Sujet : Re: Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 03. Aug 2024, 16:26:08
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8lem0$3ftpo$2@dont-email.me>
References : 1 2 3 4
User-Agent : Mozilla Thunderbird
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.
If the simulation is ever aborted the simulated DDD never reaches
its own return instruction.
When we construe the halt state of DDD as its "return"
instruction then the simulated DD never halts.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer