Re: Can D simulated by H terminate normally?

On 5/1/24 11:51 AM, olcott wrote:

On 5/1/2024 4:51 AM, Mikko wrote:

On 2024-04-30 15:50:50 +0000, olcott said:
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On 4/30/2024 5:27 AM, Mikko wrote:

On 2024-04-29 14:20:20 +0000, olcott said:
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On 4/29/2024 8:44 AM, Mikko wrote:

On 2024-04-28 18:52:06 +0000, olcott said:
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On 4/28/2024 1:39 PM, Richard Damon wrote:

On 4/28/24 2:19 PM, olcott wrote:

On 4/28/2024 1:06 PM, Richard Damon wrote:

On 4/28/24 1:50 PM, olcott wrote:

On 4/28/2024 11:08 AM, Richard Damon wrote:

On 4/28/24 11:33 AM, olcott wrote:

On 4/28/2024 10:08 AM, Richard Damon wrote:

On 4/28/24 9:52 AM, olcott wrote:

On 4/28/2024 8:19 AM, Richard Damon wrote:

On 4/28/24 8:56 AM, olcott wrote:

On 4/28/2024 3:23 AM, Mikko wrote:

On 2024-04-28 00:17:48 +0000, olcott said:
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Can D simulated by H terminate normally?

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One should not that "D simulated by H" is not the same as
"simulation of D by H". The message below seems to be more
about the latter than the former. In any case, it is more
about the properties of H than about the properties of D.
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D specifies what is essentially infinite recursion to H.
Several people agreed that D simulated by H cannot possibly
reach past its own line 03 no matter what H does.

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Nope, it is only that if H fails to be a decider.
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*We don't make this leap of logic. I never used the term decider*
*We don't make this leap of logic. I never used the term decider*
*We don't make this leap of logic. I never used the term decider*
*We don't make this leap of logic. I never used the term decider*

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You admit that people see that as being a claim about the Halting Problem, and thus the implied definitons of the terms apply.
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The only way to get people to understand that I am correct
and thus not always ignore my words and leap to the conclusion
that I must be wrong is to insist that they review every single
detail of all of my reasoning one tiny step at a time.
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No, the way to get people to understand what you are saying is to use the standard terminology, and start with what people will accept and move to what is harder to understand.
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People have no obligation to work in the direction you want them to.
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Yes, when you speak non-sense, people will ignore you, because what you speak is non-sense.
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You are just proving that you don't understand how to perform logic, or frame a persuasive arguement.
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That fact that as far as we can tell, your "logic" is based on you making up things and trying to form justifications for them, just makes people unwilling to attempt to "accept" your wild ideas to see what might make sense.
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Linguistic determinism is the concept that language and its structures
limit and determine human knowledge or thought, as well as thought
processes such as categorization, memory, and perception.
https://en.wikipedia.org/wiki/Linguistic_determinism

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So? Since formal logic isn't based on Linguistics, it doesn't directly impact it. IT might limit the forms we
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Some of the technical "terms of the art" box people into misconceptions
for which there is no escape. Some of the technical "terms of the art"
I perfectly agree with.
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*Important technical "term of the art" that I totally agree with*
Computable functions are the formalized analogue of the intuitive notion
of algorithms, in the sense that a function is computable if there
exists an algorithm that can do the job of the function, i.e. given an
input of the function domain it can return the corresponding output. https://en.wikipedia.org/wiki/Computable_function

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But you seem to miss that Halting isn't a "Computable Function", as Turing Proved.
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Even the term "halting" is problematic.
For 15 years I thought it means stops running for any reason.

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And that shows your STUPIDITY, not an error in the Theory.
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Now I know that it means reaches the final state. Half the
people here may not know that.

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No, I suspect most of the people here are smarter than that.
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Yet again only rhetoric wit no actual reasoning.
Do you believe:
(a) Halting means stopping for any reason.
(b) Halting means reaching a final state.
(c) Neither.

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The simplest way to define halting is (s): neither. Instead, it means
that it is not possible to continue the computation to an infinite
number of steps.
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Wrong answer.

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The word "you" in the question did not refer to me, so I didn't answer,
just commented.
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computation that halts… “the Turing machine will halt whenever it enters a final state” (Linz:1990:234)
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[5] Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (317-320)

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That, together with other definitions by Linz, means exactly the same
as the definition I proposed.
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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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That does not work. If you want to use the term "abnormal termination"
you must first define it.

 *I HAVE SAID THIS AT LEAST 10,000 TIMES NOW*
Every D simulated by H that cannot possibly stop running unless
aborted by H does specify non-terminating behavior to H. When
H aborts this simulation that does not count as D halting.

Which is just meaningless gobbledygook by your definitions.
It means that
int H(ptr m, ptr d) {
    return 0;
}
is always correct, because THAT H can not possible simulate the input to the end before it aborts it, and that H is all that that H can be, or it isn't THAT H.
Unless you clarify your altered definitions, H is what H is and that just becomes the conclusion.

 

Then you can compare the definitions and try
to determine whether "abnormal termination" implies halting or non-halting
or neither. Note that "halting" is a freature of a Turing machine (a Turing
machine halts or does not halt) but "abnormal termination" seems to be
a feature of a particlar simulation (a simulation of a Truing machine
is or is not abnormally terminated).
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