Re: Defining a correct halting decidability decider

On 8/4/24 3:33 PM, olcott wrote:

On 8/4/2024 2:05 PM, Richard Damon wrote:

On 8/4/24 2:49 PM, olcott wrote:

On 8/4/2024 1:38 PM, Richard Damon wrote:

On 8/4/24 10:46 AM, olcott wrote:

When we define an input that does the opposite of whatever
value that its halt decider reports there is a way for the
halt decider to report correctly.
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
HHH returns false indicating that it cannot
correctly determine that its input halts.
True would mean that its input halts.
>

>
But false indicates that the input does not halt, but it does.
>

>
I made a mistake that I corrected on a forum that allows
editing: *Defining a correct halting decidability decider*
1=input does halt
0=input cannot be decided to halt

>
And thus, not a halt decider.
>
Sorry, you are just showing your ignorance.
>
And, the problem is that a given DD *CAN* be decided about halting, just not by HHH, so "can not be decided" is not a correct answer.

 A single universal decider can correctly determine whether
or not an input could possibly be denial-of-service-attack.
0=yes does not halt or pathological self-reference
1=no  halts
 

Which isn't halt deciding, so you are just admitting you have been lying about working on the Halting Problem.
For instance, if the input is a program to try and find an even number that isn't the sum of two primes, to try and prove or disprove the Goldbach's conjecture, saying it is "possibly non-halting" is a worthless statement. (proof of non-halting would be proof of the conjecture as long as the program was "efficient" enough in its scan).
THAT is the sort of question that Halt Deciders were hoped to be able to answer, until Turing showed that Halting was Undecidable, and thus the logic system might be incomplete.