Re: DDD correctly emulated by HHH is correctly rejected as non-halting.

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Sujet : Re: DDD correctly emulated by HHH is correctly rejected as non-halting.
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theory
Date : 12. Jul 2024, 10:15:24
Autres entêtes
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Message-ID : <v6qoms$2ukg7$1@dont-email.me>
References : 1 2 3 4 5 6 7 8
User-Agent : Unison/2.2
On 2024-07-11 14:12:15 +0000, olcott said:

On 7/11/2024 1:28 AM, Mikko wrote:
On 2024-07-10 18:58:14 +0000, olcott said:
 
On 7/10/2024 1:55 PM, Alan Mackenzie wrote:
Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:
Op 10.jul.2024 om 20:12 schreef Alan Mackenzie:
[ Followup-To: set ]
 
In comp.theory Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:
 
[ .... ]
 
Proving that the simulation is incorrect. Because a correct simulation
would not abort a halting program halfway its simulation.
 
Just for clarity, a correct simulation wouldn't abort a non-halting
program either, would it?  Or have I misunderstood this correctness?
 
[ .... ]
 
A non-halting program cannot be simulated correctly in a finite time.
So, it depends whether we can call it a correct simulation, when it does
not abort. But, for some meaning of 'correct', indeed, a simulator
should not abort a non-halting program either.
 OK, thanks!
 
 In other words he is saying that when you do
1 step correctly you did 0 steps correctly.
 That is possible as "correctly" has different meaning when talking
about steps from when talking about simulations.
 
 *No that is always false*
When you did one anythings correctly then you did
more than zero anythings correctly.
If I only correcly do one thing that is not a part of my routine then
I don't do my routine correctly. If I do correctly every part of my routine
but do them in a wrong order I don't do my routine correctly.
--
Mikko

Date Sujet#  Auteur
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