Re: HHH maps its input to the behavior specified by it --- never reaches its halt state ---natural number mapping

Op 10.aug.2024 om 14:06 schreef olcott:

On 8/10/2024 6:57 AM, Richard Damon wrote:

On 8/10/24 7:30 AM, olcott wrote:

On 8/10/2024 3:29 AM, Mikko wrote:

On 2024-08-09 14:51:51 +0000, olcott said:
>

On 8/9/2024 4:03 AM, Mikko wrote:

On 2024-08-08 13:18:34 +0000, olcott said:
>

void DDD()
{
   HHH(DDD);
   return;
}
>
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.

>
The ranges of "each HHH" and "every HHH" are not defined above
so that does not really mean anything.

>
Here is something that literally does not mean anything:
"0i34ine ir m0945r (*&ubYU  I*(ubn)I*054 gfdpodf["

>
Looks like encrypted text that might mean something.
>

"Colorless green ideas sleep furiously"

>
This could be encrypted text, too, or perhaps refers to some
inside knowledge or convention.
>

I defined an infinite set of HHH x86 emulators.

>
Maybe somewnete but not in the message I commented.
>

I stipulated that each member of this set emulates
zero to infinity instructions of DDD.

>
That doesn't restrict much.
>

*I can't say it this way without losing 90% of my audience*
Each element of this set is mapped to one element of the
set of non-negative integers indicating the number of
x86 instructions of DDD that it emulates.

>
It is easier to talk about mapping if is given a name.
>

*This one seems to be good*
Each element of this set corresponds to one element of
the set of positive integers indicating the number of
x86 instructions of DDD that it emulates.

>
That would mean that only a finite number (possibly zero) of
instructions is emulated. But the restriction to DDD does not
seem reasonable.
>

>
*The set of HHH x86 emulators are defined such that*

>
I thopught HHH was a deider?
>

>
Each element of this set corresponds to one element of
the set of positive integers indicating the number of
x86 instructions of DDD that it correctly emulates.

>
And only those element of the set that either reach the final state, or simulate forever are "correct" emulators of the whole program, suitable to show halting.
>

 void DDD()
{
   HHH(DDD);
   return;
}
 In other words even though it is dead obvious to
us that a complete simulation of DDD simulated by HHH

is impossible, because HHH is programmed to abort and, therefore, it is unable to do a complete simulation.
Or are you dreaming again of the HHH that does not abort, not only as simulator, but also as input.
So, any conclusion about what it would do if it does not do what it is programmed to do, but it would do something that it is not programmed to do, namely, a complete simulation, is just speculation.
Even a beginner in software development understands that a program does what it is programmed to do. Therefore, it it is a ridiculous dream to assume that HHH would do a complete simulation and

will never halt algorithms will forever be ridiculously
more stupid and never be able to see this?
 

All HHH that abort, halt and then DDD halts. It is proved by the direct execution, by the simulation with another simulator, such as HHH1.
But HHH cannot possibly simulate *itself* correctly.
That HHH cannot reach the end, because HHH that abort cannot do a complete simulation, because it is programmed to abort, proves that the simulation is incomplete and incorrect.