I do not see any attempt in the question at generalizing—so technically the answer would be a number.
But I expect that easy to describe (but accurate) general approaches would fit right in... Thanks, — Raul On Tuesday, August 14, 2018, Jose Mario Quintana < jose.mario.quint...@gmail.com> wrote: > After I saw your message I searched for the particular Quora problem and > this is what I found, > > https://www.quora.com/What-number-must-be-subtracted- > from-21-38-55-and-106-each-so-that-the-remainders- > technically-differences-are-proportional > > It seems to me that the shape of the array is restricted to be exactly 4 > and the numbers do not have to be integers. Am I wrong? > > > On Tue, Aug 14, 2018 at 7:00 PM, 'Mike Day' via Programming < > programm...@jsoftware.com> wrote: > > > It should work on arrays of the form > > a + b * i, where a and b are integer scalars, and i is an integer > > vector. > > So, if > > q=: 6 + 7*1 2 5 11 23, > > applying this function yields > > (-/ .* % -/ .+)@:(2 2&$) q > > 7.4 > > I think you need instead something like > > Difference’s greatest common divisor, > > dgcd =: +./@:(2 -~/\ ]) > > So > > dgcd q > > 7 > > > > It’s a bit more complicated to recover the factors, 1 2 5 ..., which seem > > to be required in the Quora problem: > > (](([-|~)%])dgcd) q > > 1 2 5 11 23 > > This works for the original array, too, > > (](([-|~)%])dgcd) 21 38 55 106 > > 1 2 3 6 > > > > And > > dgcd 21 38 55 106 > > 17 > > > > Sorry for any formatting problems - typing on iPad, > > > > Mike > > > > > > Please reply to mike_liz....@tiscali.co.uk. > > Sent from my iPad > > > > > On 14 Aug 2018, at 23:18, Jose Mario Quintana < > > jose.mario.quint...@gmail.com> wrote: > > > > > > (-/ .* % -/ .+)@:(2 2&$)21 38 55 106 > > ---------------------------------------------------------------------- > > 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
