Re: The execution trace of HHH1(DDD) shows the divergence

Am Sat, 07 Jun 2025 14:06:33 -0500 schrieb olcott:

On 6/7/2025 1:59 PM, joes wrote:

Am Sat, 07 Jun 2025 12:20:42 -0500 schrieb olcott:

On 6/7/2025 11:43 AM, joes wrote:

Am Sat, 07 Jun 2025 11:02:55 -0500 schrieb olcott:

On 6/7/2025 10:54 AM, wij wrote:

On Sat, 2025-06-07 at 10:35 -0500, olcott wrote:

On 6/7/2025 10:31 AM, wij wrote:

On Sat, 2025-06-07 at 09:57 -0500, olcott wrote:

On 6/7/2025 9:54 AM, wij wrote:

On Sat, 2025-06-07 at 09:32 -0500, olcott wrote:

>

The execution trace of HHH1(DDD) shows the divergence of DDD
emulated by HHH from DDD emulated by HHH1.
Shows that DDD emulated by HHH and DDD emulated by HHH1
diverges as soon as HHH begins emulating itself emulating DDD.
>
*From the execution trace of HHH1(DDD) shown below*
DDD emulated by HHH1              DDD emulated by HHH
[00002183]
push ebp               [00002183] push ebp [00002184] mov
ebp,esp            [00002184] mov ebp,esp [00002186] push
00002183 ; DDD    [00002186] push 00002183 ; DDD [0000218b]
call 000015c3 ; HHH    [0000218b] call 000015c3 ; HHH *HHH1
emulates DDD once then HHH emulates DDD once, these match*

>
What do you mean by "then" and "once"? That implies completion and
succession, however we are only ever simulating a single call, and
only HHH1 simulates any returns of DDD/HHH.
HHH1 simulates DDD completely, HHH recurses and aborts *inside*. HHH
does not simulate DDD once, it only enters the call, but never exits.
>

The next instruction of DDD that HHH emulates is at the
machine address of 00002183.
The next instruction of DDD that HHH1 emulates is at the
machine address of 00002190.

>
At those addresses we have the first instruction of DDD and the one
after the call, respectively.
>
>
[main -> HHH1(DDD) -> HHH(DDD) -> HHH(DDD)]

New slave_stack at:198d21  DDD emulated by HHH *This is the
beginning of the divergence of the behavior*
*HHH is emulating itself emulating DDD, HHH1 never does that*

>
HHH1 does simulate HHH simulating DDD.
>

That is not the question being answered.

You say: HHH1 doesn't simulate itself. That is true. Also true is that
HHH1 simulates the same thing that HHH does,

 
The emulation of DDD by HHH(DDD) contained within the execution trace of
HHH1(DDD) diverges from the execution trace of DDD by HHH1 as soon as
HHH emulates itself emulating DDD.

No, the instructions are the same.

HHH(DDD) ALWAYS emulates itself emulating DDD. HHH1(DDD) NEVER emulates
itself emulating DDD.

Dude, I already agreed with that. It doesn't matter.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.