Sujet : Re: Defining a correct simulating halt decider
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 11. Sep 2024, 18:31:56
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vbsglu$3mme2$5@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
User-Agent : Mozilla Thunderbird
Op 11.sep.2024 om 13:41 schreef olcott:
On 9/11/2024 2:35 AM, Mikko wrote:
On 2024-09-11 00:21:36 +0000, olcott said:
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On 9/10/2024 3:52 AM, Mikko wrote:
On 2024-09-09 18:19:26 +0000, olcott said:
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On 9/8/2024 9:53 AM, Mikko wrote:
On 2024-09-07 13:57:00 +0000, olcott said:
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On 9/7/2024 3:29 AM, Mikko wrote:
On 2024-09-07 05:12:19 +0000, joes said:
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Am Fri, 06 Sep 2024 06:42:48 -0500 schrieb olcott:
On 9/6/2024 6:19 AM, Mikko wrote:
On 2024-09-05 13:24:20 +0000, olcott said:
On 9/5/2024 2:34 AM, Mikko wrote:
On 2024-09-03 13:00:50 +0000, olcott said:
On 9/3/2024 5:25 AM, Mikko wrote:
On 2024-09-02 16:38:03 +0000, olcott said:
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A halt decider is a Turing machine that computes the mapping from
its finite string input to the behavior that this finite string
specifies.
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A halt decider needn't compute the full behaviour, only whether
that behaviour is finite or infinite.
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New slave_stack at:1038c4 Begin Local Halt Decider Simulation
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Local Halt Decider: Infinite Recursion Detected Simulation Stopped
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Hence HHH(DDD)==0 is correct
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Nice to see that you don't disagree with what said.
Unvortunately I can't agree with what you say.
HHH terminates,
os DDD obviously terminates, too. No valid
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DDD emulated by HHH never reaches it final halt state.
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If that iis true it means that HHH called by DDD does not return and
therefore is not a ceicder.
The directly executed HHH is a decider.
What does simulating it change about that?
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If the simulation is incorrect it may change anything.
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PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
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However, a correct simultation faithfully imitates the original
behaviour.
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_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]
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A correct emulation obeys the x86 machine code even
if this machine code catches the machine on fire.
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It is impossible for an emulation of DDD by HHH to
reach machine address 00002183 AND YOU KNOW IT!!!
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A correct emulation of DDD does reach the machine address 0000217f and
a little later 00002183.
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*That is counter-factual and you cannot possibly show otherwise*
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A halt decider is required to predict about the actual execution,
not a couterfactual assumption.
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False assumption.
A halt decider must compute the mapping that its input
finite string specifies.
And the input, a finite string that describes a program based on the aborting HHH, describes a halting program, as proven by the direct execution, by the unmodified world class simulator and even by HHH1. The semantics of the x86 language allows only one behaviour for the finite string. Any program claiming another behaviour violates the semantics of the x86 language,
It is ridiculously stupid to assume that the fact
that DDD calls its own emulator does not change
its behavior relative to not calling its own emulator.
It ridiculous to assume that the semantics of the x86 language allows another behaviour for the finite string.
Why do you have a religious conviction to this stupid
mistake?