Thanks. I haven’t got time for a considered reply right now. Might get round to
it tomorrow afternoon.
Cheers,
Mike
Sent from my iPad
> On 27 Oct 2019, at 12:01, R.E. Boss <r.e.b...@outlook.com> wrote:
>
> I don't think your S3 is overkill, actually it simplifies the
> prefixes-suffixes machinery.
>
> ;<@(2018 (#-1+I.)@e.~ (-~{:))\ 6%~ (* (>: * >:@+:)) i.>.%:2018
> 12
>
> /:~;(3 : '<@(y |:@(](#~,: (#-1+I.)@])e.~) (-~{:))\ 6%~ (* (>:* >:@+:))
> i.>.%:>./y')2019+i:50
> 1969 11
> 1985 2
> 2010 5
> 2018 12
> 2025 1
> 2030 3
> 2030 4
> 2051 7
> 2054 13
>
>
> R.E. Boss
>
>
>> -----Oorspronkelijk bericht-----
>> Van: Programming <programming-boun...@forums.jsoftware.com>
>> Namens 'Mike Day' via Programming
>> Verzonden: donderdag 24 oktober 2019 11:08
>> Aan: programm...@jsoftware.com
>> Onderwerp: Re: [Jprogramming] Quora problem
>>
>> (I see RE Boss has just posted, while I was drafting - sorry if it's the
>> same!)
>>
>> I thought to exploit the property that
>> S3(n) = sum i^2, i <: n, is n(n+1)(2n+1)%6 but solving S(n) - S(m) = 2018
>> isn't
>> that easy!
>> Anyway, we could define
>> S3 =: 6 <.@%~ (* (>: * >:@+:))
>> but as we need a string of squares it's overkill.
>> Either way, it's useful to be able to identify the indices of ones in a
>> boolean
>> matrix:
>> ixbin =: $#.inv I.@,
>>
>> So we get, accprding to taste:
>>
>> -/"1 ixbin 2018 = -/~ +/\ *: i. >: 20
>> 12
>> -/"1 ixbin 2018 = -/~ S3 i. 20 NB. not as efficient as it looks!
>> 12
>>
>>
>> ts'-/"1 ixbin 2018 = -/~ S3 i. 20'
>> 1.53e_5 8768
>> ts'-/"1 ixbin 2018 = -/~ +/\ *: i. >: 20'
>> 8.7e_6 9856
>>
>> MIke
>>
>>
>>
>>
>>> On 24/10/2019 07:36, Skip Cave wrote:
>>> I occasionally try to solve Quora math problems to polish my J skills.
>>> Here's a problem that
>>>
>>> * "The integer 2018 can be written uniquely as the sum of consecutive
>>> squared integers. How many squares does the sum contain?"*
>>>
>>> a=.2^~i.100
>>>
>>> NB. Find how many consecutive squared integers it takes to sum to
>>> 2018. Try each length until I get a hit.
>>>
>>> +/2018=+/"1]5,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]6,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]7,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]8,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]9,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]10,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]11,\a
>>>
>>> 0
>>>
>>> +/2018=+/"1]12,\a
>>>
>>> 1 NB. A hit!
>>>
>>> +/2018=+/"1]13,\a
>>>
>>> 0
>>>
>>>
>>> So the answer is 12
>>>
>>> List them:
>>>
>>> b#~2018=+/"1 b=.12,\a
>>>
>>> 49 64 81 100 121 144 169 196 225 256 289 324
>>>
>>>
>>> Is there a way to do this search without explicit looping?
>>>
>>>
>>> Skip
>>>
>>>
>>> Skip Cave
>>> Cave Consulting LLC
>>> ----------------------------------------------------------------------
>>> For information about J forums see
>> http://www.jsoftware.com/forums.htm
>>
>>
>>
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