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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sat, 10 May 2025 11:01:13 -0400
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On 5/10/2025 10:51 AM, 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.
> 
>>>>> 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 redefined the halting problem should be about the behavior of D
>> decided by POOH, not the 'direct' behavior of D?
>>
> 
> Not at all. HHH(DDD) reports on the behavior that DDD
> actually specifies. 

Which is the behavior of the algorithm described by the input, which halts.

> All my critics require HHH to report
> on behavior that DDD does not actually specify.
> 
> The actual behavior that DDD specifies is measured
> by 

Executing DDD directly, as per the requirements:


Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
described as <X> with input Y:

A solution to the halting problem is an algorithm H that computes the 
following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly




> int sum(int x, int y) { return x + y ;}
> sum must apply the rules of arithmetic to
> its inputs thus sum(3,2) cannot correctly
> derive the sum of 5 + 7.
> 
> HHH must apply the rules of the x86 language
> to its input DDD. 

And those rules say that when DDD is actually run on an actual x86 
processor that it will halt.

> 
> When HHH1(DDD) applies these same rules to its input
> there is no recursive emulation because DDD does
> not call HHH1(DDD).
>