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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Correcting the definition of the halting problem --- Computable functions
Date: Tue, 25 Mar 2025 10:55:42 +0200
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On 2025-03-24 20:11:12 +0000, olcott said:

> On 3/24/2025 12:35 PM, dbush wrote:
>> 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:
> 
> The whole point of this post is to prove that
> no Turing machine ever reports on the behavior
> of the direct execution of another Turing machine.

Why should that be proven?

-- 
Mikko