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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Turing computable function for sum of two integers
Date: Fri, 2 May 2025 11:18:51 +0300
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On 2025-04-30 19:57:00 +0000, dbush said:

> On 4/30/2025 1:36 PM, olcott wrote:
>> On 4/29/2025 4:50 AM, Mikko wrote:
>>> On 2025-04-28 19:55:35 +0000, olcott said:
>>> 
>>>> On 4/28/2025 11:01 AM, dbush wrote:
>>>>> On 4/28/2025 11:52 AM, olcott wrote:
>>>>>> On 4/28/2025 4:01 AM, Mikko wrote:
>>>>>>> On 2025-04-16 17:36:31 +0000, olcott said:
>>>>>>> 
>>>>>>>> On 4/16/2025 7:29 AM, Richard Heathfield wrote:
>>>>>>>>> On 16/04/2025 12:40, olcott wrote:
>>>>>>>>>> sum(3,2) IS NOT THE SAME AS sum(5,2).
>>>>>>>>>> IT IS EITHER STUPID OR DISHONEST FOR YOU TO TRY TO
>>>>>>>>>> GET AWAY FOR CLAIMING THIS USING THE STRAW DECEPTION
>>>>>>>>>> INTENTIONALLY INCORRECT PARAPHRASE OF MY WORDS.
>>>>>>>>> 
>>>>>>>>> Whether sum(3,2) is or is not the same as sum(5,2) is not the question. 
>>>>>>>>> The question is whether a universal termination analyser can be 
>>>>>>>>> constructed, and the answer is that it can't.
>>>>>>>>> 
>>>>>>>>> This has been rigorously proved. If you want to overturn the proof 
>>>>>>>>> you've got your work cut out to persuade anyone to listen, not least 
>>>>>>>>> because anyone who tries to enter into a dialogue with you is met with 
>>>>>>>>> contempt and scorn.
>>>>>>>>> 
>>>>>>>>> The proof stands.
>>>>>>>>> 
>>>>>>>> 
>>>>>>>> *corresponding output to the input*
>>>>>>>> *corresponding output to the input*
>>>>>>>> *corresponding output to the input*
>>>>>>>> *corresponding output to the input*
>>>>>>>> *corresponding output to the input*
>>>>>>>> 
>>>>>>>> Not freaking allowed to look at any damn thing
>>>>>>>> else besides the freaking input. Must compute whatever
>>>>>>>> mapping ACTUALLY EXISTS FROM THIS INPUT.
>>>>>>> 
>>>>>>> A halt decider is is not allowed to compute "whatever" mapping. It is
>>>>>>> required to compute one specific mapping: to "no" if the computation
>>>>>>> described by the input can be continesd forever without halting, to
>>>>>>> "no" otherwise.
>>>>>>> 
>>>>>> 
>>>>>> It must do this by applying the finite string transformation
>>>>>> rules specified by the x86 language to the input to HHH(DD).
>>>>>> 
>>>>>> This DOES NOT DERIVE THE BEHAVIOR OF THE DIRECTLY EXECUTED DD.
>>>>>> It DOES DERIVE DD EMULATED BY HHH AND ALSO DERIVES THE RECURSIVE
>>>>>> EMULATION OF HHH EMULATING ITSELF EMULATING DD.
>>>>>> 
>>>>> 
>>>>> 
>>>>> In other words, no H exists that satisfies the following requirements,
>>>> 
>>>> BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
>>>> BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
>>>> BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
>>>> BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
>>>> BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
>>> 
>>> You have not proven that the requirements are wrong in any sense.
>>> 
>> 
>> int sum(int x, int y) { return 5; }
>> Is NOT an algorithm for the sum of two integers.
>> 
>> int sum(int x, int y) { x + y; }
>> Is an algorithm for the sum of two integers.
> 
> Obviously, but that has nothing to do with a solution to the halting function:
> 
> Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
> described as <X> with input Y:

It does. The first "sum" is analogous to Olcott's halting deciders that are
like a real halting decider except that they compute another function.

-- 
Mikko