Sujet : Re: Liar detector: Fred, Richard, Joes and Alan --- Ben's agreement
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 11. Jul 2024, 16:40:50
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6oqti$2fuva$7@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
User-Agent : Mozilla Thunderbird
On 7/11/2024 1:56 AM, Mikko wrote:
On 2024-07-10 18:27:27 +0000, joes said:
Am Wed, 10 Jul 2024 08:37:30 -0500 schrieb olcott:
On 7/10/2024 2:18 AM, Mikko wrote:
On 2024-07-09 14:14:16 +0000, olcott said:
On 7/9/2024 1:14 AM, Mikko wrote:
On 2024-07-08 17:36:58 +0000, olcott said:
On 7/8/2024 11:16 AM, Fred. Zwarts wrote:
Op 08.jul.2024 om 18:07 schreef olcott:
>
Try to show how infinity is one cycle too soon.
>
You believe that two equals infinity.
>
void Infinite_Loop()
{
HERE: goto HERE;
}
void Infinite_Recursion()
{
Infinite_Recursion();
}
void DDD()
{
HHH(DDD);
}
Two cycles is enough to correctly determine that none of the above
functions correctly emulated by HHH can possibly halt.
That you don't see this is ignorance or deception.
>
There is an important detail that determines whether an infinite
execution can be inferred. That is best illustrated by the following
examples:
void Finite_Loop()
{
int x = 10000;
HERE:
if (x > 0) {
x--;
goto HERE;
}
}
void Finite_Recursion(int n)
{
if (n > 0) {
Finite_Recursion(n + 1);
}
}
void DDD()
{
HHH(DDD); // HHH detects recursive simulation and then simulates
no more }
The important difference is that in my examples there is a
conditional instruction that can (and does) prevent infinite
exectuion.
>
When we ask:
Does the call from DDD emulated by HHH to HHH(DDD) return?
>
Why would anyone ask that? A question should make clear its topic.
Instead one could ask whether HHH can fully emulate DDD if that is what
one wants to know. Or one may think that HHH and DDD are so
unimteresting that there is no point to ask anyting about them.
>
A correct emulator can correctly any correct x86 instructions.
When it emulates non-halting code then itself does not halt.
Oh? Maybe you should give your simulator and decider different names
so they don't get confused.
A charlatan doesn't want clarity but confusion. A good charlatan just
dont what them so much that they would be noticed for that might expose
the charlatan.
It is a hierarchy of prerequisites of knowledge.
Before anyone can understand a simulating termination
analyzer based on an x86 emulator they must understand
(1) x86 emulation
(2) Termination Analysis.
So far no-one besides Ben Bacarisse has sufficiently
understood (1) "x86 emulation" well enough so that we can
move on to the (2) "Termination Analysis" aspect of simulating
termination analyzer HHH.
Once they fully understand (1) "x86 emulation" then:
*Next they must understand when this criteria have been met*
<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
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
So far of all of my reviewers only Ben Bacarisse has
understood this criteria has been met.
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
> I don't think that is the shell game. PO really /has/ an
> H (it's trivial to do for this one case) that correctly
> determines that P(P) *would* never stop running *unless* aborted.
...
> But H determines (correctly) that D would not halt if
> it were not halted. That much is a truism.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer