Re: How do computations actually work?

Liste des GroupesRevenir à c theory 
Sujet : Re: How do computations actually work?
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theory
Date : 26. May 2025, 09:46:01
Autres entêtes
Organisation : -
Message-ID : <10119o9$1u1o3$1@dont-email.me>
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
User-Agent : Unison/2.2
On 2025-05-25 14:53:06 +0000, olcott said:

On 5/25/2025 4:14 AM, Mikko wrote:
On 2025-05-24 15:18:57 +0000, olcott said:
 
On 5/24/2025 2:47 AM, Mikko wrote:
On 2025-05-23 02:47:40 +0000, olcott changed the subject to
How do computations actually work?
 Each computation works differently. It does not matter how it works as
long as there are instructions that fully specify how that computation
shall be performed.
 All termination analyzers are required to report on the
behavior that their input finite string specifies.
 To report correctly. Though the input string to a termination analyzer
usially is incomlete: the input string usually specifies different
behavours depending on the input that is not shown to the termination
analyzer, and the analyzer's report must cover all of them.
 A partial termination analyzer may fail to report but is not allowed
to report incorrectly.
 void DDD()
{
   HHH(DDD);
   return;
}
 DDD simulated by HHH cannot possibly reach its
"return" statement final halt state, only liars
will disagree.
Again a straw man deception. Where reasonable people disgraee is the
relevance of that claim. The behavour specified by DDD does reach the
final return from DDD. Whether HHH can simulate that part of the
behaviour is irrelevant. Even without the simjlation HHH decides
correctly if and only if it determines and reports that DDD halts.
--
Mikko

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