Re: This function proves that only the outermost HHH examines the execution trace

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Sujet : Re: This function proves that only the outermost HHH examines the execution trace
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 27. Jul 2024, 19:20:19
Autres entêtes
Organisation : A noiseless patient Spider
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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.
The key difference between a partial decider and a decider is that
the former case only needs to get at least one input correctly.

 An aborted C program certainly doesn't
correspond with a running turing machine - so it must be a halted turing
machine.
 
If you take the view all all new ideas are inherently incorrect
then you would be right within this false assumption.

So aborted programs are halted programs.  If you disagree, perhaps you
could point out where in my arguments above I'm wrong.
 
You seem to have the above false assumption.

-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
 
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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