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

On 6/18/2024 9:16 PM, Richard Damon wrote:

On 6/18/24 1:25 PM, olcott wrote:

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.

 No "normally" as Turing Machine have no "abnormal terminatiom"
 You just don't understand what they are.
 

>
We can derive a notion of abnormal termination for Turing
machines from the standard terms-of-the-art.

 How?
 

>
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.
>

 And then are no longer UTMs, and YES, if a machine based on such am modifed UTM (so it is no long a UTM) when the UTM stops simulating, we can not say the input halted, nor can we say it didn't halt.
 

When such a UTM has been adapted to only simulate
the first ten states of its input TMD, then every
simulated TMD with more than ten states did not
terminate normally.

The not-a-UTM just came to a no-answer state.
 

I have to go one-step-at-a-time with everyone or
they get overwhelmed and leap to the conclusion
that I am wrong.

The answer will be provided by useing an ACTUAL UTM that keeps on going, or the direct execution of the machine,
 You are just stuck in your idea that Lies are sometimes ok.

You are stuck on the idea that repeating states cannot
be recognized in a finite number of steps.
void Infinite_Loop()
{
   HERE: goto HERE;
}
void Infinite_Recursion()
{
   Infinite_Recursion();
}
void DDD()
{
   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.
This was recently confirmed in the C group.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer