Sujet : Re: No decider is accountable for the computation that itself is contained within
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 30. Jul 2024, 20:52:34
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <00e25e8f7bb0af364c2bad26b5a1ebeb76fee34d@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 30 Jul 2024 11:24:35 -0500 schrieb olcott:
On 7/30/2024 2:24 AM, joes wrote:
Am Mon, 29 Jul 2024 15:32:44 -0500 schrieb olcott:
On 7/29/2024 3:17 PM, joes wrote:
Am Mon, 29 Jul 2024 11:32:00 -0500 schrieb olcott:
On 7/28/2024 3:40 AM, Mikko wrote:
On 2024-07-27 14:21:50 +0000, olcott said:
On 7/27/2024 2:46 AM, Mikko wrote:
On 2024-07-26 16:28:43 +0000, olcott said:
Halt deciders are not allowed to report on the behavior of the
actual computation that they themselves are contained within. They
are only allowed to compute the mapping from input finite strings.
What if the input is the same as the containing computation?
It always is except in the case where the decider is reporting on the
TM description that itself is contained within.
I don't understand. "The input is not the same as the containing
computation when deciding on the description of the containing
computation"?
I mean: is that an accurate paraphrase?
An executing Turing machine is not allowed to report on its own
behavior. Every decider is only allowed to report on the behavior that
its finite string input specifies.
And what happens when those are 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.
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