Sujet : Re: Any honest person that knows the x86 language can see... predict correctly
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 01. Aug 2024, 05:24:26
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8f2lq$20ius$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14
User-Agent : Mozilla Thunderbird
On 7/31/2024 11:03 PM, wij wrote:
On Wed, 2024-07-31 at 22:51 -0500, 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.
>
>
That's right, HHH(DDD) as shown should never halt.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never
stop running unless aborted then
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
But The Halting Problem asks HHH to return 1 or 0 (so to speak, because you
don't know the detail).
Since HHH does not return 1 or 0 to answer the question, it is not a decider.
You are dealing with POO Problem.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer