Re: Can you see that D correctly simulated by H remains stuck in recursive simulation?

Op 24.mei.2024 om 18:57 schreef olcott:

On 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?