Re: Any honest person that knows the x86 language can see... predict correctly

On 7/30/24 5:13 PM, olcott wrote:

On 7/30/2024 4:07 PM, joes wrote:

Am Tue, 30 Jul 2024 15:05:54 -0500 schrieb olcott:

On 7/30/2024 1:48 PM, Fred. Zwarts wrote:

Op 30.jul.2024 om 17:14 schreef olcott:

On 7/30/2024 9:51 AM, Fred. Zwarts wrote:

Op 30.jul.2024 om 16:21 schreef olcott:

On 7/30/2024 1:52 AM, Mikko wrote:

On 2024-07-29 14:07:53 +0000, olcott said:
>

HHH(Infinite_Recursion) and HHH(DDD) show the same non-halting
behavior pattern in their derived execution traces of their
inputs.

Hard to believe as their behaviour is so different and you don't
say what pattern the see.

>
*Its all in the part that you erased*

>

We all see the differences between these two.

>
They both correctly predict behavior that must be aborted to prevent
the infinite execution of the simulating halt decider.
>

Except that the prediction for the second one is wrong. The simulation
of an aborting and halting function, like HHH, does not need to be
aborted.

I proved otherwise. When the abort code is commented out then it keeps
repeating again and again, thus conclusively proving that is must be
aborted or HHH never halts.

But the abort is not commented out in the running code!
>

 I modified the original code by commenting out
the abort and it does endlessly repeat just like
HHH correctly predicted.

But then it is a different input, as the code for the HHH that it calls MUST be considered as part of the input.
So, you are just proving yourself to be a stupid liar.

 

This is proved when it is simulate by HHH1. HHH aborts after two
recursions, which is not an infinite execution.
We know you really, really wants it to be correct. So, you are cheating
by suppressing part of the trace, in order to hide the conditional
branch instructions in the second case. But no matter how much olcott
wants it to be correct, or how many times olcott repeats that it is
correct, it does not change the fact that such a simulation is
incorrect

>