intersection

问题 I'd like to be able to drag one imageview onto another, such that when they overlap, the overlapping area changes colour. To help visualise the problem, these imageviews are circular: when they overlap, it should form a Venn diagram-style image as shown below. I know you can detect whether the intersection has occurred by using: if (self.imageview1.bounds.contains(self.imageview2.bounds)) { ... } But do not really know how to colour the area in between! 回答1: so I figured out a way to do this

问题 I have three intersecting circles with inaccurate radii. How can I determine three out of six intersection points which form the intersection area? I was initially thinking of simply getting the cluster points - points which have smallest distances between them. But since the radii are not always correct, there might be cases where the cluster points are not the points forming the intersection area. Any ideas? 回答1: For each pair of circles, find the two intersections (if they exist) on their

问题 This question already has answers here : Best way to find the intersection of multiple sets? (5 answers) Closed 12 days ago . list(set(a[0]) & set(a[1]) & set(a[2]) & set(a[3]) & set(a[4])) Does anyone know how to write this in a way such that we dont need to know apriori how many lists we will be given? (ie 5 not hard coded in)? Each a is a list of varying size. 回答1: As long as you have at least one set, you can do this: list(set(a[0]).intersection(*a[1:])) If there might be no sets, you

问题 I am trying to intersect more lists in Scheme and I need a little help. The lists look like this: The first two: (((?x john) (?city new-york)) ((?x mike) (?city chicago)) ((?x mary) (?city london))) and (((?city chicago)) ((?city new-york))) I need to look in every list (say A) from the first list and see if there is a list (say B) in the second one so that A has at least one element in common with B. If there is no such element, the resulting list will not contain A. The result for the two

问题 I am writing a raytracer in java, and I was able to get tracing of spheres working, but I believe I have something wrong with how I am tracing triangles. Here is the basic algorithm, as I understand it: First determine if the ray even intersects the plane that the triangle is on. Clip all points so they are on the same plane as the triangle (so to the xy plane, as an example). Determine if the potential intersection point falls inside or out of the triangle, based on the number of polygon

问题 I want to use intersection() by key or filter() in spark. But I really don't know how to use intersection() by key. So I tried to use filter() , but it's not worked. example - here is two RDD: data1 //RDD[(String, Int)] = Array(("a", 1), ("a", 2), ("b", 2), ("b", 3), ("c", 1)) data2 //RDD[(String, Int)] = Array(("a", 3), ("b", 5)) val data3 = data2.map{_._1} data1.filter{_._1 == data3}.collect //Array[(String, Int] = Array() I want to get a (key, value) pair with the same key as data1 based

问题 Given a set S = {s i : {z j : z ∈ N } }, what is a time-efficient algorithm for computing the unique sets of intersections of the subsets of S ? For background, I am dealing with several versions of this problem, some larger than others. In the smallest one | S | ≈ 1,000, |s i | ≈ 10,000 and the values are zip codes. Tiny example for clarity: Input: S = {{},{1},{2,3},{3,4,5,6,7,8,9,10}} Output: {{},{1},{2,3},{3,4,5,6,7,8,9,10},{3}} | S | = 4 and there are 2 4 = 16 subsets of S . However,

问题 I'm working on a Java project witch is really killing me. After several days of researching on different forums, looking for what I really need, I come to ask your help. My data : A .ply file (containing a 3D shape made of a lot of triangles) A point (3D coordinates) I would like to know if this point is contained inside the complex 3D shape. I have split this problem in 2 smaller problems : How can I represent the complex 3D shape in memory? (I found several libraries, but it seems really

问题 given a 3D grid, a 3d point as sphere center and a radius, i'd like to quickly calculate all cells contained or intersected by the sphere. Currently i take the the (gridaligned) boundingbox of the sphere and calculate the two cells for the min anx max point of this boundingbox. then, for each cell between those two cells, i do a box-sphere intersection test. would be great if there was something more efficient thanks! 回答1: There's a version of the Bresenham algorithm for drawing circles.

问题 I have to create a house like structure where user will be able to add windows on wall. The approach I am thinking is to first create a separate mesh for window which user can drag over the selected wall of the house mesh, and drops where they feel it suitable, but within the same wall or same side of the house mesh. After that I will create the whole scene again but draw the window in the wall of the house mesh instead of creating a separate window mesh. Following is what it will look like