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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: 195 page execution trace of DDD correctly simulated by HH0
Date: Thu, 27 Jun 2024 12:07:20 -0500
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On 6/27/2024 1:34 AM, Mikko wrote:
> On 2024-06-26 12:55:43 +0000, olcott said:
> 
>> On 6/26/2024 3:10 AM, Mikko wrote:
>>> On 2024-06-25 17:29:12 +0000, olcott said:
>>>
>>>> On 6/25/2024 9:13 AM, Fred. Zwarts wrote:
>>>>> Op 25.jun.2024 om 15:12 schreef olcott:
>>>>>> On 6/25/2024 7:08 AM, Fred. Zwarts wrote:
>>>>>>> Op 24.jun.2024 om 23:04 schreef olcott:
>>>>>>>> On 6/24/2024 2:36 PM, joes wrote:
>>>>>>>>> Am Mon, 24 Jun 2024 08:48:19 -0500 schrieb olcott:
>>>>>>>>>> On 6/24/2024 2:37 AM, Mikko wrote:
>>>>>>>>>>> On 2024-06-23 13:17:27 +0000, olcott said:
>>>>>>>>>>>> On 6/23/2024 3:22 AM, Mikko wrote:
>>>>>>>>>>>>> That code is not from the mentined trace file. In that file 
>>>>>>>>>>>>> _DDD()
>>>>>>>>>>>>> is at the addresses 2093..20a4. According to the trace no 
>>>>>>>>>>>>> instruction
>>>>>>>>>>>>> at the address is executed (because that address points to 
>>>>>>>>>>>>> the last
>>>>>>>>>>>>> byte of a three byte instruction.
>>>>>>>>>>>>
>>>>>>>>>>>> In order to make my examples I must edit the code and this 
>>>>>>>>>>>> changes the
>>>>>>>>>>>> addresses of some functions.
>>>>>>>>>>>
>>>>>>>>>>> Why do you need to make an example when you already have one 
>>>>>>>>>>> in the
>>>>>>>>>>> file mentioned in the subject line?
>>>>>>>>>>>
>>>>>>>>>> I had to make a few more examples such as HH1(DD,DD)
>>>>>>>>> AFACT HH1 is the same as HH0, right? What happens when HH1 
>>>>>>>>> tries to
>>>>>>>>> simulate a function DD1 that only calls HH1?
>>>>>>>>>
>>>>>>>>
>>>>>>>> typedef uint32_t u32;
>>>>>>>> u32 H(u32 P, u32 I);
>>>>>>>>
>>>>>>>> int P(u32 x)
>>>>>>>> {
>>>>>>>>    int Halt_Status = H(x, x);
>>>>>>>>    if (Halt_Status)
>>>>>>>>      HERE: goto HERE;
>>>>>>>>    return Halt_Status;
>>>>>>>> }
>>>>>>>>
>>>>>>>> int main()
>>>>>>>> {
>>>>>>>>    H(P,P);
>>>>>>>> }
>>>>>>>>
>>>>>>>> I am going to have to go through my code and standardize my names.
>>>>>>>> H(P,P) was the original name. Then I had to make a one parameter
>>>>>>>> version, a version that is identical to H, except P does not call
>>>>>>>> it and then versions using different algorithms. People have never
>>>>>>>> been able to understand the different algorithm.
>>>>>>>>
>>>>>>>> typedef void (*ptr)();
>>>>>>>> typedef int (*ptr2)();
>>>>>>>> int  HH(ptr2 P, ptr2 I); // used with int D(ptr2 P) that calls HH
>>>>>>>> int HH1(ptr2 P, ptr2 I); // used with int D(ptr2 P) that calls HH
>>>>>>>> int  HHH(ptr P);         // used with void DDD() that calls HHH
>>>>>>>> int HHH1(ptr P);         // used with void DDD() that calls HHH
>>>>>>>>
>>>>>>>> *The different algorithm version has been deprecated*
>>>>>>>> int  H(ptr2 , ptr2 I);  // used with int D(ptr2 P) that calls H
>>>>>>>> int H1(ptr2 P, ptr2 I); // used with int D(ptr2 P) that calls H
>>>>>>>>
>>>>>>>> *It is much easier for people to see the infinite recursion*
>>>>>>>> *behavior pattern when they see it actually cycle through the*
>>>>>>>> *same instructions twice*
>>>>>>>
>>>>>>> Twice is not equal to infinitely. When will you see that?
>>>>>>> It is strange that you call that an infinite recursion, when H 
>>>>>>> aborts after two cycles and the simulated H cannot reach its own 
>>>>>>> abort operation, because it is aborted when it had only one more 
>>>>>>> cycle to go.
>>>>>>> None of the aborted simulations would cycle more than twice, so 
>>>>>>> infinite recursion is not seen for an H that aborts the 
>>>>>>> simulation of itself.
>>>>>>
>>>>>> typedef void (*ptr)();
>>>>>> int H0(ptr P);
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    H0(DDD);
>>>>>> }
>>>>>>
>>>>>> int main()
>>>>>> {
>>>>>>    H0(DDD);
>>>>>> }
>>>>>>
>>>>>> _DDD()
>>>>>> [00002172] 55               push ebp      ; housekeeping
>>>>>> [00002173] 8bec             mov ebp,esp   ; housekeeping
>>>>>> [00002175] 6872210000       push 00002172 ; push DDD
>>>>>> [0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
>>>>>> [0000217f] 83c404           add esp,+04
>>>>>> [00002182] 5d               pop ebp
>>>>>> [00002183] c3               ret
>>>>>> Size in bytes:(0018) [00002183]
>>>>>>
>>>>>> The call from DDD to H0(DDD) when DDD is correctly emulated
>>>>>> by H0 cannot possibly return.
>>>>>
>>>>> Contradictio in terminis. The fact that the simulated H0 does not 
>>>>> return shows that the simulation is incorrect.
>>>>
>>>> void Infinite_Recursion()
>>>> {
>>>>    Infinite_Recursion();
>>>> }
>>>>
>>>> Ah so you simply *DON'T BELIEVE IN* infinite recursion where a
>>>> correct simulating termination analyzer would be required to
>>>> abort its simulation to correctly report non-terminating behavior.
>>>> That seems quite dumb of you.
>>>>
>>>>> The simulated H0 does not return, because it is aborted one cycle 
>>>>> too soon. One cycle later it would return.
>>>>
>>>> Complete lack of sufficient software engineering skill.
>>>
>>> The relevant area of software engineering is testing. The usual 
>>> attitude of
>>> software engineers is that a program is accpted when it has been 
>>> sufficiently
>>> tested and passed all tests. Consequently, an important part of 
>>> sofware work
>>> is the design of tests.
>>>
>>> In the current context the program to be tested is a halting decider.
>>
>> *NO IT IS NOT. H0 IS ONLY AN X86 EMULATOR*
>> After you quit lying about the behavior of DDD correctly
>> emulated by H0 then we can move on to the next point.
> 
> This discussion is about HH0. The larger context of this discussion is
> halting deiders and proofs of non-existence of halting deciders.
> 
> You have not disagreed with anyting I said, so if you say that I lied
> then you reveal that you lied.
> 
Until you agree with this we cannot move on to the next
and final point that proves I am correct. Proving that
point may possibly take longer than the rest of my life
so let's not delay this OK?

[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]

The call from DDD to H0(DDD) when DDD is correctly emulated
by x86 emulator H0 cannot possibly return.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer