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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: What it would take... People to address my points with reasoning
 instead of rhetoric -- RP
Date: Wed, 14 May 2025 10:52:10 -0500
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On 5/14/2025 6:02 AM, Fred. Zwarts wrote:
> Op 14.mei.2025 om 06:35 schreef olcott:
>> On 5/13/2025 10:48 PM, dbush wrote:
>>> On 5/13/2025 11:38 PM, olcott wrote:
>>>> On 5/13/2025 10:28 PM, dbush wrote:
>>>>> On 5/13/2025 11:06 PM, olcott wrote:
>>>>>> On 5/13/2025 9:44 PM, dbush wrote:
>>>>>>> On 5/13/2025 10:41 PM, olcott wrote:
>>>>>>>> On 5/13/2025 8:56 PM, dbush wrote:
>>>>>>>>> On 5/13/2025 9:52 PM, olcott wrote:
>>>>>>>>>> On 5/13/2025 8:38 PM, dbush wrote:
>>>>>>>>>>> On 5/13/2025 9:35 PM, olcott wrote:
>>>>>>>>>>>> On 5/13/2025 8:26 PM, dbush wrote:
>>>>>>>>>>>>> On 5/13/2025 9:16 PM, olcott wrote:
>>>>>>>>>>>>>> On 5/13/2025 8:03 PM, dbush wrote:
>>>>>>>>>>>>>>> Nope.  Russell's Paradox was derived from the base axioms 
>>>>>>>>>>>>>>> of naive set theory, proving the whole system was 
>>>>>>>>>>>>>>> inconsistent.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> In contrast, there is nothing in existing computation 
>>>>>>>>>>>>>>> theory that requires that a halt decider exists.
>>>>>>>>>>>>>
>>>>>>>>>>>>> I see you made no attempt to refute the above statement. 
>>>>>>>>>>>>> Unless you can show from the axioms of computation theory 
>>>>>>>>>>>>> that the following requirements can be met, your argument 
>>>>>>>>>>>>> has no basis:
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> 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
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> A halt decider doesn't exist
>>>>>>>>>>>>>>>> for the same reason that the set of all sets
>>>>>>>>>>>>>>>> that do not contain themselves does not exist.
>>>>>>>>>>>>>>>> *As defined both were simply wrong-headed ideas*
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> There's nothing wrong-headed about wanting to know if any 
>>>>>>>>>>>>>>> arbitrary algorithm X with input Y will halt when 
>>>>>>>>>>>>>>> executed directly. 
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Yes there is. I have proven this countless times.
>>>>>>>>>>>>>
>>>>>>>>>>>>> That requirements are impossible to satisfy doesn't make 
>>>>>>>>>>>>> them wrong. It just makes them impossible to satisfy, which 
>>>>>>>>>>>>> is a perfectly reasonable conclusion.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> It did with Russell's Paradox.
>>>>>>>>>>>> ZFC rejected the whole foundation upon which
>>>>>>>>>>>> RP was built.
>>>>>>>>>>>>
>>>>>>>>>>>> ZFC did not solve some other Russell's Paradox
>>>>>>>>>>>> it rejected the whole idea of RP as nonsense.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Unless you can show from the axioms of computation theory 
>>>>>>>>>>> that the following requirements can be met, your argument has 
>>>>>>>>>>> no basis:
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Alternatively I can do what ZFC did and over-rule
>>>>>>>>>> the whole foundation upon which the HP proofs are build.
>>>>>>>>>
>>>>>>>>> You mean the assumption that the following requirements (which 
>>>>>>>>> are *not* part of the axioms of computation theory) can be 
>>>>>>>>> satisfied? The assumption that Linz and other proved was false 
>>>>>>>>> and that you *explicitly* agreed with?
>>>>>>>>>
>>>>>>>>
>>>>>>>> The conventional halting problem proofs have your
>>>>>>>> requirements as its foundation.
>>>>>>>>
>>>>>>>
>>>>>>> They have the *assumption* that the requirements can be met, and 
>>>>>>> via proof by contradiction show the assumption to be false.
>>>>>>>
>>>>>>> And the fact that the requirements can't be met is fine, just 
>>>>>>> like the the fact that these requirements can't be met is fine:
>>>>>>>
>>>>>>> A mythic number is a number N such that N > 5 and N < 2.
>>>>>>
>>>>>> We can also say that no computation can compute
>>>>>> the square root of a dead rabbit. In none of these
>>>>>> cases is computation actually limited.
>>>>>>
>>>>>> We could equally say that no whale can give
>>>>>> birth to a pigeon. This places no actual limit
>>>>>> on the behavior of whales. Whales were never
>>>>>> meant to give birth to pigeons.
>>>>>>
>>>>>
>>>>> And as was said before:
>>>>>
>>>>> On 5/5/2025 5:39 PM, olcott wrote:
>>>>>  > On 5/5/2025 4:31 PM, dbush wrote:
>>>>>  >> Strawman.  The square root of a dead rabbit does not exist, but 
>>>>> the
>>>>>  >> question of whether any arbitrary algorithm X with input Y 
>>>>> halts when
>>>>>  >> executed directly has a correct answer in all cases.
>>>>>  >>
>>>>>  >
>>>>>  > It has a correct answer that cannot ever be computed
>>>>>
>>>>> This qualifies as both a non-rebuttal and your confirmation you 
>>>>> agree that Linz and others are correct that no algorithm exists 
>>>>> that satisfies the below 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
>>>>>
>>>>
>>>> It is true that a TM either halts or does not halt.
>>>>
>>>> None-the-less the above requirements simply ignore
>>>> that some inputs specify behavior that differs
>>>> from the behavior of their direct execution.
>>>
>>> What you think the input specifies is irrelevant. 
>>
>> What it actually specifies rules the computation.
>>
> And the input includes the code to abort, so that should be included in 
> the computation of the mapping. If it doesn't use the full 
> specification, the mapping is incorrect

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     would never stop running unless aborted then

The spec only requires HHH to determine what the behavior
of DDD would be if it never aborted. If it never aborted
then there would be no abort code for HHH to see.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer