Sujet : Re: Everyone here seems to consistently lie about this
De : abc (at) *nospam* def.com (olcott)
Groupes : comp.theoryDate : 06. Aug 2024, 13:12:05
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8t3ul$1j02h$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10
User-Agent : Mozilla Thunderbird
On 8/6/2024 6:54 AM, Richard Damon wrote:
On 8/6/24 7:35 AM, olcott wrote:
On 8/6/2024 3:07 AM, Mikko wrote:
On 2024-08-05 12:45:11 +0000, olcott said:
>
On 8/5/2024 2:27 AM, Mikko wrote:
On 2024-08-04 12:33:20 +0000, olcott said:
>
On 8/4/2024 2:15 AM, Mikko wrote:
On 2024-08-03 13:48:12 +0000, olcott said:
>
On 8/3/2024 3:06 AM, Mikko wrote:
On 2024-08-02 02:09:38 +0000, olcott said:
>
*This algorithm is used by all the simulating termination analyzers*
<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>
>
DDD is correctly emulated by HHH according to the x86
language semantics of DDD and HHH including when DDD
emulates itself emulating DDD
>
*UNTIL*
>
HHH correctly determines that never aborting this
emulation would cause DDD and HHH to endlessly repeat.
>
The determination is not correct. DDD is a halting computation, as
correctely determined by HHH1 or simly calling it from main. It is
not possible to correctly determine that ha haling computation is
non-halting, as is self-evdent from the meaning of the words.
>
>
[Who here is too stupid to know that DDD correctly simulated
by HHH cannot possibly reach its own return instruction?]
>
Who here is too stupid to know that whether DDD can reach its
own return instruction depends on code not shown below?
>
>
void DDD()
{
HHH(DDD);
return;
}
>
It is stipulated that HHH is an x86 emulator the emulates
N instructions of DDD where N is 0 to infinity.
>
That is not stipulated above. Anyway, that stipulation would not
alter the correctness of my answer.
>
>
typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
HHH(DDD);
return;
}
>
int main()
{
HHH(DDD);
}
>
In other words you do not know C well enough to comprehend
that DDD correctly simulated by any HHH cannot possibly reach
its own "return" instruction halt state.
>
You are lying again.
that
>
I am hypothesizing. If you do know C well enough to agree then
simply agree. What I said is a tautology thus disagreement <is> error.
>
And, if it *IS* a tautoligy, then you agree that "Correct Simulation" means a simulation that doesn't abort,
Your attention deficit order continues to cause you to forget
most of the details that I told you hundreds of times.
A correct emulation of 0 to infinity lines of DDD is
a correct emulation of 0 to infinity lines of DDD.
A correct simulation of N lines of DDD is sufficient to
determine that no correct simulation of DDD by any HHH
will ever stop running unless aborted.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Everyone here seems to consistently lie about this
Re: Peter Olcott seems to consistently lie about this ---