Re: What is the correct halt status for HHH(DDD) ? --- Tautology

On 7/13/24 3:45 PM, olcott wrote:

On 7/13/2024 2:40 PM, Richard Damon wrote:

On 7/13/24 3:23 PM, olcott wrote:

On 7/13/2024 2:10 PM, Richard Damon wrote:

On 7/13/24 2:53 PM, olcott wrote:

On 7/13/2024 1:33 PM, Richard Damon wrote:

On 7/13/24 2:15 PM, olcott wrote:

On 7/13/2024 12:25 PM, Richard Damon wrote:

On 7/13/24 12:48 PM, olcott wrote:

What is the correct halt status for an input to
a simulating termination analyzer that calls its
own termination analyzer?
>
typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
   HHH(DDD);
}
>
int main()
{
   HHH(DDD);
}
>

>
Halting.
>
Since HHH defined to be a termination analyzer, by that definition it must return to its caller.
>
Since DDD has no inputs, its behavior isn't affected by any inputs, and thus DDD will halt for ALL input conditions, so

>
You are stupidly saying that Infinite_Loop() will halt because
it has no inputs.
>

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Where did I say no input means halting?
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I said that since DDD has no inputs, a Termination analyizer doesn't need to look over all inputs, as there are no inputs to affect it.
>
Maybe you forget that a Termination Analyzer is not the same thing as a Halt Decider.
>
A Halt Decider determines if the given program will halt for the given input, and needs to be given both (if the program takes an input).
>
A Termination Analyzer determines if a given program will halt for every possible input, and thus is only given the program, and not the input to test.
>
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Note, for Infinite_Loop below, it IS possible for a simulator to detect a condition in a finite amount of time that an unbounded emulation would never halt, so can answer correctly non-halting.
>
The problem with trying to apply that to DDD is it isn't true, give the DDD that uses the HHH that tries to think it has determined it to never halt to a actual pure emulator (and thus DDD still calls that HHH that thinks it has determined that DDD will not halt and aborts its emulation and returns) then the pure emulatior will see that HHH make that decisioin and return to DDD which will return.
>
Thus HHH can never actually make that same conclusion. You logic is incorrrect as you presume the input changes with the emulator, but it actually doesn't, it changes with the emulator you have stated is the one that gives the correct answer, which you can't change in the middle of the problem.
>

void Infinite_Loop()
{
   HERE: goto HERE;
}
>

for HHH to be a correct termination analysizer it needs to return the value to indicate Halting.
>

Yes

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And thus, it WILL return that value to the DDD that calls it, and that DDD will halt.
>

>

Your version

>
I am asking What is the correct halt status for HHH(DDD)
that at least one element of every possible pure function
HHH can provide.
>

>
The only correct answer it is able to give is Halt.
>

>
Yet any input that must be aborted to prevent the non
termination of HHH necessarily specifies non-halting
behavior or it would never need to be aborted.

>
But the DDD that this HHH is given doesn't need to be aborted.
>
That is proven by HHH1(DDD).
>

>
That is not the same DDD instance and you know it.
That is the DDD after HHH has aborted its DDD.
>
My code shows HHH1 creating a process context and
HHH creating two more different contexts.
>
HHH1 depends on HHH aborting its own different DDD
for the DDD of HHH1 to halt.
>

>
And you ignore the fundamental fact that ALL DDDs built on the same code in HHH will behave the same.
>

 *You are just going to lie about this into perpetuity*
Any input that must be aborted to prevent the non
termination of HHH necessarily specifies non-halting
behavior or it would never need to be aborted.
 

But *THIS* input doesn't need to be aborted, because the HHH that this DDD calls DOES abort its simulation (because the one you claimed to be correct did) and thus THIS HHH could be replaced with a fully correct emulator (done without changing the input, so is at a different memory address) and that will simulate it to the end,
Your problem is your setup make you perform incorrect logic because you don't notice you are changing your input when you aren't allowed to in order to prove "need" for THIS input.
You just don't understand the fundamental of programming to know that you can't change the input when you are arguing about the properties of THAT input.