XPost: comp.theory, comp.ai.philosophy, sci.math
From: news.dead.person.stones@darjeeling.plus.com
On 19/11/2025 18:26, dart200 wrote:
> On 11/18/25 3:47 PM, Mike Terry wrote:
>> On 18/11/2025 03:10, dart200 wrote:
>>> On 11/17/25 7:07 PM, Kaz Kylheku wrote:
>>>> On 2025-11-18, dart200 wrote:
>>>>> On 11/17/25 4:31 PM, olcott wrote:
>>>>>> On 11/17/2025 6:06 PM, dart200 wrote:
>>>>>>> On 11/17/25 3:35 PM, olcott wrote:
>>>>>>>> The halting problem is requiring deciders to
>>>>>>>> compute information that is not contained in
>>>>>>>> their input.
>>>>>>>
>>>>>>> ur agreeing with turing and the halting problem:
>>>>>>>
>>>>>>> one cannot compute whether a machine halts or not from the string
>>>>>>> describing the machine
>>>>>>>
>>>>>>
>>>>>> That the halting problem limits computation
>>>>>> is like this very extreme example:
>>>>>>
>>>>>> Predict who the next president of the United States
>>>>>> will be entirely on the basis of √2 (square root of 2).
>>>>>> That cannot be derived from the input.
>>>>>
>>>>> bruh, ur agreeing with the halting problem:
>>>>>
>>>>> one cannot take the string describing the machine, and use it to compute
>>>>> whether the machine described halts
>>>>
>>>> But that isn't true; you certainly can do that. Just not using one
>>>> unified algorithm that works for absolutely all such strings.
>>>>
>>>> When it /does/ work, it's certainly not based on any input other than
>>>> the string.
>>>
>>> yes i meant generally
>>>
>>> you also can't compute generally whether you can or cannot compute whether
a an machine
>>> description halts or not
>>
>> What does that mean though?
>>
>> It sounds like you're asking for a /single/ TM that given /any/ machine
description D, must
>> compute "whether or not D's halting is computable". [And saying no such
single TM exists?]
>
> yes, it takes /single/ machine input a outputs whether /any/ other machine
could compute the input
> machine's halting semantics.
Have you read Kaz's response to my post? That explains why for any given
machine, there is always
some other machine that computes the halting status of that machine.
Basically there are only two
possible behaviours: halts or neverhalts. We just need two machines H1 and H0
that straight away
return halts/neverhalts respectively. For any machine M, either H1 or H0
correctly computes M's
halting status, so assuming normal terminology use, any single M is
decideable. (And by extension,
halting for any finite set of machines is decidable.)
Sometimes people attempt to come up with reasons why H1 and H0 don't count.
That was certainly PO's
response, and his explanation was that H1 and H0 are disqualified as halt
deciders because they
"aren't even trying". He has never explained what it means for a TM to "not
really try" to do
something; of course, TMs are just what they are, without "trying" to do
anything. We're not
talking about an olympic sport where there are points awarded for
effort/artistic interpretation
etc., it's all just "whether they work".
[Also, people like PO often confuse what the halting problem says, believing
that it is implying
that there is some machine M which "cannot be decided". That's a
misunderstanding...]
Anyhow, all of that is completely missing the point of the halting problem -
that is to find /one/
machine H that can decide /any/ input M_desc. Finding a machine that can
decide one specific input
is trivial.
Mike.
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