Re: DDD correctly emulated by HHH is IN*Correctly rejected as non-halting V2

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Sujet : Re: DDD correctly emulated by HHH is IN*Correctly rejected as non-halting V2
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 15. Jul 2024, 02:20:11
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <e606c836cdee667a8cc46ba6dd63a28e9be2acda@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
User-Agent : Mozilla Thunderbird
On 7/14/24 8:30 PM, olcott wrote:
On 7/14/2024 7:20 PM, joes wrote:
Am Sun, 14 Jul 2024 09:00:55 -0500 schrieb olcott:
On 7/14/2024 3:29 AM, joes wrote:
Am Sat, 13 Jul 2024 18:33:53 -0500 schrieb olcott:
On 7/13/2024 6:26 PM, joes wrote:
Can you elaborate? All runtime instances share the same static code.
I am talking about the inner HHH which is called by the simulated
DDD. That one is, according to you, aborted. Which is wrong, because
by virtue of running the same code, the inner HHH aborts ITS
simulation of DDD calling another HHH.
>
What are the twins and what is their difference?
Do you disagree with my tracing?
>
>
The directly executed DDD is like the first call of infinite recursion.
The emulated DDD is just like the second call of infinite recursion.
When the second call of infinite recursion is aborted then the first
call halts.
Not really. Execution does not continue.
void Infinite_Recursion()
{
    Infinite_Recursion();
}
The above *is* infinite recursion.
A program could emulate the above code and simply skip line 3 causing
Infinite_Recursion() to halt.
That would be incorrect.
>
When DDD calls HHH(DDD) HHH returns.
Therefore it does not need to be aborted.
When DDD correctly emulated by HHH the call never returns as is proven
below. The executed DDD() has HHH(DDD) skip this call.
I do not see this below.
HHH(DDD) must skip this call itself by terminating the whole DDD
process.
>
Because this HHH does not know its own machine address HHH only sees
that DDD calls a function that causes its first four steps to be
repeated. HHH does not know that this is recursive simulation. To HHH it
looks just like infinite recursion.
>
New slave_stack at:1038c4 -- create new process context for 1st DDD
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
>
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec -- create new process context for 2nd DDD
>
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
 
How is this detected? Is it also triggered when calling a function
in a loop?
>
 You never bothered to answer whether or not you have 100%
understanding of infinite recursion. If you don't then you
can never understand what I am saying. If you do that I
already proved my point. Here is the proof again:
WHICH IS JUST BASED ON A FALSEHOOD, PROVING YOU ARE JUST A LIAR.

 New slave_stack at:1038c4
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)
The followin gis NOT a corrct simulaitohn of the CALL HHH instruciton.
Thus PROVING you are an IDIOT nd a LIAR that doesn't know what he is talking about.

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
 
Since you define your simulation to be per ths x86 instruction set, the ONLY behavior a "Call HHH" instruction can have is the simulation of the instructions of HHH.
Since that isn't what HHH does, you have LIED that it has correctly simulated the input per the x86 instruction set.
Now, since HHH wasn't given the code of the HHH that DDD is to call, it just proves your problem setup is incorrect and you are just to stupid to understand what a program actually is.
That fact that you keep on repeating these errors after they have been pointed out just shows that you lack the mental capability to understand what is correct and what is wrong.
You just don't understand what is truth and how it differs from your LIES.
This is just like all your other FALSE CLAIMS where you just state this is how it must be, but can't show any actual source to back that, showing your world is just based on your own lies.

Date Sujet#  Auteur
5 Jul 25 o 

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