Re: The actual truth is that ... Turing computability issues have been addressed --- marathon winner

Am Wed, 16 Oct 2024 08:43:42 -0500 schrieb olcott:

On 10/16/2024 8:10 AM, joes wrote:

Am Wed, 16 Oct 2024 07:52:00 -0500 schrieb olcott:

On 10/16/2024 1:32 AM, joes wrote:

Am Tue, 15 Oct 2024 22:52:00 -0500 schrieb olcott:

On 10/15/2024 9:11 PM, Richard Damon wrote:

On 10/15/24 8:39 AM, olcott wrote:

On 10/15/2024 4:58 AM, joes wrote:

Am Mon, 14 Oct 2024 20:12:37 -0500 schrieb olcott:

On 10/14/2024 6:50 PM, Richard Damon wrote:

On 10/14/24 12:05 PM, olcott wrote:

On 10/14/2024 6:21 AM, Richard Damon wrote:

On 10/14/24 5:53 AM, olcott wrote:

On 10/14/2024 3:21 AM, Mikko wrote:

On 2024-10-13 12:49:01 +0000, Richard Damon said:

On 10/12/24 8:11 PM, olcott wrote:

>

Trying to change to a different analytical framework than the
one that I am stipulating is the strawman deception.
*Essentially an intentional fallacy of equivocation error*

But, you claim to be working on that Halting Problem,

I quit claiming this many messages ago and you didn't bother to
notice.

Can you please give the date and time? Did you also explicitly
disclaim it or just silently leave it out?

Even people of low intelligence that are not trying to be as
disagreeable as possible would be able to notice that a specified
C function is not a Turing machine.

But it needs to be computationally equivalent to one to ask about
Termination.

Not at all. A termination analyzer need not be a Turing computable
function.

It definitely does. An uncomputable analyser is useless.

It is true that a termination analyzer is not required to work
correctly for all inputs.

But it should work for DDD.

That there is one way that HHH can consistently catch the
non-terminating pattern of its input proves that this can be done.

DDD does terminate. Otherwise it would contradict the HP.
 

Mike suggested some different ways that would seem to be Turing
computable yet too convoluted to be time consuming for me to implement
in practice.

What are those ways?

The basic approach involves the idea that every state change of the
emulations of emulations is data that belongs to the outermost
directly executed HHH.

Nothing special. They aren't even running on the hardware proper.
 

It is too convoluted for me to provide a way for HHH to look inside
all of the emulations of emulations and pull out the data that it
needs, so knowing that this is possible is enough to know that it is
Turing computable.

It cannot dig into an infinite simulation chain.

It need not dig into an infinite emulation chain. It merely needs to see
that DDD calls HHH(DDD) twice in sequence having no termination
condition within DDD.

If it sees no termination condition in itself (it's a recursion), being
the same program, it also doesn't have one, so cannot abort.

When HHH is an x86 emulation based termination analyzer then each DDD
*correctly_emulated_by* any HHH that it calls never returns.
Each of the directly executed HHH emulator/analyzers that returns 0
correctly reports the above *non_terminating _behavior* of its input.

If HHH doesn't return, DDD doesn't either, not even the directly
executed one.

The emulated HHH is merely data to the executed termination analyzer.

When HHH is an x86 emulation based termination analyzer then each
DDD *correctly_emulated_by* any HHH that it calls never returns.

Only because the nested HHH doesn't abort.

Every nested HHH has seen one less execution trace than the next outer
one. The outermost one aborts its emulation as soon as it has seen
enough. Thus each inner HHH cannot possibly abort its own emulation.

Each inner HHH must abort if the outer does,

In the same way that each person in a marathon that are running at the
same speed and ten feet behind the person in front of them must win the
race.
The first person does not win the marathon they merely tie with everyone
else.
The fact that they start out ahead and remain ahead in the whole
marathon has no effect on whether they reach the finish line first.
 

since they are the same program. Of course, the outer doesn't simulate
the inner abort, because it has already aborted. Therefore the outer
HHH doesn't need to abort, because the inner HHHs already halt by
themselves. Reverse the if(Root) check on line 500 or what in Halt7.c.
As long as at least one of the infinitely many HHHs aborts, the whole
chain terminates (but the nested HHHs don't get simulated completely).
But they are all the same program,
so all of them must have the abort.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.