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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD emulated by HHH diverges from DDD emulated by HHH1
Date: Tue, 3 Jun 2025 23:20:23 -0400
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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:

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
 >
========== REMAINDER OF ARTICLE TRUNCATED ==========