Re: Simulating termination analyzers by dummies --- criteria is met

On 6/23/2024 2:57 AM, Mikko wrote:

On 2024-06-22 14:11:28 +0000, olcott said:
 

On 6/22/2024 8:27 AM, Richard Damon wrote:

On 6/22/24 9:04 AM, olcott wrote:

>
I am the sole inventor of the simulating halt decider.
>
Ben Bacarisse contacted professor Sipser to verify that he
really did says this. The details are in this forum about
the same date.
>
https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
   If simulating halt decider H correctly simulates its input D
   until H correctly determines that its simulated D would never
   stop running unless aborted then
>
   H can abort its simulation of D and correctly report that D
   specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

>
And, as I remember, he also verified that he disagrees with your definition of correct simulation.
>

>
*Ben also verified that the criteria have been met*
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.

>
Right, Ben was willing to do what I am not that you can prove that, by your definition, H can show that it "must" abort its simulation or the input will run forever.
>
But, just like me, he also agrees that this is NOT the defintion of Halting, so H is just shown to be a correct (partial) POOP decider but ot a Halt Decider, not even for that one input.
>

>
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.
 >
 > He knows and accepts that P(P) actually does stop. The
 > wrong answer is justified by what would happen if H
 > (and hence a different P) where not what they actually are.
 >
*Ben agrees that the criteria is met for the input*
>
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
>
*Ben disagrees that the criteria is met for the non-input*
Yet no one here can stay focused on the fact that non-inputs
*DO NOT COUNT*

 In particular, you can't. You have insisted that your "decider"
or "anlyzer" (or whatever word you happen to use) H or HH (or
hwatever name you happen to use) must return false because a
non-input (where instead of the actually called function another
function that does not halt is called) does not halt.
 

You said it backwards. When I say that I am not guilty and did
not rob the liquor store you cannot paraphrase this as he admitted
that he robbed the liquor store.
H performs a sequence of finite string transformations on
its finite input of x86 machine code. These transformations
include that D calls H(D,D) while being simulated by H. In
such a case the call from D to H(D,D) cannot possibly return.

void DDD()
{
   HHH0(DDD);
}
>
int main()
{
   Output("Input_Halts = ", HHH0(DDD));
   Output("Input_Halts = ", HHH1(DDD));
}
>
It is a verified fact that the behavior that finite string DDD
presents to HH0 is that when DDD correctly simulated by HH0
calls HH0(DDD) that this call DOES NOT RETURN.
>
It is a verified fact that the behavior that finite string DDD
presents to HH1 is that when DDD correctly simulated by HH1
calls HH0(DDD) that this call DOES RETURN.

 

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