Sujet : Re: Simulating termination analyzers by dummies --- criteria is met
De : noreply (at) *nospam* example.com (joes)
Groupes : comp.theoryDate : 24. Jun 2024, 20:52:04
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v5cip4$10816$3@i2pn2.org>
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User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Mon, 24 Jun 2024 08:46:56 -0500 schrieb olcott:
On 6/24/2024 2:22 AM, Mikko wrote:
On 2024-06-23 13:13:42 +0000, olcott said:
On 6/23/2024 2:57 AM, Mikko wrote:
On 2024-06-22 14:11:28 +0000, olcott said:
On 6/22/2024 8:27 AM, Richard Damon wrote:
On 6/22/24 9:04 AM, olcott wrote:
In particular, you can't. You have insisted that your "decider" or
"anlyzer" (or whatever word you happen to use) H or HH (or hwatever
name you happen to use) must return false because a non-input (where
instead of the actually called function another function that does
not halt is called) does not halt.
Which is all we need to know about H in ordet to determine that it is
not a decider.
void DDD()
{
H0(DDD);
}
The call from DDD to H0(DDD) when DDD is correctly emulated by H0 cannot
possibly return.
Why not? H0 is a decider AND simulator, so it can simulate itself
terminating.
-- Man kann mit dunklen Zahlen nicht rechnen. Für die eigentliche Mathematik sind sie vollkommen nutzlos. --Wolfgang Mückenheim
Re: Simulating termination analyzers by dummies --- criteria is met
Re: Simulating termination analyzers by dummies --- criteria is met