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On 5/7/2025 7:01 AM, dbush wrote:In other words, what UTM(DD) would do, which is halt.On 5/7/2025 6:16 AM, Fred. Zwarts wrote:A simulating halt decider must correctlyOp 06.mei.2025 om 21:15 schreef olcott:>None-the-less it is the words that the best selling>
author of theory of computation textbooks agreed to:
*would never stop running unless aborted*
>
is the hypothetical HHH/DD pair where the same HHH
that DD calls does not abort the simulation of its input.
>
>
Nevertheless, this change makes it fundamentally different.
I can't believe that you are so stupid to think that modifying a program does not make a program different. Are you trolling?
Given that he's shown he doesn't understand (and this list is by no means exhaustive):
>
* what requirements are
* what correct means
* what true means
* what a proof is
* how proof by contradiction works
>
I wouldn't put it past him that he actually believes it. He'll say anything to avoid admitting to himself that he wasted that last 22 years not understanding what he was working on.
>
(Anyone else that wants to add to this list, feel free)
predict *what the behavior would be* if it
did not abort its simulation.
The best selling author of theory of computation textbooksAnd you *continue* to lie that he agrees with you when it's been proven that he does not:
Professor Sipser agreed with this.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its *simulated D*
*would never stop running unless aborted* then
*simulated D would never stop running unless aborted*
means that HHH examines what the behavior of DD *would be*
if it never aborted its simulation. This is a different
hypothetical HHH/DD pair than the actual HHH/DD pair.
If it did not do this and simply kept simulatingAnd instead, it break the requirement that the input cannot be changed.
a non-terminating input it would break the requirement
that itself must halt.
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