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On 8/22/2024 11:59 AM, joes wrote:What happens when your main(HHH(DDD)) is simulated by HHH?Am Thu, 22 Aug 2024 08:36:33 -0500 schrieb olcott:On 8/22/2024 8:21 AM, joes wrote:But still emulating a D that calls an aborting H.Am Thu, 22 Aug 2024 07:59:59 -0500 schrieb olcott:On 8/22/2024 3:16 AM, Fred. Zwarts wrote:Op 22.aug.2024 om 06:22 schreef olcott:then H can abort its simulation
of D and correctly report that D specifies a non-halting
sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words
10/13/2022>
The second half proves that this is the H that aborts that is making
the prediction of the behavior of D when emulated by a hypothetical
version of itself then never aborts.
It is also not the simulator (since they are the same).THIS EXACTLY MATCHES THE SIPSER APPROVED CRITERIABut that different hypothetical HHH is a non-input.
The finite HHH(DDD)
emulates itself emulating DDD exactly once and this is sufficient
for this HHH to predict what a different HHH(DDD) do that never
aborted its emulation of its input.
How do you justify the use of a static variable?It <is> emulating the exact same code at the exact same machine addressNo; only if the same goes for the outermost one (but that doesn’tI don't know how you twist words to get that. HHH is required toHHH is supposed to predict what the behavior of DDD would be if itIf IT didn’t abort DDD calling its aborting self.
did not abort its emulation of DDD that is what the words that
Professor agreed to mean.
predict the behavior of DDD as if every HHH had its abort code
removed.
halt).
Otherwise it is not simulating itself.
exactly twice.
--Do you still not understand that HHH should predict the behaviour
of its input? Why does the HHH have an input, if it is correct to
predict the behaviour of a non-input?
Are you still cheating with the Root variable to change the input
in a non-input?
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