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On 7/3/24 10:05 PM, olcott wrote:Because the machine instructions provided to HHH specify thatOn 7/3/2024 8:47 PM, Richard Damon wrote:OF course it can. HHH might not be able to simulate it getting there,On 7/3/24 9:26 PM, olcott wrote:>On 7/3/2024 8:17 PM, Richard Damon wrote:>On 7/3/24 8:40 PM, olcott wrote:>On 7/3/2024 6:18 PM, Richard Damon wrote:>On 7/3/24 10:19 AM, olcott wrote:>On 7/3/2024 9:11 AM, joes wrote:>Am Tue, 02 Jul 2024 22:55:12 -0500 schrieb olcott:>On 7/2/2024 10:50 PM, joes wrote:>Am Tue, 02 Jul 2024 14:46:38 -0500 schrieb olcott:On 7/2/2024 2:17 PM, Fred. Zwarts wrote:Op 02.jul.2024 om 21:00 schreef olcott:On 7/2/2024 1:42 PM, Fred. Zwarts wrote:Op 02.jul.2024 om 14:22 schreef olcott:On 7/2/2024 3:22 AM, Fred. Zwarts wrote:Op 02.jul.2024 om 03:25 schreef olcott:WTF? It only calls HHH, which you just said halts.HHH halts on input DDD.Whatever HHH does, it does not run forever but aborts.HHH repeats the process twice and aborts too soon.>
DDD is correctly emulated by any HHH that can exist which calls this
emulated HHH(DDD) to repeat the process until aborted (which may be
never).
>
DDD correctly simulated by HHH cannot possibly halt.
>
An aborted simulation does not count as halting.
And doesn't show non-halting either.
>Reaching it own machine address 00002183 counts as halting.>
DDD correctly simulated by HHH cannot possibly do that.
But HHH doesn't DO a "Correct Simulation" that can show that, it only does a PARTIAL simulation.
>
<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>
>
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
Which it doesn't.
>>>
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
>
Nope, just double talk.
>
H never CORRECTLY determined that a CORRECT SIMULATION (which means one that matchs the behavior of the machine represented by the input) would never halt, sinc ehta tmachine halts.
>
_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
>
OK so it is not your ADD you continue to insist that
you can disagree with the x86 language that conclusively
proves that DDD correctly simulated by HHH cannot
possibly get past machine instruction 0000217a.
No, the claim is that it isn't the simulation by HHH that determines the actual behavior of the input, but the detailed semantics of the x86 instruciton set of the WHOLE input (which includes this particular HHH which you say DOES abort its simulation and return).
You are not paying close enough attention to the exact words
that I am saying. DDD cannot possibly reach past its own
machine address 0000217a no matter what the Hell that HHH does.
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