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I am creating a script to generate cylinders in a 3D space, however, I would like for them to not occupy the same region in space (avoid overlapping).
The cylinders are defined by a start and end point, and all have a fixed radius.
I am storing the existing cylinder in an array called listOfCylinders which is an nDim array of shape (nCylinders, 2Points [start, end], {x,y,z} coordinates of each point)
I was able to cook up:
def detect_overlap(new_start, new_end, listOfCylinders):
starts = listOfCylinders[:, 0]
ends = listOfCylinders[:, 1]
radius = 0.1
# Calculate the distance between the new cylinder and all the existing cylinders
dists = np.linalg.norm(np.cross(new_end - new_start, starts - new_start), axis=1) / np.linalg.norm(new_end - new_start)
# Check if any of the distances are less than the sum of the radii
if np.any(dists < (2*radius)):
return True
# If no overlap or intersection is found, return False
return False
But this is not accountting for situations where there is lateral overlaping.
Does anyone have a good algorithm for this?
Best Regards
WLOG one of the cylinders is vertical (otherwise rotate space). If you look at the projections of the apparent outline onto XY, you see a circle and a rectangle ended with ellipses. (For simplicity of the equations, you can also make the second cylindre parallel to XZ.)
If these 2D shapes do not overlap, your are done. Anyway, the intersection of a circle and an ellipse leads to a quartic equation.
You can repeat this process, exchanging the roles of the two cylinders. This gives a sufficient condition of non-overlap. Unfortunately, I am not sure it is necessary, though there is a direct connection to the plane separation theorem.
For a numerical approach, you can proceed as follows:
move the cylindre in the canonical position;
generate rectangles on the oblique cylindre, by rotation around the axis and using an angular parameter;
for all sides of the rectangles, detect interference with the cylindre (this involves a system of a quadratic inequation and two linear ones, which is quite tractable);
sample the angular parameter densely enough to check for no valid intersection.
I guess that a complete analytical solution is possible, but complex, and might anyway lead to equations that need to be solved numerically.
I have a 2x2 matrix of distances from a depth sensor.
The matrix is cropped so only the points we are interested in is in the frame(All the points in the cropped image contains the object).
My question is how can we determine if this object is flat or not?
The depth image is acquired from Realsense d435. I read the depth image and then multiply it by depth_scale.
The object is recognized using AI for the rgb image that is aligned with the depth image.
And I have 4 points on the object. So, all the distances in that rectangle contains the distance of the object from the sensor.
My first idea was standard deviation of all the points. But then this falls apart if the image is taken from an angle. (since the standard deviation won't be 0)
From an angle the distance of a flat object is changing uniformly on the y axis. Maybe somehow, we can use this information?
The 2x2 matrix is a numpy array in python. Maybe there are some libraries which do this already.
After reprojecting your four depth measurements to the 3D space, it becomes a problem of deciding if your set of points is coplanar. There are several ways you can go about it.
One way to do it is to reproject the points to 3D and fit a plane to all four of them there. Since you're fitting a plane to four points in three dimensions, you get an over-determined system, and it's very unlikely that all points would lie exactly on the estimated plane. At this stage, you could prescribe some tolerance to determine "goodness of fit". For instance, you could look at the R^2 coefficient.
To fit the plane you can use scipy.linalg.lstsq. Here's a good description of how it can be done: Fit plane to a set of points in 3D.
Another way to approach the problem is by calculating the volume of a tetrahedron spanned by the four points in 3D. If they are coplanar (or close to coplanar), the volume of such a tatrahedron should be equal to (or close to) 0. Assuming your pointa reprojected to 3D can be described by (x_0, y_0, z_0), ..., (x_3, y_3, z_3), the volume of the tetrahedron is equal to:
volume = abs(numpy.linalg.det(tetrahedron)) / 6, where
tetrahedron = np.array([[x_0, y_0, z_0, 1], [x_1, y_1, z_1, 1], [x_2, y_2, z_2, 1], [x_3, y_3, z_3, 1]])
To check if your points are on the same plane, (equivalently - if the tetrahedron has a small enough volume), it is now sufficient to check if
volume < TOL
for some defined small tolerance value, which must be determined experimentally.
You can define a surface by choosing three of the four 3D points.
Evaluate the distance from the remaining point to the surface.
How to choose the three points is... it may be good to choose the pattern that maximizes the area of the triangle.
This is a question that I was asked on a job interview some time ago. And I still can't figure out sensible answer.
