For a first effort, I would go with F=:1 :'(u@#~ #:@i.@(2^#))@~.' f=: /F
Hopefully that makes the issues obvious - the specification here calls for a result which grows exponentially with the size of the argument set. Also: The ~. might be extra work, but for typical cases the effort of ensuring that the argument is a set is trivial compared to the effort of constructing the result. You did not include the empty combination in your example results, but given your specification my initial inclination is to treat that as an oversight. I defined F instead of going straight for f because for testing purposes I want to be able to do (<F a), and perhaps similar things. Thanks, -- Raul On Mon, Oct 2, 2017 at 12:49 PM, Skip Cave <[email protected]> wrote: > Given a set of integers, what is the most concise and or efficient way to > list the numbers along with the sum of all combinations of the numbers? the > products of all combinations? > > for example: > > a =. 2 5 7 > + f a NB. 2, 5, 7, (2+5), (2+7), (5+7), (2+5+7) > 2 5 7 7 9 12 14 > > * f a NB. 2, 5, 7, (2*5), (2*7), (5*7), (2*5*7) > 2 5 7 10 14 35 70 > > The function 'f' should work for any verb and any size right argument noun > vector. > > Skip > > Skip Cave > Cave Consulting LLC > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
