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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: 195 page execution trace of DDD correctly simulated by HH0
Date: Tue, 25 Jun 2024 13:26:14 -0500
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On 6/25/2024 1:19 PM, Fred. Zwarts wrote:
> Op 25.jun.2024 om 19:29 schreef olcott:
>> 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.
> 
> Change of subject ignored.
> 
>>
>>> 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.
> 
> Maybe you should study some software engineering to get over it.
> 
>> Unless the outermost directly executed H0 aborts its
>> simulation after a fixed number of recursive invocations
>> NONE OF THEM DO.
> 
> Change of subject. We are talking about an H0 that aborts, so dreaming 
> of one that does not abort is irrelevant.

No one here can possibly handle more than one single point
at a time without leaping to the conclusion that I must
be incorrect. Because of this I will not tolerate moving
beyond one single point at a time.

> H0 aborts after two cycles. Then it aborts the simulated H0 which at 
> that moment has run only one cycle. One cycle later the simulated H0 
> would also return, if not aborted.
> 
>>
>> This did baffle me for three days 3.5 years ago until
>> I took the time to THINK IT ALL THE WAY THROUGH.
> 
> Apparently, your thinking went completely wrong.
> 

No the actual truth is that you are one of my least competent
reviewers.

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