Re: Simulating termination analyzers by dummies --- What does halting mean?

On 2024-06-19 13:11:10 +0000, olcott said:

On 6/19/2024 3:18 AM, Fred. Zwarts wrote:

Op 18.jun.2024 om 19:25 schreef olcott:

On 6/18/2024 12:06 PM, joes wrote:
 void DDD()
{
   H0(DDD);
}
 DDD correctly simulated by any H0 cannot possibly halt.
 

DDD halts iff H0 halts.

 Halting is a technical term-of-the-art that corresponds
to terminates normally. Because Turing machines are
abstract mathematical objects there has been no notion
of abnormal termination for a Turing machine.
 We can derive a notion of abnormal termination for Turing
machines from the standard terms-of-the-art.
 Some TM's loop and thus never stop running, this is classical
non-halting behavior. UTM's simulate Turing machine descriptions.
This is the same thing as an interpreter interpreting the
source-code of a program.
 A UTM can be adapted so that it only simulates a fixed number
of iterations of an input that loops. When this UTM stops
simulating this Turing machine description we cannot correctly
say that this looping input halted.
 

 If the code specifies 5 iterations and the simulator simulates only 3 iterations, it is incorrect to conclude that the repetition show non-halting behaviour.

 It is correct do say that the simulated input did not terminate
normally, thus defining the notion of abnormal termination
within Turing machines.

No, it is not. It is correct to say that the input was not simulated to its
normal termination. The input to the simulator is not a simulated input, it
is a real input.
You have not defined "abnormal termination" of a Turing machine, nor
presented any reason to do so.
--
Mikko