Sujet : Re: 197 page execution trace of DDD correctly simulated by HHH
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 29. Jun 2024, 21:44:18
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v5po6i$1h5u1$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
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
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.
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? If it doesn’t halt, the outer one doesn’t either
(since it is an instance of the same code, with the same input).
If the outer one halts, it doesn’t need to abort the inner one (which
then also halts).
Alternatively HHH(DDD)
here:
simulates DDD that calls HHH(DDD) that goto here: until out-of-memory
error.
So it does not abort?
-- Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:Objectively I am a genius.
…