I have a set of points (500 points), each has its own coordinate. These points are represented by two different 2-D coordinate systems (e.g., (X, Y) and (x, y)). I have a list of 500 (X,Y)-type of coordinates and 500 (x,y)-type of coordinates. My goal is to match them pairwise. There should be 1-1 mapping between the two sets. I can also convert a distance from one coordinate system to the other.
I am thinking about using pairwise distances as a means to match. Please let me know if there is a simple way to do that. Thank you.
Related
I want to create a set of N random convex disjoint polygons on a plane where each polygon must have at most V vertices, where N and V are parameters for my function, and I'd like to obtain a distribution as close as possible to uniform (every possible set being equally probable). Also I need to randomly select two points on the plane that either match with one of the vertices in the scene or are in empty space (not inside a polygon).
I already implemented for other reasons in the same programming language an AABB tree, Separating Axis Theorem-based collision detection between convex polygons and I can generate a random convex polygon with arbitrary amount of vertices inside a circle of given radius. My best bet thus far:
Generate a random polygon using the function I have available.
Query the AABB tree to test for interception with existing polygons.
If the AABB tree query returns empty set I push the generated polygon into it, otherwise I test with SAT against all the other polygons whose AABB overlaps with the generated one's. If SAT returns "no intersection" I push the polygon, otherwise I discard it.
Repeat from 1 until N polygons are generated.
Generate a random number in {0,1}
If the generated number is 1 I pick a random polygon and a random vertex on it as a point
If the generated number is 0 I generate a random position in (x,y) and test if it falls within some polygon (I might create a tiny AABB around it and exploit the AABB tree to reduce the required number of PiP tests). In case it's not inside any polygon I approve it as a valid point, otherwise I repeat from 5.
Repeat from 5 once more to get the second point.
I think the solution would possibly work, but unfortunately there's no way to guarantee that I can generate N such polygons for very large N, or find two good points in an acceptable time, and I'm programming in React, where long operations run on the main thread blocking the UI till they end. I could circumvent the issue by ejecting from create-react-app and learn Web Workers, which would require probably more time than it's worth for me.
This is definitely non-uniform distribution, but perhaps you could begin by generating N points in the plane and then computing the Voronoi diagram for those points. The Voronoi diagram can be computed in O(n log n) time with Fortune's algorithm. The cells of the Voronoi diagram are convex, so you can then construct a random polygon of the desired number of vertices that lies within each cell of the diagram.
By Balu Ertl - Own work, CC BY-SA 4.0, Link
Ok, here is another proposal. I have little knowledge of js, but could cook up something in Python.
Use Poisson disk sampling with distance parameter d to generate N samples of the centers
For a given center make a circle with R≤d.
Generate V angles using Dirichlet distribution such that sum of angles is equal to 2π. Sort them.
Place vertices on the circle using angles generate at step#3 and connect them. This would be be your polygon
UPDATE
Instead using Poisson disk sampling for step 1, one could use Sobol quasi-random sequences. For N points sampled in the 1x1 box (well, you have to scale it afterwards), least distance between points would be
d = 0.5 * sqrt(D) / N,
where D is dimension of the problem, 2 in your case. So radius of the circle for step 2 would be 0.25 * sqrt(2) / N. This ties nicely N and d.
https://www.sciencedirect.com/science/article/abs/pii/S0378475406002382
What is the Difference between Geodist(sfield,x,y) and dist(2,x,y,a,b) in Apache Solr for Geo-Spacial Searches ??
dist(2,x,y,0,0) :- calculates the Euclidean distance between (0,0) and (x,y) for each document. Return the Distance between two Vectors (points) in an n-dimensional space.
I was earlier using geodist() distance function for Geo-Spatial searches on my website but its response time was large. so have done a POC(proof of concept) for different distance functions and found that dist(2,x,y,0,0) distance function is relatively taking half of the time. But I want to know the reason behind this and the algorithms which both functions are using to calculate the distance.
I have to make a difference matrix for the same to convey it further.
The main difference is that geodist() is intended to work with spatial field types.
