Sujet : Re: Can D simulated by H terminate normally?
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 05. May 2024, 16:38:48
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v185lo$1t4hn$5@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 32 33
User-Agent : Mozilla Thunderbird
On 5/5/2024 3:14 AM, Mikko wrote:
On 2024-05-04 13:56:27 +0000, olcott said:
On 5/4/2024 4:47 AM, Mikko wrote:
On 2024-05-03 11:55:15 +0000, olcott said:
>
On 5/3/2024 4:33 AM, Mikko wrote:
On 2024-05-02 18:35:19 +0000, olcott said:
>
On 5/2/2024 4:39 AM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:
On 4/30/2024 5:46 PM, Richard Damon wrote:
On 4/30/24 12:15 PM, olcott wrote:
On 4/30/2024 10:44 AM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:
On 4/30/2024 3:46 AM, Fred. Zwarts wrote:
Op 29.apr.2024 om 21:04 schreef olcott:
>
[ .... ]
>
When we add the brand new idea of {simulating termination analyzer} to
the existing idea of TM's then we must be careful how we define halting
otherwise every infinite loop will be construed as halting.
>
>
Why?
>
That doesn't mean the machine reached a final state.
>
>
Alan seems to believe that a final state is whatever state that an
aborted simulation ends up in.
>
Only through your twisted reasoning. For your information, I hold to the
standard definition of final state, i.e. one which has no state following
it. An aborted simulation is in some state, and that state is a final
one, since there is none following it.
>
On 4/30/2024 10:44 AM, Alan Mackenzie wrote:
You are thus mistaken in believing "abnormal" termination
isn't a final state.
>
Only if you try to define something that is NOT related to Halting, do
you get into that issue.
>
"The all new ideas are wrong" assessment.
Simulating termination analyzers <are> related to halting.
>
Except you cannot define what such a thing is, and that relationship is
anything but clear.
>
When a simulating termination analyzer matches one of three
non-halting behavior patterns
(a) Simple Infinite loop
(b) Simple Infinite Recursion
(c) Simple Recursive Simulation
>
Simple recursive simulation is not a non-halting behaviour
if the recursion is not infinite.
>
In other words the only way that we can tell that an infinite
loop never halts is to simulate it until the end of time?
>
The phrase "in other words" is not correct here as it means that
what follows means the same as what precedes, and that is not
true here.
>
For same loops the only wha to detect non-termination may be
to simulate to infinity but they can be considered exluded by
the term "simple" in (a).
>
There are repeating state non-halting behavior patterns
that can be recognized. These are three more functions
where H derives the correct halt status:
>
void Infinite_Recursion(u32 N)
{
Infinite_Recursion(N);
}
>
Per (b) that is non-halting and indeed it is (though the
execution may crash for "out of memeory").
>
>
It is not actually infinite though because H recognizes the non-halting
behavior pattern, aborts the simulation and reports non-halting.
The recursion is infinite. The simulation by H is incomplete and finite.
Do you understand that it is ridiculously stupid for a simulating
termination analyzer to simulate a non-terminating input forever?
It is the exact same thing with D simulated by H on the basis
of the directly executed H(D,D).
>
void Infinite_Loop()
{
HERE: goto HERE;
}
>
Per (a) that is non-halting and indeed it is.
>
It is not actually infinite though because H recognizes the non-halting
behavior pattern, aborts the simulation and reports non-halting.
The loop is infinite. The simulation by H is incomplete and finite.
Do you understand that it is ridiculously stupid for a simulating
termination analyzer to simulate a non-terminating input forever?
It is the exact same thing with D simulated by H on the basis
of the directly executed H(D,D).
>
>
int factorial(int n)
{
if (n >= 1)
return n*factorial(n-1);
else
return 1;
}
>
Per (c) that is non-halting but in reality it is not.
Ergo, the rule (c) is wrong.
>
That was an example of an input that H correctly determines
does halt.
Maybe but it is an example of your non-halting pattern (c) as
presented above.
No it is not. A simulating termination analyzer must get at least one
non-terminating input correctly and one terminating input correctly.
You couldn't even tell that ordinary factorial halts?
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Can D simulated by H terminate normally?
Re: Can D simulated by H terminate normally? POE