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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Correcting the definition of the halting problem --- Computable
 functions
Date: Mon, 24 Mar 2025 13:35:23 -0400
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On 3/24/2025 12:44 PM, olcott wrote:
> On 3/24/2025 10:14 AM, dbush wrote:
>> On 3/24/2025 11:03 AM, olcott wrote:
>>> On 3/24/2025 6:23 AM, Richard Damon wrote:
>>>> On 3/23/25 11:09 PM, olcott wrote:
>>>>> It is impossible for HHH compute the function from the direct
>>>>> execution of DDD because DDD is not the finite string input
>>>>> basis from which all computations must begin.
>>>>> https://en.wikipedia.org/wiki/Computable_function
>>>>
>>>> WHy isn't DDD made into the correct finite string?i
>>>>
>>>
>>> DDD is a semantically and syntactically correct finite
>>> stirng of the x86 machine language.
>>
>> Which includes the machine code of DDD, the machine code of HHH, and 
>> the machine code of everything it calls down to the OS level.
>>
>>>
>>>> That seems to be your own fault.
>>>>
>>>> The problem has always been that you want to use the wrong string 
>>>> for DDD by excluding the code for HHH from it.
>>>>
>>>
>>> DDD emulated by HHH directly causes recursive emulation
>>> because it calls HHH(DDD) to emulate itself again. HHH
>>> complies until HHH determines that this cycle cannot
>>> possibly reach the final halt state of DDD.
>>>
>>
>> Which is another way of saying that HHH can't determine that DDD halts 
>> when executed directly.
>>
> 
> given an input of the function domain it can
> return the corresponding output.
> https://en.wikipedia.org/wiki/Computable_function
> 
> Computable functions are only allowed to compute the
> mapping from their input finite strings to an output.
> 


The HHH you implemented is computing *a* computable function, but it's 
not computing the halting function:


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