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

On 3/25/25 6:49 PM, olcott wrote:

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.

No, YOUR question is incorrect as regards a Halt Decider, as it is just a strawman.
The question for a Halt Decider isn't what the decider can see in its own (partial) emulation, but what would be seen giving that identical COMPLETE input (thus must includeing the code for the called function) to a COMPLETE emulator.

 _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.
 

I guess you are admitting you don't understand what a correct emulation is, especially per the defintion of the x86 langugage.
The ONLY correct emulation of the call EEE instruction would be to continue the emulation at the address of EEE, that is 000015d2.
IF that isn't part of the full input, then the input can not be correctly emulated, as it wasn't actually a program.
If that is part of the full input, then changing it, will of course have the possiblity of changing the behavior, and since HHH and EEE are not the exact same program, this says nothing about the DDD that was given to HHH and which calls that same version of HHH.
Sorry, you are just proving your utter stupidity.