Re: DDD emulated by HHH diverges from DDD emulated by HHH1

On 6/3/2025 9:42 PM, dbush wrote:

On 6/3/2025 10:29 PM, olcott wrote:

On 6/3/2025 8:57 PM, dbush wrote:

On 6/3/2025 5:14 PM, olcott wrote:

On 6/3/2025 3:48 PM, joes wrote:

Am Tue, 03 Jun 2025 14:47:23 -0500 schrieb olcott:

On 6/3/2025 3:28 AM, Fred. Zwarts wrote:

Op 02.jun.2025 om 17:52 schreef olcott:

>

DDD correctly emulated by HHH diverges from DDD correctly emulated by
HHH1 as soon as HHH begins emulating itself emulating DDD, marked
below.
*HHH1 never emulates itself emulating DDD*

>

*This is the beginning of the divergence of the behavior*
*of DDD emulated by HHH versus DDD emulated by HHH1*

>

Misleading words when you change the meaning of diverging.
Mike showed the traces side by side. Even after many requests, you
still cannot show the first instruction that is interpreted differently
by HHH and HHH1. The only difference is that HHH gives up the
simulation too early.

>
As soon as HHH begins emulating itself and HHH1 NEVER begins emulating
itself THIS IS THE DIVERGENCE.

Yes, that is exactly the point where HHH aborts.

>
Both the divergence and the abort are shown below.
>
_DDD()
[00002183] 55             push ebp
[00002184] 8bec           mov ebp,esp
[00002186] 6883210000     push 00002183 ; push DDD
[0000218b] e833f4ffff     call 000015c3 ; call HHH
[00002190] 83c404         add esp,+04
[00002193] 5d             pop ebp
[00002194] c3             ret
Size in bytes:(0018) [00002194]
>
_main()
[000021a3] 55             push ebp
[000021a4] 8bec           mov ebp,esp
[000021a6] 6883210000     push 00002183 ; push DDD
[000021ab] e843f3ffff     call 000014f3 ; call HHH1
[000021b0] 83c404         add esp,+04
[000021b3] 33c0           xor eax,eax
[000021b5] 5d             pop ebp
[000021b6] c3             ret
Size in bytes:(0020) [000021b6]
>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  ========== =============
[000021a3][0010382d][00000000] 55         push ebp      ; main()
[000021a4][0010382d][00000000] 8bec       mov ebp,esp   ; main()
[000021a6][00103829][00002183] 6883210000 push 00002183 ; push DDD
[000021ab][00103825][000021b0] e843f3ffff call 000014f3 ; call HHH1
New slave_stack at:1038d1
>
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138d9
[00002183][001138c9][001138cd] 55         push ebp      ; DDD of HHH1
[00002184][001138c9][001138cd] 8bec       mov ebp,esp   ; DDD of HHH1
[00002186][001138c5][00002183] 6883210000 push 00002183 ; push DDD
[0000218b][001138c1][00002190] e833f4ffff call 000015c3 ; call HHH
New slave_stack at:14e2f9
>
Begin Local Halt Decider Simulation   Execution Trace Stored at:15e301
[00002183][0015e2f1][0015e2f5] 55         push ebp      ; DDD of HHH[0]
[00002184][0015e2f1][0015e2f5] 8bec       mov ebp,esp   ; DDD of HHH[0]
[00002186][0015e2ed][00002183] 6883210000 push 00002183 ; push DDD
[0000218b][0015e2e9][00002190] e833f4ffff call 000015c3 ; call HHH
New slave_stack at:198d21
>
THIS IS WHERE THE DIVERGENCE OF DDD EMULATED BY HHH
AND DDD EMULATED BY HHH1 BEGINS

>
So how exactly do HHH and HHH1 emulate the first instruction of HHH differently?
>

>
The question is incorrect.
HHH emulates DDD two times and HHH1 emulates DDD one time
the whole second time is the divergence.

 There is no divergence if the instructions are emulated exactly the same in both cases. 

HHH1(DDD) emulates DDD exactly one time.
HHH(DDD) emulates DDD exactly two times.
The whole second time that HHH emulates DDD is
divergence.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer