Sujet : Re: Every HHH(DDD) is correct to reject its input
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 13. Jun 2025, 17:45:39
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <102hkjj$3i2uo$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
User-Agent : Mozilla Thunderbird
On 6/13/2025 5:14 AM, Mikko wrote:
On 2025-06-11 13:59:00 +0000, olcott said:
On 6/11/2025 8:04 AM, Fred. Zwarts wrote:
Op 10.jun.2025 om 18:49 schreef olcott:
On 6/10/2025 2:07 AM, Mikko wrote:
On 2025-06-08 06:04:45 +0000, olcott said:
>
On 6/8/2025 12:54 AM, Mikko wrote:
On 2025-06-07 13:53:53 +0000, olcott said:
>
On 6/7/2025 3:19 AM, Mikko wrote:
On 2025-06-06 17:15:10 +0000, olcott said:
>
On 6/6/2025 2:58 AM, Mikko wrote:
On 2025-06-05 16:01:46 +0000, olcott said:
>
On 6/5/2025 2:42 AM, Mikko wrote:
On 2025-06-04 15:00:07 +0000, olcott said:
>
On 6/4/2025 2:39 AM, Mikko wrote:
On 2025-06-02 05:12:26 +0000, olcott said:
>
On 6/1/2025 6:20 AM, Mikko wrote:
On 2025-05-31 19:21:10 +0000, olcott said:
>
On 5/31/2025 2:11 PM, Mr Flibble wrote:
Olcott is doing this:
>
int main()
{
DDD(); // DDD calls HHH
}
>
This is incorrect as it is a category (type) error in the form of
conflation of the EXECUTION of DDD with the SIMULATION of DDD: to
completely and correctly simulate/analyse DDD there must be no execution
of DDD prior to the simulation of DDD.
>
Olcott should be doing this:
>
int main()
{
HHH(DDD);
}
>
I would have left it there except that many dozens of
reviewers have pointed out that they believe that HHH
is supposed to report on the behavior of its caller.
>
A halt decider is required to report on the computation it is asked
about. There is no requirement that a halt decider knows or can find
out whether it is called by the program about which is required to
report. Consequently, whether the computaton asked about calls the
decider is irrelevant.
>
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*
>
If it does then the "input" is not DDD, which specifies a halting
behaviour if HHH is a decider.
>
You can say these things only by making
sure to ignore the verified facts.
>
We can ignore irrelevant facts. But if you ignore relevant requirements
you can't prove that your soliution is correct.
>
As long as DDD emulated by HHH cannot possibly reach
its own "return" instruction final halt state then
DDD is non halting even if it is never simulated.
>
That is not what "non-halting" means. Anything said about "DDD emulated
by HHH" is irrelevant. Wikipedia says: "In computability theory, the
halting problem is the problem of determining, from a description of an
arbitrary computer program and an input, whether the program will finish
running, or continue to run forever." Your HHH(DDD) does not do that.
>
<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
>
That is not a definition of the meaning of halting. That is a diagnostic
>
It is one definition of non-halting in that
it logically entails never reaching the
simulated "return" statement final halt state.
>
It is an incorrect definition,
>
It *is* the industry standard definition.
There is no indusstry standard definition. Industry uses the common
language meaning that "non-halting" is what does not halt.
If you run an infinite loop on your computer
and then smash the computer with a sledge hammer
to make the infinite loop stop running,
THIS DOES NOT COUNT AS HALTING.
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Every HHH(DDD) is correct to reject its input
Re: Every HHH(DDD) is correct to reject its input