Sujet : Re: A computable function that reports on the behavior of its actual self is not allowed
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logicDate : 13. May 2024, 01:12:40
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v1rlto$324ln$3@dont-email.me>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
On 5/12/2024 6:57 PM, Richard Damon wrote:
On 5/12/24 7:14 PM, olcott wrote:
On 5/12/2024 5:40 PM, Richard Damon wrote:
On 5/12/24 6:21 PM, olcott wrote:
On 5/12/2024 4:40 PM, Richard Damon wrote:
On 5/12/24 5:18 PM, olcott wrote:
On 5/12/2024 2:27 PM, olcott wrote:
>
Computable functions are the basic objects of study in computability
theory. Computable functions are the formalized analogue of the
intuitive notion of algorithms, in the sense that a function is
computable if there exists an algorithm that can do the job of the
function, i.e. given an input of the function domain it can return the
corresponding output.
>
https://en.wikipedia.org/wiki/Computable_function
>
A computable function that reports on the behavior of its actual
self (or reports on the behavior of its caller) is not allowed.
>
A decider must halt whereas simulating a pathological input
that would never halt unless aborted can only halt by aborting.
>
This causes the direct execution of this input after it has been aborted
to have different behavior than the simulated input that cannot possibly
stop running unless aborted.
>
>
*MORE PRECISE WORDING* (this may take a few more rewrites)
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
It is a verified fact that the directly executed Ĥ ⟨Ĥ⟩ cannot possibly
stop running unless simulating partial halt decider embedded_H aborts
its simulation of its input.
>
But since embedded_H implements a specific algorithm, either it will or it won't. "unless" is a meaningless word here, it implies a case that can't happen.
>
We can look at the two possible cases.
>
First, if embedded_H doesn't ever abort its simulation, then, as you have desceribed, THAT embedded_H creates a H^ that will never halt, but the H that was based on will also never abort its simulation (or you lied that embedded_H is the needed copy of H) and thus never answer and fail to be a decider.
>
>
It can answer without halting by transitioning to its own internal
non-final state of embedded_H.qn without ever reaching Ĥ.qn. Every
simulated instance of embedded_H would do this same thing and then
continue simulating its input.
So, you just don't understand how algorithms work, and how compuations are DEFINED.
>
>
If you want to try to define a new system of compuation that allows giving answer without the algorithm ending, but still allows all the composition operations that are included in computation theory, go ahead and try.
>
You then need to show that it is Turing Complete, which means that you can't outlaw any computation allowed in a Turing Machine, like H^.
>
>
*It <is> a way for embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to get the correct answer*
*It <is> a way for embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to get the correct answer*
*It <is> a way for embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to get the correct answer*
Nope.
Claiming to do something that isn't the way defined doesn't make it work.
It is not a halt decider. It is not even a termination analyzer.
It <is> an huge improvement over both YES and NO from H are the
wrong answer. NO from H IS THE CORRECT ANSWER.
That it like claiming you can make you cat bark, by trying to call your dog a cat.
>
Every prior work that I have ever seen and probably every prior
work that exists essentially concludes that both YES and NO are the
wrong answer for H to provide for every H/D pair where H and D have
the HP pathological relationship.
Which ever answer H gives, will be wrong.
I JUST PROVED OTHERWISE.
YES IS THE CORRECT ANSWER AND AS LONG AS H PROVIDES
THIS ANSWER WITHOUT STOPPING NOTHING CONTRADICTS IT.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer