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

On 6/4/2025 8:48 PM, Richard Damon wrote:

On 6/4/25 10:52 AM, olcott wrote:

On 6/4/2025 1:54 AM, Mikko wrote:

On 2025-06-03 19:57:09 +0000, olcott said:
>

On 6/3/2025 2:37 AM, Mikko wrote:

On 2025-06-02 15:52:53 +0000, olcott said:
>

DDD correctly emulated by HHH diverges from DDD
correctly emulated by HHH1 as soon as HHH begins
emulating itself emulating DDD, marked below.

>
The DDD emulated (correctly or otherwise) by HHH is the same DDD as
the one emulated (correctly or otherwise) so both specify the same
behaviour.

>
No they do not. When DDD calls its own emulator its behavior
is different than when DDD calls another different emulator.

>
If the input string does not unambiguously specify one and only one
behaviour it is incorrectly encoded and not a valid input string.
The halting problem of Truing machines requires that every pair of
a Turing macnine and input is descibed so that the behaviour to
be decided about is the only behaviour that meets to the description.
>

>
The code proves what it proves.
>
_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

 So what "simulation" is the above? It seems that you are showing a trace from x86, not what HHH is doing.
 

What I am showing is DDD emulated by HHH1
side-by-side with DDD emulated by HHH
*They initially match up*
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
*The matching is now all used up*
*Then DDD emulated by HHH does something*
*that DDD emulated by HHH1 never does*
*it emulates DDD all over again*
[00002183] push ebp      ;
[00002184] mov ebp,esp   ;
[00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer