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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sat, 10 May 2025 11:44:31 -0500
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On 5/10/2025 11:21 AM, wij wrote:
> On Sat, 2025-05-10 at 11:15 -0500, olcott wrote:
>> On 5/10/2025 11:00 AM, wij wrote:
>>> On Sat, 2025-05-10 at 10:43 -0500, olcott wrote:
>>>> On 5/10/2025 10:14 AM, wij wrote:
>>>>> On Sat, 2025-05-10 at 09:51 -0500, olcott wrote:
>>>>>> On 5/10/2025 1:19 AM, wij wrote:
>>>>>>> On Sat, 2025-05-10 at 01:06 -0500, olcott wrote:
>>>>>>>> On 5/10/2025 1:00 AM, wij wrote:
>>>>>>>>> On Sat, 2025-05-10 at 00:41 -0500, olcott wrote:
>>>>>>>>>> On 5/10/2025 12:27 AM, wij wrote:
>>>>>>>>>>> On Sat, 2025-05-10 at 00:19 -0500, olcott wrote:
>>>>>>>>>>>> On 5/10/2025 12:13 AM, wij wrote:
>>>>>>>>>>>>> On Sat, 2025-05-10 at 00:06 -0500, olcott wrote:>>
>>>>>>>>>>>>>> When mathematical mapping is properly understood
>>>>>>>>>>>>>> it will be known that functions computed by models
>>>>>>>>>>>>>> of computation must transform their input into
>>>>>>>>>>>>>> outputs according to the specific steps of an
>>>>>>>>>>>>>> algorithm.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _DDD()
>>>>>>>>>>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>>>>>>>>>>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>>>>>>>>>>>>>> [00002175] 6872210000 push 00002172 ; push DDD
>>>>>>>>>>>>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>>>>>>>>>>>>> [0000217f] 83c404     add esp,+04
>>>>>>>>>>>>>> [00002182] 5d         pop ebp
>>>>>>>>>>>>>> [00002183] c3         ret
>>>>>>>>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> For example HHH(DDD) only correctly map to the
>>>>>>>>>>>>>> behavior that its input actually specifies by correctly
>>>>>>>>>>>>>> emulating DDD according to the rules of the x86 language.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> This causes the first four instructions of DDD
>>>>>>>>>>>>>> to be emulated followed by HHH emulating itself
>>>>>>>>>>>>>> emulating the first three instructions of DDD.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It is right at this recursive simulation just
>>>>>>>>>>>>>> before HHH(DDD) is called again that HHH recognizes
>>>>>>>>>>>>>> the repeating pattern and rejects DDD.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Yes, but you still did not answer the question: Is POOH exactly about HP?
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>        >>>>> H(D)=1 if D() halt.
>>>>>>>>>>>>        >>>>> H(D)=0 if D() not halt.
>>>>>>>>>>>>
>>>>>>>>>>>> Right now it is mostly about proving the
>>>>>>>>>>>> above requirements are is mistaken.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Why is the requirement invalid?
>>>>>>>>>>>
>>>>>>>>>>> H(D)=1 if D() halt.
>>>>>>>>>>> H(D)=0 if D() not halt.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>>> The notion that the behavior specified by the finite
>>>>>>>>>> string input to a simulating termination analyzer
>>>>>>>>>
>>>>>>>>> POOH reads(takes) its input as a function, not 'finite string'.
>>>>>>>>> Are you talking about POOH now? There is no POOH that takes
>>>>>>>>> 'finite string'.
>>>>>>>>>
>>>>>>>>
>>>>>>>> It <is> a finite string of x86 bytes.
>>>>>>>
>>>>>>> Disagree.
>>>>>>> The D in Halt7.c (I just saw once) does not treat H as 'finite string',
>>>>>>> D calls H. H also does not treat D as 'finite string'.
>>>>>>>
>>>>>>
>>>>>> HHH and DDD and DD are the most recent functions.
>>>>>> HHH does emulate its finite strings of x86 machine code
>>>>>> according to the rules of the x86 language.
>>>>>
>>>>> This is from a copy of Halt7.c:
>>>>>
>>>>> void P(ptr x)
>>>>> {
>>>>>      int Halt_Status = H(x, x);
>>>>>      if (Halt_Status)
>>>>>        HERE: goto HERE;
>>>>>      return;
>>>>> }
>>>>>
>>>>> int main()
>>>>> {
>>>>>      Output("Input_Halts = ", H(P, P));
>>>>> }
>>>>>
>>>>> H reads a *pointer*.
>>>>> In P, P *calls* H.
>>>>>
>>>>> Both do not process 'finite string'.
>>>>>
>>>>
>>>> What it is it a pointer to a box of chocolates?
>>>> finite strings are passed as pointers to finite
>>>> string in C.
>>>
>>> Nope. I don't believe it is a pointer to chocolates, even if you say so.
>>> It is about the code of D/H itself. They do not process string, the fact says
>>> the author of the program does not intend to process 'string'.
>>>
>>
>> The input to HHH(DDD) is a pointer to a finite string
>> of machine code. HHH applies an x86 emulator to this
>> finite string.
> 
> Do you read English? Do D/H read and process 'finite string'?
> 

Like I have always said... DD, and DDD are finite
strings of x86 machine code and HHH emulates these
according to the x86 language.


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