Re: The input to HHH(DDD) specifies a non-halting sequence of configurations +++

On 6/15/2025 3:47 AM, Fred. Zwarts wrote:

Op 14.jun.2025 om 16:17 schreef olcott:

On 6/13/2025 6:28 AM, Mikko wrote:

On 2025-06-11 14:11:32 +0000, olcott said:
>

On 6/11/2025 3:29 AM, Mikko wrote:

On 2025-06-10 16:10:49 +0000, olcott said:
>

On 6/10/2025 7:01 AM, Mikko wrote:

On 2025-06-09 14:46:30 +0000, olcott said:
>

On 6/9/2025 6:24 AM, Richard Damon wrote:

On 6/8/25 10:50 PM, olcott wrote:

void DDD()
{
   HHH(DDD);
   return;
}
>
The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*
>
*Every rebuttal to this changes the words*
>

>
>
So, you think a partial simulation defines behavior?
>
Where do you get that LIE from?
>

>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
>
I am no so stupid that I require a complete
simulation of a non-terminating input.

>
Yes you are. You just express your stupidity in another way.
>

>
It only takes two simulations of DDD by HHH for HHH
to correctly recognize a non-halting behavior pattern.

>
Either the pattern or the recognition is incorrect.

>
DDD correctly simulated by HHH cannot possibly reach its
own "return" statement final halt state. This by itself
*is* complete proof that the input to HHH(DDD) specifies
non-halting behavior.

>
No, it is not. The words "cannot possibly" are not sufficiently
meaningful to prove anything. HHH does what it does and does
not what it does not. But what it can or cannot do, possiby or
otherwise?
>

>
It is required that one have the technical competence of
a first year CS student that knows C to understand that
it is self-evident that the input to HHH(DDD) specifies
behavior such that DDD correctly simulated by HHH cannot
possibly reach its simulated "return" statement.

 Indeed, even a beginner will see that HHH fails to reach the end of the simulation of a halting program.

<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>
The criteria agreed to by the best selling author of theory
of computation textbooks disagrees.

>
It is also required that one know that in computer science
halting means reaching a final halt state.

 But preventing a program to halt (e.g. by turning off a computer, or aborting a simulation) does not make a program non-halting.
 

>
If you have less technical competence than this then the
problem is your lack of technical competence.
>

 So, why do you continue if you do not even understand this?

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer