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

On 10/31/2024 5:49 AM, Mikko wrote:

On 2024-10-29 14:35:34 +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.
>

>
*That people fail to agree with this and also fail to*
*correctly point out any error seems to indicate dishonestly*
*or lack of technical competence*
>
DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.

 - irrelevant

100% perfectly relevant within the philosophy of computation
*THE TITLE OF THIS THREAD*
[The philosophy of computation reformulates existing ideas on a new basis ---]

- couterfactual
 

You can baselessly claim that verified facts are counter-factual
you cannot show this.
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
If you don't even understand the x86 language and claim
that I am wrong that would make you a liar.
https://chatgpt.com/share/67158ec6-3398-8011-98d1-41198baa29f2
ChatGPT explains all of the details of how and why I am correct
and will vigorously argue against anyone that says otherwise.
The key to getting correct reasoning from ChatGPT is to exhaustively
explain all of the details of an algorithm and its input such that
your explanation and its analysis fits within 4000 words. When you go
over that limit it simply forgets key details and makes big mistakes.
ChatGPT totally understands Simulating termination analyzer HHH
apply to input DDD: (as proven by the above link)
void DDD()
{
   HHH(DDD);
   return;
}
ChatGPT gets overwhelmed by this same HHH applied to DD
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer