Sujet : Re: Hypothetical possibilities --- stupid rebuttal
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theoryDate : 01. Aug 2024, 10:14:15
Autres entêtes
Organisation : -
Message-ID : <v8fg4n$22utp$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
User-Agent : Unison/2.2
On 2024-07-30 21:11:45 +0000, olcott said:
On 7/30/2024 1:44 AM, Mikko wrote:
On 2024-07-29 16:17:32 +0000, olcott said:
On 7/28/2024 3:04 AM, Mikko wrote:
On 2024-07-27 13:55:56 +0000, olcott said:
On 7/27/2024 1:54 AM, Mikko wrote:
On 2024-07-26 13:58:54 +0000, olcott said:
On 7/26/2024 3:05 AM, Mikko wrote:
On 2024-07-24 13:38:08 +0000, olcott said:
On 7/24/2024 4:01 AM, Mikko wrote:
On 2024-07-23 14:41:11 +0000, olcott said:
On 7/23/2024 2:32 AM, Mikko wrote:
On 2024-07-22 15:05:41 +0000, olcott said:
On 7/22/2024 6:05 AM, Mikko wrote:
On 2024-07-20 15:28:31 +0000, olcott said:
void DDD()
{
HHH(DDD);
}
int main()
{
DDD();
}
(a) Termination Analyzers / Partial Halt Deciders must halt
this is a design requirement.
For a partial analyzer or deciders this is not always required.
*You can't even get my words correctly*
A termination analyzer must report on the behavior of at least
one input for all of the inputs of this one input. This is
met when a termination analyzer analyzes an input having no inputs.
A partial halt decider must correctly determine the halt status
of at least one input and its specific input (if any).
HHH is both a partial halt decider and a termination analyzer
for DDD and a few other inputs having no input.
(b) Every simulating termination analyzer HHH either
aborts the simulation of its input or not.
This must be interpreted to mean that a simulating termination analyzer
may abort its simulation for some simulated abort and simulate others
to the termination.
I am talking about hypothetical possible ways that HHH could be encoded.
(a) HHH(DDD) is encoded to abort its simulation.
(b) HHH(DDD) is encoded to never abort its simulation.
(c) Within the hypothetical case where HHH does not abort
the simulation of its input {HHH, emulated DDD and executed DDD}
never stop running.
The case is not very hypothetical. Given the HHH you already have,
it is fairly easy to construct the "hypothetical" HHH and see what
it actually does.
(a) HHH(DDD) is encoded to abort its simulation.
(b) HHH(DDD) is encoded to never abort its simulation.
This violates the design requirement of (a) therefore HHH must
abort the simulation of its input.
The violation simply means that the "hypothetical" HHH is not a
termination analyzer of partial halt decider in sense (a). What
it "must" be or do depends on the requirements.
Therefore (a) is correct and (b) is incorrect according to the
design requirements for HHH that it must halt.
It is also a truism that any input that must be aborted
is a non-halting input.
No, it is not. The "must" and "non-halting" belong to different worlds.
The word "must" blongs to requirements. The word "non-halting" is a
feature of a program. They are unrelated, so one cannot be inferred
from the other.
When-so-ever there are two hypothetical possible way to encode
a simulating halt decider for a specific input
(a) one aborts its simulation of DDD
(b) never aborts its simulation of DDD
Does the simulator that simulates the beginning and end of the
simulated computation but skips a part in ghe middle belong to
class (a) or class (b)?
That is off topic. I am only referring to a sequence of
1 to N x86 machine language instructions simulated according
to the x86 semantic meaning of these instructions.
No, it isn't. Abortion of simulation is a deviation form x86 macine
language semantics. What I ask about does not deviate more.
In other words you are saying that it is absolutely impossible
to make an x86 program that is an x86 emulator that correctly
emulates a finite number of instructions of non-terminating
input x86 machine code.
You are lying again. That is not the same in other words, and I am
not saying what you falsely claim.
I am not lying I am paraphrasing so that we can come to a mutual
understanding.
It is lying to paraphrase so that the original meaning is not preserved.
I make my point mere clearly here:
[Any honest person that knows the x86 language can see...]
That does not alter the fact that you lied above. Therefore the term
"honest person" does not include you. So what is your evidence that
any (or even some) honest person can see...?
*I clarified what I meant responding to your original message here*
http://al.howardknight.net/?STYPE=msgid&MSGI=%3Cv8bc6j%24159av%241%40dont-email.me%3E
That does not really claify beyond what you said here, and what it
clarifies is irrelevant to the fact that you lied above.
-- Mikko