Sujet : Re: Defining a correct simulating halt decider
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 05. Sep 2024, 21:41:28
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <f4ca32c31d70ce51426c3a731a55012c94b836a5@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Thu, 05 Sep 2024 13:10:13 -0500 schrieb olcott:
On 9/5/2024 12:22 PM, joes wrote:
Am Thu, 05 Sep 2024 12:17:01 -0500 schrieb olcott:
On 9/5/2024 11:56 AM, joes wrote:
Am Thu, 05 Sep 2024 11:52:04 -0500 schrieb olcott:
On 9/5/2024 11:34 AM, joes wrote:
Am Thu, 05 Sep 2024 11:10:40 -0500 schrieb olcott:
On 9/5/2024 10:57 AM, joes wrote:
The directly executed HHH correctly determines that its emulated DDD
must be aborted because DDD keeps *THE EMULATED HHH* stuck in
recursive emulation.
Why doesn’t the simulated HHH abort?
The first HHH cannot wait for its HHH to abort which is waiting for
its HHH to abort on and on with no HHH ever aborting.
But why does HHH halt and return that itself doesn’t halt?
First agree that you understand the first part so that we don't
endlessly digress away from the point.
I smell evasion but fine, I understand that HHH cannot wait.
Now that I think about it, HHH could recognise itself (wouldn’t it
need to be a quine?)… no, the copy would do the same.
-- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:It is not guaranteed that n+1 exists for every n.
…