Question is:
you are given set of points (x,y). Find 2 most distant points. Distant from each other.
For example, for points: (0,0), (1,1), (-8, 5) - the most distant are: (1,1) and (-8,5) because the distance between them is larger from both (0,0)-(1,1) and (0,0)-(-8,5).
The obvious approach is to calculate all distances between all points, and find maximum. The problem is that it is O(n^2), which makes it prohibitively expensive for large datasets.
There is approach with first tracking points that are on the boundary, and then calculating distances for them, on the premise that there will be less points on boundary than "inside", but it's still expensive, and will fail in worst case scenario.
Tried to search the web, but didn't find any sensible answer - although this might be simply my lack of search skills.
For this specific problem, with just a list of Euclidean points, one way is to find the convex hull of the set of points. The two distant points can then be found by traversing the hull once with the rotating calipers method.
Here is an O(N log N) implementation:
http://mukeshiiitm.wordpress.com/2008/05/27/find-the-farthest-pair-of-points/
If the list of points is already sorted, you can remove the sort to get the optimal O(N) complexity.
For a more general problem of finding most distant points in a graph:
Algorithm to find two points furthest away from each other
The accepted answer works in O(N^2).
Boundary point algorithms abound (look for convex hull algorithms). From there, it should take O(N) time to find the most-distant opposite points.
From the author's comment: first find any pair of opposite points on the hull, and then walk around it in semi-lock-step fashion. Depending on the angles between edges, you will have to advance either one walker or the other, but it will always take O(N) to circumnavigate the hull.
You are looking for an algorithm to compute the diameter of a set of points, Diam(S). It can be shown that this is the same as the diameter of the convex hull of S, Diam(S) = Diam(CH(S)). So first compute the convex hull of the set.
Now you have to find all the antipodal points on the convex hull and pick the pair with maximum distance. There are O(n) antipodal points on a convex polygon. So this gives a O(n lg n) algorithm for finding the farthest points.
This technique is known as Rotating Calipers. This is what Marcelo Cantos describes in his answer.
If you write the algorithm carefully, you can do without computing angles. For details, check this URL.
A stochastic algorithm to find the most distant pair would be
Choose a random point
Get the point most distant to it
Repeat a few times
Remove all visited points
Choose another random point and repeat a few times.
You are in O(n) as long as you predetermine "a few times", but are not guaranteed to actually find the most distant pair. But depending on your set of points the result should be pretty good. =)
This question is introduced at Introduction to Algorithm. It mentioned 1) Calculate Convex Hull O(NlgN). 2) If there is M vectex on Convex Hull. Then we need O(M) to find the farthest pair.
I find this helpful links. It includes analysis of algorithm details and program.
http://www.seas.gwu.edu/~simhaweb/alg/lectures/module1/module1.html
Wish this will be helpful.
Find the mean of all the points, measure the difference between all points and the mean, take the point the largest distance from the mean and find the point farthest from it. Those points will be the absolute corners of the convex hull and the two most distant points.
I recently did this for a project that needed convex hulls confined to randomly directed infinite planes. It worked great.
See the comments: this solution isn't guaranteed to produce the correct answer.
Just a few thoughts:
You might look at only the points that define the convex hull of your set of points to reduce the number,... but it still looks a bit "not optimal".
Otherwise there might be a recursive quad/oct-tree approach to rapidly bound some distances between sets of points and eliminate large parts of your data.
This seems easy if the points are given in Cartesian coordinates. So easy that I'm pretty sure that I'm overlooking something. Feel free to point out what I'm missing!
Find the points with the max and min values of their x, y, and z coordinates (6 points total). These should be the most "remote" of all the boundary points.
Compute all the distances (30 unique distances)
Find the max distance
The two points that correspond to this max distance are the ones you're looking for.
Here's a good solution, which works in O(n log n). It's called Rotating Caliper’s Method.
https://www.geeksforgeeks.org/maximum-distance-between-two-points-in-coordinate-plane-using-rotating-calipers-method/
Firstly you find the convex hull, which you can make in O(n log n) with the Graham's scan. Only the point from the convex hull can provide you the maximal distance. This algorithm arranges points of the convex hull in the clockwise traversal. This property will be used later.
