Re: Simulating termination analyzers for dummies

On 6/17/24 9:16 PM, olcott wrote:

On 6/17/2024 5:42 PM, Richard Damon wrote:

On 6/17/24 8:20 AM, olcott wrote:

On 6/17/2024 3:31 AM, Fred. Zwarts wrote:

Op 17.jun.2024 om 05:33 schreef olcott:

To understand this analysis requires a sufficient knowledge of
the C programming language and what an x86 emulator does.
>
Unless every single detail is made 100% explicit false assumptions
always slip though the cracks. This is why it must be examined at
the C level before it is examined at the Turing Machine level.
>
typedef void (*ptr)();
int H0(ptr P);
>
void Infinite_Loop()
{
   HERE: goto HERE;
}
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void DDD()
{
   H0(DDD);
   return;
}
>
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}
>
Every C programmer that knows what an x86 emulator is knows that when H0
emulates the machine language of Infinite_Loop, Infinite_Recursion, and
DDD that it must abort these emulations so that itself can terminate
normally.
>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as non-
halting.
>

>
For Infinite_Loop and Infinite_Recursion that might be true, because there the simulator processes the whole input.
>
The H0 case is very different. For H0 there is indeed a false assumption, as you mentioned. Here H0 needs to simulate itself, but the simulation is never able to reach the final state of the simulated self. The abort is always one cycle too early, so that the simulating H0 misses the abort. Therefore this results in a false negative.
(Note that H0 should process its input, which includes the H0 that aborts, not a non-input with an H that does not abort.)
>
This results in a impossible dilemma for the programmer. It he creates a H that does not abort, it will not terminate.

>
*Therefore what I said is correct*
When every input that must be aborted is construed as non-halting
then the input to H0(DDD) is correctly construed as non-halting.

>
In other words, if you allow yourself to LIE, you can claim the wrong answer is right.
>
Since your "Needing to abort" is NOT the same as halting, all you are doing is admitting that your whole logic system is based on the principle that LIES ARE OK.
>

 "Needing to abort" <is> the same as a NOT halting input.
You are simply too ignorant to understand this.
 

Nope, not if you are comparing DIFFERENT version of the input.

When I explain this in terms of of mathematical mappings
from finite strings to behaviors this simply leaps over
everyone's head.
 

As has been shown, you can apply the input to H0 (when you don't change it, so the call to H0 still goes to this H0), to a UTM and it will reach the final end, so *THIS* H0 did not "Need" to abort its input, but did because it was programmend to.
Change the H0 to H1, which simulates longer, and keep the input DDD calling H0, and we see that if H1 simulates long enough, it will see the end to.
The problem is you try to justify changing the input when you change 'H', which is just a lie, as the input needs to be a COMPLETE PROGRAM to even HAVE behavior, and that complere program includes the exact version of the decider that it is designed to refute,
Yes, your verskon where the input changes shows something, and that is that you don't understand the meaning of theory.
We can only ask for behavior of specific programs, your idea of a "template" as in input is flawed, as the template doesn't HAVE a definied behavior, only specific instances do.
It is like being asked the sum of 2 + and then not giving the second number, there is no valid answer to that.
Thus, your logic is just shown to be based on invalid principles and LIES.