Re: Can D simulated by H terminate normally?

On 5/1/2024 7:16 PM, Richard Damon wrote:

On 5/1/24 11:26 AM, olcott wrote:

On 5/1/2024 4:42 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 4/30/2024 5:46 PM, Richard Damon wrote:

On 4/30/24 11:50 AM, olcott wrote:

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Since the notion of abnormal termination could not exist prior to my
creation of a simulating halt decider and does exist within this
frame-of-reference we must construe abnormal termination as not
halting.  If we don't do this we end up with actual infinite loops
that halt.

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Except that Turing Machine do not have a concept of "Abnormal
Termination",

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Indeed, not.
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They do now, otherwise simulating termination analyzers are defined
to report that infinite loops always halt because they abort their
simulation of this infinite loop to report not halting.

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Balderdash.  "Simulating termination analyzers" aren't defined at all.
Until we have some definition of them, it is impossible to discuss their
properties sensibly.
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When you say that I have not defined them at all you are ignoring the
10,000 times that I have defined them in this forum.
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Any simulated input that does not need to be aborted to prevent
its own infinite execution is an input that terminates normally.
This counts as halting.

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Except that without a functioning halting decider, it is impossible to
know whether a simulated input "needs to be aborted" or not.  We know
there are no functioning halting deciders.
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It is a termination analyzer thus is not required to be infallibly
correct on every possible input. It must get at least one input
correctly.

 In other words, it is a TOY.
 By your definition:
 H(ptr m, ptr d) {
     return 1;
}
 is a correct termination analyzer, as it will get at least one input correctly.
 

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D simulated by H cannot possibly reach past its own line 3 even
in an infinite number of simulated steps.

 WRONG. see my other post.
 

Bring up the details of your rebuttal and make sure
to not diverge from the above two lines AT ALL WHAT-SO-EVER.

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All inputs that must be aborted terminate abnormally, thus does
not count as halting.

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I have shown, in the last two days, that "terminating abnormally",
whatever that might mean in a turing machine, is indeed halting.  You
chose not to respond to those parts of my posts.
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If we take your definition then all infinite loops halt therefore
you are wrong.
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 No, because the Turing Machine, in an infinite loop, doesn't terminate abnormally.
 Only the SIMULATION of the machine terminated abnormally.
 

None-the-less the notion of terminating abnormally is entailed
by a simulating termination analyzer.

 

you are just showing that your system isn't actually the
equivlent to the Turing Problem.

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You don't even understand that no infinite loop halts.

 Right, not even abnormally.
 Only simulations terminate abnormally, but (partial) simulations are not actual machine behavior.
 You are just showing all the more that you don't understand the difference between the REALITY of the machine behavior shown by the actual running of the machine, and the "imaginary" world of a simulation of it. The imaginary only become reality if it is taken to completion.
 

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yes, we can define that some "final states" are to be considered
"abnormal terminations" and some "Normal Termination", but that
doesn't change the nature of the problem.

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*The step that corrects the error of the halting problem comes last*

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