Secondly, for all the points on the convex hull, you'll need to find the most distant point on this hull (it's called the antipodal point here). You don't have to find all the antipodal points separately (which would give quadratic time). Let's say the points of the convex hall are called p_1, ..., p_n, and their order corresponds to the clockwise traversal. There is a property of convex polygons that when you iterate through points p_j on the hull in the clockwise order and calculate the distances d(p_i, p_j), these distances firstly don't decrease (and maybe increase) and then don't increase (and maybe decrease). So you can find the maximum distance easily in this case. But when you've found the correct antipodal point p_j* for the p_i, you can start this search for p_{i+1} with the candidates points starting from that p_j*. You don't need to check all previously seen points. in total p_i iterates through points p_1, ..., p_n once, and p_j iterates through these points at most twice, because p_j can never catch up p_i as it would give zero distance, and we stop when the distance starts decreasing.
A solution that has runtime complexity O(N) is a combination of the above
answers. In detail:
(1) One can compute the convex hull with runtime complexity O(N) if you
use counting sort as an internal polar angle sort and are willing to
use angles rounded to the nearest integer [0, 359], inclusive.
(2) Note that the number of points on the convex hull is then N_H which is usually less than N.
We can speculate about the size of the hull from information in Cormen et al. Introduction to Algorithms, Exercise 33-5.
For sparse-hulled distributions of a unit-radius disk, a convex polygon with k sides, and a 2-D normal distribution respectively as n^(1/3), log_2(n), sqrt(log_2(n)).
The furthest pair problem is then between comparison of points on the hull.
This is N_H^2, but each leading point's search for distance point can be
truncated when the distances start to decrease if the points are traversed
in the order of the convex hull (those points are ordered CCW from first point).
The runtime complexity for this part is then O(N_H^2).
Because N_H^2 is usually less than N, the total runtime complexity
for furthest pair is O(N) with a caveat of using integer degree angles to reduce the sort in the convex hull to linear.
Given a set of points {(x1,y1), (x2,y2) ... (xn,yn)} find 2 most distant points.
My approach:
1). You need a reference point (xa,ya), and it will be:
xa = ( x1 + x2 +...+ xn )/n
ya = ( y1 + y2 +...+ yn )/n
2). Calculate all distance from point (xa,ya) to (x1,y1), (x2,y2),...(xn,yn)
The first "most distant point" (xb,yb) is the one with the maximum distance.
3). Calculate all distance from point (xb,yb) to (x1,y1), (x2,y2),...(xn,yn)
The other "most distant point" (xc,yc) is the one with the maximum distance.
So you got your most distant points (xb,yb) (xc,yc) in O(n)
For example, for points: (0,0), (1,1), (-8, 5)
1). Reference point (xa,ya) = (-2.333, 2)
2). Calculate distances:
from (-2.333, 2) to (0,0) : 3.073
from (-2.333, 2) to (1,1) : 3.480
from (-2.333, 2) to (-8, 5) : 6.411
So the first most distant point is (-8, 5)
3). Calculate distances:
from (-8, 5) to (0,0) : 9.434
from (-8, 5) to (1,1) : 9.849
from (-8, 5) to (-8, 5) : 0
So the other most distant point is (1, 1)
It might seem a bit odd that I am asking for python code to calculate the area of a polygon with a list of (x,y) coordinates given that there have been solutions offered in stackoverflow in the past. However, I have found that all the solutions provided are sensitive to the order of the list of (x,y) coordinates given. For example, with the code below to find an area of a polygon:
def area(p):
return 0.5 * abs(sum(x0*y1 - x1*y0
for ((x0, y0), (x1, y1)) in segments(p)))
def segments(p):
return zip(p, p[1:] + [p[0]])
coordinates1 = [(0.5,0.5), (1.5,0.5), (0.5,1.5), (1.5,1.5)]
coordinates2 = [(0.5,0.5), (1.5,0.5), (1.5,1.5), (0.5,1.5)]
print "coordinates1", area(coordinates1)
print "coordinates2", area(coordinates2)
This returns
coordinates1 0.0
coordinates2 1.0 #This is the correct area
For the same set of coordinates but with a different order. How would I correct this in order to get the area of the non-intersecting full polygon with a list of random (x,y) coordinates that I want to make into a non-intersecting polygon?
EDIT: I realise now that there can be multiple non-intersecting polygons from a set of coodinates. Basically I am using scipy.spatial.Voronoi to create Voronoi cells and I wish to calculate the area of the cells once I've fed the coordinates to the scipy Voronoi function - unfortunately the function doesn't always output the coordinates in the order that will allow me to calculate the correct area.
Several non-intersecting polygons can be created from a random list of coordinates (depending on its order), and each polygon will have a different area, so it is essential that you specify the order of the coordinates to build the polygon (see attached picture for an example).
