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

On 10/30/24 8:24 AM, olcott wrote:

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.

And thus is no longer a UTM, so doesnt have the [properties of  UTM.

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

Nope, since the "Halt Status" of the machine represented by ⟨Ĥ⟩ ⟨Ĥ⟩ is BY DEFININTION the behavior of Ĥ ⟨Ĥ⟩, which halts since you admit that embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ -> Ĥ.qn and thus so does  Ĥ ⟨Ĥ⟩.
You are just proving your utter stupidity.
Remember, you don't get to change the meaning of the words of the problem you claim to be doing, no matter how much you want to.