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

Op 10.aug.2024 om 20:41 schreef olcott:

On 8/10/2024 1:35 PM, Fred. Zwarts wrote:

Op 10.aug.2024 om 16:24 schreef olcott:

On 8/10/2024 9:00 AM, Fred. Zwarts wrote:

Op 10.aug.2024 om 15:37 schreef olcott:

On 8/10/2024 8:21 AM, Fred. Zwarts wrote:

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.

>
A complete simulation of DDD by a pure x86 emulator
named HHH cannot possibly reach its own "return"
instruction halt state.

>
Indeed, HHH fails to reach its own halt state. HHH cannot possibly simulate itself up to its halt state.
Which proves that the simulation is incomplete and, therefore, incorrect.
>

>
That an emulation of an input is necessary correct no matter
what-the-Hell it does as long as it conforms to the semantics
of the x86 language is either over your head or you persistently
lie about it.
>

>
No evidence for these claims. We know that a simulation of a halting program is correct only when it reaches the halt state,

 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 void Infinite_Loop()
{
   HERE: goto HERE;
}
 You seem to be an ignoramus. Ordinary ignorance can be corrected.

Completely irrelevant. There is no infinite loop in HHH because HHH aborts after N recursions, not after an Infinite_Recursion.
Your trace showed that HHH runs only two recursions. Two is not infinite.
Apparently your memory lasts only one or two days.
We have shown this before, but you forget it again and again.
Therefore, HHH looks more like:
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
Th whole discussion makes me think of a discussion with someone with dementia. After one round we are back at the beginning.