Re: My reviewers think that halt deciders must report on the behavior of their caller

On 6/6/2025 2:29 PM, joes wrote:

Am Fri, 06 Jun 2025 12:28:50 -0500 schrieb olcott:

On 6/6/2025 3:13 AM, Fred. Zwarts wrote:

Op 05.jun.2025 om 17:46 schreef olcott:

On 6/5/2025 2:01 AM, Fred. Zwarts wrote:

Op 05.jun.2025 om 06:33 schreef olcott:

On 6/4/2025 10:41 PM, dbush wrote:

On 6/4/2025 11:32 PM, olcott wrote:

On 6/4/2025 9:56 PM, dbush wrote:

On 6/4/2025 10:44 PM, olcott wrote:

On 6/4/2025 9:13 PM, dbush wrote:

On 6/4/2025 10:09 PM, olcott wrote:

On 6/4/2025 8:43 PM, Richard Damon wrote:

On 6/4/25 11:50 AM, olcott wrote:

 

The only possible way that HHH can report on the direct
execution of DDD() is for HHH to report on the behavior of
its caller:

It doesn't need to. It has all the code of DDD to analyse.
 

int main()
{
     DDD();     // this HHH(DDD);  // is not the caller of
     this: this
is }              // asking what the above will do

>
That is just not the way that computation actually works.

 Yes it is. Your comments are correct: running HHH(DDD) is asking what
running DDD() will do. You can do so by simulating it *uninterrupted*.
You won't of course always get an answer, not because DDD wouldn't halt,
but because HHH doesn't. In preventing itself non-halting, HHH gets
the answer wrong. All would be well if it just returned that DDD halts.
 

Sure it is.  We don't care how the mapping is generated, only that
it is generated.

There is not enough information in the input to know how the caller
works.

 There definitely is. The code of DDD (including HHH) completely determines
its path.
 

Counterfactual. The input is a pointer to the start of a function.

>
Prove it.
>

It seems you have no idea of the C language.
In HHH(DDD), DDD is a parameter. This parameter is the input for HHH.
DDD is also a function. In C a function, when used as a parameter, is a
pointer.
I assume this is another clever way to distract the attention from the
fact that your claims are counterfactual.

>
DDD emulated by HHH specifies executed HHH emulates DDD that calls
emulated HHH(DDD)
that emulates DDD that calls emulated HHH(DDD)
that emulates DDD that calls emulated HHH(DDD)...

No, DDD specifies that it calls HHH to simulate itself, whereupon that
HHH will stop simulating and return to DDD,

void DDD()
{
   HHH(DDD);
   return;
}
The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*
*Every rebuttal to this changes the words*
*YOU JUST CHANGED THESE WORDS*

which would then halt and
the outer HHH would report so - if the outer HHH wouldn't abort before.
 

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer