olcott <polcott333@gmail.com> wrote:To be fair to the LLMs, they're only parroting what he tells them to parrot. If you ask one neutrally about HHH, it'll wobble a bit because of the expression "correctly simulates" before deducing that HHH is not what Olcott claims for it:
[ .... ]
Basically everyone that disagrees with me isThat is a barefaced lie. You have never dealt in verified facts. You
disagreeing with verified facts and they
simply refuse to look at the verification.
have continually called what you would like to be true a "verified fact".
When challenged to cite a name, date and source for such verification you
have never come up with one.
There is no verification to look at. The output of LLMs is not
verification, particularly as they were likely trained largely on the
contents of comp.theory over the last 20 years. They're just echoing
back the falsehoods you've unwittingly fed them with.
--------------------------------------------
I am testing the assumption that simulating termination analyzer HHH correctly simulates its input until it:
(a) Detects a non-terminating behavior pattern: abort simulation and return 0.
(b) Simulated input reaches its simulated "return" statement: return 1.
typedef int (*ptr)();
int HHH(ptr P);
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
What value should HHH(DD) correctly return?
What does that tell me about my assumption?
--------------------------------------------
Answer: HHH(DD) should correctly return 0.
Reasoning:
Definition of HHH behavior:
Returns 0 if it detects non-termination.
Returns 1 if the simulated program terminates.
Simulate DD:
int DD() {
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
The control flow depends on Halt_Status = HHH(DD).
If Halt_Status == 1: DD enters an infinite loop (goto HERE).
If Halt_Status == 0: DD returns 0 (terminates).
Self-reference paradox:
HHH(DD) must predict DD’s behavior.
If HHH(DD) predicts termination → returns 1. Then DD loops forever → contradiction.
If HHH(DD) predicts non-termination → returns 0. Then DD terminates → consistent with definition.
Conclusion:
The only consistent value is HHH(DD) = 0.
Implication for your assumption:
Your assumption that HHH correctly simulates its input program until detecting termination or non-termination fails in general.
This DD example shows no algorithm can simulate or decide termination correctly for all inputs—it's the halting problem in disguise.
Summary:
HHH(DD) = 0.
That result proves your assumption leads to contradiction, so a correct universal HHH cannot exist.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within
Les messages affichés proviennent d'usenet.