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

Am Thu, 14 Nov 2024 07:22:12 -0600 schrieb olcott:

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.
The DDD executed in main() is never pushed onto the stack of any HHH
thus <is not> the input DDD.

It starts at the same address, however. In what sense is the input not
the DDD with that entry point?

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.

remember that the measure
never has been: Does the emulated DDD ever stop running?
The measure is: Does the emulated DDD ever reach its "return"
instruction final halt state?

Under what circumstances does DDD stop running without returning
or return without stopping?

If you were technically competent in the x86 language you would know
that the answer to this has always been no.

Which is a failure of HHH, since DDD itself returns and stops running.

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.

That is imprecise: there is no single DDD that loops. But I agree that
HHH can't simulate itself aborting and returning. Why is that?

--
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.