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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: The execution trace of HHH1(DDD) shows the divergence
Date: Sat, 7 Jun 2025 19:24:53 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sat, 07 Jun 2025 14:06:33 -0500 schrieb olcott:
> On 6/7/2025 1:59 PM, joes wrote:
>> Am Sat, 07 Jun 2025 12:20:42 -0500 schrieb olcott:
>>> On 6/7/2025 11:43 AM, joes wrote:
>>>> Am Sat, 07 Jun 2025 11:02:55 -0500 schrieb olcott:
>>>>> On 6/7/2025 10:54 AM, wij wrote:
>>>>>> On Sat, 2025-06-07 at 10:35 -0500, olcott wrote:
>>>>>>> On 6/7/2025 10:31 AM, wij wrote:
>>>>>>>> On Sat, 2025-06-07 at 09:57 -0500, olcott wrote:
>>>>>>>>> On 6/7/2025 9:54 AM, wij wrote:
>>>>>>>>>> On Sat, 2025-06-07 at 09:32 -0500, olcott wrote:
>>>>
>>>>>>>>>>> The execution trace of HHH1(DDD) shows the divergence of DDD
>>>>>>>>>>> emulated by HHH from DDD emulated by HHH1.
>>>>>>>>>>> Shows that DDD emulated by HHH and DDD emulated by HHH1
>>>>>>>>>>> diverges as soon as HHH begins emulating itself emulating DDD.
>>>>>>>>>>>
>>>>>>>>>>> *From the execution trace of HHH1(DDD) shown below*
>>>>>>>>>>> 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 *HHH1
>>>>>>>>>>> emulates DDD once then HHH emulates DDD once, these match*
>>>>
>>>> What do you mean by "then" and "once"? That implies completion and
>>>> succession, however we are only ever simulating a single call, and
>>>> only HHH1 simulates any returns of DDD/HHH.
>>>> HHH1 simulates DDD completely, HHH recurses and aborts *inside*. HHH
>>>> does not simulate DDD once, it only enters the call, but never exits.
>>>>
>>>>>>>>>>> The next instruction of DDD that HHH emulates is at the
>>>>>>>>>>> machine address of 00002183.
>>>>>>>>>>> The next instruction of DDD that HHH1 emulates is at the
>>>>>>>>>>> machine address of 00002190.
>>>>
>>>> At those addresses we have the first instruction of DDD and the one
>>>> after the call, respectively.
>>>>
>>>>
>>>> [main -> HHH1(DDD) -> HHH(DDD) -> HHH(DDD)]
>>>>>>>>>>> New slave_stack at:198d21  DDD emulated by HHH *This is the
>>>>>>>>>>> beginning of the divergence of the behavior*
>>>>>>>>>>> *HHH is emulating itself emulating DDD, HHH1 never does that*
>>>>
>>>> HHH1 does simulate HHH simulating DDD.
>>>>
>>> That is not the question being answered.
>> You say: HHH1 doesn't simulate itself. That is true. Also true is that
>> HHH1 simulates the same thing that HHH does,
> 
> The emulation of DDD by HHH(DDD) contained within the execution trace of
> HHH1(DDD) diverges from the execution trace of DDD by HHH1 as soon as
> HHH emulates itself emulating DDD.
No, the instructions are the same.

> HHH(DDD) ALWAYS emulates itself emulating DDD. HHH1(DDD) NEVER emulates
> itself emulating DDD.
Dude, I already agreed with that. It doesn't matter.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.