Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

On 2/9/25 2:57 PM, olcott wrote:

On 2/9/2025 1:42 PM, joes wrote:

Am Sun, 09 Feb 2025 13:00:05 -0600 schrieb olcott:

On 2/9/2025 12:47 PM, Fred. Zwarts wrote:

Op 09.feb.2025 om 17:49 schreef olcott:

On 2/9/2025 10:43 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 17:37 schreef olcott:

On 2/9/2025 9:53 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 16:15 schreef olcott:

On 2/9/2025 2:09 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 07:04 schreef olcott:

On 2/8/2025 3:49 PM, Fred. Zwarts wrote:

Op 08.feb.2025 om 15:43 schreef olcott:

On 2/8/2025 3:54 AM, Fred. Zwarts wrote:

Op 08.feb.2025 om 00:13 schreef olcott:

>

The input to HHH(DD) cannot possibly terminate normally. Referring
to some other DD does not change this verfied fact.
>

That DD halts is a verified fact.

The input to HHH(DD) DOES NOT HALT !!!

>
It is a verified fact that the finite string describes a halting
program. Du to a bug, HHH does not see that, because it investigates
only the first few instructions of DD. HHH is unable to process the
call from DD to HHH correctly.

>
DD simulated by HHH cannot possibly terminate normally.

>
Indeed, because HHH fails to simulate itself up to the end.
This is verified with:
         int main() {
           return HHH(main);
         }
>

There is no simulating itself to the end with the above example either.
Apparently you do not understand the basic notion of recursion very
well.

 

That’s the point. HHH doesn’t terminate.
>

Counter-factual.
 

Then it is wrong, as if HHH returns 0, it returns 0 when DD calls it so DD will terminate and HHH will just be wrong.
You need to decide if HHH does a correct simulation or is a decider.
It can't be both, as they are mutually exclusive.