I would like to compute the mean and standard deviation of the areas of a set of Voronoi regions in 2D (if the region extends to infinity, I'll just clip it to the unit square).
However if possible I would like to do this computation from the Delaunay Triangulation without explicitly computing the Voronoi regions? Is this even possible, or is it better to just compute the Voronoi diagram explicitly?
In order to calculate the voronoi region of a vertex you need to iterate the 1-ring around it. Then the area of the region is defined as:
A = 1/8 * (sum for every adjacent vertex p_i) { (cot alpha_i + cot beta_i) * (p_i - c).Length² }
In the image you can see the whole voronoi region in light red. A part of it is shown in dark red. This is one of the parts accumulated by the sum. alpha and beta are the angles as visible in the image. c is the center vertex position. p_i is the opposite vertex_position. alpha, beta and p_i change while iterating. c keeps its value.
If you calculate those parts for every adjacent vertex, you get 8 times the area of the voronoi region.
Related
I need some hints on this one:
A polygon P is star-shaped if there exists a point p in the interior of P such that any other point (vertex) on the boundary is visible to p.
Given a polygon P, how can i determine if P is a star shaped polygon?
Time complexity should be o(n) on average.
Ive been sitting on this for a while now, Any help will be appericiated.
very weird definition of star according to that circle and pie are also stars ...
First simple and O(n) possibility I can think of is to render visibility map:
compute BBOX of the shape
create 2D map of the BBOX and clear it with zero
so map 2D array (texture) to the BBOX of some resolution xs*ys
for each convex vertex increment visibility map
simply by rendering "infinite" triangle/quad onto the map
You can use winding rule to chose if vertex is convex or concave by simply checking the sign of z coordinate of the adjacent edges cross product against the winding rule of your shape.
scan the 2D map for cells containing number of convex vertexes
all the cells/pixels containing number of convex vertexes are your possible Z so if any found your shape is a "star".
This is O(n*xs*ys) where n is number of (convex) vertexes and xs*ys is resolution of the visibility map. Note if your resolution is too low due to inaccuracies you might produce false negatives/positives ... if (max) resolution of the map is constant then the complexity will turn to O(n).
The rendering can be done simply for example with OpenGL and STENCIL buffer which directly has operation to increment STENCIL pixel however that one will limit the n to 255 as STENCIL is only 8 bit these days (after changes in OpenGL)... However you can workaround this by seting the BBOX to 1 and clear the exterior of the triangle/quad instead of incrementing its interrior. then the pixels holding 1 are your Z this might be used with any rendering engine no need for STENCIL
I need a way to compute the distance beetween a point and the bounding edge of a polygon.
If the point is outside the polygon, the distance will be posivite
If the point is inside the polygon, the distance will be negative
This is called SDF for Signed Distance Field/Function
The polygon itself is composed of multiple paths, can be concave, with holes, but not self intersecting, and with a lot of clockwise ordered points (10000+).
I've found some existing solutions, but they require to test the point against each polygon edge, which is not efficient enough.
Here is the visual result produced (green is positive, red is negative):
So I've tried the following:
Put the polygon edges in a quadtree
To compute the distance, find the closest edge to the point and change the sign depending on which side of the edge the point is.
Sadly, it doesn't works when the point is at the same distance of multiple edges, such as corners.
I've tried adding condition so a point is outside the polygon if it's on the exterior side of all the edges, but it doesn't solve the interior problem, and the other way around.
Can't wrap my head around it...
If anyone curious, the idea is to later use some shader to produce images like this :
EDIT
To clarify, here is a close up of the problem arising at corners :
For all the points in area A, the closest segment is S1, so no problem
For all the points in area E, the closest segment is S2, so no problem either
All points in area B, C and D are at the same distance of S1 and S2
Points in area B are on the exterior side of S1 and interior side of S2
Points in area D are on the interior side of S1 and exterior side of S2
Points in area C are on the exterior side of both segments
One might think that a point has to be on the interior side of both segments in order to be considered "in". It solves the problem for angles < 180°, but the problem is mirrored for angles > 180°
Worst, two or more corners can share the same position (like the four way corner in the low part of first image)...
I hope this solves your problem.
This is implemented in Python.
First, we use imageio to import the image as an array. We need to use a modified version of your image (I filled the interior region in white).
Then, we transform the RGBA matrix into a binary matrix with a 0 contour defining your interface (phi in the code snippet)
Here's phi from the snippet below (interior region value = +0.5, exterior region value = -0.5):
import imageio
import numpy as np
import matplotlib.pyplot as plt
import skfmm
# Load image
im = imageio.imread("0WmVI_filled.png")
ima = np.array(im)
# Differentiate the inside / outside region
phi = np.int64(np.any(ima[:, :, :3], axis = 2))
# The array will go from - 1 to 0. Add 0.5(arbitrary) so there 's a 0 contour.
phi = np.where(phi, 0, -1) + 0.5
# Show phi
plt.imshow(phi)
plt.xticks([])
plt.yticks([])
plt.colorbar()
plt.show()
# Compute signed distance
# dx = cell(pixel) size
sd = skfmm.distance(phi, dx = 1)
# Plot results
plt.imshow(sd)
plt.colorbar()
plt.show()
Finally, we use the scikit-fmm module to compute the signed distance.
Here's the resulting signed distance field:
For closed, non-intersecting and well oriented polygons, you can speed up the calculation of a signed distance field by limiting the work to feature extrusions based on this paper.
The closest point to a line lies within the edge extrusion as you have shown in your image (regions A and E). The closest feature for the points in B, C and D are not the edges but the vertex.
The algorithm is:
for each edge and vertex construct negative and positive extrusions
for each point, determine which extrusions they are in and find the smallest magnitude distance of the correct sign.
The work is reduced to a point-in-polygon test and distance calculations to lines and points. For reduced work, you can consider limited extrusions of the same size to define a thin shell where distance values are calculated. This seems the desired functionality for your halo shading example.
While you still have to iterate over all features, both extrusion types are always convex so you can have early exits from quad trees, half-plane tests and other optimisations, especially with limited extrusion distances.
The extrusions of edges are rectangles in the surface normal direction (negative normal for interior distances).
From 1:
The extrusions for vertices are wedges whose sides are the normals of the edges that meet at that vertex. Vertices have either positive or negative extrusions depending on the angle between the edges. Flat vertices will have no extrusions as the space is covered by the edge extrusions.
From 1:
I'm looking for an algorithm that converts a regular grid of heights (e.g. 1024x1024) to a triangular irregular network. Here is an image showing an example of a triangular irregular network:
I've looked in the internet for an algorithms to convert it, but I just couldn't find one. Basically the triangle density depends on the roughness and/or pixel error (when rasterized), or something like that.
Here's an idea for a two-step algorithm: Do a Delaunay triangulation based on a rough mesh first, then smoothe out the triangles recursively until a certain error criterion is met.
For the first step, identify a set of vertices for the Delaunay triangulation. These vertices coincide with pixel coordinates. Extreme points that are either higher or lower than all four neighbouring pixels should be in the set as should ridge points on the borders where the adjacent pixels along the border are either lower or higher. This should give a coarse triangular mesh. You can get a finer mesh by also including pixels that have a high curvature.
In the second step, iterate through all triangles. Scan through the triangle along the pixel grid and accumulate an error square for each pixel inside the triangle and also identify the points of maximum and minimum signed error. If the average error per pixel does not meet your criterion, add the points of lowest and highest error to your triangulation. Verify the new triangles and re-triangulate as necessary.
Notes:
The coarse triangulation in step one should be reasonably fast. If the height map is ragged, you might end up with too many vertices in the ragged area. In that case, the hight map might be smoothed with a Gaussian filter before applying the algorithm.
The recursive re-triangulation is probably not so fast, because determining the error requires scanning the triangles over and over. (The process should get faster as the triangle size decreases, but still.) A good criterion for finding vertices in step 1 might speed up step 2.
You can scan a triangle by finding the bounding box of pixels. Find the barycentric coordinates s, t of the lower left point of the bounding box and also the barycentric increments (dsx, dtx) and (dsy, dty) that correspond to a pixel move in the x and y directions. You can then scan the bounding box in two loops over the included pixels (x, y), calculate the barycentric coordinates (s, t) from your delta vectors and accumulate the error if you are inside the triangle, i.e. when s > 0, t > 0 and s + t < 1.
I haven't implemented this algorithm (yet - it is an interesting task), but I imagine that finding a good balance between speed and mesh quality is a matter of tailoring error criteria and vertex selection to the current height map.
I have a height map of NxN values.
I would like to find, given a point A (the red dot), whose x and y coordinates are given (and z is known from the data, so A is a vertex of the surface) a set of points that lie on the circumference of the circle with center in A and radius R that are a good approximation of a circular "cloth" (in grey) draped on the imaginary surface described by the data points.
The sampling, the reciprocal distances between the set of points that I am trying to find, doesn't need to be uniform, but still I would like to find at least all the points that are an intersection of the edges of the mesh with the circle at distance R from A.
How to find this set of points?
Is this a known problem?
(source: keplero.com)
-- edit
The assumption that Jan is using is right: the samples form a regular rectangular or square grid (in the X-Y plane) aligned with [0,0]. But I would like to take the displacement in the Z direction into account to compute the distance. you can see the height map as a terrain, and the algorithm I am looking for as the instructions to give to an explorer that, traveling just on paths of given latitude or longitude, mark the points that are at distance R from A. Walking distance, that is taking into account all the Z displacements done so far. The explorer climbs and go down in the valleys too.
