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

Op 26.mei.2024 om 15:20 schreef olcott:

On 5/26/2024 6:01 AM, Fred. Zwarts wrote:

Op 26.mei.2024 om 06:19 schreef olcott:

On 5/25/2024 2:32 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.
>

>
Olcott's own words are that the simulation of D never reaches past line 03. So the lines following line 03 do not play a role and, therefore, can be removed without changing the claim. This leads to:
>
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         return H(p, p);
04       }
05
06       int main()
07       {
08         H(D,D);
09         return 0;
10       }
>
>
What we see is that the only property of D that is used is that it is a parameter duplicator. (Is that why it is called D?). H needs 2 parameters, but it can be given only one input parameter, so the parameter duplicator is required to allow H to decide about itself.
>
>
>
Of the infinite set of H that simulate at least one step, none of them, when simulated by H, halts, because none of them reaches its final state. Olcott's claim is equivalent to the claim of non-halting behaviour of H.
This means that a simulating halt-decider is a bad idea, because the decider itself does not halt.

>
Not at all.
    A simulator is an x86 emulator that correctly emulates 1 to N of the
    x86 instructions of D in the order specified by the x86 instructions
    of D. This may include M recursive emulations of H emulating itself
    emulating D.
>
    This means that D cannot possibly reach its own line 06 and halt
    in any finite steps of correct simulation. H is free to halt at
    any time after these N finite steps of correct simulation.
>
>

>
D does not reach it own line 04 because the simulation of H does not return to D. So, it shows that the simulation of H does not reach it final state, which proves that H does not halt.

 Your transformation would have been acceptable if you retained the
fact that H is a pure function that always halts and returns some value.
 

Since H is required to halt, but H shows that H does not halt (because the simulation of H never reaches its final state), the conclusion must be that there is no such H.