Sujet : Re: HHH maps its input to the behavior specified by it --- partial simulation never reaches its halt state, but the program does.
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 10. Aug 2024, 03:57:30
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <69f5a47bf03695a1d24a44c807a8ae6234b8cf78@i2pn2.org>
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On 8/9/24 1:02 PM, olcott wrote:
On 8/9/2024 10:42 AM, Fred. Zwarts wrote:
Op 09.aug.2024 om 17:04 schreef olcott:
>>>> void Infinite_Loop()
{
HERE: goto HERE;
}
>
The correct simulation of the above never halts.
>
void DDD()
{
HHH(DDD);
return;
}
A correct simulation of N instructions of DDD <is>
A correct simulation of N instructions of DDD.
You are trying to get away with denying a truism.
But not a correct simulation of ALL the instruction of DDD, so doesn't indicate if it halts.
I guess you think if you drive one mile on a highway, and then get off because you think the road will never stop, that proves that the road will never stop.
Sorry, it doesn't work that way, The only simulation that shows, by itself, that an input to be non-halting, is a simulation that runs forever.
Remember, EVERY DDD that calls a different HHH is a different input and has a different correct simulation, and all of the ones that use a HHH that aborts and returns will halt themselves even though HHH's simulation didn't reach there.