Liste des Groupes | Revenir à s logic |
On 3/11/24 6:39 PM, olcott wrote:There is no correct answer for this (program/input pair): (H/⟨Ĥ⟩ ⟨Ĥ⟩)On 3/11/2024 7:08 PM, Richard Damon wrote:So, you admit you were wrong, and are just stupid.On 3/11/24 4:00 PM, olcott wrote:>On 3/11/2024 12:13 PM, Richard Damon wrote:>On 3/11/24 6:19 AM, olcott wrote:>On 3/11/2024 1:02 AM, Richard Damon wrote:>On 3/10/24 10:44 PM, olcott wrote:>
>*Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
>
Yes, you have rotely repeated that many times, not knowing what that implies, or doesn't imply.
>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
>
*Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
*Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
*Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
Really?
>
A "submachine" doesn't have a requirement except to act as the machine it is a copy of.
>
There is no such thing as a sub-machine.
OF course there is.
>
What do you think H^.H is?
>
You are just showing how stupid you are.
An an abstract concept that humans can talk about that
cannot actually be implemented as any actual aspect of
any actual Turing machine.
Why isn't H^.H the implementation in the Turing Machine H^ of a submachine that is a copy of H?
>So seems most of your ideas.
The ghost of Bug Bunny exists as a fictional idea.
>If H (H^) (H^) says Qn, as you normally say, then it is a FACT that>>>So, what you are REALLY Saying is that:>
>
H (H^) (H^) gets the wrong answer for every implementation of H,
>
Both answers of YES and NO are incorrect for any H ⟨Ĥ⟩ ⟨Ĥ⟩
that is inconsistent with the behavior of Ĥ ⟨Ĥ⟩ when an
exact copy of this same H is embedded within Ĥ.
No, there IS a 'Correct answer', it just isn't the one that that H gives.
>
You are bullshitting yourself, yet not bullshitting me.
There is no correct answer for the H/⟨Ĥ⟩ ⟨Ĥ⟩ program/input pair.
H^ (H^) will go to Qn and Halt,
And thus the CORRECT ANSWER is Qy.
If you changed your mind, an now H (H^) (H^) says Qy, then it is a fact that H^ (H^) will also goto Qy and loop forever.
And thus, the CORRECT ANSWER is Qn
Les messages affichés proviennent d'usenet.