Yes. Has anyone investigated the relative strengths of these idioms? What’s the overhead of setting up the sparse representation? If I always use my version, it’s just that I’m lazy,
Cheers Mike Sent from my iPad > On 24 Oct 2019, at 10:16, R.E. Boss <r.e.b...@outlook.com> wrote: > > Your ixbin is equal to (4&$.$.) > > > 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 >> >> >> >> -- >> This email has been checked for viruses by Avast antivirus software. >> https://www.avast.com/antivirus >> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
