Re: Halting Problem: What Constitutes Pathological Input

On 5/5/25 10:29 PM, olcott wrote:

On 5/5/2025 8:06 PM, Richard Damon wrote:

On 5/5/25 11:51 AM, olcott wrote:

On 5/5/2025 10:17 AM, Mr Flibble wrote:

What constitutes halting problem pathological input:
>
Input that would cause infinite recursion when using a decider of the
simulating kind.
>
Such input forms a category error which results in the halting problem
being ill-formed as currently defined.
>
/Flibble

>
I prefer to look at it as a counter-example that refutes
all of the halting problem proofs.
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

>
Which isn't a program until you include the SPECIFIC HHH that it refutes, and thus your talk about correctly emulated by HHH is just a lie.
>

>
https://github.com/plolcott/x86utm
>
The x86utm operating system includes fully
operational HHH and DD.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
>
When HHH computes the mapping from *its input* to
the behavior of DD emulated by HHH this includes
HHH emulating itself emulating DD. This matches
the infinite recursion behavior pattern.
>

>
And *ITS INPUT*, for the HHH that answers 0, is the representation of a program

 Not at all. This has always been stupidly wrong.
The input is actually a 100% perfectly precise
sequence of steps. With pathological self-reference
some of these steps are inside the termination analyzer.
 

Can't be, as the input needs to be about a program, which must, by the definition of a program, include all its algorithm.
Yes, there are steps that also occur in the termination analyzer, but they have been effectively copied into the program the input describes.
Note, nothing says that the representation of the program has to be an assembly level description of it. It has to be a complete description, that 100% defines the results the code will generate (and if it will generate) but it doesn't need to be the exact assembly code,
YOU even understand that, as you present the code as "C" code, which isn't assembly.
What you forget is that the input program INCLUDES as its definiton, all of the code it uses, and thus the call to the decider it is built on includes that code into the decider, and that is a FIXED and DETERMINDED version of the decider, the one that THIS version of the input is designed to make wrong.
This doesn't change when you hypothosize a different decider looking at THIS input.