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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 08:12:45 -0500
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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.

Until you acknowledge this is true, this is the
only thing that I am willing to talk to you about.

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