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On 6/9/2025 5:31 AM, Fred. Zwarts wrote:But the abort is coded in the input. Dreaming of a HHH that does not abort is not a valid argument, and certainly not showing technical competence.Op 09.jun.2025 om 06:15 schreef olcott:This is merely a lack of sufficient technical competenceOn 6/8/2025 10:42 PM, dbush wrote:>On 6/8/2025 11:39 PM, olcott wrote:>On 6/8/2025 10:32 PM, dbush wrote:>On 6/8/2025 11:16 PM, olcott wrote:>On 6/8/2025 10:08 PM, dbush wrote:>On 6/8/2025 10:50 PM, olcott wrote:>void DDD()>
{
HHH(DDD);
return;
}
>
The *input* to simulating termination analyzer HHH(DDD)
No it's not, as halt deciders / termination analyzers work with algorithms,
That is stupidly counter-factual.
>
That you think that shows that
My understanding is deeper than yours.
No decider ever takes any algorithm as its input.
But they take a description/specification of an algorithm,
There you go.
>which is what is meant in this context.>
It turns out that this detail makes a big difference.
>And because your HHH does not work with the description/ specification of an algorithm, by your own admission, you're not working on the halting problem.>
>
HHH(DDD) takes a finite string of x86 instructions
that specify that HHH simulates itself simulating DDD.
And HHH fails to see the specification of the x86 instructions. It aborts before it can see how the program ends.
>
on your part. It is a verified fact that unless the outer
HHH aborts its simulation of DDD that DDD simulated by HHH
the directly executed DDD() and the directly executed HHH()
would never stop running. That you cannot directly see this
is merely your own lack of sufficient technical competence.
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