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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 22:54:46 -0400
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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.

>
>> 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.