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On 2024-05-24 18:52:31 +0000, Fred. Zwarts said:Correct simulated is fully specified above: *Correct Simulation Defined*
Op 24.mei.2024 om 18:57 schreef olcott:This has the problem that it is not clear what the corrct simulationOn 5/24/2024 10:01 AM, Fred. Zwarts wrote:>Op 23.mei.2024 om 18:52 schreef olcott:>typedef int (*ptr)(); // ptr is pointer to int function in C>
00 int H(ptr p, ptr i);
01 int D(ptr p)
02 {
03 int Halt_Status = H(p, p);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 int main()
10 {
11 H(D,D);
12 return 0;
13 }
>
The above template refers to an infinite set of H/D pairs where D is
correctly simulated by pure function H. This was done because many
reviewers used the shell game ploy to endlessly switch which H/D was
being referred to.
>
*Correct Simulation Defined*
This is provided because every reviewer had a different notion of
correct simulation that diverges from this notion.
>
In the above case a simulator is an x86 emulator that correctly emulates
at least one of the x86 instructions of D in the order specified by the
x86 instructions of D.
>
This may include correctly emulating the x86 instructions of H in the
order specified by the x86 instructions of H thus calling H(D,D) in
recursive simulation.
>
*Execution Trace*
Line 11: main() invokes H(D,D); H(D,D) simulates lines 01, 02, and 03 of
D. This invokes H(D,D) again to repeat the process in endless recursive
simulation.
The case can be simplified even more (D is not needed):
>
We are ONLY asking about whether D correctly simulated by pure function H can possibly reach its own final state at line 06 and halt.
>
Because H is a pure function we know that H halts.
https://en.wikipedia.org/wiki/Pure_function#
Every H of the above H/D pairs returns the meaningless value of 56.
Maybe if you simplify your question, the answer is easier to find:
The case can be simplified by eliminating the complexity of the template D:
>
typedef int (*ptr)(); // ptr is pointer to int function in C
00 int H(ptr p, ptr i);
01 int main()
02 {
03 H(H,H);
04 return 0;
05 }
>
Of the infinite set of H that simulate at least one step of its input, none of them, when simulated by H, halts, because none of them possibly reaches its final state.
So, does H correctly recognize non-halting behaviour in H?
>
D is an unneeded complexity, because the only property of D that is needed is that it calls H, so why not using H as input directly?
of H(H) would be, considering that H requires two arguments. The purpose
of D is to avoid that problem.
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