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

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.
_III()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push III
[0000217a] e853f4ffff call 000015d2 ; call EEE(III)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
The first four instructions of the finite string
of machine code at machine address 00002172 are
repeated until EEE reaches its finite limit.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer