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

On 2024-10-29 13:56:19 +0000, olcott said:

On 10/29/2024 2:57 AM, Mikko wrote:

On 2024-10-29 00:57:30 +0000, olcott said:
 

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,
 

 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.

 Nope, read the problem you have quoted in the past.
 

 Ultimately I trust Linz the most on this:
 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
 Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 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:
 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.

 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.

 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.

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.
--
Mikko