Re: Halting Problem: What Constitutes Pathological Input

On 5/7/2025 10:08 PM, Richard Damon wrote:

On 5/7/25 9:51 AM, olcott wrote:

On 5/7/2025 4:48 AM, Mikko wrote:

On 2025-05-06 18:40:16 +0000, olcott said:
>

On 5/6/2025 10:53 AM, joes wrote:

Am Tue, 06 May 2025 10:29:59 -0500 schrieb olcott:

On 5/6/2025 4:35 AM, Mikko wrote:

On 2025-05-05 17:37:20 +0000, olcott said:

>

The above example is category error because it asks HHH(DD) to report
on the direct execution of DD() and the input to HHH specifies a
different sequence of steps.

>
No, it does not. The input is DD specifides exactly the same sequence
of steps as DD. HHH just answers about a different sequence of steps
instead of the the seqeunce specified by its input.

As agreed to below:
>

<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
>
*input D* is the actual input *would never stop running unless aborted*
is the hypothetical H/D pair where H does not abort.

>

H should simulate its actual input D that calls the aborting H, not a
hypothetical version of D that calls a pure simulator.
>

>
*would never stop running unless aborted*
refers to the same HHH that DD calls yet
this hypothetical HHH does not abort.
>

You cannot possibly show the exact execution trace where DD is correctly
emulated by HHH and this emulated DD reaches past its own machine
address [0000213c].

>

Duh, no simulator can simulate itself correctly. But HHH1 can simulate
DD/HHH.
>

>
HHH does simulate itself correctly yet must create

>
No, it cannot simulate itself to the point where it returns.
>

>
HHH(DD) does return.

 And thus so does the HHH(DD) that DD calls.
 

In other words HHH caught DD trying to cause itself to halt.

>
<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
>
HHH only need simulate DD until it correctly determines
*simulated D would never stop running unless aborted*
This refers to a hypothetical HHH/DD pair where HHH never aborts.
>

Right, that a correct simulation of D would never stop running,
 

No because that would make the termination analyzer
fail to terminate and that is not allowed dumbell bee.

That includes using the fact that H does what it does, and the input is what it is.
 And thus NOT the hypothetical HHH/DD pair, but the hypothetical HHH looking at the actual DD which still calls the origianl HHH
 

It is an inherent aspect of the nature of simulating
termination analyzer is that they must always correctly
predict what the behavior of their input would be
IF THEY NEVER ABORTED. If they don't always do this
they will sometimes get stuck in non-termination.

You just don't understand what (correct) simulation means in the field, or what a program actually is.
 

Everyone here seems to think they they get to stipulate
what correct it, rather what it really is.

And, your whole system has an improper intertwining of the TWO programs that are mentioned (H and D).
 

They are essentially a pair of C functions that
would be in infinite recursion with each other
except that one of them has the power to stop it.

Please point out anywhere in the proof where he implies that H or D are NOT actual complete programs, it is sort of a requirement as Turing Machies always are complete, that is part of their beauty, you can't make a non-computation/program Turing Machine.
 Sorry, you really are showing that you are THAT Stupid.

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