It is a little complicated, but you can draw all the objects by the following code: from mpl_toolkits.mplot3d import Axes3D import matplotlib.pyplot as plt import numpy as np from itertools import product, combinations fig = plt.figure() ax = fig.gca(projection=’3d’) ax.set_aspect(“equal”) # draw cube r = [-1, 1] for s, e in combinations(np.array(list(product(r, r, r))), 2): … Read more
I had the same problem, but it was fixed by running following code CREATE EXTENSION postgis; In detail, open pgAdmin select (click) your database click “SQL” icon on the bar run “CREATE EXTENSION postgis;” code
To flip the triangle you really just need to change the direction of iteration. Instead of going from i = 0 to i < lines you need to go down from i = lines-1 to i >= 0 You also need to change the c to how many spaces and symbols you want to start … Read more
find the coin (green bounding box rectangle) either manually or by some search for specific color,pattern,hough transform,segmentation… This will limit the area to search for next steps find the boundary (distinct red edge in color intensity) so create a list of points that are the coin boundary (be careful with shadows) just scan for high … Read more
I found this answer and just modified some minor things to work with dart, I ran a test on a hardcoded polygon. The list _area is my polygon and _polygons is required for my mapcontroller. final Set<Polygon> _polygons = {}; List<LatLng> _area = [ LatLng(-17.770992200, -63.207739700), LatLng(-17.776386600, -63.213576200), LatLng(-17.778348200, -63.213576200), LatLng(-17.786848100, -63.214262900), LatLng(-17.798289700, -63.211001300), LatLng(-17.810547700, … Read more
That really depends on how the lines are represented. I’m going to assume that you have them represented in the parametric form x0(t) = u0 + t v0 x1(t) = u1 + t v1 Here, the x‘s, u‘s, and v‘s are vectors (further denoted in bold) in ℜ2 and t ∈ [0, 1]. These two … Read more
The answers based on angles are cumbersome to implement and can’t be easily generalized to data in higher dimensions. A better approach is that mentioned in my and WimC’s answers here: given the distance matrix D(i, j), define M(i, j) = 0.5*(D(1, j)^2 + D(i, 1)^2 – D(i, j)^2) which should be a positive semi-definite … Read more
You can apply an object’s transform to the object’s geometry directly, and then reset the position, rotation, and scale like so: object.updateMatrix(); object.geometry.applyMatrix4( object.matrix ); object.position.set( 0, 0, 0 ); object.rotation.set( 0, 0, 0 ); object.scale.set( 1, 1, 1 ); object.updateMatrix(); three.js r.150
Perform line intersection tests for each pair of lines, one from each polygon. If no pairs of lines intersect and one of the line end-points of polygon A is inside polygon B, then A is entirely inside B. The above works for any type of polygon. If the polygons are convex, you can skip the … Read more
Eric Meyer’s book “Eric Meyer on CSS” talks about this (Project 10) and the text wrap solutions that you found use the same principle. Using a simple border-radius: 50% does not affect the shape of the content box, which are rectangular at this time. For example, see the demo by Kyle Sevenoaks. There is a … Read more