The trivial algorithm for this would be something like this. We know that given R, the maximum displacement on the x and y axis corresponds to a completely flat surface. If there is no slope, the x,y points will all be in the bounding square Ax-R < x < Ax+r and Ay-R
At this point, it would start traveling to the close cells, since if the perimeter enters the edge of one cell of the grid, it also have to exit that cell.
I reckon this is going to be quite difficult to solve in an exact fashion, so I would suggest trying the straightforward approach of simulating the paths that your explorers would take on the surface.
Given your starting point A and a travel distance d, calculate a circle of points P on the XY plane that are d from A.
For each of the points p in P, intersect the line segment A-p with your grid so that you end up with a sequence of points where the explorer crosses from one grid square to the next, in the order that this would happen if the explorer were travelling from A. These points should then be given a z-coordinate by interpolation from your grid data.
You can thus advance through this point sequence and keep track of the distance travelled so far. Eventually the target distance will be reached - adjust p to be at this point.
P now contains the perimeter that you're looking for. Adjust the sample fidelity (size of P) according to your needs.
Just to clarify - You have a triangulated surface in 3d and, for a given starting vertex Vi in the mesh you would like to find the set of vertices U that are reachable via paths along the surface (i.e. geodesics) with length Li <= R.
One approach would be to transform this to a graph-based problem:
Form the weighted, undirected graph G(V,E), where V is the set of vertices in the triangulated surface mesh and E is the set of edges in this mesh. The edge weight should be the Euclidean (3d) length of each edge. This graph is a discrete distance map - the distance "along the surface" between each adjacent vertex in the mesh.
Run a variant of Dijkstra's algorithm from the starting vertex Vi, only expanding paths with length Li that satisfy the constraint Li <= R. The set of vertices visited U, will be those that can be reached by the shortest (geodesic) path with Li <= R.
The accuracy of this approach should be related to the resolution of the surface mesh - as long as the surface curvature within each element is not too high the Euclidean edge length should be a good approximation to the actual geodesic distance, if not, the surface mesh should be refined in that area.
Hope this helps.
How can I find the largest circle that can fit inside a concave polygon?
A brute force algorithm is OK as long as it can handle polygons with ~50 vertices in real-time.
The key to solving this problem is first making an observation: the center of the largest circle that will fit inside an arbitrary polygon is the point that is:
Inside the polygon; and
Furthest from any point on the edges of the polygon.
Why? Because every point on the edge of a circle is equidistant from that center. By definition, the largest circle will have the largest radius and will touch the polygon on at least two points so if you find the point inside furthest from the polygon you've found the center of the circle.
This problem appears in geography and is solved iteratively to any arbitrary precision. Its called the Poles of Inaccessibility problem. See Poles of Inaccessibility: A Calculation Algorithm for the Remotest Places on Earth.
The basic algorithm works like this:
Define R as a rectilinear region from (xmin,ymin) to (xmax,ymax);
Divide R into an arbitrary number of points. The paper uses 21 as a heuristic (meaning divide the height and width by 20);
Clip any points that are outside the polygon;
For the remainder find the point that is furthest from any point on the edge;
From that point define a new R with smaller intervals and bounds and repeat from step 2 to get to any arbitrary precision answer. The paper reduces R by a factor of the square root of 2.
One note, How to test if a point is inside the polygon or not: The simplest solution to this part of the problem is to cast a ray to the right of the point. If it crosses an odd number of edges, it's within the polygon. If it's an even number, it's outside.
Also, as far as testing the distance to any edge there are two cases you need to consider:
The point is perpendicular to a point on that edge (within the bounds of the two vertices); or
It isn't.
(2) is easy. The distance to the edge is the minimum of the distances to the two vertices. For (1), the closest point on that edge will be the point that intersects the edge at a 90 degree angle starting from the point you're testing. See Distance of a Point to a Ray or Segment.
In case anyone is looking for a practical implementation, I designed a faster algorithm that solves this problem for a given precision and made it a JavaScript library. It's similar to the iterative grid algorithm described by #cletus, but it's guaranteed to obtain global optimum, and is also 20-40 times faster in practice.
Check it out: https://github.com/mapbox/polylabel
An O(n log(n)) algorithm:
Construct the Voronoi Diagram of the edges in P. This can be done with, for example, Fortunes algorithm.
For Voronoi nodes (points equidistant to three or more edges) inside P;
Find the node with the maximum distance to edges in P. This node is the centre of the maximum inscribed circle.
Summary: In theory, this can be done in O(n) time. In practice you can do it in O(n log n) time.
