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On 7/11/2024 2:20 AM, Fred. Zwarts wrote:And since the only "correct simulation" would be of ALL the stes,Op 10.jul.2024 om 20:58 schreef olcott:What you said logically entails that a correct simulationOn 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 not what I said.
of 1 step counts as a correct simulation of 0 steps.
But the correct simulation of N steps is NOT the correct simulation, which implies of ALL steps.What I said is that if a program needs two steps for a simulation, it is incorrect to simulate only one step and then abort and report it will never halt.I am talking about the correct simulation of N steps and you
English seems to be a difficult language for you.
are trying to get away with saying there is no such thing
as the correct simulation of N steps. That is either terribly
confused or dishonest, yet incorrect either way.
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