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On 6/6/2025 3:29 AM, Fred. Zwarts wrote:And if the outer one aborts, they all abort.Op 05.jun.2025 om 17:50 schreef olcott:You seem to remain unqualified to make this assessmentOn 6/5/2025 2:14 AM, Fred. Zwarts wrote:But HHH does not *detect* that. The abort is based on an *invalid assumption*, not on what HHH sees. HHH sees only the first few recursive recursions of a finite recursion and misses the fact that the next cycle if the simulated HHH will abort and halt.Op 04.jun.2025 om 18:31 schreef olcott:>On 6/4/2025 4:13 AM, Fred. Zwarts wrote:There we see that HHH aborts after a finite recursion.Op 03.jun.2025 om 21:53 schreef olcott:>On 6/3/2025 7:45 AM, dbush wrote:>On 6/2/2025 10:58 PM, Mike Terry wrote:>Even if presented with /direct observations/ contradicting his position, PO can (will) just invent new magical thinking that only he is smart enough to understand, in order to somehow justify his busted intuitions.>
My favorite is that the directly executed D(D) doesn't halt even though it looks like it does:
>
>
On 1/24/24 19:18, olcott wrote:
> The directly executed D(D) reaches a final state and exits normally.
> BECAUSE ANOTHER ASPECT OF THE SAME COMPUTATION HAS BEEN ABORTED,
> Thus meeting the correct non-halting criteria if any step of
> a computation must be aborted to prevent its infinite execution
> then this computation DOES NOT HALT (even if it looks like it does).
>
If the second call of otherwise infinite recursion had
to be aborted to prevent actual infinite recursion then
this call always was non-halting even when it was forced
to stop running.
>
But since there is no infinite recursion, no abort is needed.
*You just contradicted yourself*
>
void DDD()
{
HHH(DDD);
return;
}
>
HHH simulates DDD that calls HHH(DDD)
that simulates DDD that calls HHH(DDD)
that simulates DDD that calls HHH(DDD)
that simulates DDD that calls HHH(DDD)...
>
HHH is supposed to abort as soon as it detects that
the simulated input cannot possibly reach its own
simulated "return" instruction final state.
>
by continuing to fail to understand that because every
instance of HHH has the exact same machine code that
unless the outermost HHH aborts then none of them abort.
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