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On 4/29/2025 5:03 PM, Richard Heathfield wrote:Then it is attacking not the Halting Problem but the Olcott Problem, which is of interest to nobody but you.On 29/04/2025 22:38, olcott wrote:The domain of HHH is DD.
>
<snip>
>>>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
HHH is correct DD as non-halting BECAUSE THAT IS
WHAT THE INPUT TO HHH(DD) SPECIFIES.
You're going round the same loop again.
>
Either your HHH() is a universal termination analyser or it isn't.
No, it doesn't. To do that, you would have to devise a program to correctly analyse *any* of *all* possible programs, given any of *all* possible inputs.If it isn't, it's irrelevant to the Halting Problem,It correctly refutes the conventional proof of the
Halting Problem proofs.
The "impossible" input specifiesIt's not an impossible input. There's nothing impossible about being able to analyse /one/ program in line with your expectations. The impossibility comes from HHH being able to correctly determine halting behaviour for *any* program (although I would grant that the program can at least be expected to be syntactically correct).
non-halting behavior and the contradictory part of DD
is unreachable code.
Have you ever done any actual programming?No, mate. I paint potatoes. There's good money in spud daubs.
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