Ah.. hmm... Is this what you were thinking of?
(i. >./)(+ * *@[)/\. 0 1 1 1 0 1 0 1 Note that there could be multiple blocks the same length, though, so maybe instead: (I.@:= >./)(+ * *@[)/\. 0 1 1 1 0 1 0 1 That's an interesting approach you've got going here, and it might turn out to be faster than mine. I'm not certain it will always give you the best answer though, for the general case. For the general case I think you'd have to consider a variety of possibilities (for example: given the same perimeter, a square will have a larger area than some other rectangle). Still, I think you're onto something worth pursuing, here. Thanks, -- Raul On Tue, Jun 20, 2017 at 5:30 PM, Martin Kreuzer <[email protected]> wrote: > @Raul > > Beg your pardon for taking your time ... > > This is what I did (as an approach); maybe I learn something from you (or > somebody else) proving me wrong. > > Anyway, this seems to work on the given set of data: > > Taking the mean of the column heights > > ] v=. 2 6 7 4 1 7 > 2 6 7 4 1 7 > am=. +/ % # > am v > 4.5 > > gives a "compacted" rectangle of 4.5 times 6 (base) = 27 units of area. > > IIf I 'floor' the mean > > ] mf=. <. am v > 4 > > and take a look at which columns exceed (the floor of ) the mean I get this > matrix > > v,:mf<: v > 2 6 7 4 1 7 > 0 1 1 1 0 1 > > Here I got stuck with my coding, so I will describe in words only what I > intend to do further on: > > * Extract the indices of the largest (longest) block of 1s (1..3) and > calculate its length (3) > * Within that block search for the lowest column value (4) > * Multiply block length (3) with lowest column value within that range (4) > to obtain a rectangle area of 3*4 = 12 units. > > Thanks for your patience. > -M > > > At 2017-06-20 16:07, you wrote: > >> Both of those are valid answers. >> >> The problem is simply to determine the largest area. >> >> Thanks, >> >> -- >> Raul >> >> >> On Tue, Jun 20, 2017 at 11:58 AM, 'Pascal Jasmin' via Programming >> <[email protected]> wrote: >> > is the answer to your example, 6x2 or 4x3? >> >> >> >> > ________________________________ >> > From: Raul Miller <[email protected]> >> > To: Programming forum <[email protected]> >> > Sent: Tuesday, June 20, 2017 11:30 AM >> > Subject: [Jprogramming] largest rectangle problem >> >> >> >> > 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 > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
