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

On 6/3/2025 11:53 PM, olcott wrote:

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

On 6/3/2025 11:26 PM, olcott wrote:

On 6/3/2025 10:20 PM, dbush wrote:

On 6/3/2025 11:07 PM, olcott wrote:

On 6/3/2025 9:54 PM, dbush wrote:

On 6/3/2025 10:51 PM, olcott wrote:

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.
>

>
Let the record show that Peter Olcott has failed to identify an instruction that HHH and HHH1 emulated differently.
>

>
When HHH emulates itself emulating DDD and emulates
DDD a second time this second emulation of DDD begins
at its own address 00002183.
>
HHH1 only emulates DDD exactly once.
>

>
>
Irrelevant, as that does not change the fact that the emulations performed by HHH and HHH1 are exactly the same up to the point that HHH aborts, as you have just admitted on the record:
>

>
No matter what anyone ever said or misconstrued the
actual execution trace proves:
>

>
That the emulations performed by HHH and HHH1 are identical up to the point that HHH aborts,

 Counter-factual as anyone that understands
the x86 language can clearly see.
 

False, as the side-by-side trace show exactly that and as you have admitted on the record *multiple times*:
On 5/6/2025 5:17 PM, dbush wrote:
 > On 5/6/2025 5:03 PM, olcott wrote:
 >> On 5/6/2025 3:51 PM, dbush wrote:
 >>> On 5/6/2025 4:46 PM, olcott wrote:
 >>>> On 5/6/2025 3:31 PM, dbush wrote:
 >>>>> Then what is the first instruction emulated by HHH that differs
 >>>>> from the emulation performed by UTM?
 >>>>>
 >>>>
 >>>> HHH1 is exactly the same as HHH except that DD
 >>>> does not call HHH1. This IS the UTM emulator.
 >>>> It does not abort.
 >>>
 >>> Last chance:
 >>>
 >>> What is the first instruction emulated by HHH that differs from the
 >>> emulation performed by HHH1?
 >>
 >> Go back and read the part you ignored moron.
 >
 > Let the record show that Peter Olcott has neglected to identify an
 > instruction that HHH emulates differently from HHH1.
 >
 >>> Failure to provide this in your next message or within one hour of
 >>> your next post in this newsgroup will be taken as your official on-
 >>> the-record admission that the emulations performed by HHH and HHH1
 >>> are in fact exactly the same up until the point that HHH aborts, at
 >>> which point HHH did not correctly simulate the last instruction it
 >>> simulated as you are previously on record as admitting.
 >
 > Therefore, as per the above requirements:
 >
 > LET THE RECORD SHOW
 >
 > That Peter Olcott
 >
 > Has *officially* admitted
 >
 > That the emulations performed by HHH and HHH1 are in fact exactly the
 > same up until the point that HHH aborts, at which point HHH did not
 > correctly simulate the last instruction it simulated as he is previously
 > on record as admitting.
On 6/3/2025 10:54 PM, dbush wrote:
 > On 6/3/2025 10:51 PM, olcott wrote:
 >> 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:
 >>>>> 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.
 >>
 >
 > Let the record show that Peter Olcott has failed to identify an
 > instruction that HHH and HHH1 emulated differently.
 >
 >>
 >>> What happened before either emulation started is irrelevant.
 >>>
 >>> The only way for the emulations to diverge is if there is a
 >>> particular instruction such that X happens if HHH emulates it and Y
 >>> happens if HHH1 emulates it.  Again, what happened before either
 >>> emulation started is irrelevant.
 >>>
 >>>
 >>> So I'll ask one more time:  how exactly do HHH and HHH1 emulate the
 >>> first instruction of HHH, or *any* instruction that is part of the
 >>> emulation of DDD, differently?
 >>>
 >>> Failure to provide the above explanation in your next reply or within
 >>> one hour of your next post in this newsgroup will be taken as your
 >>> official on-the-record admission that the emulations of DDD performed
 >>> by HHH and HHH1 do *not* diverge but are in fact the same up to the
 >>> point that HHH aborts.
 >
 > Therefore, as per the above criteria:
 >
 > Let The Record Show
 >
 > That Peter Olcott
 >
 > Has *officially* admitted:
 >
 > That the emulations of DDD by HHH and HHH1 in fact do *not* diverge but
 > are in fact the same up to the point that HHH aborts.