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On 8/1/2024 11:11 AM, joes wrote:Sipser agreed only to a correct simulation.Am Thu, 01 Aug 2024 09:30:00 -0500 schrieb olcott:On 8/1/2024 9:23 AM, Fred. Zwarts wrote:>Op 01.aug.2024 om 15:29 schreef olcott:On 8/1/2024 8:12 AM, Fred. Zwarts wrote:Op 01.aug.2024 om 14:20 schreef olcott:On 8/1/2024 3:10 AM, Fred. Zwarts wrote:Op 31.jul.2024 om 23:23 schreef olcott:On 7/31/2024 3:01 PM, Fred. Zwarts wrote:Op 31.jul.2024 om 17:14 schreef olcott:On 7/31/2024 3:44 AM, Fred. Zwarts wrote:Op 31.jul.2024 om 06:09 schreef olcott:If you weren't a clueless wonder you would understand that DDD correctlyIt also shows that HHH when simulating itself, does not reach the endThe trace stops and hides what happens when 000015d2 is called.These next lines conclusively prove that DDD is being correctly
Olcott is hiding the conditional branch instructions in the
recursion.
emulated by HHH after DDD calls HHH(DDD).
of its own simulation.
emulated by HHH including HHH emulating itself emulated DDD has no end
of correct emulation.It does if the simulated HHH aborts, but its simulating copy preempts<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
that. Indeed, it has no choice, but if it didn't abort, the simulation
wouldn't abort either. Therefore it can't simulate itself.
>
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>
I spent two years carefully composing the above before I evenThe never ending pattern is there only in your dreams. The HHH that halts after two cycles has a halting pattern. Only HHH, when simulating itself, cannot see that after two cycles, because it needs one more cycle.
asked professor Sipser to review it.
DDD is correctly emulated by HHH until HHH sees the same
never ending pattern that anyone else can see.
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