Re: DD simulated by HHH cannot possibly halt (Halting Problem)

On 05/04/2025 19:11, olcott wrote:

On 4/5/2025 11:25 AM, dbush wrote:

On 4/5/2025 11:59 AM, olcott wrote:

On 4/5/2025 2:42 AM, Richard Heathfield wrote:

On 05/04/2025 07:14, olcott wrote:

On 4/4/2025 10:49 PM, Richard Heathfield wrote:

On 05/04/2025 00:41, olcott wrote:

*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination. The
only rebuttal to this is rejecting the notion that
deciders must always halt.

>

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>

In other words, you operate on the principle that deciders don't have to (and indeed can't) always make a correct decision on whether an input program halts.
>

>
The termination analyzer HHH would be correct
to determine that it must stop simulating DD to
prevent its own non-termination

>
Fine, but then it fails to do its job. What you are learning (albeit slowly) is that the termination analyser HHH can't analyse whether DD terminates. It is therefore not a general purpose termination analyser.
>

>
Introduction to the Theory of Computation 3rd Edition
by Michael Sipser (Author) (best selling textbook)
>
<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>

>
But not what you think he agreed to:
>
>

 You have to show that by showing the details of how
what he agreed to is not accurately paraphrased by
*Simulating termination analyzer Principle*

No, you have to show firstly that your H determines anything at all about D's behaviour. Then you have to show that its determination is correct. And then you have to show that it can do this for any input.
Alternatively, you could stop arguing with all these naysayers and simply submit a paper to the Journal of the ACM.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
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