https://www.jsoftware.com/help/dictionary/d631.htm
x u&.v y ↔ vi (v x) u (v y) Here: u is + v is *: vi is %: (or *:inv) x is 3 y is 4 So these are equivalent 3 +&.*: 4 %: (*:3) + (*: 4) *:inv (*:3) + (*: 4) I hope this makes sense. -- Raul On Sat, Oct 23, 2021 at 3:03 PM More Rice <[email protected]> wrote: > > Hello, > > (Sorry for the previous empty email - web page problem) > > please excuse another newbie question ... > > Ref: https://code.jsoftware.com/wiki/Vocabulary/starco > > pythag =: +&.*: > 3 pythag 4 > 5 > > + operated dyadically and acted on both x and y - ok. > > but how does *: know to act on x as well? Isn't pythag using the monadic > definition of *: to square y only? > > so magical ... > > thank you for the pointer and have a great weekend. > > > Maurice > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
