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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 01:06:48 -0500
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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.

>> does sometimes differ from the behavior of its direct
>> execution. It is a provably different sequence of steps.
> 

This is a verified fact.
The pathological relationship that inputs can have
with their simulating termination analyzer changes
the behavior of these inputs relative to their direct
execution.

> So, you are talking about the behavior of the 'simulating termination analyzer'
> i.e. POOH? (not really about the HP)
> 
> 
> 
> 


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