HHH(DDD)==0 is correct

On 7/1/2025 11:37 AM, Mr Flibble wrote:

On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
 

On 6/30/25 2:30 PM, Mr Flibble wrote:
>
PO just works off the lie that a correct simulation of the input is
different than the direct execution, even though he can't show the
instruction actually correctly simulated where they differ, and thus
proves he is lying.
>
The closest he comes is claiming that the simulation of the "Call HHH"
must be different when simulated then when executed, as for "some
reason" it must be just because otherwise HHH can't do the simulation.
>
Sorry, not being able to do something doesn't mean you get to redefine
it,
>
You ar4e just showing you are as stupid as he is.

 No. A simulator does not have to run a simulation to completion if it can
determine that the input, A PROGRAM, never halts.
 /Flibble

The most direct way to analyze this is that
HHH(DDD)==0 and HHH1(DDD)==1 are both correct
because DDD calls HHH(DDD) in recursive simulation and
DDD does not call HHH1(DDD) in recursive simulation.
*I can't imagine how Mike does not get this*
*Context of above dialogue*
*Context of above dialogue*
*Context of above dialogue*
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
   HHH(DDD);
   return;
}
int main()
{
   HHH(DDD);
}
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0. (HHH1 has identical code)
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer