On 3/15/2025 5:12 AM, Mikko wrote:That can't be right. Otherwise my simulator could just not simulateOn 2025-03-14 14:39:30 +0000, olcott said:YES.On 3/14/2025 4:03 AM, Mikko wrote:On 2025-03-13 20:56:22 +0000, olcott said:>On 3/13/2025 4:22 AM, Mikko wrote:>On 2025-03-13 00:36:04 +0000, olcott said:>
>
>void DDD()>
{
HHH(DDD);
return;
}
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
When HHH correctly emulates N steps of the above functions none of
them can possibly reach their own "return" instruction and
terminate normally.
Nevertheless, assuming HHH is a decider, Infinite_Loop and
Infinite_Recursion specify a non-terminating behaviour, DDD
specifies a terminating behaviour
What is the sequence of machine language instructions of DDD
emulated by HHH such that DDD reaches its machine address 00002183?
Irrelevant off-topic distraction.
Proving that you don't have a clue that Rice's Theorem is anchored in
the behavior that its finite string input specifies.
Another irrelevant off-topic distraction, this time involving a false
claim.
One can be a competent C programmer without knowing anyting about
Rice's Theorem.
>Rice's Theorem is about semantic properties in general, not justDoes THE INPUT TO simulating termination analyzer HHH encode a C
behaviours.
The unsolvability of the halting problem is just a special case.
function that reaches its "return"
instruction [WHEN SIMULATED BY HHH] (The definition of simulating
termination analyzer) ???
at all and say that no input halts.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>key word "correctly"
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.
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| 21 Apr 26 |  … |
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