Re: Correcting the definition of the halting problem --- Computable functions

Am Tue, 25 Mar 2025 17:49:06 -0500 schrieb olcott:

On 3/25/2025 4:29 PM, joes wrote:

Am Tue, 25 Mar 2025 08:01:14 -0500 schrieb olcott:

On 3/25/2025 3:47 AM, joes wrote:

 

A pure simulator can not limit the number of steps. Also III doesn't
halt in, say, 3 steps. Why should III call a different instance that
doesn't abort, when it is being simulated?
>

The fact that the same states in the program-under-test keep repeating
such that the program-under-test cannot possibly reach its own final
halt state proves that program-under-test does not halt.

 

They don't repeat, though, not in the same stack frame. And the test
program is part of the program under test. Can you answer my question?
 

Your question was incorrect.

How so? You said III calls a non-aborting simulator instead of the one
it is being simulated by. Those are not the same.

The first four instructions of the finite string of machine code at
machine address 00002172 are repeated until EEE reaches its finite
limit.

A simulator shouldn't be limited. And again, the instructions are not
repeated in the same stack frame.

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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.