Liste des Groupes | Revenir à theory |
On 5/20/2025 2:06 AM, Mikko wrote:That is a wild guess of HHH, not a correct determination. When it sees the call to HHH and we know that HHH halts, we know that there is only a finite recursion, so the 'would never stop running' exists only in your dreams.On 2025-05-20 04:20:54 +0000, olcott said:*H correctly simulates its input D until*
><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
>
Do you understand that we are only evaluating whether
or not HHH/DDD meets this above criteria?
I do understand that the meaning of the behaviour is not mentioned
in the creteria and is therefore irrelevant, an obvious consequence
of which is that your "WRONG!" above is false.
>
specifies that HHH must simulate DDD according
to the meaning of the rules of the x86 language.
The meaning of every step of the behavior is
precisely specified by the x86 language.
_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]
*H correctly simulates its input D*
00002172 00002173 00002175 0000217a
H correctly simulates itself simulating DDD
00002172 00002173 00002175 0000217a
*until H correctly determines that its simulated D*
*would never stop running unless aborted*
H sees DDD call the same function with the sameHHH does not even see a full cycle, so it cannot know that there are no conditional branches in the cycle. You can view a full cycle in different ways:
parameter and there are no conditional branch
instructions from the beginning of DDD to calling
HHH(DDD) again. This repeating pattern proves
non-termination.
Les messages affichés proviennent d'usenet.