Most spatial implementation are based on Lucene's Points API, which is a BKD Index. This field type is strictly limited to coordinates in lat/lon decimal degrees. Behind the scenes, latitude and longitude are indexed as separate numbers. Four main field types are available for spatial search :
LatLonPointSpatialField
LatLonType (now deprecated) and its non-geodetic twin PointType
SpatialRecursivePrefixTreeFieldType (RPT for short), including RptWithGeometrySpatialField, a derivative
BBoxField (for areas, 4 instances of another field type referred to by numberType)
In geodist (sfield, x, y), sfield is a spatial field type that represents two points (lat,lon), so the direct equivalent using dist() would be to implement dist (2, sfieldX, sfieldY, x, y) with sfieldX and sfieldY being respectively the (lat,lon) coordinates of sfield.
Using dist (power, a, b, ...) you can't query a spatial field type. In order to perform the same spatial search, you would have to specify every point's dimension separately. It would require 2 indexed fields (or values per field at least) for 2 dimensions, 3 for 3d, and so on. That makes a huge difference because you would have to index every coordinates of each point separately.
Besides, you can also use geodist() as is with the BBoxField field type that indexes a single rectangle per document field and supports searching via a bounding box. To do the same with dist() you would have to compute the center point of the box to input each one of its coordinates as a function argument, so it would be too much hassle to yield the same result if you want to use an area as parameter.
Lastly, LatLonPointSpatialField for example does distance calculations based on Haversine formula (Great Circle), BBoxField does it a little faster because the rectangular shape is faster to compute. It's true that dist() may be even faster but remember that requires more field to be indexed, a lot of preprocess at query time to be able to yield the same calculated distance, and, as mentioned by Mats, it wouldn't take the earth' curvature into account.
An euclidean distance doesn't account for the curvature of the earth. If you're only sorting by the distance, the behavior can be OK - but only if your hits are within a small geographical area (the value of a unit compared to meters greatly change when you're getting closer to the poles).
There's an extensive and good answer that explains the difference between a Euclidean distance and a proper geographical distance (usually calculated using haversine) available at the GIS Stack Exchange.
Although at small scales any smooth surface looks like a plane, the accuracy of the Pythagorean formula depends on the coordinates used. When those coordinates are latitude and longitude on a sphere (or ellipsoid), we can expect that
Distances along lines of longitude will be reasonably accurate.
Distances along the Equator will be reasonably accurate.
All other distances will be erroneous, in rough proportion to the differences in latitude and longitude.
To calculate the dihedral angles between two planes, one needs four points: two lie on the intersecting edge and two lie on each corresponding plane. The full mathematical formulation can be found here.
Now my question is concerned with data structure and how to efficiently calculate all the dihedral angles in a hexahedron. Suppose I have a data structure as followed
vertices[8] // Contains all the vertices of the hexahedral
edges[12] = {{vertices[i], vertices[k]}, {vertices[i], vertices[j]}...} // Each cell contain an edge formed by the two vertices.
face[6] = { {vertices[i], vertices[j], vertices[k], vertices[l]}, {..} ...} // Each face contains the four vertices that form a face of the hexahedral.
Supposed all the faces of this hexahedron is flat (i.e. all four vertices of a face is coplanar), what is a good strategy to calculate all the dihedral angles of a hexahedral being defined in this way?
At the moment, my pseudocode looks like
for all edges
loop through the face list to find all faces that contain the edges
for the face that both contain the vertices of the sharing edge, find the other points
then used the formulation proposed above.
which appears quite clumsy and slow. Any better suggestions?
For each face, find the normal vector. Take the cross product of the two diagonal vectors and normalize to unit length; then
For each pair of adjacent faces with normal vectors A and B, the dihedral angle between them is acos(A \dot B)
I'm currently working on a project where I want to draw different mathematical objects onto a 3D cube. It works as it should for Points and Lines given as a vector equation. Now I have a plane given as a parametric equation. This plane can be somewhere in the 3D space and may be visible on the screen, which is this 3D cube. The cube acts as an AABB.
First thing I needed to know was whether the plane intersects with the cube. To do this I made lines who are identical to the edges of this cube and then doing 12 line/plane intersections, calculating whether the line is hit inside the line segment(edge) which is part of the AABB. Doing this I will get a set of Points defining the visible part of the plane in the cube which I have to draw.
