Sujet : Re: This function proves that only the outermost HHH examines the execution trace
De : acm (at) *nospam* muc.de (Alan Mackenzie)
Groupes : comp.theoryDate : 27. Jul 2024, 22:45:50
Autres entêtes
Organisation : muc.de e.V.
Message-ID : <v83pqe$2nhr$4@news.muc.de>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : tin/2.6.3-20231224 ("Banff") (FreeBSD/14.1-RELEASE (amd64))
olcott <
polcott333@gmail.com> wrote:
On 7/27/2024 4:16 PM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:
On 7/27/2024 3:20 PM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:
On 7/27/2024 1:14 PM, Alan Mackenzie wrote:
olcott <polcott333@gmail.com> wrote:
Stopping running is not the same as halting.
DDD emulated by HHH stops running when its emulation has been aborted.
This is not the same as reaching its ret instruction and terminating
normally (AKA halting).
I think you're wrong, here. All your C programs are a stand in
for turing machines. A turing machine is either running or
halted. There is no third state "aborted".
Until you take the conventional ideas of
(a) UTM
(b) TM Description
(c) Decider
and combine them together to become a simulating partial halt decider.
Where does the notion of "aborted", as being distinct from halted, come
from?
After all of these years and you don't get that?
"Aborted" being distinct from halted is an incoherent notion. It isn't
consistent with turing machines. I was hoping you could give a
justification for it.
A simulating partial halt decider can stop simulating
its input when it detects a non-halting behavior pattern.
This does not count as the input halting.
Says who? Well, OK, it would be the machine halting, not the input, but
that's a small point.
void Infinite_Recursion()
{
Infinite_Recursion();
}
[ .... ]
Do you understand that HHH(Infinite_Recursion) correctly
implements this criteria for the above input?
There's nothing wrong with my understanding, but I'm not sure what
"implementing a criterion (not "criteria")" means, any more than
"listening to the height of a wall". You don't make it clear which
criterion you're talking about.
<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
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If you can't understand that Infinite_Recursion() doesn't
halt then you don't know the subject matter nearly well enough.
That appears to have nothing to do with my point, which is that "aborted"
is the same turing machine state as "halted".
I would be grateful if one of the group's experts would post here to
clear up this point for me.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
-- Alan Mackenzie (Nuremberg, Germany).