Re: The philosophy of computation reformulates existing ideas on a new basis ---SUCCINCT

On 11/14/2024 2:56 AM, joes wrote:

Am Wed, 13 Nov 2024 17:11:30 -0600 schrieb olcott:

On 11/13/2024 4:58 AM, Mikko wrote:

On 2024-11-12 13:58:03 +0000, olcott said:

On 11/12/2024 1:12 AM, joes wrote:

Am Mon, 11 Nov 2024 10:35:57 -0600 schrieb olcott:

On 11/11/2024 10:25 AM, joes wrote:

Am Mon, 11 Nov 2024 08:58:02 -0600 schrieb olcott:

On 11/11/2024 4:54 AM, Mikko wrote:

On 2024-11-09 14:36:07 +0000, olcott said:

On 11/9/2024 7:53 AM, Mikko wrote:

>

The actual computation itself does involve HHH emulating itself
emulating DDD. To simply pretend that this does not occur seems
dishonest.

Which is what you are doing: you pretend that DDD calls some other
HHH that doesn’t abort.

DDD emulated by HHH does not reach its "return" instruction final
halt state whether HHH aborts its emulation or not.

When DDD calls a simulator that aborts, that simulator returns to
DDD, which then halts.

It is not the same DDD as the DDD under test.

 What, then, is the DDD "under test"?

The machine code address that is passed to HHH on the stack
is the input to HHH thus the code under test. It specifies
that HHH emulates itself emulating DDD.
HHH is required to abort the emulation of any input that
would otherwise result in its own non-termination. DDD
is such an input.
The DDD executed in main() is never pushed onto the stack
of any HHH thus <is not> the input DDD.

 

If the DDD under the test is not the same as DDD then the test is
performed incorrectly and the test result is not valid.

The DDD under test IS THE INPUT DDD

Yes, exactly. In particular, the one that calls the aborting HHH.
 

void DDD()
{
   HHH(DDD);
   return;
}
If you are not a brain dead moron you could remember that the
measure never has been: Does the emulated DDD ever stop running?
after be told that this dozens of times.
The measure is: Does the emulated DDD ever reach its "return"
instruction final halt state?
_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]
If you were technically competent in the x86 language you
would know that the answer to this has always been no.
The emulated DDD cycles through its first four instructions
never reaching its "ret" instruction final halt state no
matter how many times it is emulated.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer