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

On 6/30/2024 3:42 AM, joes wrote:

Am Sat, 29 Jun 2024 15:03:02 -0500 schrieb olcott:

On 6/29/2024 2:44 PM, joes wrote:

Am Fri, 28 Jun 2024 14:28:20 -0500 schrieb olcott:

On 6/28/2024 2:18 PM, joes wrote:

Am Fri, 28 Jun 2024 12:53:46 -0500 schrieb olcott:

On 6/28/2024 12:41 PM, joes wrote:

Thanks for leaving the unanswered questions in place, though I’d
rather have you answer them.
Am Fri, 28 Jun 2024 12:05:18 -0500 schrieb olcott:

On 6/28/2024 11:26 AM, joes wrote:

Am Fri, 28 Jun 2024 10:25:36 -0500 schrieb olcott:

On 6/28/2024 8:14 AM, joes wrote:

Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:

>

To the caller DDD, which then returns to its own caller H0,
which returns „halting” to main… hold on.

Why doesn’t the first recursive H return?

Question still not answered.

It should abort, just like the outer one.
>

The outer HHH meets its abort criteria one execution trace
sooner than the next inner one because HHH needs to see
two complete execution traces before its abort criteria
has been met.
 As soon as the outer HHH sees the inner one complete one
full execution trace then the outer one has its abort
criteria.
 

HHH(DDD)
simulates DDD that calls HHH(DDD) that simulates DDD that calls
HHH(DDD)
that proves to the outer directly executed HHH that it must abort and
reject.

How does it do that?

Over your head. I have explained it too many times and you just can't
get it.

 

No, I mean: why does the inner simulator repeat instead of aborting,
the same as the outer one does?
>

Technically it is called detecting a repeating state.

Yeah, I know. My point is: all recursive calls both enter and detect
a repeating state.
>

 The inner ones always see one less execution trace
than the next outer one, thus could only meet their
abort criteria after they have already been aborted.