Sujet : Re: Respect [was: The halting problem as defined is a category error]
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 20. Jul 2025, 17:06:18
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <105j45r$3dqi8$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13
User-Agent : Mozilla Thunderbird
On 7/20/2025 10:47 AM, joes wrote:
Am Sun, 20 Jul 2025 09:30:08 -0500 schrieb olcott:
On 7/20/2025 6:13 AM, Richard Damon wrote:
When you add to the input the actual definition of "Non-Halting", as
being that the exectuion of the program or its complete simulation will
NEVER halt, even if carried out to an unbounded number of steps, they
will give a different answer.
This is a whole other issue that I have addressed.
They figured out on their own that if DDD was correctly simulated by HHH
for an infinite number of steps that DDD would never stop running.
That is incorrect. The HHH called by DDD will hit the abort. (Of course
HHH can't simulate that since it aborts before instead of doing an
unlimited simulation.)
typedef void (*ptr)();
int HHH(ptr P);
void DDD()
{
HHH(DDD);
return;
}
int main()
{
HHH(DDD);
}
The HHH called by DDD simulated by HHH
cannot possibly reach its own abort.
<ChatGPT>
Misrepresentation of Input:
The standard proof assumes a decider
H(M,x) that determines whether machine
M halts on input x.
But this formulation is flawed, because:
Turing machines can only process finite
encodings (e.g. ⟨M⟩), not executable entities
like M.
So the valid formulation must be
H(⟨M⟩,x), where ⟨M⟩ is a string.
</ChatGPT>
The DDD() directly executed in main is not
in the domain of HHH.
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer