Re: Flat out dishonest or totally ignorant?

On 7/5/24 8:15 AM, olcott wrote:

On 7/5/2024 2:32 AM, Mikko wrote:

On 2024-07-04 15:56:10 +0000, olcott said:
>

On 7/4/2024 10:07 AM, joes wrote:

Am Thu, 04 Jul 2024 08:43:22 -0500 schrieb olcott:

On 7/4/2024 8:38 AM, joes wrote:

Am Thu, 04 Jul 2024 07:50:51 -0500 schrieb olcott:

On 7/4/2024 5:38 AM, joes wrote:

Am Wed, 03 Jul 2024 11:21:01 -0500 schrieb olcott:

On 7/3/2024 11:09 AM, Fred. Zwarts wrote:

Op 03.jul.2024 om 17:55 schreef olcott:

On 7/3/2024 10:52 AM, Fred. Zwarts wrote:

Op 03.jul.2024 om 15:24 schreef olcott:

On 7/3/2024 3:42 AM, Fred. Zwarts wrote:

Op 03.jul.2024 om 05:55 schreef olcott:

On 7/2/2024 10:50 PM, joes wrote:

Am Tue, 02 Jul 2024 14:46:38 -0500 schrieb olcott:

On 7/2/2024 2:17 PM, Fred. Zwarts wrote:

Op 02.jul.2024 om 21:00 schreef olcott:

On 7/2/2024 1:42 PM, Fred. Zwarts wrote:

Op 02.jul.2024 om 14:22 schreef olcott:

On 7/2/2024 3:22 AM, Fred. Zwarts wrote:

Op 02.jul.2024 om 03:25 schreef olcott:

>

Similarly, if you think that HHH can simulate itself correctly,
you are wrong.
        int H(ptr p, ptr i);
        int main()
        {
          return H(main, 0);
        }
You showed that H returns, but that the simulation thinks it does
not return.
DDD is making it unnecessarily complex, but has the same problem.

main correctly emulated by H never stops running unless aborted.

HHH is unable to simulate main correctly, because it unable to
simulate itself correctly.
The 'unless phrase' is misleading, because we are talking about a H
*does* abort. Dreaming of one that does not abort, is irrelevant.
The correctly simulated main would stop, because the simulated H is
only one cycle away from its return when its simulation is aborted.

HHH is required to report on what would happen if HHH did not abort.
HHH is forbidden from getting its own self stuck in infinite
execution. Emulated instances of itself is not its actual self.

No. HHH is simulating itself, not a different function that does not
abort. All calls are instances of the same code with the same
parameters. They all do the same thing: aborting.

HHH always meets its abort criteria first because it always sees at
least one fully execution trace of DDD before the next inner one. It
is stupidly incorrect to think that HHH can wait on the next one.

Stupidly incorrect is thinking that the next one wouldn’t abort just
because that part isn’t simulated.

Unless the outermost one aborts none of them do.

Since the outermost aborts, all of them do.
>

>
This the same same as saying the when everyone in
a foot race is in single file and 15 feet behind
the one in front of them that everyone will come
in first place. No you are wrong.

>
That "first place" is not in "the same" but your own lie.
In a turthful paraphrase there would be "to the fininsh line"
or something like that instead.
>

 _DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
 When HHH(DDD) simulates its input that calls HHH(DDD)
to simulate its input that simulates its input the outer
HHH has seen the first four instructions repeated, the
inner HHH has not seen the first four instructions repeated.
The outer one has met its abort criteria, the inner one
has not met its abort criteria.
 If you now the x86 language you would already know that.
 

And DDD() will always reach its end if HHH(DDD) returns.
HHH's simulaition may not reach that point, but DDD does.
You just don't understand that HHH can not abort the actualol DDD, only its emulation of it,
You logic is just stuck in your own fantasy world, and can't understand what REALITY actually i, because you have decided that you get to make the rules of the shared system, when you don't.