Re: 197 page execution trace of DDD correctly simulated by HHH

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Sujet : Re: 197 page execution trace of DDD correctly simulated by HHH
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theory
Date : 30. Jun 2024, 09:42:12
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v5r5p4$1irrf$1@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Sat, 29 Jun 2024 15:03:02 -0500 schrieb olcott:
On 6/29/2024 2:44 PM, joes wrote:
Am Fri, 28 Jun 2024 14:28:20 -0500 schrieb olcott:
On 6/28/2024 2:18 PM, joes wrote:
Am Fri, 28 Jun 2024 12:53:46 -0500 schrieb olcott:
On 6/28/2024 12:41 PM, joes wrote:
Thanks for leaving the unanswered questions in place, though I’d
rather have you answer them.
Am Fri, 28 Jun 2024 12:05:18 -0500 schrieb olcott:
On 6/28/2024 11:26 AM, joes wrote:
Am Fri, 28 Jun 2024 10:25:36 -0500 schrieb olcott:
On 6/28/2024 8:14 AM, joes wrote:
Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:
 
To the caller DDD, which then returns to its own caller H0,
which returns „halting” to main… hold on.
Why doesn’t the first recursive H return?
Question still not answered.
It should abort, just like the outer one.

HHH(DDD)
simulates DDD that calls HHH(DDD) that simulates DDD that calls
HHH(DDD)
that proves to the outer directly executed HHH that it must abort and
reject.
How does it do that?
Over your head. I have explained it too many times and you just can't
get it.
No, I mean: why does the inner simulator repeat instead of aborting,
the same as the outer one does?

Technically it is called detecting a repeating state.
Yeah, I know. My point is: all recursive calls both enter and detect
a repeating state.

I don't mean "New Jersey, New Jersey".
Please don’t regard me as dumber than I am.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.

Date Sujet#  Auteur
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