Sujet : Re: Any honest person that knows the x86 language can see... predict correctly
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 01. Aug 2024, 15:10:42
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8g513$26s53$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
User-Agent : Mozilla Thunderbird
Op 01.aug.2024 om 15:04 schreef olcott:
On 8/1/2024 7:56 AM, Fred. Zwarts wrote:
Op 01.aug.2024 om 13:51 schreef olcott:
On 8/1/2024 2:46 AM, Fred. Zwarts wrote:
Op 01.aug.2024 om 05:51 schreef olcott:
On 7/31/2024 10:08 PM, wij wrote:
On Tue, 2024-07-30 at 18:50 -0500, olcott wrote:
>
It is not supposed to be a general solution to the halting problem.
it only shows how the "impossible" input is correctly determined
to be non halting.
>
>
But how do you determine it is non-halting?
>
As I know you are even unable to define what 'halt' mean !!!
>
I have done this thousands of times and after someone
has read these thousands of times they say that I never
said it once.
>
void DDD()
{
HHH(DDD);
return;
}
>
int main()
{
HHH(DDD);
}
>
If DDD correctly emulated by HHH cannot possibly
reach its return instruction then it never halts.
>
>
>
But a correct simulation is impossible.
>
When HHH does what-ever-the-hell the x86 semantics specifies
then HHH is correct.
>
>
But since HHH deviates from the semantics of the x86 language (by skipping instructions of a halting program) it is incorrect.
>
Only a freaking moron would believe that a non
terminating input should be simulated forever.
We are not talking about your dreams of a non aborting HHH. We are talking about an aborting HHH which terminates.
So, who is the freaking moron talking about non-terminating input?
Dreams are no substitute for fact or for logic.
Skipping the last cycle of the simulation of HHH, which aborts after two cycles, is a deviation from the semantics of the x86 language.
But, you are a slow learner. Probably you will repeat the same error another thousand times.