Re: A simulating halt decider applied to the The Peter Linz Turing Machine description ⟨Ĥ⟩

On 5/26/24 10:53 PM, olcott wrote:

On 5/26/2024 9:30 PM, Richard Damon wrote:

On 5/26/24 10:06 PM, olcott wrote:

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
  Ĥ copies its own Turing machine description: ⟨Ĥ⟩
  then invokes embedded_H that simulates ⟨Ĥ⟩ with ⟨Ĥ⟩ as input.
>
It is an easily verified fact that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
embedded_H cannot possibly reach its own simulated final state of
⟨Ĥ.qn⟩ in any finite sequence of steps such as Googolplex ^ Googolplex
number of steps. https://en.wikipedia.org/wiki/Googolplex

>
So, it can verify that if H / embedded_H was programmed not to halt it simulation, then the H^ built on that H will be non-halting, but that doesn't say anything about the DIFFERENT H^ built on an H that does abort its simulation and returns 0.
>

 You really do have an actual problem paying attention.
It is not a mere trollish ruse. I poured engine oil
into my power steering fluid Friday night so I can relate.

Shows your lack of paying attention to details, as you argument also shows.

 As far as paying attention to words that have been written
I read and reread many times until I am sure that I get it.
Even then I get it wrong once in a while.

You mean most of time, Rereading doesn't help when you don't know the meaning of the words. Thats the problem with talking about a field you haven't studied.

 When embedded_H correctly simulates a Googolplex ^ Googolplex steps
of ⟨Ĥ⟩ ⟨Ĥ⟩ then embedded_H halts.

And thus embedded_H is NOT a UTM, and its simulation say NOTHING about the behavior of the machine it is simulating.

 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its
own simulated final state of ⟨Ĥ.qn⟩ in any finite number of steps.
 

Which is a statment that means nothing, as you have implicitly agreed.
An aborted simulation does NOT indicate that the input is a non-halting machine.
By the exact same logic, a machine that simulated ONE step and then aborted can say that it was unable to reach the final state.
Remember, in all the machine you looked at, each H^ is only looked at by ONE decider, which simulated for some number of steps and then gave up and aborted.
NONE of the other machine/input pairs have ANY connection to that input, as they are simulations of DIFFERENT inputs.
You may be figuring out how to word your statements to get the answer you want, but in doing so you are removing all the meaning out of that statement that you need to go on.
All you have shown is that for this type of input, you simulator aborts its simulation before it gets to the end, and tries to find logic, which doesn't exist, to say it was right in aborting.
Since for every input, based on a simulator that does decide to abort and return 0, we can simulate that input with a UTM (or a variation of H that just simulates long enough) and reach the final state, your claim that embedded_H "can't" reach the final state is incorrect. It DOESN'T reach the final state, but for the input it was given, if it simulated longer, it would have. Note the key, the input it was given, as that is the problem in view, the specific input given to the specific H you want to present as your correct halt decider.