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• Open source jigsaw piece generator

  - by codecowboy

  Hi, Does anyone know of a C-based open source class / framework which can generate a random jigsaw puzzle from an image or generate a random puzzle template which could then be applied to an image? The puzzle pieces must have male and female notches/holes. There should be more than one template so that the puzzle does not become too easy. The target system is iOS / Mac. If not, how would you approach this problem? The puzzle pieces should be as close to a real jigsaw piece shape as possible and the system must be dynamic so that the user can use their own photos or download photos. thanks!

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• Reconstructing simple 3d enviroment(room) from photo

  - by Riz

  I have photo of a room with three walls and floor/ceiling or both. I am trying to reconstruct this room in 3d asking user for minimal input. Right now I use 8 points defined by user, angles of left and right wall(they can be quite different from 90) and one size "InLeftBottom-InRightBottom"(I need to have real size of this room for later use). I have no info about user's camera(I can read EXIF to get FOV and use constant height but this can be only used as additional info). Is this possible to ask user for less info? Maybe it's possible to get wall angles without user interaction? Or maybe I am completly wrong and should use different approach?

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• WPF, convert Path.DataProperty to Segment objects

  - by user275587

  I was wondering if there was a tool to convert a path data like "M 0 0 l 10 10" to it's equivalent line/curve segment code. Currently I'm using: string pathXaml = "<Path xmlns=\"http://schemas.microsoft.com/winfx/2006/xaml/presentation\" xmlns:x=\"http://schemas.microsoft.com/winfx/2006/xaml\" Data=\"M 0 0 l 10 10\"/>"; Path path = (Path)System.Windows.Markup.XamlReader.Load(pathXaml); It appears to me that calling XamlParser is much slower than explicitly creating the line segments. However converting a lot of paths by hand is very tedious.

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• How to know the line joining two points?

  - by dafero

  I have two points and I want to know the line which is joining them. I don't want to draw the line. I want to create a matrix with all the points which formed the line. In the future, I want to solve if two points belong or not to a shape. And this is the first part.

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• Generate a polygon from line.

  - by VOX

  I want to draw a line with thickness in j2me. This can easily be achieved in desktop java by setting Pen width as thickness value. However in j2me, Pen class does not support width. My idea is to generate a polygon from the line I have, that resembles the line with thickness i want to draw. In picture, on the left is what I have, a line with points. On the right is what I want, a polygon that when filled, a line with thickness. Could anyone know how to generate a polygon from line? http://www.freeimagehosting.net/uploads/140e43c2d2.gif

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• Surface Area of a Spheroid in Python

  - by user3678321

  I'm trying to write a function that calculates the surface area of a prolate or oblate spheroid. Here's a link to where I got the formulas (http://en.wikipedia.org/wiki/Prolate_spheroid & http://en.wikipedia.org/wiki/Oblate_spheroid). I think I've written them wrong, but here is my code so far; from math import pi, sqrt, asin, degrees, tanh def checkio(height, width): height = float(height) width = float(width) lst = [] if height == width: r = 0.5 * width surface_area = 4 * pi * r**2 surface_area = round(surface_area, 2) lst.append(surface_area) elif height > width: #If spheroid is prolate a = 0.5 * width b = 0.5 * height e = 1 - a / b surface_area = 2 * pi * a**2 * (1 + b / a * e * degrees(asin**-1(e))) surface_area = round(surface_area, 2) lst.append(surface_area) elif height < width: #If spheroid is oblate a = 0.5 * height b = 0.5 * width e = 1 - b / a surface_area = 2 * pi * a**2 * (1 + 1 - e**2 / e * tanh**-1(e)) surface_area = round(surface_area, 2) lst.append(surface_area, 2) return lst

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• Determining if two rays intersect

  - by Faken

  I have two rays on a 2D plane that extend to infinity but both have a starting point. They are both described by a starting point and a vector in the direction of the ray extending to infinity. I want to find out if the two rays intersect but i don't need to know where they intersect (its part of a collision detection algorithm). Everything i have looked at so far describes finding the intersection point of two lines or line segments. Anyone know a fast algorithm to solve this?

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• Determine mouse click for all screen resolutions

