Re: No decider is accountable for the computation that itself is contained within, unless that is its input

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Sujet : Re: No decider is accountable for the computation that itself is contained within, unless that is its input
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 31. Jul 2024, 04:18:59
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <a611c93efbf6087987c1d61103780171f70868f8@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
User-Agent : Mozilla Thunderbird
On 7/30/24 10:41 PM, olcott wrote:
On 7/30/2024 9:13 PM, Mike Terry wrote:
On 30/07/2024 22:22, olcott wrote:
On 7/30/2024 4:09 PM, joes wrote:
Am Tue, 30 Jul 2024 15:13:34 -0500 schrieb olcott:
On 7/30/2024 2:52 PM, joes wrote:
Am Tue, 30 Jul 2024 11:24:35 -0500 schrieb olcott:
On 7/30/2024 2:24 AM, joes wrote:
Am Mon, 29 Jul 2024 15:32:44 -0500 schrieb olcott:
On 7/29/2024 3:17 PM, joes wrote:
Am Mon, 29 Jul 2024 11:32:00 -0500 schrieb olcott:
On 7/28/2024 3:40 AM, Mikko wrote:
On 2024-07-27 14:21:50 +0000, olcott said:
On 7/27/2024 2:46 AM, Mikko wrote:
On 2024-07-26 16:28:43 +0000, olcott said:
>
Halt deciders are not allowed to report on the behavior of the
actual computation that they themselves are contained within. They
are only allowed to compute the mapping from input finite strings.
What if the input is the same as the containing computation?
It always is except in the case where the decider is reporting on
the TM description that itself is contained within.
>
I don't understand. "The input is not the same as the containing
computation when deciding on the description of the containing
computation"?
I mean: is that an accurate paraphrase?
>
An executing Turing machine is not allowed to report on its own
behavior. Every decider is only allowed to report on the behavior that
its finite string input specifies.
And what happens when those are the same?
>
That is always the case except in the rare exception that I discovered
where a simulating halt decider is simulating the input that calls
itself.
>
Always? Most TMs don't get themselves as input. OTOH that is one of
the most interesting cases.
The description of a TM specifies the behaviour of that machine
when it is running.
>
>
The x86 code of DDD when correctly emulated by HHH according
to the semantics of the x86 code of DDD and HHH does have
different behavior that the directly executed DDD as a matter
of verified fact for three years.
>
People deny this as if a smash a Boston Cream pie in the face
and they deny that there ever was any pie even while their
voice is incoherent because they are talking through the pie
smashed on their face.
>
Hehe, when you go on like this I can't help thinking of "head crusher":
>
<https://www.youtube.com/watch?v=8t4pmlHRokg>
>
Mike.
>
 It has always been ridiculously stupid to say that DDD
is not correctly emulated by HHH because how the hell
would would get to the first instruction of DDD if HHH
did not correctly emulate DDD ???
ILLOGICAL STATMENT.
Doing part of something right doesn't mean you did all of it right,
IF HHH can't get even start its emulation, that would be a very bad emulation,

 The same thing applies when DDD calls HHH(DDD).
Which MUST be emulated as the instructions of HHH and NOT what you shpw below.
Below just proves that you are lying.,

 Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
 Is the real issue that no one here besides me has any clue
about the x86 language and they are all just faking it?
No, you are just PROVING that you are lying.

 I have no idea what you are saying and on this basis I am
sure that you must be wrong?
 
So you admit that if you don't understand something that means the other person is lying,
That is just admitting that you are stupid AND a pathological liar.
That you don't understand something doesn't make it wrong, but you using that as a basic to call something wrong proves YOU to be just a liar.
Sorry, you are too stupid for you own good and are just sinking your reputation under a mile deep pile of your shitty lies.

Date Sujet#  Auteur
5 Jul 25 o 

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