Re: Any honest person that knows the x86 language can see... predict INcorrectly

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Sujet : Re: Any honest person that knows the x86 language can see... predict INcorrectly
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 02. Aug 2024, 00:33:31
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <4ad177a17f83c3e8fbc5f745e026427d491cb81c@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
User-Agent : Mozilla Thunderbird
On 8/1/24 7:51 AM, olcott wrote:
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 when it doesn't, it is wrong.
Since HHH doesn't know what the call instruction will ACTUALLY do, we prove that it is the later.
You prove that a call to HHH with the parameter DDD will return 0, so the fact that HHH decides that isn't what it does means it is just wrong, and you claiming the wrong answer is right just makes you a pathological liar.

Date Sujet#  Auteur
7 Jul 25 o 

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