  - by Hallik

  I have some simple javascript that determines where a click happens within a browser here: var clickDoc = (document.documentElement != undefined && document.documentElement.clientHeight != 0) ? document.documentElement : document.body; var x = evt.clientX; var y = evt.clientY; var w = clickDoc.clientWidth != undefined ? clickDoc.clientWidth : window.innerWidth; var h = clickDoc.clientHeight != undefined ? clickDoc.clientHeight : window.innerHeight; var scrollx = window.pageXOffset == undefined ? clickDoc.scrollLeft : window.pageXOffset; var scrolly = window.pageYOffset == undefined ? clickDoc.scrollTop : window.pageYOffset; params = '&x=' + (x + scrollx) + '&y=' + (y + scrolly) + '&w=' + w + '&random=' + Date(); All of this data gets stored in a DB. Later I retrieve it and display where all the clicks happened on that page. This works fine if I do all my clicks in one resolution, and then display it back in the same resolution, but this not the case. there can be large amounts of resolutions used. In my test case I was clicking on the screen with a screen resolution of 1260x1080. I retrieved all the data and displayed it in the same resolution. But when I use a different monitor (tried 1024x768 and 1920x1080. The marks shift to the incorrect spot. My question is, if I am storing the width and height of the client, and the x/y position of the click. If 3 different users all with different screen resolutions click on the same word, and a 4th user goes to view where all of those clicks happened, how can I plot the x/y position correctly to show that everyone clicked in the same space, no matter the resolution? If this belongs in a better section, please let me know as well.

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• Find most right and left point of a horizontal circle in 3d Vector environment

  - by Olivier de Jonge

  I'm drawing a 3D pie chart that is rendered with in 3D vectors, projected to 2D vectors and then drawn on a Graphics object. I want to calculate the most left and right point of the circle The method to create a vector, draw and project to a 2d vector are below. Anyone knows the answer? public class Vector3d { public var x:Number; public var y:Number; public var z:Number; //the angle that the 3D is viewed in tele or wide angle. public static var viewDist:Number = 700; function Vector3d(x:Number, y:Number, z:Number){ this.x = x; this.y = y; this.z = z; } public function project2DNew():Vector { var p:Number = getPerspective(); return new Vector(p * x, p * y); } public function getPerspective():Number{ return viewDist / (this.z + viewDist); } }

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• resampling a series of points

  - by clamp

  hello, i have an array of points in 3d (imagine the trajectory of a ball) with X samples. now, i want to resample these points so that i have a new array with positions with y samples. y can be bigger or smaller than x but not smaller than 1. there will always be at least 1 sample. how would an algorithm look like to resample the original array into a new one? thanks!

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• Algorithm to Group All the Cycles Together

  - by Ngu Soon Hui

  I have a lot of cycles ( indicated by numeric values, for example, 1-2-3-4 corresponds to a cycle, with 4 edges, edge 1 is {1:2}, edge 2 is {2:3}, edge 3 is {3,4}, edge 4 is {4,1}, and so on). A cycle is said to be connected to another cycle if they share one and only one edge. For example, let's say I have two cycles 1-2-3-4 and 5-6-7-8, then there are two cycle groups because these two cycles are not connecting to each other. If I have two cycles 1-2-3-4 and 3-4-5-6, then I have only one cycle group because these two cycles share the same edge. What is the most efficient way to find all the cycle groups?

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• Determining if two lines intersect

  - by Faken

  I have two lines that extend to infinity but both have a starting point. They are both described by a starting point and a vector in the direction of the line extending to infinity. I want to find out if the two lines intersect but i don't need to know where they intersect (its part of a collision detection algorithm). Everything i have looked at so far describes finding the intersection point of two lines or line segments. Anyone know a fast algorithm to solve this?

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• Uniform distance between points

  - by Reonarudo

  Hello, How could I, having a path defined by several points that are not in a uniform distance from each other, redefine along the same path the same number of points but with a uniform distance. I'm trying to do this in Objective-C with NSArrays of CGPoints but so far I haven't had any luck with this. Thank you for any help. EDIT I was wondering if it would help to reduce the number of points, like when detecting if 3 points are collinear we could remove the middle one, but I'm not sure that would help.

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• Is a point inside or outside a polygon which is on the surface of a globe

  - by richard

  How do I determine if a point is inside or outside a polygon that lies on the the surface of the earth? The inside of the polygon can be determined via the right hand rule, ie. the inside of the polygon is on your right hand side when you walk around the polygon. The polygon may Circle either pole Cross the 180 longitude Cover more than 50% of the globe As the globe is a sphere the normal ray crossing algorithms do not work correctly.

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• QT Question (general math really): How can I tell if an Ellipse and a Line have an intersection?

  - by asdsa

  The ellipse is actually a circle. I know all the coordinates of both things, I just need to know whether or not a line goes through an ellipse. Is this built in anywhere? SPecifically I am using QGraphicsEllipseIem and QLine but i can convert them to any other type thanks

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• Centre of a circle that intersects two points

  - by Jason

  Given two points in a 2D plane, and a circle of radius r that intersects both of those points, what would be the formula to calculate the centre of that circle? I realise there would two places the circle can be positioned. I would want the circle whose centre is encountered first in a clockwise direction when sweeping the line that joins the two points around one of those points, starting from an arbitrary angle. I guess that is the next stage in my problem, after I find an answer for the first part. I'm hoping the whole calculation can be done without trigonometry for speed. I'm starting with integer coordinates and will end with integer coordinates, if that helps.

