On 1/30/2023 6:10 AM, Richard Damon wrote:
> On 1/30/23 12:08 AM, olcott wrote:
>> On 1/29/2023 8:52 PM, Richard Damon wrote:
>>> On 1/29/23 9:27 PM, olcott wrote:
>>>> On 1/29/2023 8:06 PM, Richard Damon wrote:
>>>>> On 1/29/23 8:48 PM, olcott wrote:
>>>>>> On 1/29/2023 7:41 PM, Richard Damon wrote:
>>>>>>> On 1/29/23 8:28 PM, olcott wrote:
>>>>>>>> On 1/29/2023 7:10 PM, Richard Damon wrote:
>>>>>>>>> On 1/29/23 7:28 PM, olcott wrote:
>>>>>>>>>> On 1/28/2023 9:47 PM, Richard Damon wrote:
>>>>>>>>>>> On 1/28/23 4:29 PM, olcott wrote:
>>>>>>>>>>>> On 1/28/2023 3:13 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 1/28/23 3:59 PM, olcott wrote:
>>>>>>>>>>>>>> On 1/28/2023 2:42 PM, Richard Damon wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Yes, you can build SOME Deciders from a Sumation.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Not All Deciders can be built from a Sumation.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> One TM is *ALL* deciders.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Really?
>>>>>>>>>>>>>
>>>>>>>>>>>>> It can be both a is_prime and is_perfect decider?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Maybe your problem is you don't understand at all what a
>>>>>>>>>>>>> decider is?
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You just don't understand the basics of category theory,
>>>>>>>>>>>>>>> and just fall into the fallacy of Proof by Example.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Just because one subset of a set happens to have a
>>>>>>>>>>>>>>> property doesn't mean that property applies to the whole
>>>>>>>>>>>>>>> class.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Try and provide a single counter-example where IS_SUM gets
>>>>>>>>>>>>>> the wrong answer.
>>>>>>>>>>>>>
>>>>>>>>>>>>> If the question is Is the number prime?
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If there are no counter-examples that prove that IS_SUM
>>>>>>>>>>>>>> gets the wrong
>>>>>>>>>>>>>> answer then this logically entails that IS_SUM always gets
>>>>>>>>>>>>>> the correct
>>>>>>>>>>>>>> answer.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> But only to the one question it was built for.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Likewise for DOES_HALT, there are zero countability issues with
>>>>>>>>>>>> DOES_HALT. DOES_HALT merely needs to compute the mapping
>>>>>>>>>>>> from any
>>>>>>>>>>>> arbitrary input pair of finite strings to its accept or
>>>>>>>>>>>> reject state.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Can you PROVE that this is doable?
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> We can change IS_SUM to allow any arbitrary finite set of
>>>>>>>>>> finite string
>>>>>>>>>> inputs. If any of these finite strings contains a character
>>>>>>>>>> that is not
>>>>>>>>>> an ASCII digit then IS_SUM rejects, otherwise IS_SUM is as it was
>>>>>>>>>> previously specified.
>>>>>>>>>
>>>>>>>>> Ok, how do you use IS_SUM to answer the IS_PRIME question?
>>>>>>>>>
>>>>>>>>
>>>>>>>> IS_PRIME would have a single finite string of ASCII digits as
>>>>>>>> its only input and you already know this is computable.
>>>>>>>>
>>>>>>>>> IS_SUM takes the countably infinite number of inputs
>>>>>>>>
>>>>>>>> IS_SUM takes a finite set of finite string inputs
>>>>>>>> IS_SUM takes a finite set of finite string inputs
>>>>>>>> IS_SUM takes a finite set of finite string inputs
>>>>>>>> IS_SUM takes a finite set of finite string inputs
>>>>>>>> IS_SUM takes a finite set of finite string inputs
>>>>>>>
>>>>>>> And computes ONE mapping, out of the uncountable infinte many
>>>>>>> mappings of strings -> the set of answers for each input.
>>>>>> IS_SUM need not ever count any uncountable set. All that IS_SUM
>>>>>> must do is compute each mapping that it is presented with, one
>>>>>> finite set of finite strings at a time.
>>>>>>
>>>>>
>>>>>
>>>>> So????
>>>>>
>>>>>
>>>>> We aren't talking about IS_SUM being non-computable,
>>>>
>>>> Then the "you" of "we" is off topic. Countability was supposed to be
>>>> an alternate proof that halting is undecidable.
>>>>
>>> Right, countablility of mappings.
>>>
>>> The computability of a single mapping show nothing about the
>>> countability arguement.
>>>
>>> No one says that NO mappings are computable, just that most are not.
>>>
>>
>> No one can show even one mapping of a finite set of string inputs to
>> IS_SUM that is not computable, thus the conclusion that there are more
>> than one counter-example is totally refuted.
>
> ????
>
> You seem to think that is_sum computes multiple mappings.
>
> *A* mapping, as being talked about here, is a complete listing of the
> required output for every possible input.
>
> IS_SUM computes *ONE* mapping of input to output.
>
> Yes, it can computer the
>
>>
>>> You clearly don't understand categorical logic.
>>>
>>
>> Categorically exhaustive reasoning is my key innovation it utterly
>> eliminates the error of omission.
>
> No, it is your key fallicy. You seem to think that a detail analysis of
> one case tells you all you need about every case that you don't look at.
>
> That is just
>
>
>>
>> If it is impossible to show as many as a single counter-example then we
>> can correctly generalize this to say that more than one counter example
>> also does not exist.
>
> No, you are just using the fallacy of the Red Herring, because you
> demand a counter-example of something that isn't the problem.
>

If no counter example can be provided that a decision problem is
undecidable then this proves that it is decidable.

> How does your "IS_SUM" machine answer both the question of "is 8 the sum
> of two primes?" and "is 8 a perfect number?" (assuming the first is what
> you are defininig IS_SUM to be, not sure what else you mean, sometime it
> seems like IS_SUM is computing the sum of two numbers give to is, which
> isn't an "IS" question.
>

IS_SUM determines if its first finite string of ASCII digits is the sum
of its remaining finite list of finite strings of ASCII digits. Maybe
you need to take notes. There is also an adaptation of IS_SUM that takes
arbitrary finite strings.

>
> Note, these two question both take the SAME input (8), but have
> different outputs,
>
> For IS_SUM_OF_PRIMES, the answer is yes, because 8 = 3 + 5
>
> For IS_PERFECT, the answer is no, because 8 has factors to 1, 2, 4 which
> only total to 7, not 8.
>
> You claim same machine, in is the same, output will be the same, thus
> must give the wrong answer to one of them.

One decision problem per TM. A finite number of different decision
problems could be selected for computation by a single TM when the
ordinal number of the decision problem is the first input parameter.

Every decision problem that operates on a finite set of finite strings
is not uncomputable for any countability reasons.

The most straight forward to validate the Goldbach Conjecture is to
simply test every element of the set of natural numbers. Since this is
not a finite list of inputs it is not computable. Mathematical induction
can be used to algorithmically compress the analysis of some infinite
sets. https://www.britannica.com/science/Goldbach-conjecture

>>
>>
>>> Again, your use of the fallacy of proof by example.
>>
>

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer