Re: Defining a correct simulating halt decider

On 9/5/24 12:10 PM, olcott wrote:

On 9/5/2024 10:57 AM, joes wrote:

Am Thu, 05 Sep 2024 08:24:20 -0500 schrieb olcott:

On 9/5/2024 2:34 AM, Mikko wrote:

On 2024-09-03 13:00:50 +0000, olcott said:

On 9/3/2024 5:25 AM, Mikko wrote:

On 2024-09-02 16:38:03 +0000, olcott said:
>

A halt decider is a Turing machine that computes the mapping from
its finite string input to the behavior that this finite string
specifies.

>
A halt decider needn't compute the full behaviour, only whether that
behaviour is finite or infinite.

>

Nice to see that you don't disagree with what said. Unvortunately I
can't agree with what you say.
HHH terminates, so DDD obviously terminates, too. No valid
C interpretaion of allows DDD to continue forever after HHH jas
terminated.

DDD emulated by HHH never reaches it final halt state. It looks like I
have to repeat this 10,000 times before anyone ever notices that I said
it at least once.

We have noticed.
>

Show the details of how DDD emulated by HHH reaches its own machine
address 0000217f.

 

By HHH returning, which we are guaranteed from its definition as a
decider.
>

 Are you trying to get away with a lie by erasing the context?
 _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]
 Show the details of how DDD emulated by HHH
reaches its own machine address 0000217f.
 00002172, 00002173, 00002175, 0000217a calls HHH(DDD)
then
00002172, 00002173, 00002175, 0000217a calls HHH(DDD)...
 How the F--- Does the emulated HHH return?
Show the exact freaking steps of how the emulated HHH returns !!!
 00002172, 00002173, 00002175, 0000217a calls HHH(DDD)
then what comes next???
 

000015d2, which is part of the program DDD, as that includes all the code used by DDD.
What, are you still that stupid to not be able to learn the definitions?
Look at your 200 page trace.
Replace the 4 lines at the begining with the addresses of DDD insteand of the addresss of main, and similarly the end, and you get the exact emulation that a correctly emulating HHH would do of THIS DDD (that calls the HHH that it does).
VOLA, your trace of DDD correctly emulated reaching the final state.
The fact that the HHH that it calls doesn't do it just proves that your HHH fails to meet its requirement.
Remember, the correct emulation of the DDD emulated by HHH is NOT the same thing as the emulation of DDD done by this HHH.
You are just proving your ignorance by repeating that error.