Re: DDD emulated by HHH --- (does not refer to prior posts)

On 8/28/2024 2:21 PM, joes wrote:

Am Wed, 28 Aug 2024 11:44:53 -0500 schrieb olcott:

On 8/28/2024 11:31 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 18:21 schreef olcott:

On 8/28/2024 11:11 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 17:13 schreef olcott:

On 8/28/2024 9:57 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 14:59 schreef olcott:

On 8/28/2024 7:46 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 14:12 schreef olcott:

On 8/28/2024 4:09 AM, Fred. Zwarts wrote:

Op 27.aug.2024 om 14:44 schreef olcott:

On 8/27/2024 3:38 AM, Fred. Zwarts wrote:

Op 27.aug.2024 om 04:33 schreef olcott:

 

When we assume that:
(a) HHH is an x86 emulator that is in the same memory space
as DDD.
(b) HHH emulates DDD according to the semantics of the x86
language.
then we can see that DDD emulated by HHH cannot possibly get
past its own machine address 0000217a.

>
Yes, we see. In fact DDD is not needed at all.

Or are trying to distract the attention from the fact that DDD
is not needed is a simple truism, a tautology in your terms?
>

When 100% of the whole point is for HHH to correctly determine
whether or not DDD would stop running if not aborted *IT IS
RIDICULOUSLY STUPID TO SAY THAT DDD IS NOT NEEDED*

Like Fred has been saying for a month, what is HHH(HHH,HHH)?
 

When without DDD it is clear as crystal that HHH cannot possibly
simulate itself correctly:

 

You may repeat it many more times, but HHH violated the semantics
of the x86 language by skipping the last few instructions of a
halting program. This finite string, when given for direct
execution, shows a halting behaviour. This is the proof what the
semantics of the x86 language means for this finite string: a
halting program.

It is very telling to see where these exchanges peter out (haha).
 

A dishonest dodge way form the subject of DDD emulated by HHH.

And when the x86 string tells the computer that there is a halting
program and the simulator decides that there is a non-halting
program, this proves that the simulation is incorrect.
Clear as crystal: the semantics of the x86 string is proved by its
direct execution.
This is shown in the example below, where the direct execution of
HHH halts, but HHH decides that it does not halt.
>

By this same reasoning that fact that you are no longer hungry AFTER
you have eaten proves that you never needed to eat.
The behavior of DDD before HHH aborts its simulation (before it has
eaten) it not the same behavior after DDD has been aborted (after it
has eaten).

I do not understand this. There is no „after having been aborted”.
 

Then this is your lack of sufficient software engineering skill.
The directly executed DDD() has different behavior than DDD
emulated by HHH because DDD() benefits from HHH having already
aborted its emulation of DDD. HHH itself does not receive this
benefit.

If hungry stands for fear for infinite recursion

>
hungry stands for will not stop running unless aborted just like will
remain hungry until eating is always true whenever hungry

Your HHH will see a 'special condition' after a few recursions, abort
and halt.

Why to do dishonestly try to get away with the strawman deception and
change the subject to HHH?
>
It is a design requirement that HHH halts if it doesn't halt it is
wrong.

Then why does it report itself as nonterminating? (There is nothing
else in DDD that would cause that.)
 

How could it do that? IT MUST TERMINATE TO REPORT ANYTHING.

When DDD emulated by HHH according to the semantics of the x86 language
cannot possibly reach its own machine address of 00002183, then HHH is
correct to reject DDD as non-halting even of HHH does this entirely by
wild guess.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer