Re: Can D simulated by H terminate normally?

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:

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The ONLY way that we can determine if any computation is correct is
when it meets its specification. When a TM is specified to calculate
the sum of a pair of decimal integers and it derives any decimal
integer other than 5 from inputs 2,3 then it is incorrect.

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Changing the subject. The question is not whether it is correct, but
whether it halts. Incorrect programs exist and even those program may
halt.

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I had to address this:

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On 4/29/2024 11:17 AM, Alan Mackenzie wrote:

There is no notion of "correct" in a turing machine.  It is either
running, or has reached a final state.  In the TM equivalent of "core
dump", a final state has most definitely been reached.

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I would indeed be charmed if you would address it, but you have evaded
it, as you have evaded most of the points I made yesterday.
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Note that I said there is no correctness _IN_ a turing machine.  This is
independent of whether or not that turing machine is useful for some
external purpose.
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Note also that you wilfully distorted my meaning by trimming.  The full
context was:
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Core dump abnormal termination does not count as the program
correctly finished its processing.

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There is no notion of "correct" in a turing machine.  It is either
running, or has reached a final state.  In the TM equivalent of "core
dump", a final state has most definitely been reached.

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Your use of the word "correctly" in "correctly finished its processing"
is wrong.  A turing machine is either running or it's finished its
processing.  From the point of view of the tm, there is no "correct" or
"incorrect" associated with the latter condition; it's simply reached a
final state.
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You are thus mistaken in believing "abnormal" termination isn't a final
state.
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 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.
Only if you try to define something that is NOT related to Halting, do you get into that issue.
Your problem is you want to try to define "Halting" as something that can ALWAYS be measured finitely. Yes, that makes it decidable, but makes it also not the same as the actual halting.
It also gets very hard (if not impossible) to define it in terms that can ACTUALLY be measured finitely, and also not have a trivial decay path.

-- 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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