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

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.
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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.
>

>
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*
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.
I have shown a weird yet not impossible way for H say say NO
correctly.

>
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.
>

 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*
>
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?
 

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer