Re: Simulating termination analyzers for dummies

On 6/17/2024 8:24 PM, Richard Damon wrote:

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.
 

It is ALWAYS the exact same sequence of bytes.

 

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.
 

Not so much.
void DDD()
{
   UTM(DDD);
}
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer