
the correct measure of the behavior of the actual input.
 

But not the D that he was talking about, which is FIXED to always use the original decider.
Sorry, you don't get to put words in peoples mouths, and ar just showing how much your world is based on lying.