Liste des Groupes | Revenir à c theory |
On Sat, 2025-07-19 at 16:05 -0500, olcott wrote:I have proven that I do and you only deny thisOn 7/19/2025 3:57 PM, wij wrote:Yes, there is type mismatch problems in nearly all discussions.On Sat, 2025-07-19 at 15:41 -0500, olcott wrote:>On 7/19/2025 3:14 PM, wij wrote:>>>
HP is very simple: H(D)=1 if D halts, H(D)=0 if D does not halt.
>
The standard proof assumes a decider
H(M,x) that determines whether machine
M halts on input x.
>
But this formulation is flawed, because:
Whatever the 'formulation' is, the HP result is a fact that no H can decide
the halting status of any given D.
>
And that is wrong because H(⟨D⟩) is correctly determined.
It has always been a type mismatch error when H(D) was
assumed.
But I don't think you will understand what it is.
DD points to the finite string machineA type mismatch: HHH(DD) or HHH(<DDD>)?>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.
Halting Problem::= H(D)=1 if D halts, H(D)=0 if D does not halt.
The conclusion is, no such H exists.
>
And that is wrong because H(⟨D⟩) is correctly determined.
It has always been a type mismatch error when H(D) was
assumed.
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
Not at all. The HP proof claims that DDDD correctly simulated by HHH cannot reach pastThat is roughly what HP proof says.
the "if" statement thus cannot reach the "return"
statement. T
You chopped up my statement in the middle of a word.his makes HHH(DD)==0 correct.How is this statement from?
HHH(DD) above shows it cannot return to report 0.Factually incorrect.
(I guess you might say something and doing another, again)
--'formulation' does not really matter.>
If 'formulation' matters, it is another problem.
>
>
Les messages affichés proviennent d'usenet.