Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

Op 08.feb.2025 om 15:43 schreef olcott:

On 2/8/2025 3:54 AM, Fred. Zwarts wrote:

Op 08.feb.2025 om 00:13 schreef olcott:

Experts in the C programming language will know that DD
correctly simulated by HHH cannot possibly reach its own
"if" statement.

>
Yes, it demonstrates the incapability of HHH to correctly determine the halting behaviour of DD
>

>
The finite string DD specifies non-terminating recursive
simulation to simulating termination analyzer HHH. This
makes HHH necessarily correct to reject its input as
non-halting.

>
The finite string defines one behaviour. This finite string, when given to an X86 processor shows halting behaviour. This finite string,when given to a world class simulator, shows halting behaviour. Only HHH fails to see this proven halting behaviour. So it proves the failure of HHH.
HHH aborts the simulation on unsound grounds one cycle before the simulation would terminate normally.
>

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>
https://github.com/plolcott/x86utm/blob/master/Halt7.c
has fully operational HHH and DD
>
The halting problem has always been a mathematical mapping
from finite strings to behaviors.

>
Yes. And the behaviour of this finite string has been proven to show halting behaviour. Only Olcott's HHH fails to see it.
His misunderstanding is that he thinks that the behaviour defined by the finite string depends on the simulator.

>
When DD calls HHH(DD) in recursive simulation it is a
verified fact that DD cannot possibly halt.

 Which proves the failure of HHH. It does not reach the end of a halting program. All other methods show that DD halts.