On 8/7/24 9:54 AM, olcott wrote:
On 8/7/2024 2:29 AM, Mikko wrote:
On 2024-08-05 13:49:44 +0000, olcott said:
On 8/5/2024 2:39 AM, Mikko wrote:
On 2024-08-04 18:59:03 +0000, olcott said:
On 8/4/2024 1:51 PM, Richard Damon wrote:
On 8/4/24 9:53 AM, olcott wrote:
On 8/4/2024 1:22 AM, Fred. Zwarts wrote:
Op 03.aug.2024 om 18:35 schreef olcott:
∞ instructions of DDD correctly emulated by HHH[∞] never >>>>>>>>> reach their own "return" instruction final state.
So you are saying that the infinite one does?
Dreaming again of HHH that does not abort? Dreams are no
substitute for facts.
The HHH that aborts and halts, halts. A tautology.
void DDD()
{
HHH(DDD);
return;
}
That is the right answer to the wrong question.
I am asking whether or not DDD emulated by HHH
reaches its "return" instruction.
But the "DDD emulated by HHH" is the program DDD above,
When I say DDD emulated by HHH I mean at any level of
emulation and not and direct execution.
If you mean anything other than what the words mean you wihout
a definition in the beginning of the same message then it is
not reasonable to expect anyone to understand what you mean.
Instead people may think that you mean what you say or that
you don't know what you are saying.
If you don't understand what the word "emulate" means look it up.
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?
No, you have to admit that you will get the wrong answer.
Remember, HHH is a fixed defined piece of code, so it has a fixed
defined answer that it WILL give based on the algorithm that HHH used.
What ever that answer is, it will be wrong, as can be easily determined
by many other deciders.
DDD does specify non-halting behavior to HHH and HHH must
report on this non-halting behavior that DDD specifies.
No, because halting is an OBJECTIVE criteria, and either HHH will return
and that makes DDD a halting program (so HHH would have needed to have
guessed 1 to be right), or HHH will never return, making DDD a
non-halting program (and ONLY that makes it non-halting) but HHH turns
out not to be a decider.
All halt deciders compute the mapping from their input
finite string to the behavior that this finite string
specifies.
Which, for a halt decide is DEFINED to be the mapping based on the
direct execution of the program, and whether it reaches a final state
iin a finite number of steps or not.
That HHH can't compute this does not make that not the mapping it NEEDS
to compute to be a Halt Decider, that just means there is no such thing
as a universally correct Halt Decider.
No halt decider is ever allowed to report on the behavior
of any computation that itself is contained within unless
this is the same behavior that its finite string input
specifies.
Says WHAT? This is just you stating you lie again.
ALL Halt deciders are REQUIRED (to be a halt decider) that answer about
*ANY* program described by their input, even if it includes a copy of
itself.
You just proved yourself to be a ingnoran LIAR that reckless repeats
lies that he has been informed are lies because you stupidly refuse to
even look at the actual definition, because you don't "beleive" in them.
Sorry, you don't get to not believe in the rules of the game that have
been establish.
You just canceled yourself out of the game, and booked your ticket to
the garbage heap of life.
void DDD()
{
HHH(DDD);
return;
}
Any expert in the C language that knows what x86 emulators
are knows that DDD correctly emulated by HHH specifies what
is essentially equivalent to infinite recursion that cannot
possibly reach its "return" instruction halt state.
It seems that no one here has that degree of expertise.
That they know that they don't understand these things
and still say that I am wrong is dishonest.
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