Yes. One way of thinking about that would be that the elements have orthogonal sides, and slanted "rectangles" would thus include empty space on the edges.
Thanks, -- Raul On Tue, Jun 20, 2017 at 4:49 PM, 'Mike Day' via Programming <[email protected]> wrote: > I've only just seen this. > I expect it's obvious, but you ARE looking only for rectangles with > "vertical" and "horizontal" sides, i.e. no rotations other than multiples of > pi/2 involved? > Thanks, > Mike > > Please reply to [email protected]. > Sent from my iPad > >> On 20 Jun 2017, at 16:29, Raul Miller <[email protected]> wrote: >> >> Something I stumbled over today. >> >> If we have a series of bars of varying height, what's the largest >> rectangle that can be drawn over the bars without covering any empty >> space. >> >> For example: >> >> '*'#"0~2 6 7 4 1 7 >> ** >> ****** >> ******* >> **** >> * >> ******* >> >> I'll post a solution later, and I'll be interested in seeing if it's >> basically the only obvious approach or if there's a variety of good >> approaches. (I have reason to believe, though, that there's a better >> way than what I came up with.) >> >> Thanks, >> >> -- >> 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 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