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• Finding whether a point lies inside a rectangle or not

  - by avd

  The rectangle can be oriented in any way...need not be axis aligned. Now I want to find whether a point lies inside the rectangle or not. One method I could think of was to rotate the rectangle and point coordinates to make the rectangle axis aligned and then by simply testing the coordinates of point whether they lies within that of rectangle's or not. The above method requires rotation and hence floating point operations. Is there any other efficient way to do this??

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• How to draw a filled circle in Java?

  - by Roman

  I have a JPanel with a Grid Layout. In the "cells" of the grid I can put different elements (for example JButtons). There is no problems with that. But now I want to put a filled circle in some of the cells. I also would like to relate an ActionListener with these circles. In more details, if I click the circle it disappears from the current cell and appears in another one. How can I do it in Java? I am using Swing.

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• Shortest distance between points on a toroidally wrapped (x- and y- wrapping) map?

  - by mstksg

  I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean rectangle, but when a point moves to the right boundary, it will appear at the left boundary (at the same y value), given by x_new = x_old % width Basically, points are plotted based on: (x_new, y_new) = ( x_old % width, y_old % height) Think Pac Man -- walking off one edge of the screen will make you appear on the opposite edge. What's the best way to calculate the shortest distance between two points? The typical implementation suggests a large distance for points on opposite corners of the map, when in reality, the real wrapped distance is very close. The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula. But that would involve many checks, calculations, operations -- some that I feel might be unnecessary. Is there a better way?

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• Given the lat/lon of 2 close points on earth (<10m), How do I calculate the distance in metres?

  - by Rory

  I have the lat/lon of 2 points on the earth. They are really close together, <10m. Let's assume the earth is flat. How do I calculate the distance between them in metres? I know about tools (PostGIS, etc.) that can do this correctly, however I'm just doing a rough and ready type, and I'm OK with low accuracy. At such small sizes a difference of 1% is only 10cm, which is fine for me. I'm doing this in stock python. I'm OK with a standard Euclidean distance thing.

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• Graphics Question: How do I restrict the mouse cursor to within a circle?

  - by Dan

  I'm playing with XNA. When I click the left mouse button, I record the X,Y co-ordinates. Keeping the mouse button held down, moving the mouse draws a line from this origin to the current mouse position. I've offset this into the middle of the window. Now, what I'd like to do is restrict the mouse cursor to within a circle (with a radius of N, centred on the middle of the screen). Restricting the mouse to a rectangular region is easy enough (by adjusting the origin by the difference of the mouse position and the size of the region), but I haven't a clue on how to start doing it for a circular region. Can anyone explain how to do this? Any advice on where to start would be helpful.

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• Is there any algorithm for calculating area of a shape given co-ordinates that define the shape ?

  - by Superior0

  So I have some function that receives N random 2D points. Is there any algorithm to calculate area of the shape defined by the input points?

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• Python point lookup (coordinate binning?)

  - by Rince

  Greetings, I am trying to bin an array of points (x, y) into an array of boxes [(x0, y0), (x1, y0), (x0, y1), (x1, y1)] (tuples are the corner points) So far I have the following routine: def isInside(self, point, x0, x1, y0, y1): pr1 = getProduct(point, (x0, y0), (x1, y0)) if pr1 >= 0: pr2 = getProduct(point, (x1, y0), (x1, y1)) if pr2 >= 0: pr3 = getProduct(point, (x1, y1), (x0, y1)) if pr3 >= 0: pr4 = getProduct(point, (x0, y1), (x0, y0)) if pr4 >= 0: return True return False def getProduct(origin, pointA, pointB): product = (pointA[0] - origin[0])*(pointB[1] - origin[1]) - (pointB[0] - origin[0])*(pointA[1] - origin[1]) return product Is there any better way then point-by-point lookup? Maybe some not-obvious numpy routine? Thank you!

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• 3D coordinate of 2D point given camera and view plane

  - by Myx

  I wish to generate rays from the camera through the viewing plane. In order to do this, I need my camera position ("eye"), the up, right, and towards vectors (where towards is the vector from the camera in the direction of the object that the camera is looking at) and P, the point on the viewing plane. Once I have these, the ray that's generated is: ray = camera_eye + t*(P-camera_eye); where t is the distance along the ray (assume t = 1 for now). My question is, how do I obtain the 3D coordinates of point P given that it is located at position (i,j) on the viewing plane? Assume that the upper left and lower right corners of the viewing plane are given. NOTE: The viewing plane is not actually a plane in the sense that it doesn't extend infinitely in all directions. Rather, one may think of this plane as a widthxheight image. In the x direction, the range is 0--width and in the y direction the range is 0--height. I wish to find the 3D coordinate of the (i,j)th element, 0

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• Given 3 pts, how do I calculate the normal vector? [closed]

  - by WaggingSiberian

  Given three 3D points (A,B, & C) how do I calculate the normal vector? The three points define a plane and I want the vector perpendicular to this plane. Can I get sample C# code that demonstrates this?

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