Re: Any honest person that knows the x86 language can see... predict correctly is to say Halting,

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Sujet : Re: Any honest person that knows the x86 language can see... predict correctly is to say Halting,
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 01. Aug 2024, 13:30:47
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8fv5n$25l0a$4@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14
User-Agent : Mozilla Thunderbird
On 8/1/2024 6:34 AM, Richard Damon wrote:
On 7/31/24 11:51 PM, olcott wrote:
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 only *IF*  HHH *DOES* correctly emulate its input, which means it can't abort its emulation,
*No stupid it has never meant that*
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     *simulating halt decider H correctly simulates its input D*
     *until H correctly determines that its simulated D would never*
     *stop running unless aborted*
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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