Sujet : Re: HHH maps its input to the behavior specified by it --- never reaches its halt state
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 08. Aug 2024, 13:37:49
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <f6769b3a6eb0025227cf1e37abd8bd6670eb7f0c@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 8/8/24 12:41 AM, 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.
Nope, and none of the HHH's that you have shown a trace actually showed a correct x86 emulation in their output, as none of them have shown the instrucitons of HHH that DDD called.
The only trace that shows that wasn't generated by HHH, but was a trace of HHH doing its steps, which shows that the needed trace CAN be made, but you first need to make HHH follow the requirements.
*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*
Yes there are, that is all you seem to use.
Everyone can tell that Richard is trying to get way
with disagreeing with a tautology.
Nope, just with falacy.
The only errors are yours each error was pointed
out separately in the Preceding posts.
Your continued lying and deceit and failure to actually answer the errors pointed out is just proving that you have secured your place in the trash heap of eternity.
You are just projecting your own lies and deciet on others, proving you utter incapability to form a logica l thought
Sorry, your aren't worth debating any more, as you have just proved by never actually responding the the error, that you are just to stupid to understand what you are doing.
YOU are the prototype of problem with this country, so stuck in your own ideas that you will not even look at the truth.
Sorry, but it is hard to imagine someone making themselves a stupdid as you have done to yourself.