Sujet : Re: Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 04. Aug 2024, 08:16:43
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8n6cc$3tv07$2@dont-email.me>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
Op 03.aug.2024 om 18:03 schreef olcott:
On 8/3/2024 10:33 AM, Richard Damon wrote:
On 8/3/24 10:26 AM, olcott wrote:
On 8/3/2024 9:04 AM, Fred. Zwarts wrote:
Op 03.aug.2024 om 15:50 schreef olcott:
On 8/3/2024 3:14 AM, Fred. Zwarts wrote:
Op 02.aug.2024 om 22:57 schreef olcott:
Who here is too stupid to know that DDD correctly simulated
by HHH cannot possibly reach its own return instruction?
>
void DDD()
{
HHH(DDD);
return;
}
>
>
Which proves that the simulation is incorrect.
>
When are you going to understand that you are not allowed
to disagree with the semantics of the x86 language?
>
>
>
I do not disagree.
When are you going to understand that it is a deviation of the semantics of the x86 language to skip instructions of a halting program,
>
HHH(DDD) simulates DDD that calls HHH(DDD) to repeat the process.
>
If it does this an infinite number of times the simulated DDD
never reaches its own return instruction.
>
If it does this a googolplex number of times the simulated DDD
never reaches its own return instruction.
>
Nope, the PARTIAL SIMULATION of DDD never reaches the return instruction.
>
For N = 0; while N <= googolplex; N++
N instructions of DDD correctly emulated by HHH[N] never
reach their own "return" instruction final state.
Proving that for each N the simulation is incomplete and therefore incorrect.
∞ instructions of DDD correctly emulated by HHH[∞] never
reach their own "return" instruction final state.
Proving that HHH cannot possibly simulate *itself* correctly.
Thus any HHH that takes a wild guess that DDD emulated
by itself never halts is always correct.
No, it proves that HHH cannot possibly simulate itself correctly. All HHH that abort are incorrect because they abort one cycle before the simulation of a halting program would halt. The HHH that does not abort is incorrect because it does not halt.
…