Re: 195 page execution trace of DDD correctly simulated by HH0

Op 26.jun.2024 om 15:30 schreef olcott:

On 6/26/2024 3:01 AM, Fred. Zwarts wrote:

Op 25.jun.2024 om 21:30 schreef olcott:

On 6/25/2024 2:17 PM, Fred. Zwarts wrote:

>
It might be true, but it is irrelevant, because the simulated H0 is aborted prematurely. The simulating H0 aborts after two cycles,

>
*I am not even talking about a simulating halt decider yet dumbo*

>
Neither am I. Why do you mention a simulating halt decider? (Who is the dumbo?)
>

 Simulated H0 is never aborted.
Only termination analyzers do that.

So, you change subject again. We agreed to discuss one point at a time: the H0 that aborts after two cycles. Now you claim that H0 is never aborted.

 

If you can't begin to comprehend x86 emulators then our conversation
is dead right here.

>
Fortunately, I am very well able to do so.
But it seems that you have to learn a few basic facts about simulation.
>

>
For every x86 emulator Ho that can possibly exist
at machine address 0000217a...
>
_DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
>
The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.

>
So, you repeat your claim without showing any error in my reasoning.
Therefore, I repeat again:
>

 It is like I say 2 + 3 = 5 and you simply say no I am wrong.
My reasoning is that the first four lines of DDD necessary must
repeat for DDD correctly simulated by any x86 emulator at machine
address 0000217a.
 If you reasoning is that a magic fairy sprinkles magic
fairy dust on line four then you are wrong.

If ... But that if is not satisfied. So each programmer knows that the 'then' is not relevant.
Your reasoning that an H0 that aborts after two cycles magically never aborts itself is the problem. That is like magically changing 2+3 into 7-8.
It might be true hat H0 cannot return, but it is irrelevant, because the simulated H0 is aborted prematurely. The simulating H0 aborts after two cycles, when the simulated H0 has one cycle to go before it would return.
So, the only reason that the simulated H0 does not return is that it is aborted prematurely. A correct simulation would show that one cycle later it would return.
It seems that it is over your head that a prematurely aborted simulation is not a correct simulation.
There are a few reasons why a simulation can be incorrect:
1) Some instructions are incorrectly simulated.
2) Instructions are simulated out of order.
3) Some instructions are not simulated at all.
Your problem is 3).
You only prove that your H0 is unable to simulate itself correctly, because it aborts too soon and fails to simulate the end of itself.