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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
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Date: Thu, 20 Jun 2024 09:53:13 +0200
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Op 19.jun.2024 om 17:46 schreef olcott:
> On 6/19/2024 10:39 AM, Fred. Zwarts wrote:
>> 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.
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