My first line select the two roots: r=: 13 :'> }. p. y' r 0 100 _2 50 0
These are the roots: Now find the axis of symmetry x value (the average of the roots, a=: 13 :'(+/y)%#y' a 50 0 25 So if I apply a to r 0 100 _2 I expect 25. I get 1 2 -----OriginalaMessage----- From: programming-boun...@forums.jsoftware.com [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km Sent: Saturday, February 23, 2013 9:53 PM To: programm...@jsoftware.com Subject: Re: [Jprogramming] The farmer's fence Linda, if you enter 50 0 and then enter 1 2 $ 50 0 the printed results appear the same, but the first has two items and the second only one item. What do you expect the average of a one-item array to be? Can you figure out how to use rank to get what you want from r 0 100 _2 ? Sent from my iPad On Feb 23, 2013, at 6:21 PM, "Linda Alvord" <lindaalv...@verizon.net> wrote: > A should give an average. The roots are 50 0 . > > Shouldn't the average be 25 rather than 1 2 ? > > Linda > > -----Original Message----- > From: programming-boun...@forums.jsoftware.com > [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km > Sent: Saturday, February 23, 2013 6:45 PM > To: programm...@jsoftware.com > Subject: Re: [Jprogramming] The farmer's fence > > r 0 100 _2 > 50 0 > $r 0 100 _2 > 1 2 > > Sent from my iPad > > > On Feb 23, 2013, at 4:35 PM, "Linda Alvord" <lindaalv...@verizon.net> wrote: > >> The roots of a polynomial: >> >> r=: 13 :'> }. p. y' >> r 0 100 _2 : >> 50 0 >> r >> [: > [: }. p. >> >> >> The average of the roots or x coordinate of axis of symmetry: >> a=: 13 :'(+/y)%#y' >> a 50 0 >> 25 >> a >> +/ % # >> >> Find the maximum: >> >> 0 100 _2 p. 25 >> 1250m >> >> Sadly this doesn't work: >> >> a r 0 100 _2 >> 50 0 >> >> Any idea why not? >> >> Linda >> >> >> -----Original Message----- >> From: programming-boun...@forums.jsoftware.com >> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Alex >> Giannakopoulos >> Sent: Saturday, February 23, 2013 3:28 PM >> To: J Programming forum >> Subject: Re: [Jprogramming] The farmer's fence >> >> Nice little gotcha there, assuming that the shape will be a square, >> since a square maximizes the contained area for a rectangle, while >> forgetting that the wall gives you extra perimeter for free, >> depending on > the shape. >> >> By the same analogy I'd tackle Roger's version of the problem, i.e. >> find ANY shape that will maximize the area: >> Again, I suspect that going for a (semi)circle might be essentially >> the same gotcha. >> >> I haven't got time to code it at the moment, but I'd investigate an >> (half) ellipse and also a parabola. >> Will need some integration though, to find the expression for the >> length of their curves. >> --------------------------------------------------------------------- >> - 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 > > ---------------------------------------------------------------------- > 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
