(x&f)^:n y is the same as x f^:n y, by definition.
----- Original Message ----- From: Bo Jacoby <[email protected]> Date: Saturday, March 20, 2010 8:46 Subject: Re: [Jprogramming] recursive composition To: Programming forum <[email protected]> > Hi John > but > (x&f)^:3 y > _2 > - Bo > > --- Den lør 20/3/10 skrev John Randall > <[email protected]>: > Fra: John Randall <[email protected]> > Emne: Re: [Jprogramming] recursive composition > Til: "Programming forum" <[email protected]> > Dato: lørdag 20. marts 2010 16.17 > > Raul Miller wrote: > > On Fri, Mar 19, 2010 at 1:43 PM, David Ward Lambert > > <[email protected]> wrote: > >> NB. j programmers MUST understand data flow. > >> NB. x f^:3 y differs from (f&y)^:3 x > > > > To illustrate this we should find an operation which > > is not associative and which produces a plausible result > > for one of those two cases. > > It is easy to show by example that these are different, for example > > x=:2 > y=:4 > f=:- > x f^:3 y > _2 > (f&y)^:3 x > _10 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