I now have up to 6 points A, B, C, D, E and F defining the polygon ABCDEF I would like to draw. To do this I want to split the polygon into triangles for example: ABC, ACD, ADE, AED. I would draw this triangles like described here. The problem I am currently facing is, that I (believe I) need to order the points to get correct triangles and then a correctly drawn polygon. I found out about convex hulls and found QuickHull which works in three dimensional space. There is just one problem with this algorithm: At the beginning I need to create a three dimensional simplex to have a starting point for the algorithm. But as all my points are in the same plane they simply form a two dimensional plane. Thus I think this algorithm won't work.
My question is now: How do I order these 3D points resulting in a polygon that should be a 2D convex hull of these points? And if this is a limitation: I need to do this in C.
Thanks for your help!
One approach is to express the coordinates of the intersection points in the space of the plane, which is 2D, instead of the global 3D space. Depending on how exactly you computed these points, you may already have these (say (U, V)) coordinates. If not, compute two orthonormal vectors that belong to the plane and take the dot products with the (X, Y, Z) intersections. Then you can find the convex hull in 2D.
The 8 corners of the cube can be on either side of the plane, and have a + or - sign when the coordinates are plugged in the implicit equation of the plane (actually the W coordinate of the vertices). This forms a maximum of 2^8=256 configurations (of which not all are possible).
For efficiency, you can solve all these configurations once for all, and for every case list the intersections that form the polygon in the correct order. Then for a given case, compute the 8 sign bits, pack them in a byte and lookup the table of polygons.
Update: direct face construction.
Alternatively, you can proceed by tracking the intersection points from edge to edge.
Start from an edge of the cube known to traverse the plane. This edge belongs to two faces. Choose one arbitrarily. Then the plane cuts this face in a triangle and a pentagon, or two quadrilaterals. Go to the other the intersection with an edge of the face. Take the other face bordered by this new edge. This face is cut in a triangle and a pentagon...
Continuing this process, you will traverse a set of faces and corresponding segments that define the section polygon.
In the figure, you start from the intersection on edge HD, belonging to face DCGH. Then move to the edge GC, also in face CGFB. From there, move to edge FG, also in face EFGH. Move to edge EH, also in face ADHE. And you are back on edge HD.
Complete discussion must take into account the case of the plane through one or more vertices of the cube. (But you can cheat by slightly translating the plane, constructing the intersection polygon and removing the tiny edges that may have been artificially created this way.)
I know this may be a duplicate, but it seems like a variation on the 'Closest pair of Points' algorithm.
Given a Set of N points (x, y) in the unit square and a distance d, find all pair of points such that the distance between them is at most d.
For large N the brute force method is not an option. Besides the 'sweep line' and 'divide and conquer' methods, is there a simpler solution? These pair of points are the edges of an undirected graph, that i need to traverse it and say if it's connected or not (which i already did using DFS, but when N = 1 million it never finishes!).
Any pseudocode, comments or ideas are welcome,
Thanks!
EDIT: I found this on Sedgewick book (i'm looking at the code right now):
Program 3.18 uses a two-dimensional array of linked lists to improve the running time of Program 3.7 by a factor of about 1/d2 when N is sufficiently large. It divides the unit
square up into a grid of equal-sized smaller squares. Then, for each square, it builds a linked list of all the
points that fall into that square. The two-dimensional array provides the capability to access immediately
the set of points close to a given point; the linked lists provide the flexibility to store the points where
they may fall without our having to know ahead of time how many points fall into each grid square.
We are really looking for points that are inside of a circle of center (x,y) and radius d.
The square that encloses circle is a square of center (x,y) and sides 2d. Any point out of this square does not need to be checked, it's out. So, a point a (xa, ya) is out if abs(xa - x) > d or abs (ya -yb) > d.
Same for the square that is enclosed by that circle is a square of center (x,y) and diagonals 2d. Any point out of this square does not need to be checked, it's in. So, a point a (xa, ya) is in if abs(xa - x) < (d * 1.412) or abs(ya -yb) < (d * 1.412).
Those two easy rules combined reduce a lot the number of points to be checked. If we sort the point by their x, filter the possible points, sort by their y, we come to the ones we really need to calculate the full distance.
For any given point, you can use a Manhattan distance (x-delta plus y-delta) heuristic to filter out most of the points that are not within the distance "d" - filter out any points whose Manhattan distance is greater than (sqrt(2) * d), then run the expensive-and-precise distance test on the remaining points.