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

Am Wed, 19 Jun 2024 08:31:24 -0500 schrieb olcott:

On 6/19/2024 4:08 AM, joes wrote:

Am Tue, 18 Jun 2024 21:51:56 -0500 schrieb olcott:

On 6/18/2024 9:41 PM, Richard Damon wrote:

On 6/18/24 10:30 PM, olcott wrote:

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:

Terminating is a property of the actual machine, and not a simulation
of it.

Still true.

Thus according to your faulty reasoning when the source-code of a C
program is simulated by interpreter this is mere nonsense gibberish
having nothing to do what the behavior that this source-code
specifies.

YOUR partial decider makes everything halt, even that which doesn't.
So yes, it can't simulate infinite loops.

My partial decider makes everything stop running this is not at all the
same as halting. Novices get very confused about this.

Your nonsimulator halts even when given nonterminating input.

You could say the SIMULATION didn't terminate normally, but you can't
say the machine didn't or even the Turing Machine Description, as you
could give that exact same TMD to a real UTM and find out the actual
behaviof or the input.

Sure you can otherwise interpreters of source-code would be a bogus
concept.

When I write an infinite loop, I want it to be interpreted as an
infinite loop. Your H0 is bogus.

[deflection snipped]

You just have lost track of the defintions of what is REALITY (the
actual behavior of the machine) and what is just imagination.

(a) If the simulation of the x86 machine language of the
function does not prove the actual behavior that this finite string
specifies then source-code interpreters are a bogus concept.
(b) source-code interpreters are NOT a bogus concept.

Your H0 is not an interpreter. It aborts infinite loops.

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.

Which doesn't mean the program DDD needs to be abort to have it halt.

The verified that that it does need to be aborted contradicts your
nonsense to the contrary.

[The verification that it ... ?]
If H0 halts, so does DDD (which only calls it).

My partial decider makes everything stop running this is not at all the
same as halting. Novices get very confused about this.

I was talking about DDD. It calls H0, which shall halt. Then DDD returns,
having terminated.

--
joes