Liste des Groupes | Revenir à s logic |
On 3/10/24 6:24 PM, olcott wrote:In computability theory and computational complexity theory,On 3/10/2024 7:40 PM, immibis wrote:And when you remember that when we posse that ACTUAL question, the input is a FIXED machine, (not a template that changes by the decide that it trying to decide it) then there are a LOT of machines that get the right answer. The key is we know that there is ONE that doesn't, the one that particular decider was built to foil. Thus, the problem isn't an invalid question.On 10/03/24 20:16, olcott wrote:>On 3/10/2024 2:09 PM, immibis wrote:>On 10/03/24 19:32, olcott wrote:>On 3/10/2024 1:08 PM, immibis wrote:>On 10/03/24 18:17, olcott wrote:>ZFC simply tossed out the Russell's Paradox question as unsound.>
So you are saying that some Turing machines are not sound?
>>ZFC simply tossed out the Russell's Paradox question as unsoundBoth H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:>
(a) Their input halts H.qy
(b) Their input fails to halt or has a pathological
relationship to itself H.qn.
But the "Pathological Relationship" is ALLOWED.
>
expressly disallowing the "Pathological Relationship".
So you are saying that some Turing machines are not real Turing machines?
>I am only claiming that both H and Ĥ.H correctly say YES>
when their input halts and correctly say NOT YES otherwise.
well the halting problem requires them to correctly say NO, so you haven't solved it
All decision problem instances of program/input such that both
yes and no are the wrong answer toss out the input as invalid.
>
all decision problems are defined so that all instances are valid or else they are not defined properly
>
Not in the case of Russell's Paradox.
And now we are back to: Every Turing machine and input pair defines an execution sequence. Every sequence is either finite or infinite. Therefore it is well-defined and there is no paradox.
>
Can you show me a Turing machine that specifies a sequence of configurations that is not finite or infinite?
When we construe every yes/no question that cannot possibly
have a correct yes/no answer as an incorrect question
>
then we must correspondingly construe every decider/input
pair that has no correct yes/no answer as invalid input.
>
Les messages affichés proviennent d'usenet.