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 : 30. Apr 2025, 19:11:28
Autres entêtes
Organisation : Fix this later
Message-ID : <vutp4h$q23t$1@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 31
User-Agent : Mozilla Thunderbird
On 30/04/2025 18:28, olcott wrote:
On 4/30/2025 10:46 AM, Richard Heathfield wrote:
On 30/04/2025 16:15, olcott wrote:
On 4/29/2025 5:03 PM, Richard Heathfield wrote:
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.
>
The domain of HHH is DD.
>
Then it is attacking not the Halting Problem but the Olcott Problem, which is of interest to nobody but you.
>
Because you don't pay any attention at all
you did not bother to notice that I have never been
attacking the Halting Problem only the conventional
Halting Problem proof.
HHH(DD) doesn't attack the Halting Problem proof any more than a bucketful of marshmallows attacks Fort Knox. It fails to model the problem correctly.
THE IMPOSSIBLE INPUT IS REJECTED AS NON-HALTING.
If HHH rejects DD, it fails to meet the spec for a decider, which doesn't have licence to reject programs.
If it reports on DD as halting, DD fails to meet the spec for an input that you appear to agree is impossible to report on.
-- Richard HeathfieldEmail: rjh at cpax dot org dot uk"Usenet is a strange place" - dmr 29 July 1999Sig line 4 vacant - apply within
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