Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT

On 2025-06-25 14:14:07 +0000, olcott said:

On 6/25/2025 1:27 AM, Mikko wrote:

On 2025-06-24 14:06:12 +0000, olcott said:
 

On 6/24/2025 2:54 AM, Fred. Zwarts wrote:

Op 23.jun.2025 om 16:50 schreef olcott:

On 6/23/2025 3:24 AM, Fred. Zwarts wrote:

Op 22.jun.2025 om 21:27 schreef olcott:

On 6/22/2025 11:01 AM, Fred. Zwarts wrote:

Op 20.jun.2025 om 16:53 schreef olcott:

On 6/20/2025 4:42 AM, Fred. Zwarts wrote:

Op 19.jun.2025 om 17:23 schreef olcott:

On 6/19/2025 3:55 AM, Fred. Zwarts wrote:

Op 18.jun.2025 om 17:41 schreef olcott:

On 6/18/2025 4:36 AM, Fred. Zwarts wrote:

Op 17.jun.2025 om 16:36 schreef olcott:

On 6/17/2025 4:28 AM, Fred. Zwarts wrote:

Op 17.jun.2025 om 00:26 schreef olcott:

On 6/16/2025 3:53 AM, Fred. Zwarts wrote:

Op 15.jun.2025 om 22:10 schreef olcott:

void DDD()
{
   HHH(DDD);
   return;
}
 When I challenge anyone to show the details of exactly
how DDD correctly simulated by ANY simulating termination
analyzer HHH can possibly reach its own simulated "return"
statement final halt state they ignore this challenge.

 It seems very difficult for you to read.
We clearly stated that the challenge is improper.

 Are you too stupid to understand that dogmatic
assertions that are utterly bereft of any supporting
reasoning DO NOT COUNT AS REBUTTALS ???

 No, you are too stupid to realise that challenging for a recipe to draw a square circle does not count as a proof that square circles exist.
 

 Claiming that I made a mistake with no ability to
show this mistake is DISHONEST.
 

 Indeed, but irrelevant,

 That alternative is that you are dishonest.
When you claim that I am wrong and have
no ability to show how and where I am wrong
this would seem to make you a liar.
 No one has ever even attempted to show the details
of how this is not correct:
 void DDD()
{
   HHH(DDD);
   return;
}
 When one or more instructions of DDD are correctly
simulated by ANY simulating termination analyzer HHH
then this correctly simulated DDD never reaches its
simulated "return" statement final halt state.

 Indeed, HHH fails to reach the end of the simulation, even though the end is only one cycle further from the point where it gave up the simulation.
 

 That is counter-factual and over-your-head.
 

 No evidence presented for this claim. Dreaming again?
Even a beginner understands that when HHH has code to abort and halt, the simulated HHH runs one cycle behind the simulating HHH, so that when the simulating HHH aborts, the simulated HHH is only one cycle away from the same point.

 Proving that you do not understand what unreachable code is.
First year CS students and EE majors may not understand this.
All CS graduates would understand this.

 That you do not understand what I write makes it difficult for you to learn from your errors.
It is not that difficult. Try again and pay full attention to it.
Even a beginner understands that when HHH has code to abort and halt,
the simulated HHH runs one cycle behind the simulating HHH, so that when the simulating HHH aborts, the simulated HHH is only one cycle away from the same point.

 Yes this is factual.
 *This is only ordinary computer programming with*
*no theory of computation computer science required*
 Every simulated HHH remains one cycle behind its simulator
no matter how deep the recursive simulations go. This means
that the outermost directly executed HHH reaches its abort
criteria first.

 And it fails to see that the simulated HHH would reach exactly the same abort criteria one cycle later.
In this way, it misses the fact that it is simulating an HHH that would abort and halt.
 

 void Infinite_Loop()
{
   HERE: goto HERE;
   printf("Fred Zwarts can't understand this is never reached\n");
}
 

Another claim without any evidence.
 Olcott does not understand that his HHH does not see an infinite loop.
It aborts and halt, so the recursion is finite.

 You didn't even use the term recursion correctly.
Infinite loops have nothing to do with recursion.

 And infinite loops have nothing to do with a simulator simulating itself. Therefore, talking about infinite loops is changing the subject.
 

Mike understands that HHH could recognize an infinite
loop correctly.
     The process in which a function calls itself directly
    or indirectly is called recursion and the corresponding
    function is called a recursive function.
https://www.geeksforgeeks.org/introduction-to-recursion-2/
 Lines 987 to 992 is where infinite loops are recognized
Lines 996 to 1005 is where infinite recursion is recognized
https://github.com/plolcott/x86utm/blob/master/Halt7.c
 HHH correctly emulates the x86 machine code of its
input until one of those two patterns is matched.

 But there is a bug in the code that tries to recognise an infinite recursion.

 There is no bug. Quit your defamation.
 

It forgets to count the conditional branch instructions when simulating the simulator.

 *It does not forget them. They are irrelevant*
 The question being asked is this:
Can DDD correctly simulated by any termination analyzer
HHH that can possibly exist reach its own "return" statement
final halt state?

 Why would anyone ask that question or care about the answer?

 In computer science the only measure of halting
is reaching a final halt state. Stopping running
for any other reason does not count as halting.

In computer science the only measure of non-halting is the
possibility to execute an unlimited number of steps without
halting. An execution of a limited number of steps does not
count as non-haltign.
--
Mikko