The Voronoi cells are convex, so that the polygon is unambiguously defined.
You can compute the convex hull of the points, but as there are no reflex vertices to be removed, the procedure is simpler.
1) sort the points by increasing abscissa; in case of ties, sort on ordinates (this is a lexicographical ordering);
2) consider the straight line from the first point to the last and split the point sequence in a left and a right subsequence (with respect to the line);
3) the requested polygon is the concatenation of the left subsequence and the right one, reversed.
Is there even such a thing as a 3D centroid? Let me be perfectly clear—I've been reading and reading about centroids for the last 2 days both on this site and across the web, so I'm perfectly aware at the existing posts on the topic, including Wikipedia.
That said, let me explain what I'm trying to do. Basically, I want to take a selection of edges and/or vertices, but NOT faces. Then, I want to place an object at the 3D centroid position.
I'll tell you what I don't want:
The vertices average, which would pull too far in any direction that has a more high-detailed mesh.
The bounding box center, because I already have something working for this scenario.
I'm open to suggestions about center of mass, but I don't see how this would work, because vertices or edges alone don't define any sort of mass, especially when I just have an edge loop selected.
For kicks, I'll show you some PyMEL that I worked up, using #Emile's code as reference, but I don't think it's working the way it should:
from pymel.core import ls, spaceLocator
from pymel.core.datatypes import Vector
from pymel.core.nodetypes import NurbsCurve
def get_centroid(node):
if not isinstance(node, NurbsCurve):
raise TypeError("Requires NurbsCurve.")
centroid = Vector(0, 0, 0)
signed_area = 0.0
cvs = node.getCVs(space='world')
v0 = cvs[len(cvs) - 1]
for i, cv in enumerate(cvs[:-1]):
v1 = cv
a = v0.x * v1.y - v1.x * v0.y
signed_area += a
centroid += sum([v0, v1]) * a
v0 = v1
signed_area *= 0.5
centroid /= 6 * signed_area
return centroid
texas = ls(selection=True)[0]
centroid = get_centroid(texas)
print(centroid)
spaceLocator(position=centroid)
In theory centroid = SUM(pos*volume)/SUM(volume) when you split the part into finite volumes each with a location pos and volume value volume.
This is precisely the calculation done for finding the center of gravity of a composite part.
There is not just a 3D centroid, there is an n-dimensional centroid, and the formula for it is given in the "By integral formula" section of the Wikipedia article you cite.
Perhaps you are having trouble setting up this integral? You have not defined your shape.
[Edit] I'll beef up this answer in response to your comment. Since you have described your shape in terms of edges and vertices, then I'll assume it is a polyhedron. You can partition a polyedron into pyramids, find the centroids of the pyramids, and then the centroid of your shape is the centroid of the centroids (this last calculation is done using ja72's formula).
I'll assume your shape is convex (no hollow parts---if this is not the case then break it into convex chunks). You can partition it into pyramids (triangulate it) by picking a point in the interior and drawing edges to the vertices. Then each face of your shape is the base of a pyramid. There are formulas for the centroid of a pyramid (you can look this up, it's 1/4 the way from the centroid of the face to your interior point). Then as was said, the centroid of your shape is the centroid of the centroids---ja72's finite calculation, not an integral---as given in the other answer.
This is the same algorithm as in Hugh Bothwell's answer, however I believe that 1/4 is correct instead of 1/3. Perhaps you can find some code for it lurking around somewhere using the search terms in this description.
I like the question. Centre of mass sounds right, but the question then becomes, what mass for each vertex?
Why not use the average length of each edge that includes the vertex? This should compensate nicely areas with a dense mesh.
You will have to recreate face information from the vertices (essentially a Delauney triangulation).
If your vertices define a convex hull, you can pick any arbitrary point A inside the object. Treat your object as a collection of pyramidal prisms having apex A and each face as a base.
For each face, find the area Fa and the 2d centroid Fc; then the prism's mass is proportional to the volume (== 1/3 base * height (component of Fc-A perpendicular to the face)) and you can disregard the constant of proportionality so long as you do the same for all prisms; the center of mass is (2/3 A + 1/3 Fc), or a third of the way from the apex to the 2d centroid of the base.
You can then do a mass-weighted average of the center-of-mass points to find the 3d centroid of the object as a whole.
The same process should work for non-convex hulls - or even for A outside the hull - but the face-calculation may be a problem; you will need to be careful about the handedness of your faces.