Re: Simulating termination analyzers for dummies

On 6/17/2024 9:49 AM, Fred. Zwarts wrote:

Op 17.jun.2024 om 16:34 schreef olcott:

On 6/17/2024 9:18 AM, Fred. Zwarts wrote:

Op 17.jun.2024 om 15:47 schreef olcott:

On 6/17/2024 8:30 AM, Fred. Zwarts wrote:

Op 17.jun.2024 om 14:20 schreef olcott:

On 6/17/2024 3:31 AM, Fred. Zwarts wrote:

Op 17.jun.2024 om 05:33 schreef olcott:

To understand this analysis requires a sufficient knowledge of
the C programming language and what an x86 emulator does.
>
Unless every single detail is made 100% explicit false assumptions
always slip though the cracks. This is why it must be examined at
the C level before it is examined at the Turing Machine level.
>
typedef void (*ptr)();
int H0(ptr P);
>
void Infinite_Loop()
{
   HERE: goto HERE;
}
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void DDD()
{
   H0(DDD);
   return;
}
>
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}
>
Every C programmer that knows what an x86 emulator is knows that when H0
emulates the machine language of Infinite_Loop, Infinite_Recursion, and
DDD that it must abort these emulations so that itself can terminate
normally.
>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as non-
halting.
>

>
For Infinite_Loop and Infinite_Recursion that might be true, because there the simulator processes the whole input.
>
The H0 case is very different. For H0 there is indeed a false assumption, as you mentioned. Here H0 needs to simulate itself, but the simulation is never able to reach the final state of the simulated self. The abort is always one cycle too early, so that the simulating H0 misses the abort. Therefore this results in a false negative.
(Note that H0 should process its input, which includes the H0 that aborts, not a non-input with an H that does not abort.)
>
This results in a impossible dilemma for the programmer. It he creates a H that does not abort, it will not terminate.

>
*Therefore what I said is correct*

>
No, that is not a logical conclusion.

>
Every C programmer that knows what an x86 emulator is knows
that when H0 emulates the machine language of Infinite_Loop, Infinite_Recursion, and DDD that it must abort these emulations
so that itself can terminate normally.

>
That might be correct.

>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as non-
halting.

>
That is wrong. It only shows that H0 is unable to simulate itself. It tells nothing about the halting of the input.
>

>
*Too late you have already affirmed the words above*
Affirming the first part necessitates the second part.
>

That is not logical. If a non-aborting program is wrong, it does not follow that a program that aborts is correct.
Please, think before you reply.
>
So, I repeat:
The logical conclusion if both aborting and not aborting result in errors, is: a halt-decider cannot be based on such a simulation.

>
Your view here is merely ignorant of the fact that deciders
must report on the behavior specified by their inputs.
>

 It is incorrect to assume that a failing simulation is able to report about its input.
The simulation fails, because H0 is unable to simulate itself.
 

There is no possible way for the call to H0 by DDD
correctly simulated by any H0 to return to its caller.
_DDD()
[00001fd2] 55               push ebp
[00001fd3] 8bec             mov ebp,esp
[00001fd5] 68d21f0000       push 00001fd2 ; push address of DDD
[00001fda] e8f3f9ffff       call 000019d2 ; call  H0
[00001fdf] 83c404           add esp,+04
[00001fe2] 5d               pop ebp
[00001fe3] c3               ret
Size in bytes:(0018) [00001fe3]
*THAT THIS IS OVER YOUR HEAD DOES NOT MEAN THAT I AM INCORRECT*
DDD correctly simulated by H0 *is* the behavior that
the finite string of x86 machine code of DDD specifies.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer