On 4/4/22 6:32 PM, olcott wrote:
> On 4/4/2022 5:23 PM, wij wrote:
>> On Tuesday, 5 April 2022 at 04:47:42 UTC+8, olcott wrote:
>>> On 4/4/2022 3:36 PM, olcott wrote:
>>>> On 4/4/2022 2:51 PM, Ben Bacarisse wrote:
>>>>> olcott <No...@NoWhere.com> writes:
>>>>>
>>>>>> On 4/4/2022 11:23 AM, Ben Bacarisse wrote:
>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 4/4/2022 10:32 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 4/4/2022 5:14 AM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 4/3/2022 8:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>
>>>>>>>>>>>>> It might be time to skip ahead because the next exercise is to
>>>>>>>>>>>>> do the
>>>>>>>>>>>>> same for P, a TM that decides if a string encodes a prime
>>>>>>>>>>>>> number.  Can
>>>>>>>>>>>>> you think of how to specify that without giving an algorithm?
>>>>>>>>>>>>>           P.q0 ??? ⊦* P.qy    if ???
>>>>>>>>>>>>>           P.q0 ??? ⊦* P.qn    otherwise.
>>>>>>>>>>>>> (The three ??? won't all be the same things.)  Any idea how to
>>>>>>>>>>>>> flesh
>>>>>>>>>>>>> this out?  If you can, you will be able to do it for E very
>>>>>>>>>>>>> easily too.
>>>>>>>>>>>>
>>>>>>>>>>>> P.q0 S ⊦* P.qy    if Is-Prime-Number(S)
>>>>>>>>>>>> P.q0 S ⊦* P.qn    otherwise.
>>>>>>>>>>> That's getting close.  We know, from how the notation works,
>>>>>>>>>>> that S is a
>>>>>>>>>>> string of symbols in the TM's alphabet.  But writing
>>>>>>>>>>> Is-Prime-Number(S)
>>>>>>>>>>> suggests that is not natural language.  That's a very hard route
>>>>>>>>>>> to go
>>>>>>>>>>> down.  I'd have to ask you for the definition of
>>>>>>>>>>> Is-Prime-Number.
>>>>>>>>>>> Defining it symbolically is messy and if the definition is /not/
>>>>>>>>>>> formal,
>>>>>>>>>>> dressing the definition up with a formal-looking name is just
>>>>>>>>>>> superficial.
>>>>>>>>>>
>>>>>>>>>> It goes through some tedious steps to see if it is a prime
>>>>>>>>>> number:
>>>>>>>>>>
>>>>>>>>>> A prime number is a natural number greater than 1 that is not a
>>>>>>>>>> product of two smaller natural numbers.
>>>>>>>>> We've hit a bit of a road-block rather sooner that I had expected.
>>>>>>>>> First off, there's no need to define what a prime number is.  If
>>>>>>>>> at some
>>>>>>>>> point it turns out that your readers do not know, go ahead and
>>>>>>>>> define
>>>>>>>>> it, but it's too widely understood by comp.theory readers to
>>>>>>>>> bother
>>>>>>>>> about.
>>>>>>>>> But writing (as I think you are suggesting)
>>>>>>>>>       P.q0 S ⊦* P.qy    it goes through some tedious steps to see
>>>>>>>>> if it is a
>>>>>>>>>                         prime number
>>>>>>>>> is not really adequate.  There are two 'it'.  To what do they
>>>>>>>>> refer?
>>>>>>>>
>>>>>>>> You told me to make sure that I do not provide an algorithm.
>>>>>>> Yes, that good.  You didn't.
>>>>>>> Now, to what do the two 'it's refer?
>>>>>
>>>>> OK, maybe things have gone too far already, but why are you
>>>>> ignoring my
>>>>> questions?  Your phrase used 'it' twice.  What did you intend to refer
>>>>> to by these two pronouns?  It was not an idle question.  I think when
>>>>> you answer it, at least one problem will become clear.
>>>>>
>>>>>>>> This is somewhat algorithmic:
>>>>>>> No, no.  The non-algorithmic way is best.  You should be able to
>>>>>>> specific what a computation does even when yo have no idea how to
>>>>>>> write
>>>>>>> the an algorithm to do it.  Sometimes there isn't ad algorithm!
>>>>>>>
>>>>>>>>> and nothing in P.q0 S ⊦* P.qy is a number so there is nothing
>>>>>>>>> there to
>>>>>>>>> be a prime number.
>>>>>>>>> Can you see how you (a) make it shorter, (b) make it clearer?
>>>>>>> My reply has three questions in it (depending on how you count)
>>>>>>> but you
>>>>>>> didn't answer any of them.  This will only work if you try to answer
>>>>>>> these questions.  Sometimes the answer will be "I don't know what
>>>>>>> you
>>>>>>> mean", but that's a perfectly good answer.
>>>>>>
>>>>>> Anything besides the bare function name is somewhat algorithmic so
>>>>>> what you are asking for seems impossible.
>>>>>
>>>>> Here's a supporting exercise.  Write down at least half a dozen
>>>>> strings
>>>>> that might be passed to P.  At least one of them should be a string
>>>>> that
>>>>> P must accept and at least one must be a string the P should
>>>>> reject, but
>>>>> you should say, for each one, whether P accepts of rejects it.
>>>>>
>>>>> Be prepared for me to raise questions about what the strings
>>>>> represent.
>>>>> It's easy to assume conventions from everyday life that should be
>>>>> stated
>>>>> explicitly.  You must have come across this in software: "the manual
>>>>> said the input should be a number but it went wrong for सहस्र."
>>>>>
>>>>> Finally, don't fuss about the prime bit.  Just use the word prime.
>>>>> Everyone one knows what it means.  The key thing here is to state
>>>>> /what/
>>>>> must be prime for P to correctly accept a string.
>>>>>
>>>>
>>>> I am estimating that we will finally achieve closure on this.
>>>> Creating a common language between us will achieve the basis for mutual
>>>> understanding.
>>>>
>>> The process that we are doing looks like it will be effective on the
>>> basis of eliminating all hidden assumptions.
>>> --
>>> Copyright 2022 Pete Olcott
>>>
>>> "Talent hits a target no one else can hit;
>>> Genius hits a target no one else can see."
>>> Arthur Schopenhauer
>>
>> No use. Even all the people you find agree with you is still useless.
>> To claim
>> you solved the HP problem, you have to show ALL your H first. People
>> cannot
>> judge or teach 'claims'. But your P already showed wrong, no need to
>> publish H.
>
>
> My analysis is based on this model: If "an X is a Y" and Z says that "an
> X is a Y" then anything in the universe that disagrees is necessarily
> incorrect.

Unless Z also says that an A is a B when it isn't. That is the failicy
of proving by example.

>
> "an X is a Y" =
> The input to embedded_H specifies a non-halting sequence of
> configurations. (input is non-halting)

Except it doesn't, at least not if embeddd_H says it is.

Since if embedded_H applied to <H^> <H^> -> Qn, then H^ applied to <H^>
-> H^.Qn an Halts, and it H^ applied to <H^> Halts then BY DEFINITION,
<H^> <H^> is defined to specify a HALTING COMPUTATION.

Maybe it doesn't POOP, but it Halts.

>
> Z says that "an X is a Y" =
> embedded_H rejects its input.

Then Z is wrong, since mbedded_H rejected a Halting Input.

>
> If you can think of a case where
> "an X is a Y" is simultaneously true and false then you have a rebuttal,
> otherwise not so much.
>

Except you