Path: ...!3.eu.feeder.erje.net!feeder.erje.net!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3
 ---IGNORING ALL OTHER REPLIES
Date: Wed, 19 Jun 2024 17:39:32 +0200
Organization: A noiseless patient Spider
Lines: 275
Message-ID: <v4uu3k$1vtc8$4@dont-email.me>
References: <v4kf3h$3h3iu$7@dont-email.me> <v4m5l6$3v4ql$1@dont-email.me>
 <v4mmsd$1qt6$3@dont-email.me> <v4oo36$hnns$1@dont-email.me>
 <v4pc7t$ln46$3@dont-email.me> <v4revs$18him$1@dont-email.me>
 <v4s07h$1boeu$5@dont-email.me> <v4sb8k$1e266$2@dont-email.me>
 <v4slo4$1g4ib$1@dont-email.me> <v4u33c$1rrod$1@dont-email.me>
 <v4ukld$1vpm0$1@dont-email.me> <v4unf7$1vtc8$1@dont-email.me>
 <v4urgp$21810$1@dont-email.me>
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
Injection-Date: Wed, 19 Jun 2024 17:39:33 +0200 (CEST)
Injection-Info: dont-email.me; posting-host="f73c32778335b0321bd1b04b79c4c70c";
	logging-data="2094472"; mail-complaints-to="abuse@eternal-september.org";	posting-account="U2FsdGVkX19qUv4FYmT+HbD5p9dDj7kM"
User-Agent: Mozilla Thunderbird
Cancel-Lock: sha1:yKWVeF7N7y4iGgABCXT1rEm0z1Y=
Content-Language: en-GB
In-Reply-To: <v4urgp$21810$1@dont-email.me>
Bytes: 12964

Op 19.jun.2024 om 16:55 schreef olcott:
> On 6/19/2024 8:46 AM, Fred. Zwarts wrote:
>> Op 19.jun.2024 om 14:58 schreef olcott:
>>> On 6/19/2024 2:58 AM, Fred. Zwarts wrote:
>>>> Op 18.jun.2024 om 21:04 schreef olcott:
>>>>> On 6/18/2024 11:05 AM, Mikko wrote:
>>>>>> On 2024-06-18 12:57:21 +0000, olcott said:
>>>>>>
>>>>>>> On 6/18/2024 3:03 AM, Mikko wrote:
>>>>>>>> On 2024-06-17 13:03:56 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 6/17/2024 2:20 AM, Mikko wrote:
>>>>>>>>>> On 2024-06-16 12:47:09 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 6/16/2024 2:53 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-06-15 16:22:09 +0000, olcott said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 6/13/2024 8:24 PM, Richard Damon wrote:
>>>>>>>>>>>>>  > On 6/13/24 11:32 AM, olcott wrote:
>>>>>>>>>>>>>  >>
>>>>>>>>>>>>>  >> It is contingent upon you to show the exact steps of 
>>>>>>>>>>>>> how H computes
>>>>>>>>>>>>>  >> the mapping from the x86 machine language finite string 
>>>>>>>>>>>>> input to
>>>>>>>>>>>>>  >> H(D,D) using the finite string transformation rules 
>>>>>>>>>>>>> specified by
>>>>>>>>>>>>>  >> the semantics of the x86 programming language that 
>>>>>>>>>>>>> reaches the
>>>>>>>>>>>>>  >> behavior of the directly executed D(D)
>>>>>>>>>>>>>  >>
>>>>>>>>>>>>>  >
>>>>>>>>>>>>>  > Why? I don't claim it can.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The first six steps of this mapping are when instructions
>>>>>>>>>>>>> at the machine address range of [00000cfc] to [00000d06]
>>>>>>>>>>>>> are simulated/executed.
>>>>>>>>>>>>>
>>>>>>>>>>>>> After that the behavior of D correctly simulated by H diverges
>>>>>>>>>>>>> from the behavior of D(D) because the call to H(D,D) by D
>>>>>>>>>>>>> correctly simulated by H cannot possibly return to D.
>>>>>>>>>>>>>
>>>>>>>>>>>>> _D()
>>>>>>>>>>>>> [00000cfc](01) 55          push ebp
>>>>>>>>>>>>> [00000cfd](02) 8bec        mov ebp,esp
>>>>>>>>>>>>> [00000cff](03) 8b4508      mov eax,[ebp+08]
>>>>>>>>>>>>> [00000d02](01) 50          push eax       ; push D
>>>>>>>>>>>>> [00000d03](03) 8b4d08      mov ecx,[ebp+08]
>>>>>>>>>>>>> [00000d06](01) 51          push ecx       ; push D
>>>>>>>>>>>>> [00000d07](05) e800feffff  call 00000b0c  ; call H
>>>>>>>>>>>>> [00000d0c](03) 83c408      add esp,+08
>>>>>>>>>>>>> [00000d0f](02) 85c0        test eax,eax
>>>>>>>>>>>>> [00000d11](02) 7404        jz 00000d17
>>>>>>>>>>>>> [00000d13](02) 33c0        xor eax,eax
>>>>>>>>>>>>> [00000d15](02) eb05        jmp 00000d1c
>>>>>>>>>>>>> [00000d17](05) b801000000  mov eax,00000001
>>>>>>>>>>>>> [00000d1c](01) 5d          pop ebp
>>>>>>>>>>>>> [00000d1d](01) c3          ret
>>>>>>>>>>>>> Size in bytes:(0034) [00000d1d]
>>>>>>>>>>>>
>>>>>>>>>>>> When you put "V2" or "V3" or something similar on the 
>>>>>>>>>>>> subject line
>>>>>>>>>>>> you should tell what is different from the original version.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> I ask what are the steps
>>>>>>>>>>> I provide 6 steps and then ask what are the next steps.
>>>>>>>>>>> I provide all of the steps.
>>>>>>>>>>
>>>>>>>>>> In which version?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *This is the simplest possible version*
>>>>>>>>>
>>>>>>>>> void DDD()
>>>>>>>>> {
>>>>>>>>>    H0(DDD);
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> After six steps of DDD are correctly emulated by H0
>>>>>>>>> what machine address of DDD would it be at?
>>>>>>>>>
>>>>>>>>> _DDD()
>>>>>>>>> [00001fd2] 55               push ebp      ; housekeeping
>>>>>>>>> [00001fd3] 8bec             mov ebp,esp   ; housekeeping
>>>>>>>>> [00001fd5] 68d21f0000       push 00001fd2 ; push DDD
>>>>>>>>> [00001fda] e8f3f9ffff       call 000019d2 ; call H0
>>>>>>>>> [00001fdf] 83c404           add esp,+04   ; housekeeping
>>>>>>>>> [00001fe2] 5d               pop ebp       ; housekeeping
>>>>>>>>> [00001fe3] c3               ret           ; return to caller
>>>>>>>>> Size in bytes:(0018) [00001fe3]
>>>>>>>>
>>>>>>>> So how is this a difference between the original version and V2 
>>>>>>>> and V3?
>>>>>>>>
>>>>>>>
>>>>>>> No params thus easier to see that it pushes its own machine address.
>>>>>>
>>>>>> My question is still unanswered.
>>>>>>
>>>>>
>>>>> *The simplest possible case*
>>>>> void DDD()
>>>>> {
>>>>>    H0(DDD);
>>>>> }
>>>>>
>>>>> *The next simplest case*
>>>>> typedef void (*ptr)();
>>>>> void DDD(ptr x)
>>>>> {
>>>>>    HH(x, x);
>>>>> }
>>>>>
>>>>> *The conventional case*
>>>>> typedef int (*ptr2)();
>>>>> int P(ptr2 x)
>>>>> {
>>>>>    int Halt_Status = H(x, x);
>>>>>    if (Halt_Status)
>>>>>      HERE: goto HERE;
>>>>>    return Halt_Status;
>>>>> }
>>>>>
>>>>> I had to keep dumbing it down to make it more
>>>>> difficult to reject out-of-hand without review.
>>>>
>>>> No, this is not the simplest case. You are making it unnecessary 
>>>> complex. The simplest case is:
>>>>
>>>>         int main()
>>>>         {
>>>>           return H(main, 0);
>>>>         }
>>>>
>>>> No D, DD, or DDD is needed.
>>>> For this case you proved that main halts, whereas H reports 
>>>> non-halting, i.e. a false negative.
>>>> This shows that in your more complex (but equivalent) cases there is 
>>>> also a false negative.
>>>> Of course, you prefer the more complex cases, because there you can 
>>>> play your invalid claim that the direct execution and the simulation 
>>>> of DDD(DDD) show different behaviour., but the simplest case shows 
>>>> that it is not true. It is just a false negative.
>>>
>>> _DDD()
>>> [000020a2] 55         push ebp      ; housekeeping
>>> [000020a3] 8bec       mov ebp,esp   ; housekeeping
>>> [000020a5] 68a2200000 push 000020a2 ; push DDD
>>> [000020aa] e8f3f9ffff call 00001aa2 ; call H0
>>> [000020af] 83c404     add esp,+04   ; housekeeping
>>> [000020b2] 5d         pop ebp       ; housekeeping
>>> [000020b3] c3         ret           ; never gets here
>>> Size in bytes:(0018) [000020b3]
>>>
>>> Exactly which step of DDD emulated by H0 was emulated
>>> incorrectly such that this emulation would be complete?
>>> AKA DDD emulated by H0 reaches machine address [000020b3]
>>>
>>>
>>
>> That has been pointed out to you so many times. 
> 
> When falsehoods are pointed out an unlimited number of times
> they still remain falsehoods.
> 
>> It seems really to difficult for you. So, you prefer to forget or 
>> ignore it.
>> The 'call' instruction at 000020aa is incorrectly simulated. 
> 
> As a matter of fact it is not incorrectly simulated.
> I am showing this with HH0 instead of H0 because
> the trace provided by HH0 is easier to understand.
========== REMAINDER OF ARTICLE TRUNCATED ==========