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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: 195 page execution trace of DDD correctly simulated by HH0
Date: Thu, 27 Jun 2024 09:34:33 +0300
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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.

-- 
Mikko