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

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?