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

Op 11.aug.2024 om 13:45 schreef olcott:

On 8/11/2024 1:30 AM, Mikko wrote:

On 2024-08-10 11:30:34 +0000, olcott said:
>

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:
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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*
>
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.

>
As we onece observed, this would be clearer with incdices.
No journal woth of consideration will accept an article
that uses the same name for a specific program and a set.
>
>

 void DDD()
{
   HHH(DDD);
   return;
}
 None-the-less it is clear that of the above specified infinite
set DDD correctly emulated by each element of that set never
reaches its own "return" instruction halt state.

Since no DDD is correctly simulated by HHH, we are talking about the properties of an empty set.
But, indeed, the simulation of DDD by HHH fails to reach the halt state. It aborts one cycle before the simulated HHH would reach its 'return' instruction, after which DDD would reach its halt state.

 My words must be understandable by ordinary C programmers
and computer scientists. The latter tend to conclude that
my work is incorrect as soon as they know the subject matter
before actually seeing what I said.
 

Every C programmer understands that a simulation fails if it does not reach the end of a halting program.
Your own words are that HHH halts (that is why it aborts: it must halt). The direct execution of HHH halts. The simulation by another simulator (e.g. HHH1) halts. The trace shows that HHH halts. A lot of evidence that HHH halts.
Only HHH as a simulator fails to reach the end of the simulated HHH, which makes clear that it is not a property of HHH, or DDD, but a failure of this simulator.
HHH cannot possibly simulate itself correctly up to the end.
When HHH aborts after 100 cycles, the simulated HHH has done only 99 cycles.
We can try to abort one cycle earlier, after 99 cycles, so that the path for the simulation to the end is shorter, but then the simulator also aborts earlier, when the simulated HHH has done only 98 cycles, again one cycle too soon.
It is impossible to make the path short enough, because the simulation always aborts one cycle before the end of the simulated HHH.