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On 6/20/24 10:12 AM, olcott wrote:*This seems to be the first time that you told the truth about this*On 6/20/2024 3:09 AM, Fred. Zwarts wrote:Why, you aren't?Op 20.jun.2024 om 02:00 schreef olcott:>This shows all of the steps of HH0 simulating DDD>
calling a simulated HH0 simulating DDD
>
https://liarparadox.org/HH0_(DDD)_Full_Trace.pdf
*Some of the key instructions are color coded*
GREEN---DebugStep Address
RED-----HH Address
YELLOW--All of the DDD instructions
CYAN----Return from DebugStep to Decide_Halting_HH
>
_DDD()
[000020a2] 55 push ebp ; housekeeping
[000020a3] 8bec mov ebp,esp ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404 add esp,+04 ; housekeeping
[000020b2] 5d pop ebp ; housekeeping
[000020b3] c3 ret ; never gets here
Size in bytes:(0018) [000020b3]
>
Exactly which step of DDD emulated by H0 was emulated
incorrectly such that this emulation would be complete?
AKA DDD emulated by H0 reaches machine address [000020b3]
>
>
>
If the simulation of a program with a loop of 5 iterations is aborted after 3 iterations, all instructions are correctly simulated. Nevertheless, it is an incorrect simulation, because it should simulate up to the final state of the program.
>
It would be helpful if you answer the actual question being asked
right here and thus not answer some other question that was asked
somewhere else.
You seem to think you are God or something that gets to set the rules.
YOU ARE NOT.
>Right, but they don't do it by "Correctly Simulating" the input, but by a PARTIAL simulation that provides the needed information to prove that an ACTUAL CORRECT (and complete) simulation of that input would not halt.Similarly, if a simulator which aborts after 2 cycles of recursive simulation of it self, it simulates only 1 of the 2 cycles of itself. So, it is incorrect, not because one instruction was simulated incorrectly, but because it did not simulate up to the final state of the simulated self.>
>
void Infinite_Loop()
{
HERE: goto HERE;
}
>
It also looks like you fail to comprehend that it is possible
for a simulating termination analyzer to recognize inputs that
would never terminate by recognizing the repeating state of
these inputs after a finite number of steps of correct simulation.
Les messages affichés proviennent d'usenet.