Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/10/2025 11:06 AM, wij wrote:

On Sat, 2025-05-10 at 10:45 -0500, olcott wrote:

On 5/10/2025 10:28 AM, wij wrote:

On Sat, 2025-05-10 at 09:33 -0500, olcott wrote:

On 5/10/2025 7:37 AM, Bonita Montero wrote:

Am 09.05.2025 um 04:22 schrieb olcott:
>

Look at their replies to this post.
Not a one of them will agree that
>
void DDD()
{
     HHH(DDD);
     return; // final halt state
}
>
When 1 or more instructions of DDD are correctly
simulated by HHH then the correctly simulated DDD cannot
possibly reach its "return" instruction (final halt state).
>
They have consistently disagreed with this
simple point for three years.

>
I guess that not even a professor of theoretical computer
science would spend years working on so few lines of code.
>

>
I created a whole x86utm operating system.
It correctly determines that the halting problem's
otherwise "impossible" input is actually non halting.
>
int DD()
{
     int Halt_Status = HHH(DD);
     if (Halt_Status)
       HERE: goto HERE;
     return Halt_Status;
}
>
https://github.com/plolcott/x86utm
>

>
Nope.
  From I know HHH(DD) decides whether the input DD is "impossible" input or not.
>

>
DD has the standard form of the "impossible" input.
HHH merely rejects it as non-halting.
>

 You said 'merely' rejects it as non-halting.
So, POOH do not answer the input of any other function?
 

The input that has baffled computer scientists for 90
years is merely correctly determined to be non-halting
when the behavior of this input is measured by HHH
emulating this input according to the rules of the x86
language.
The same thing applies to the Linz proof yet cannot
be understood until after HHH(DDD) and HHH(DD) are
fully understood.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer