Re: How Partial Simulations correctly determine non-halting ---Mike Terry Error

On 6/5/2024 12:25 PM, John Smith wrote:

On 5/06/24 18:49, olcott wrote:
 

My idea was to have the executed HH pass a portion of what is
essentially its own Turing Machine tape down to the simulated
instances of HH. It does do this now.

 In other words, your idea is to have an incorrect simulation.
 DD(DD) doesn't get passed a portion of any other machine's tape. If the simulation does, then the simulation is simulating something other than DD(DD). It might be simulating DD(DD,SecretParameter).
 

The only issue left that seems to not matter is that each simulated
HH needs to see if it must initialize its own tape.

 Why shouldn't it always initialize its own tape?
   Since this

has no effect on its halt status decision I don't think it makes
any difference.
>
I will double check everything to make sure there is no data passed
from the outer simulations to the inner simulations that can possibly
be used for any halt status decision by these inner simulated
instances of HH.

 There are actually two different ways that it's possible to understand your program.
 The inner DebugStep doesn't work the same as the outer DebugStep. Depending on which kind of viewpoint we use, we can say that the inner DebugStep is using secret information so it knows to work differently, or we can say that the inner DebugStep is normal (if directly executed), but it simulates differently from its direct execution (an incorrect simulation).
 

>
I really appreciate your help on this.
>

 

You don't seem to know enough computer science about this.
Mike Terry has reviewed it and I am going by his review.
Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
On 3/1/2024 12:41 PM, Mike Terry wrote:
 >
 > Obviously a simulator has access to the internal state
 > (tape contents etc.) of the simulated machine. No problem there.
I am thinking as long as the inner simulations do not receive any
data that can possibly change their own halt status decision then
we still have a computable function.
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
There must be some way that a UTM can pass a portion of its own tape
down to the simulated instances of itself so that this outer directly
executed UTM can see what they are doing.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer