Sujet : Re: Why Peter Olcott is both right and wrong
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 16. May 2025, 20:50:52
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <10084us$3tum4$2@dont-email.me>
References : 1 2 3 4 5
User-Agent : Mozilla Thunderbird
Op 16.mei.2025 om 17:34 schreef olcott:
On 5/16/2025 1:54 AM, Mikko wrote:
On 2025-05-15 15:33:01 +0000, Mr Flibble said:
>
On Thu, 15 May 2025 10:13:50 -0500, olcott wrote:
>
On 5/15/2025 1:27 AM, Mr Flibble wrote:
Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.
>
Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.
>
/Flibble
>
Introduction to the Theory of Computation 3rd Edition by Michael Sipser
(Author)
4.4 out of 5 stars 568 ratings
>
https://www.amazon.com/Introduction-Theory-Computation-Michael- Sipser/dp/
113318779X
>
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D until
H correctly determines that its simulated D would never stop
running unless aborted then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>
HHH does correctly reject DDD and DD according to the exact meaning of
the above words. It also seems to me that those words are a truism.
>
Sipser is wrong: he is disagreeing with Turing et al that pathological
input is undecidable.
>
Which sentence of Sipser contradicts which sentence of Turing?
Why do you think that Sipser is wrong and not Turing?
>
A simulating halt decider (SHD) (as defined above)
proves that there cannot be any input that actually
does the opposite of whatever value that its SHD
returns.
int main()
{
DDD(); // The HHH called by DDD() cannot report on its caller.
}
No, it proves that a correct simulation is impossible. Either the simulation does not halt, or it aborts before it can see that the input halts. The input specifies a halting program, but the programmer by programming the abort, the programmer made HHH blind for this specification. That HHH does skip this important part of the specification, does not mean that the specification is not present in the input.
Of course, you will ignore this verifiable facts and just repeat your claims. Is that because you don't understand them, or because your dreams overrule the facts? Come out of rebuttal mode and try to think. Learning something new is not so dangerous as you seem to think.