Liste des Groupes | Revenir à s logic |
On 3/9/2024 9:33 AM, immibis wrote:HOW?On 9/03/24 16:04, olcott wrote:No it does not. Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ has some preceding steps that copy its inputOn 3/9/2024 7:11 AM, immibis wrote:>On 8/03/24 20:08, olcott wrote:>On 3/8/2024 12:58 PM, immibis wrote:>On 8/03/24 19:12, olcott wrote:>>>
Unless it is an extra parameter it has no basis for doing this.
If it is an extra parameter then it is no longer the Linz proof.
>
So it's impossible to make a Turing machine that writes 12345 onto its tape unless 12345 is a parameter?
That is a valid point.
It is impossible to get me to talk about that until
we first have full closure that the Linz H correctly
determines the halt status of the Linz Ĥ ⟨Ĥ⟩ when run
in the Olcott master UTM.
>
When run in the Olcott master UTM as Olcott machines, the Linz proof does not work because it is designed for Turing machines. A different proof works.
Because Olcott machines are more powerful than Turing Machines they can correctly determine the halt status of this input: H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>.
>
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> always calculates the same answer that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> calculates.
>
and H ⟨Ĥ⟩ ⟨Ĥ⟩ does not have these steps. Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see all of these
steps.
It turns out that this single difference is why Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation.I.E, if you lie about the system, you can claim anything you want.
I only created Olcott machines so that it is easy to see how Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩Except you don't explain HOW it does that,
would detect that it must abort its simulation and transition to Ĥ.Hqn.
Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ might still be able to do this thusYou are just proving you are too stupid to understand what yoare talking about.
nullifying your counter-example.
Your reviews are often quite good. Some of Richard's reviews
were very good and some of them are stuck in circular reasoning
anchored in very persistently false assumptions.
Les messages affichés proviennent d'usenet.