Liste des Groupes | Revenir à s logic |
On 3/10/2024 7:46 PM, immibis wrote:This is incorrect because Linz H and Ĥ do not have machine addresses. Any rebuttal based on the fact that Linz H has a machine address is refuted.On 11/03/24 01:32, olcott wrote:I am taking H1(D,D) to be Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ and H(D,D) to be Linz H ⟨Ĥ⟩.On 3/10/2024 2:16 PM, immibis wrote:>>>
I noticed that you gave up on Olcott machines and now you are back to your old bullshit ways of pretending that the same machine can produce two different execution traces on the same input. Why don't you show us an execution trace where that happens? Both traces must show the first instruction that is different in both traces and I recommend showing 20 more instructions after that, but you can abort one after that time, if it doesn't halt, to prevent the trace getting infinitely long.
Turing Machines and Olcott machines cannot properly implement
H1(D,D) and H(D,D) that know their own machine address.
"Know their own machine address" isn't an objective specification and it has nothing to do with the halting problem anyway. Turing machines don't have machine addresses and Olcott machines don't have machine addresses either.
>My C code proves these two have different behavior:>
(a) H1(D,D) + H1_machine_address
(b) H(D,D) + H_machine_address
Of course.
>Because they are different computations they are>
not required to have the same behavior.
Of course. One of them even answers the halting problem correctly in this case. But that's because this case isn't the Linz counterexample for H1. If you built the Linz counterexample for H1, which is D1, you'd find that H1 gets it wrong, so H1 doesn't solve the halting problem.
Since we are using machine addresses there is no need for copies.This makes Linz not applicable.
This simplifies Linz Ĥ down to this.
Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ haltsIn other words:
Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
The Linz proof relies on the fact that copies of a machine behave PRECISELY IDENTICALLY. If you change to a system where they do not automatically behave the same, then it is YOUR responsibility to make sure they do behave the same.Turing machines and Olcott machines cannot know their own machineThey cannot be implemented as Turing Machines or Olcott>
Machines. They can be implemented as RASP machines proven
by the fact that they are implemented as C functions.
They can be implemented as Turing machines. Each one is different from the other. They are not the same.
address in a way that cannot be circumvented. X86 virtual machines
and thus (possibly augmented) RASP machines can.
Les messages affichés proviennent d'usenet.