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On 8/5/2024 2:27 AM, Mikko wrote:You are lying again.On 2024-08-04 12:33:20 +0000, olcott said:typedef void (*ptr)();
On 8/4/2024 2:15 AM, Mikko wrote:That is not stipulated above. Anyway, that stipulation would notOn 2024-08-03 13:48:12 +0000, olcott said:void DDD()
On 8/3/2024 3:06 AM, Mikko wrote:Who here is too stupid to know that whether DDD can reach itsOn 2024-08-02 02:09:38 +0000, olcott said:[Who here is too stupid to know that DDD correctly simulated
*This algorithm is used by all the simulating termination analyzers*The determination is not correct. DDD is a halting computation, as
<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.
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.
by HHH cannot possibly reach its own return instruction?]
own return instruction depends on code not shown below?
{
HHH(DDD);
return;
}
It is stipulated that HHH is an x86 emulator the emulates
N instructions of DDD where N is 0 to infinity.
alter the correctness of my answer.
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.
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