Re: The philosophy of computation reformulates existing ideas on a new basis ---

On 10/31/24 8:36 AM, olcott wrote:

On 10/31/2024 5:45 AM, Mikko wrote:

On 2024-10-30 12:24:13 +0000, olcott said:
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On 10/30/2024 5:07 AM, Mikko wrote:

On 2024-10-29 13:56:19 +0000, olcott said:
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On 10/29/2024 2:57 AM, Mikko wrote:

On 2024-10-29 00:57:30 +0000, olcott said:
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On 10/28/2024 6:56 PM, Richard Damon wrote:

On 10/28/24 11:04 AM, olcott wrote:

On 10/28/2024 6:16 AM, Richard Damon wrote:

The machine being used to compute the Halting Function has taken a finite string description, the Halting Function itself always took a Turing Machine,
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That is incorrect. It has always been the finite string Turing Machine
description of a Turing machine is the input to the halt decider.
There are always been a distinction between the abstraction and the
encoding.

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Nope, read the problem you have quoted in the past.
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Ultimately I trust Linz the most on this:
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the problem is: given the description of a Turing machine
M and an input w, does M, when started in the initial
configuration qow, perform a computation that eventually halts?
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
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Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
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Linz also makes sure to ignore that the behavior of ⟨Ĥ⟩ ⟨Ĥ⟩
correctly simulated by embedded_H cannot possibly reach
either ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩ because like everyone else he rejects
simulation out of hand:
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We cannot find the answer by simulating the action of M on w,
say by performing it on a universal Turing machine, because
there is no limit on the length of the computation.

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That statement does not fully reject simulation but is correct in
the observation that non-halting cannot be determied in finite time
by a complete simulation so someting else is needed instead of or
in addition to a partial simulation. Linz does include simulationg
Turing machines in his proof that no Turing machine is a halt decider.

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To the best of my knowledge no one besides me ever came up with the
idea of making a simulating halt decider / emulating termination
analyzer.

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Textboods may mention the idea but there is not much to say about it,
only that it does not give a complete solution. Linz' proof covers
all Turing machines. A simulating halt decider that is not a Turing
machine is not interesting because there is no known way to make it.

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In other words you are saying that there is no such thing as a
UTM. Not a smart thing to say. embedded_H was adapted from a UTM.

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I already said that you should not use the expression "In other wordw".
It is not clear what you mean by it but you boviously don't mean what
the phrase really means.
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 The only way to verify mutual understanding is to
keep paraphrasing back and forth until there is
mutual agreement.
 If my paraphrase is inaccurate then you must point
out the exact details of the inaccuracy.
 

Besides, it is not a good idea to lie about what other people say.
Even when one says "In other words".
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 I am translating it into what I am understanding
you are saying if you don't mean that that it is
up to you to correct the vagueness and ambiguity
of your words by seeing how I misunderstood them.
 

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

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There should be an "or" between the last two lines.
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embedded_H does correctly determine the halt status of the
Linz ⟨Ĥ⟩ ⟨Ĥ⟩ when embedded_H computes the mapping from its
finite string input to the behavior this finite string actually
specifies.

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It does not matter what a non-exstent Turing machine does. If H is
a possible Turing machine then embedded_H determines incorrectly.
If not then Ĥ is not a Turing machine and therefore not relevant.
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 Within the title of this thread UTM based embedded_H does
correctly determine that its simulation of the finite string
⟨Ĥ⟩ cannot possibly reach its own states of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
Since we can see this it is plausible that embedded_H can see this.

But it embedded_H is a UTM, then so is H, so it is categoricaly impossible for it to "abort" its emulation/simulition of its input, as to do so makes it not a UTM?

 We could stupidly say that ZFC is not doing
https://en.wikipedia.org/wiki/Naive_set_theory
correctly and on this basis it incorrectly rejects
set as members of themselves.
 

And, if you want to "reject" classical computation theory, you need to FORMALLY define you new system, and FORMALLY show what it can do, and then try to persuade people that your system has something of value.
Since we don't have the Russel Paradox of Naive Set Theory, you need to do a lot of work to show that you have a good replacement for a well working system.
If your system can't handle "Turing Complete" systems, that will be a big hurdle.