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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: Tue, 25 Mar 2025 13:04:10 -0400
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On 3/25/2025 12:56 PM, olcott wrote:
> On 3/25/2025 11:46 AM, dbush wrote:
>> On 3/25/2025 12:40 PM, olcott wrote:
>>> On 3/25/2025 10:17 AM, dbush wrote:
>>>> On 3/25/2025 11:13 AM, olcott wrote:
>>>>> On 3/25/2025 10:02 AM, dbush wrote:
>>>>>> On 3/25/2025 10:53 AM, olcott wrote:
>>>>>>> On 3/25/2025 9:45 AM, dbush wrote:
>>>>>>>> On 3/24/2025 11:29 PM, olcott wrote:
>>>>>>>>> On 3/24/2025 10:12 PM, dbush wrote:
>>>>>>>>>> On 3/24/2025 10:07 PM, olcott wrote:
>>>>>>>>>>> On 3/24/2025 8:46 PM, André G. Isaak wrote:
>>>>>>>>>>>> On 2025-03-24 19:33, olcott wrote:
>>>>>>>>>>>>> On 3/24/2025 7:00 PM, André G. Isaak wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>>> In the post you were responding to I pointed out that 
>>>>>>>>>>>>>> computable functions are mathematical objects.
>>>>>>>>>>>>>
>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Computable_function
>>>>>>>>>>>>>
>>>>>>>>>>>>> Computable functions implemented using models of computation
>>>>>>>>>>>>> would seem to be more concrete than pure math functions.
>>>>>>>>>>>>
>>>>>>>>>>>> Those are called computations or algorithms, not computable 
>>>>>>>>>>>> functions.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> https://en.wikipedia.org/wiki/Pure_function
>>>>>>>>>>> Is another way to look at computable functions implemented
>>>>>>>>>>> by some concrete model of computation.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> And not all mathematical functions are computable, such as the 
>>>>>>>>>> halting function.
>>>>>>>>>>
>>>>>>>>>>>> The halting problems asks whether there *is* an algorithm 
>>>>>>>>>>>> which can compute the halting function, but the halting 
>>>>>>>>>>>> function itself is a purely mathematical object which exists 
>>>>>>>>>>>> prior to, and independent of, any such algorithm (if one 
>>>>>>>>>>>> existed).
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> None-the-less it only has specific elements of its domain
>>>>>>>>>>> as its entire basis. For Turing machines this always means
>>>>>>>>>>> a finite string that (for example) encodes a specific
>>>>>>>>>>> sequence of moves.
>>>>>>>>>>
>>>>>>>>>> False.  *All* turing machine are the domain of the halting 
>>>>>>>>>> function, and the existence of UTMs show that all turning 
>>>>>>>>>> machines can be described by a finite string.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You just aren't paying enough attention. Turing machines
>>>>>>>>> are never in the domain of any computable function.
>>>>>>>>> <snip>
>>>>>>>>>
>>>>>>>>
>>>>>>>> False.  The mathematical function that counts the number of 
>>>>>>>> instructions in a turing machine is computable.
>>>>>>>>
>>>>>>>
>>>>>>> It is impossible for an actual Turing machine to
>>>>>>> be input to any other TM.
>>>>>>>
>>>>>>
>>>>>> But a description of a turing machine can be, for example in the 
>>>>>> form of source code or a binary.  And a turing machine by 
>>>>>> definition *always* behaves the same for a given input when 
>>>>>> executing directly.
>>>>>
>>>>> IT IS COUNTER-FACTUAL THAT A MACHINE DESCRIPTION ALWAYS
>>>>> SPECIFIES
>>>>> BEHAVIOR IDENTICAL TO THE DIRECTLY EXECUTED MACHINE.
>>>>>
>>>>> _III()
>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>> [00002173] 8bec       mov  ebp,esp  ; housekeeping
>>>>> [00002175] 6872210000 push 00002172 ; push III
>>>>> [0000217a] e853f4ffff call 000015d2 ; call EEE(III)
>>>>> [0000217f] 83c404     add  esp,+04
>>>>> [00002182] 5d         pop  ebp
>>>>> [00002183] c3         ret
>>>>> Size in bytes:(0018) [00002183]
>>>>
>>>> That is not the complete description.  The complete description 
>>>> consists of the code of III
>>>
>>> and the fact that EEE 
>>
>> Is called by III makes the code of EEE part of the fixed input, as 
>> well as everything that EEE calls down to the OS level.
>>
> 
> Which is not relevant to whether or not III emulated
> by EEE reaches its own final halt state.
> 

Which is why III emulated by EEE is not relevant.

Remember, we want to know whether an algorithm exists that can do this:


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