Re: A computable function that reports on the behavior of its actual self is not allowed

On 5/12/24 8:12 PM, olcott wrote:

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:

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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.

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https://en.wikipedia.org/wiki/Computable_function
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A computable function that reports on the behavior of its actual
self (or reports on the behavior of its caller) is not allowed.
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A decider must halt whereas simulating a pathological input
that would never halt unless aborted can only halt by aborting.
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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.
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*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
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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.

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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.
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We can look at the two possible cases.
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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.
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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.
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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.
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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^.
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*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*

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Nope.
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Claiming to do something that isn't the way defined doesn't make it work.
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 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.

It isn't even a "program" per the term-of-the-art, since it generates and answer without ending, thus your "proof" is based on lies.

 

That it like claiming you can make you cat bark, by trying to call your dog a cat.
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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.

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Which ever answer H gives, will be wrong.
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 I JUST PROVED OTHERWISE.
YES IS THE CORRECT ANSWER AND AS LONG AS H PROVIDES
THIS ANSWER WITHOUT STOPPING NOTHING CONTRADICTS IT.
 

Where?
You claimed the equivalent of a cat being a 10 story office building.
embedded_H can NOT give an answer and continue to run and still be a program per the computability theory.
Your claim is that the right answer can't be the right answer because it is impossible for H to determine it. That is just throwing a two-year-olds hissy fit.
You just don't seem to understand the basic computation model., because you are too stupid.