On 15/12/2025 16:31, olcott wrote:

On 12/15/2025 3:20 AM, Mikko wrote:

On 15/12/2025 02:15, olcott wrote:

On 12/14/2025 4:46 AM, Mikko wrote:

On 11/12/2025 16:38, olcott wrote:

On 12/11/2025 2:53 AM, Mikko wrote:

olcott kirjoitti 10.12.2025 klo 18.27:

DD() executed from main() calls HHH(DD) thus is
not one-and-the-same-thing as an argument to HHH.

If the last sentence is true then this is not the counter exmaple
mentioned in certain proofs of noncomputability of halting and
therefore not relevant in that context. The halting problem reuqires >>>>>> that HHH can determine whether the counter example halts. That is, >>>>>> you must be able to replace "???" in

ÿÿ #include <stdio.h> // or your replacement
ÿÿ int main (void)
ÿÿ {
ÿÿÿÿ int Halt_Status = HHH(???); // put the correct argument here
ÿÿÿÿ printf("HHH says: %s\n", Halt_Status ? "halts" : "does not
halt");
ÿÿÿÿ return Halt_Status;
ÿÿ }

with whatever specifies the behaviour of DD to HHH. If you can't
do this then HHH is not a halt decider nor a partial halt decider.

When the halting problem requires a halt decider
to report on the behavior of a Turing machine this
is always a category error.

That you don't know what "category error" means does not justify your
claim. Apparently you can't apply definitions.

Turing machines only compute functions from finite
strings they never compute functions from Turing
machines.

True, but irrelevant to questions about category errors.

A halt decider can at best compute the behavior of
a Turing machine through the proxy of a finite
string machine description it never computes it
directly from another Turing machine.

Whenever any textbook says that a halt decider
must compute halting for machine M on input w
is it wrong.

Which textbook actually says "must"? It is not wrong to say "must" in
the sense that any decider that does not compute whether machine M
halts on input w is not a halt decider. But using "must" is not the
clearest way to say it because the word "must" other meanings.

It actually computes halting that this input pair specifies (?M?, w).

There is an unbalanced parenthesis above.

No halt decider ever computes the halt status
of a machine except through the proxy of finite
strings.

No halt decider computes anything because there are not halt deciders.
But we may compare to univerals Turing machines, which do exist.
A finite string tells to a universal Turing machine everything needed
to fully simulate the behaviour of any Turing machine with any input.

Because all computation only transforms finite
strings the result of this transformation is the
ultimate correct answer.

No, it is not the ultimate correct answer to every question. In
particular, it differes from the ultimate correct answer to at
least one halting question.

--
Mikko

--- PyGate Linux v1.5.2
* Origin: Dragon's Lair, PyGate NNTP<>Fido Gate (3:633/10)