Sujet : Re: Everyone here seems to consistently lie about this
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theoryDate : 06. Aug 2024, 10:07:58
Autres entêtes
Organisation : -
Message-ID : <v8slku$1gfnf$1@dont-email.me>
References : 1 2 3 4 5 6 7
User-Agent : Unison/2.2
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.
-- Mikko