Re: 197 page execution trace of DDD correctly simulated by HHH

Op 28.jun.2024 om 18:54 schreef olcott:

On 6/28/2024 10:48 AM, Fred. Zwarts wrote:

Op 28.jun.2024 om 17:32 schreef olcott:

On 6/28/2024 10:22 AM, Fred. Zwarts wrote:

Op 28.jun.2024 om 17:04 schreef olcott:

On 6/28/2024 9:53 AM, Fred. Zwarts wrote:

Op 28.jun.2024 om 16:27 schreef olcott:

On 6/28/2024 3:23 AM, Fred. Zwarts wrote:

Op 27.jun.2024 om 19:30 schreef olcott:

>
When you prove that you are totally overwhelmed and confused
by the original issue I break it down into simpler steps.
>
If you don't have a slight clue about the C programming
language then the first step is you must learn this language
otherwise it is like trying to talk to someone about
differential calculus that does not know how to count to ten.

>
If... But since this if does not apply, the the is irrelevant.
You keep repeating irrelevant texts to hide that you cannot show any error in my reasoning.
>

>
typedef void (*ptr)();
int H0(ptr P);
>
void Infinite_Loop()
{
   HERE: goto HERE;
}
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void DDD()
{
   H0(DDD);
}
>
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}
>
Every C programmer that knows what an x86 emulator is knows that when H0
emulates the machine language of Infinite_Loop, Infinite_Recursion, and
DDD that it must abort these emulations so that itself can terminate
normally.
>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as non-halting
by returning 0 to its caller.
>
Simulating termination analyzers must report on the behavior that their
finite string input specifies thus H0 must report that DDD correctly
emulated by H0 remains stuck in recursive simulation.
>

>
Another attempt to distract from the subject.You claim you are not talking about halt-deciders or termination analyzers, but now you bring them up again.
>

>
https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
>
I only do this because you have gotten overwhelmed.
I prove my point step-by-step and because you don't
understand any of the steps you leap to the conclusion
that I am wrong.
>

We are discussing an H0 that aborts after two cycles. I do not tolerate to go away from this point.
>

>
I updated all of my names in my code.
// HHH(DDD) and HHH1(DDD) are the standard names for DDD input
// DDD calls HHH(DDD). HHH1 is identical to HHH.
>
// HH(DD,DD) and HH1(DD,DD) are the standard names for (DD,DD) input
// DD calls HH(DD,DD) and HH1 is identical to HH.
>
>
You haven't shown that you even understand that Infinite_Recursion()
doesn't halt. You must understand this before you can understand
the more complex example of DDD.

>
We agreed to talk only about the simulator which aborts after two cycles of recursive simulation.

>
Not if you don't have the prerequisites.
>

>
I have them.
But you try to distract from the fact that you do not even understand a two cycle recursive simulation. We cannot talk about infinite recursion before you understand a two cycle recursive simulation..

>
I spent two years coming up with these precise words before
I contacted professor Sipser for his approval.
>
<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>
>
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.
 >
>

>
But there is no correct simulation, so Sipser's approval does not apply.

 The semantics of the x86 programming language specifies that
N steps of DDD were correctly emulated until the infinite
recursion behavior pattern was correctly matched.

What you say is that two equals infinite.
*We are not talking about infinite recursion. We are talking about a two cycle recursive simulation.*

 This has better color coding than the prior version.
https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
 

Let's agree about a two cycle recursive simulation, for which you have not shown any evidence that it can be done correctly.

 I have proven the verified fact that DDD is correctly emulated by
HHH and HHH correctly emulates itself emulating DDD until the outer
HHH sees that DDD meets the infinite recursion behavior pattern.
 

There is no infinite recursion pattern. The simulated HHH runs one cycle behind the simulating HHH. The simulating HHH aborts the simulated HHH only one cycle before the simulated HHH would return.
Aborting the simulation at this point makes it incorrect.
So, stop talking about infinite, because apparently do not understand what it means. Infinite is much more than two.