Extending your relationship slightly: v"a"b <=> v"(a <. b <. (v b. 0)) where a and b are three integers each to match the rank of v.
This determines the rank in which v is executed; however, the expression v"a"b will have rank b when used in an expression. v=:(<@i."0)"1 'x';v 3 2 +-+-----+---+ |x|0 1 2|0 1| +-+-----+---+ Here <@i. executes as rank 0 0 0, where when used with ; v runs as rank 1 1 1. On Fri, Jun 21, 2013 at 11:57 PM, Michal D. <[email protected]>wrote: > By taking the minimum rank I mean: > > v"a b"c d <=> v"(a b <. c d) > > Interesting Raul, thanks for pointing that one out. Expressions resulting > in errors always seem to trigger the funny corner cases. > > The other behaviour resistant to simplification that I came up would be > mixing positive and negative ranks. > > Cheers, > > Mike > > > On Thu, Jun 20, 2013 at 6:26 AM, Raul Miller <[email protected]> > wrote: > > > $(i.3)(+"0"0 _)i.4 > > 3 4 > > > > simplifying by using the minimum of all nested ranks: > > > > $(i.3)(+"0)i.4 > > |length error > > > > -- > > Raul > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
