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On 10/31/2024 5:49 AM, Mikko wrote:And, that system has UNDEFINED behavior per the semantics of the x86 language, as the code at 000015d2 has not be defined per the x86 language.On 2024-10-29 14:35:34 +0000, olcott said:100% perfectly relevant within the philosophy of computation
>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
*THE TITLE OF THIS THREAD*
[The philosophy of computation reformulates existing ideas on a new basis ---]
- couterfactualYou 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/67158ec6-3398-8011-98d1-41198baa29f2Nope, it admits that you are wrong, when you tell it to forget your lies.
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 exhaustivelyBut you lied to it, and so your arguement is based on lies.
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;
}
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