Re: DD specifies non-terminating behavior to HHH --- ONE POINT AT A TIME !!!

On 2/21/25 6:15 PM, olcott wrote:

On 2/20/2025 3:35 AM, Fred. Zwarts wrote:

Op 20.feb.2025 om 01:34 schreef olcott:

On 2/19/2025 4:55 AM, Fred. Zwarts wrote:

Op 18.feb.2025 om 17:48 schreef olcott:

On 2/18/2025 8:11 AM, Fred. Zwarts wrote:

Op 18.feb.2025 om 14:37 schreef olcott:

On 2/18/2025 6:25 AM, Richard Damon wrote:

On 2/18/25 6:26 AM, olcott wrote:

On 2/18/2025 3:24 AM, Mikko wrote:

On 2025-02-17 09:05:42 +0000, Fred. Zwarts said:
>

Op 16.feb.2025 om 23:51 schreef olcott:

On 2/16/2025 4:30 PM, joes wrote:

Am Sun, 16 Feb 2025 15:58:14 -0600 schrieb olcott:

On 2/16/2025 2:02 PM, joes wrote:

Am Sun, 16 Feb 2025 13:24:14 -0600 schrieb olcott:

On 2/16/2025 10:35 AM, joes wrote:

Am Sun, 16 Feb 2025 06:51:12 -0600 schrieb olcott:

On 2/15/2025 2:49 AM, Mikko wrote:

On 2025-02-14 12:40:04 +0000, olcott said:

On 2/14/2025 2:58 AM, Mikko wrote:

On 2025-02-14 00:07:23 +0000, olcott said:

On 2/13/2025 3:20 AM, Mikko wrote:

On 2025-02-13 04:21:34 +0000, olcott said:

On 2/12/2025 4:04 AM, Mikko wrote:

On 2025-02-11 14:41:38 +0000, olcott said:

>

DD  correctly simulated by HHH cannot possibly terminate normally.

>
That claim has already shown to be false. Nothing above shows that
HHH does not return 0. If it does DD also returns 0.

When we are referring to the above DD simulated by HHH and not
trying to get away with changing the subject to some other DD
somewhere else

such as one that calls a non-aborting version of HHH
>

then anyone with sufficient knowledge of C programming knows that no
instance of DD shown above simulated by any corresponding instance
of HHH can possibly terminate normally.

Well, then that corresponding (by what?) HHH isn’t a decider.

I am focusing on the isomorphic notion of a termination analyzer.

(There are other deciders that are not termination analysers.)
>

A simulating termination analyzer correctly rejects any input that
must be aborted to prevent its own non-termination.

Yes, in particular itself is not such an input, because we *know* that
it halts, because it is a decider. You can’t have your cake and eat it
too.

I am not even using the confusing term "halts".
Instead I am using in its place "terminates normally".
DD correctly simulated by HHH cannot possibly terminate normally.

What’s confusing about „halts”? I find it clearer as it does not imply
an ambiguous „abnormal termination”. How does HHH simulate DD
terminating abnormally, then? Why doesn’t it terminate abnormally
itself?
You can substitute the term: the input DD to HHH does not need to be
aborted, because the simulated decider terminates.
>

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
  int Halt_Status = HHH(DD);
  if (Halt_Status)
    HERE: goto HERE;
  return Halt_Status;
}
>
int main()
{
  HHH(DD);
}
>
Every simulated input that must be aborted to
prevent the non-termination of HHH is stipulated
to be correctly rejected by HHH as non-terminating.
>

A very strange and invalid stipulation.

>
It merely means that the words do not have their ordinary meaning.
>

>
Unless HHH(DD) aborts its simulation of DD itself cannot possibly terminate normally. Every expert in the C programming language
can see this. People that are not experts get confused by the loop
after the "if" statement.
>

>
So? Since it does that, it needs to presume that the copy of itself it sees called does that.
>

>
Not at all. Perhaps your technical skill is much more woefully
deficient than I ever imagined.
>
Here is the point that you just missed Unless the first HHH
that sees the non-terminating pattern aborts its simulation
none of them do because they all have the exact same code.
>
>

>
The point Olcott misses is that if the non-terminating HHH is changed to abort the simulation, the program is changed. He does not understand that a modification of a program makes a change. Such a change modifies the behaviour of the program. The non- termination behaviour has disappeared with this change and only remains in his dreams. After this change, the simulation would terminate normally and HHH should no longer abort. But it does, because the code that detects the 'special condition' has a bug, which makes that it does not see that the program has been changed into a halting program.

>
>
When I focus on one single-point:
I get two years of dodging and this point is never addressed.
>
[DD simulated by HHH cannot possibly terminate normally]
>
void DDD()
{
   HHH(DDD);
   return;
}
>
int main()
{
   HHH(Infinite_Recursion);
   HHH(DDD);
}
>

It is not true that this point has never been addressed. Olcott ignores it when it is addressed.
>
What is the point? Even if HHH fails to simulate the halting program DD up to the end because it is logically impossible for it to complete the simulation, it still fails.

>
It fails In the same way that every CAD system
will never correctly represent a geometric circle that has
four equal length sides in the same two dimensional plane.
>
>
>

Indeed. Such a CAD system fails if it is given the task to draw a square circle. Similarly, HHH fails if it is given the task to determine the termination behaviour of DD.

 It DOES NOT FAIL when it is defined coherently.
The square root of a basket of rotten eggs is also not computable.

Except you haven't shown a "coherent" definition, since your logic is based on variable constants and claims based on lies.
All you are doing is showing your utter ignorance of what you talk about, and that you are nothing but an ignorant pathological lying fraud.

 

We can't blame them, because they cannot possibly complete the impossible task. We only blame the one that is trying do something that has been proven that it cannot possibly work.