Generalized Voronoi diagrams.
If you consider the vertices and edges of the polygon as a set of sites and tessellate the interior into the "nearest neighbor cells" then you get the so-called (generalized) Voronoi diagram. The Voronoi diagram consists of nodes and edges connecting them. The clearance of a node is the distance to its defining polygon faces.
(Here the polygon even has holes; the principle still works.)
The key observation now is that the center of the maximum inscribed circle touches three faces (vertices or edges) of the polygon, and no other face can be closer. So the center has to lie on a Voronoi node, i.e, the node with the largest clearance.
In the example above the node that marks the center of the maximum inscribed circle touches two edges and a vertex of the polygon.
The medial axis, by the way, is the Voronoi diagram with those Voronoi edges removed that emanate from reflex vertices. Hence, the center of the maximum inscribed circle also lies on the medial axis.
Source: A blog article of mine that deals with generalizations of maximum inscribed circles at some point. There you can find more on Voronoi diagrams and their relation to maximum inscribed circles.
Algorithms & implementations.
You could actually compute the Voronoi diagram. A worst-case O(n log n) algorithm for points and segments is given by Fortune, A sweepline algorithm for Voronoi diagrams, SoCG'86. Held published the software package Vroni with an expected O(n log n) time complexity, which actually computes the maximum inscribed circle, too. And there seems to be an implementation in boost, too.
For simple polygons (i.e., without holes) a time-optimal algorithm that runs in O(n) time is due to Chin et al., Finding the Medial Axis of a Simple Polygon in Linear Time, 1999.
Brute force.
However, as you stated that you are fine with a brute-force algorithm: What about simply trying out all triplets of sites (vertices and edges). For each triplet you find candidate Voronoi nodes, i.e., equidistant loci to the three sites and check whether any other site would intersect the candidate maximum inscribed circle. If there is an intersection you dismiss the candidate. Take the greatest you can find over all triplets.
See chapter 3 in my Master thesis about more details on computing equidistant loci for three sites.
I implemented a piece of python code based on cv2 to get the maximum/largest inscribed circle inside mask/polygon/contours. It supports non-convex/hollow shape.
import cv2
import numpy as np
def get_test_mask():
# Create an image
r = 100
mask = np.zeros((4 * r, 4 * r), dtype=np.uint8)
# Create a sequence of points to make a contour
vert = [None] * 6
vert[0] = (3 * r // 2, int(1.34 * r))
vert[1] = (1 * r, 2 * r)
vert[2] = (3 * r // 2, int(2.866 * r))
vert[3] = (5 * r // 2, int(2.866 * r))
vert[4] = (3 * r, 2 * r)
vert[5] = (5 * r // 2, int(1.34 * r))
# Draw it in mask
for i in range(6):
cv2.line(mask, vert[i], vert[(i + 1) % 6], (255), 63)
return mask
mask = get_test_mask()
"""
Get the maximum/largest inscribed circle inside mask/polygon/contours.
Support non-convex/hollow shape
"""
dist_map = cv2.distanceTransform(mask, cv2.DIST_L2, cv2.DIST_MASK_PRECISE)
_, radius, _, center = cv2.minMaxLoc(dist_map)
result = cv2.cvtColor(mask, cv2.COLOR_GRAY2BGR)
cv2.circle(result, tuple(center), int(radius), (0, 0, 255), 2, cv2.LINE_8, 0)
# minEnclosingCircle directly by cv2
contours, _ = cv2.findContours(mask, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)[-2:]
center2, radius2 = cv2.minEnclosingCircle(np.concatenate(contours, 0))
cv2.circle(result, (int(center2[0]), int(center2[1])), int(radius2), (0, 255, 0,), 2)
cv2.imshow("mask", mask)
cv2.imshow("result", result)
cv2.waitKey(0)
Red circle is max inscribed circle
Source: https://gist.github.com/DIYer22/f82dc329b27c2766b21bec4a563703cc
I used Straight Skeletons to place an image inside a polygon with three steps:
Find the straight skeleton using the Straight Skeleton algorithm (pic 1)
Base on the straight skeleton, find the largest circle (pic 2)
Draw the image inside that circle (pic 3)
Try it at: https://smartdiagram.com/simple-infographics-3d-charts-2/
An O(n log X) algorithm, where X depends on the precision you want.
Binary search for largest radius R for a circle:
At each iteration, for a given radius r, push each edge E, "inward" by R, to get E'. For each edge E', define half-plane H as the set of all points "inside" the the polygon (using E' as the boundary). Now, compute the intersection of all these half-planes E', which could be done in O(n) time. If the intersection is non-empty, then if you draw a circle with radius r using any point in the intersection as the center, it will be inside the given polygon.