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On 5/12/2024 5:40 PM, Richard Damon wrote:Nope.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.*It <is> a way for embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to get the correct answer*So, you just don't understand how algorithms work, and how compuations are DEFINED.>>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.
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.
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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*
Every prior work that I have ever seen and probably every priorWhich ever answer H gives, will be wrong.
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.
I have shown a weird yet not impossible way for H say say NONope, just shows you don't understand how programs work.
correctly.
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In this case embedded_H <is> an actual UTM that has the extra
feature of examining all of the state transitions of its input
to see what we can all see that Ĥ ⟨Ĥ⟩ remains stuck in recursive
simulation.
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And a UTM doesn't reveal its answer until it come to a final state, just like ALL Turing Machines or equivalent computation.
>*Or we can get an actual partial halt decider as follows*>
*Or we can get an actual partial halt decider as follows*
*Or we can get an actual partial halt decider as follows*
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No decider is ever allowed to report on its own behavior thus embedded_H
as a simulating partial halt decider is NOT ALLOWED to report on the
direct execution of Ĥ ⟨Ĥ⟩ because this IS REPORTING ON ITS OWN BEHAVIOR.
WHO SAYS THIS?
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