Sujet : Re: HHH maps its input to the behavior specified by it --- HHH never reaches its halt state so never decides, or it decides wrong.
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 08. Aug 2024, 03:04:17
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <57b1a48ad47f583a80e4c85fbb61a1ad29a60d12@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Mozilla Thunderbird
On 8/7/24 9:35 PM, olcott wrote:
On 8/7/2024 8:22 PM, Richard Damon wrote:
On 8/7/24 9:12 PM, olcott wrote:
On 8/7/2024 8:03 PM, Richard Damon wrote:
On 8/7/24 2:14 PM, olcott wrote:
On 8/7/2024 1:02 PM, joes wrote:
Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:
On 8/7/2024 2:29 AM, Mikko wrote:
On 2024-08-05 13:49:44 +0000, olcott said:
>
I know what it means. But the inflected form "emulated" does not mean
what you apparently think it means. You seem to think that "DDD
emulated by HHH" means whatever HHH thinks DDD means but it does not.
DDD means what it means whether HHH emulates it or not.
>
In other words when DDD is defined to have a pathological relationship
to HHH we can just close our eyes and ignore it and pretend that it
doesn't exist?
It doesn't change anything about DDD. HHH was supposed to decide anything
and can't fulfill that promise. That doesn't mean that DDD is somehow
faulty, it's just a counterexample.
>
>
void DDD()
{
HHH(DDD);
return;
}
>
*HHH is required to report on the behavior of DDD*
Anyone that does not understand that HHH meets this criteria
has insufficient understanding.
>
But it doesn't, as a correct simulation of a DDD that calls an HHH that returns will stop running,
>
I really think that you must be a liar here because
you have known this for years:
>
On 8/2/2024 11:32 PM, Jeff Barnett wrote:
> ...In some formulations, there are specific states
> defined as "halting states" and the machine only
> halts if either the start state is a halt state...
>
> ...these and many other definitions all have
> equivalent computing prowess...
>
Anyone that knows C knows that DDD correctly simulated
by any HHH cannot possibly reach its "return" {halt state}.
>
>
But the problem is that you HHH ODESN'T correctly emulate the DDD it is given, because it aborts its emulation.
Each HHH of every HHH that can possibly exist definitely
emulates zero to infinity instructions correctly. In
none of these cases does the emulated DDD ever reach
its "return" instruction halt state.
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
*There are no double-talk weasel words around this*
Then why do you use double-talk wasel words that conflate the behavior of the program DDD that calls the HHH that emulates its input for a finite number of steps, and then returns 0 to its caller (that is DDD) making DDD halt, with the PARTIAL (and thus INCORRECT) emulation that HHH did of the DDD it was given that aborts before it gets to that point in the behavior of the program.
Yes, there is one class of HHH that actually do perform a correct emulation of their input, and never abort it, and satisfy your claim, unfortunately those HHH and the DDDs that call them never return to give an answer, and are a TOTALLY DISTINCT group of machines/inputs from the others, and becuase DDD does include its HHH as part of its definition (or it couldn't be emulated past 4 instructions) they are clearly DIFFERENT inputs so not relevent to your problem.
I KNOW, because you need that lie to make your false claim sound like it might be correct.
Sorry, but it seems everything you accuse others of doing tends to be something you actualy are doin gyour self.
Just like your buddy Trump does. You know, you use a lot of the same logic he does, ignore the truth and just keep repeating your lies.