Sujet : Re: Turing Machine computable functions apply finite string transformations to inputs VERIFIED FACT
De : rjh (at) *nospam* cpax.org.uk (Richard Heathfield)
Groupes : comp.theoryDate : 29. Apr 2025, 23:03:52
Autres entêtes
Organisation : Fix this later
Message-ID : <vuric8$2gjif$3@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
User-Agent : Mozilla Thunderbird
On 29/04/2025 22:38, olcott wrote:
<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. If it isn't, it's irrelevant to the Halting Problem, and we can ignore it. If it is, however, then we know that it doesn't work for all inputs, even if (as you claim) it works for one.
<snip>
DD <is> the Halting Problem counter-example input to HHH.
for the same reason we can't devise a universally accurate termination analyser that executes the code to see what happens.
>
There is no evidence of that.
Sure there is. Not just evidence, but an actual, rigorous, mathematical proof.
-- Richard HeathfieldEmail: rjh at cpax dot org dot uk"Usenet is a strange place" - dmr 29 July 1999Sig line 4 vacant - apply within