Re: People are still trying to get away with disagreeing with the semantics of the x86 language

Op 29.jun.2024 om 18:09 schreef olcott:

People are still trying to get away with disagreeing with
the semantics of the x86 language. That is isomorphic to
trying to get away with disagreeing with arithmetic.
 typedef void (*ptr)();
int H0(ptr P);
 void Infinite_Loop()
{
   HERE: goto HERE;
}
 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 void DDD()
{
   H0(DDD);
}
 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 by returning 0 to its caller.
 Simulating termination analyzers must report on the behavior
that their finite string input specifies thus H0 must report
that DDD correctly emulated by H0 remains stuck in recursive
simulation.
 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
      H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
 People are trying to get away with disagreeing with the semantics
of the x86 language by disagreeing that
 The call from DDD to HHH(DDD) when N steps of DDD are correctly
emulated by any pure function x86 emulator HHH cannot possibly
return.
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
  *A 100% complete and total rewrite of the prior paper*
https://www.researchgate.net/publication/381636432_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_P
 

Apparently the x86 language and simulation are to complex for you. Make it simple. What about the following halt decider:
   int Halt (ptr p) {
     return 1;
   }
It is clear that the question "Will de program described in the input halt" is incorrect, because Halt is unable to process that question, so we cannot ask that question. Therefore, we need to redefine the question to "Does the program described in the input halt or not?".
And since the expression (X || !X) always result in true for any value of X, the Halt function is always correct and we have solved the halting problem.