Re: HHH(DDD)==0 is correct

On 7/3/2025 9:16 AM, Richard Damon wrote:

On 7/2/25 10:50 PM, olcott wrote:

On 7/1/2025 11:37 AM, Mr Flibble wrote:

On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
>

On 6/30/25 2:30 PM, Mr Flibble wrote:
>
PO just works off the lie that a correct simulation of the input is
different than the direct execution, even though he can't show the
instruction actually correctly simulated where they differ, and thus
proves he is lying.
>
The closest he comes is claiming that the simulation of the "Call HHH"
must be different when simulated then when executed, as for "some
reason" it must be just because otherwise HHH can't do the simulation.
>
Sorry, not being able to do something doesn't mean you get to redefine
it,
>
You ar4e just showing you are as stupid as he is.

>
No. A simulator does not have to run a simulation to completion if it can
determine that the input, A PROGRAM, never halts.
>
/Flibble

>
The most direct way to analyze this is that
HHH(DDD)==0 and HHH1(DDD)==1 are both correct
because DDD calls HHH(DDD) in recursive simulation and
DDD does not call HHH1(DDD) in recursive simulation.

 Nope. It seems you don't understand what the question actually IS because you have just lied to yourself so much that you lost the understanding of the queiston.
 

>
*I can't imagine how Mike does not get this*

 I can't understand
 

>
*Context of above dialogue*
*Context of above dialogue*
*Context of above dialogue*

 Context of your context:
 A Halt Decider is supposed to decide if the program given to it (via some correct representation) will halt when run.
 Thus, "the input" needs to represent a program
 

>
typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
   HHH(DDD);
   return;
}
>
int main()
{
   HHH(DDD);
}

 Which, by itself, isn't a valid input, or program. as HHH is undefined.
 Each different definition of HHH, gives a different problem.
 Your "logic" seems to be based on trying to re-define what a program is, which just makes it a lie.
 "Programs" must be complete and self-contained in the field of computability theory, something you don't seem to understand.
 

>
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0. (HHH1 has identical code)
>
>

 But it CAN'T simulate the above input. as it isn't valid.
 You need to add the code of HHH to the input to let HHH simulate "the input" to get anything.
 

No I do not. The above paragraph has every detail that is needed.

And at that point, you have different inputs for different HHHs, and possibly different behaviors, which you logic forgets to take into account, which just breaks it.
 

Wrong.
It is because the what I specified does take this
into account that HHH(DDD)==0 and HHH1(DDD)==1 are correct.

Sorry, but you sre just showing your own stupidity, and ignorance of what you are talking about.

You never point out any actual mistake.
You only point out what you fail to understand.
*Every chatbot totally understands this correctly*
 >> Termination Analyzer HHH simulates its input until
 >> it detects a non-terminating behavior pattern. When
 >> HHH detects such a pattern it aborts its simulation
 >> and returns 0.
That you do not understand is your own limitation.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer