Re: Any honest person that knows the x86 language can see... predict correctly

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Sujet : Re: Any honest person that knows the x86 language can see... predict correctly
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 31. Jul 2024, 11:52:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8d527$1i7t1$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10
User-Agent : Mozilla Thunderbird
On 7/31/2024 3:54 AM, joes wrote:
Am Tue, 30 Jul 2024 16:13:55 -0500 schrieb olcott:
On 7/30/2024 4:07 PM, joes wrote:
Am Tue, 30 Jul 2024 15:05:54 -0500 schrieb olcott:
On 7/30/2024 1:48 PM, Fred. Zwarts wrote:
Op 30.jul.2024 om 17:14 schreef olcott:
On 7/30/2024 9:51 AM, Fred. Zwarts wrote:
Op 30.jul.2024 om 16:21 schreef olcott:
On 7/30/2024 1:52 AM, Mikko wrote:
On 2024-07-29 14:07:53 +0000, olcott said:
 
I proved otherwise. When the abort code is commented out then it keeps
repeating again and again, thus conclusively proving that is must be
aborted or HHH never halts.
But the abort is not commented out in the running code!

I modified the original code by commenting out the abort and it does
endlessly repeat just like HHH correctly predicted.

Yes, and that modification makes HHH not call itself
Not at all. It makes HHH stop aborting DDD.
So that HHH and DDD endlessly repeat.

but a different
program. You'd need to also comment out the outermost abort; then it
wouldn't halt, but if you change HHH to abort, you change all copies
of it at the same time (to keep the recursive call structure).
 
If your name is Charlie and your leg gets amputated you
are still yourself, you don't get renamed to Bill.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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