Sujet : Re: V5 --- Professor Sipser
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 22. Aug 2024, 14:21:24
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <e12d5d2caec39f6964f567343dad8333a92970fe@i2pn2.org>
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User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
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:
<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
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>
We swap the word "determines" for "predicts"
When we swap thew word "halt decider" for "termination analyzer" the
above is translated from computer science into software engineering.
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.
THIS EXACTLY MATCHES THE SIPSER APPROVED CRITERIA 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.
But that different hypothetical HHH is a non-input.
HHH is supposed to predict what the behavior of DDD would be if it did
not abort its emulation of DDD that is what the words that Professor
agreed to mean.
If IT didn’t abort DDD calling its aborting self.
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?
-- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:It is not guaranteed that n+1 exists for every n.