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

On 2/19/25 7:38 PM, olcott wrote:

On 2/19/2025 1:41 AM, joes wrote:

Am Tue, 18 Feb 2025 22:08:20 -0600 schrieb olcott:

On 2/18/2025 5:32 PM, Richard Damon wrote:

On 2/18/25 8:37 AM, olcott wrote:

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.
>

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.
>

And you miss, that since the first does it, all of them do it,

Unless THE FIRST ONE THAT SEES IT DOES IT NONE OF THEM DO

That’s what he said.
>

 Also when the first one does it none of the rest of them
can possibly do it because that are no longer being simulated.
 

But that is just confusing the behavior of a partial simulation of a machine with the behavior of the machine.
Which just shows that you have no idea about what you are talking about.
Someone who actually understand something about programming would understand this, but that isn't you.