area

Is there a standard for this? Algorithm name? Say: I have 10 polygons of different sizes. I have an area of specific size. I want to know how to fill the most polygons in that area, and how they are fitted. Note: Polygons may be rotated depending on the restriction set. One possible name is a Packing Problem . It is related to the Knapsack Problem . These problems tend to be NP-hard, and many require heuristics. If you can constrain the allowed forms of polygons and of the area, there may exist a more efficient algorithm for your special case. You can have a look at "Dancing Links" in

in gnuplot, when you try to fill an area under a curve, the tics of both axes are hinded behind the solid area. Is there any way to give them to the front? I am using postcript terminal, where no transparent features are allowed (i guess) Thanks There sure is! Before plotting, run this interactively or in a script: set tics front Try help tics interactively for more info. user3207862 After trying many options I realized that the following works properly: set tics front It seems important to write it before the plot command. ;-) 来源： https://stackoverflow.com/questions/21193547/gnuplot-fill-area

What is the fastest way of getting the area of any arbitrary Windows region? I know I can enumerate all points of bounding rectangle and call the PtInRegion() function but it seems not very fast. Maybe you know some faster way? When you call GetRegionData , you'll get a list of non-overlapping rectangles that make up the region. Add up their areas, something like this: function GetRegionArea(rgn: HRgn): Cardinal; var x: DWord; Data: PRgnData; Header: PRgnDataHeader; Rects: PRect; Width, Height: Integer; i: Integer; begin x := GetRegionData(rgn, 0, nil); Win32Check(x <> 0); GetMem(Data, x); try

How do I calculate the area of a polygon stored in a MySql database? The polygons' points are lat longs. So, degrees and minutes seem to be causing a problem. I've tried: SELECT AREA( my_polygon ) FROM `my_table` WHERE name = 'Newport' Because, the points are lat longs, I get weird results. (I'm not able to switch to Postgre). Is there a way to do this in MySQL? I'd like to get the results in sq. meters or sq. km or sq. miles-- any of these would be fine. You've got to transform those lats and lons into a more appropriate coordinate system. Since the earth is a sphere, you're talking about

I have used bwconvhull to detect a certain part of the image, as you can see in the image, there are number of objects with certain centroids. What I want to do is to detect the object with the largest area (1st big one from the left) and neglect the others. Which method should I follow? I will be very thankful for your help. The following is the code (it's very roughly written as I am still working on it. Sorry for any inconvenience in advance) CH_objects = bwconvhull(c,'objects'); imshow(CH_objects); title('Objects Convex Hull'); bwarea(CH_objects) Ilabel = bwlabel(CH_objects,8); stat =

I'm looking everywhere to answer this question but nowhere can I find anything about it. Can anyone tell me whether we can affect the area which received the item hover border-radius property. So that the effect of changes such as color took place after hitting a real area viewed item? Do not block that physically exists in the DOM as a square? This is simple img. and some simple fiddle: www.jsfiddle.net/nawAE Well, you can use SVG and pointer-events: https://developer.mozilla.org/en-US/docs/Web/CSS/pointer-events?redirectlocale=en-US&redirectslug=CSS%2Fpointer-events Or just use SVG, maybe

I tried to using the algorithm shown here: https://discuss.leetcode.com/topic/15733/my-java-solution-sum-of-areas-overlapped-area However, that algorithm only deals with finding the areas of only TWO overlapped rectangles. How would I go on about finding the area of the intersection of say 3, or 4 or 5, etc number of overlapping rectangles, if I know the length, breadth of each rectangle? benten Shapely is a good library for stuff like this. from shapely.geometry import box # make some rectangles (for demonstration purposes and intersect with each other) rect1 = box(0,0,5,2) rect2 = box(0.5,0

I've to write a method that has the following signature public class Position { double longitude; double latitude; } boolean isInsideTheArea(Position center, double radius, Position point); So if point is inside the area that has the center as its center and radius as its radius in miles , this should return true , false otherwise. Use the Haversine formula to compute the distance between center and point . If that distance is greater than radius , return false ; otherwise return true . Pseudocode: def haversineDistance(a, b): # snip... return foo def isInsideTheArea (center, radius, point):