There are N points on a 2D grid (x,y). I need to find the shortest path, from point A to point B, but I can only travel from one point to another and I can't travel between two points if the distance between them is farther than a distance D. I thought it might be solved by using some kind of modified Dijkstra's algorithm, but I'm not sure how, because I've never implemented it before, just studied it on Wiki.
Well, Dijkstra finds shortest paths in graphs. So just consider the grid points to be nodes in a graph with edges between each node S and all other nodes T such that dist(S, T) <= D. You don't have to actually construct the graph because the edges are easily determined as needed by Dijkstra. Just check all nodes in a square around S with radius D. A S-T edge exists iff (Sx - Tx)^2 (Sy - Ty)^2 <= D^2.
Wiki explanation is sufficient for this.
Dijkstra's algorithm takes 3 inputs. The Graph, Starting node and Ending node.
To construct the graph just do this
For i 1..n in points
For j i+1..n in points
if(dist(points[i],points[j])<=D)
add j to childs of i
add i to childs of j
After constructing the graph, perform dijkstra.
The subtlety of a question like this lies in a critical definition - what is the measure of distance in your grid?
The are many different shortest path problems and solutions, and they are studied throughout mathematics. They are each characterised by the 'topology' of the area being searched. Consider a few distinct topologies with their own solutions:
A one sided piece of paper
Suppose your grid represents coordinates on a piece of paper - the shortest path is easy to find, as it is simply a straight line between those points.
The surface of the moon
If your grid represents locations on the moon in terms of latitude and longitude, the shortest path is an arc along the moon's surface - If you drove "in a straight line" between two points on the moon, you would be travelling in an arc, because of the moon's curvature.
Road Intersections
If you want to find the distance between two intersections in a grid of roads, where the traffic on each road has a different speed, and you can only travel along the roads, then you can find the shortest path using Dijkstra's algorithm.
One way road intersections
A slight variation of the above - we only need to consider roads in one direction. There might not be any paths in this case.
Summary
To give a good solution, we need to understand the topology of your grid. If the distance is pythagerous's theorem than that indicates euclidean geometry (like in the piece of paper example), so the solution is a straight line.
Is it possible you mean that you can travel between any two points if the are closer than D - like flying a plane between airports, for example?
EDIT: I didn't see your comment because you didn't use #. In your case your grid is like the airports a plane can fly between. The shortest path is found using Dijkstra's algorithm - the immediate neighbours of a point are all points closer than D. Find them, represent it all as a graph, and use Dijkstra's algorithm.
I would suggest using the formula to find the distance between 2 points i.e sqrt((x2-x1)^2+(y2-y1)^2). This distance is always the shortest between 2 points.
Related
I am currently working on a hole detection problem in 3D point cloud data. I am referring to this paper "Detecting Holes in Point Set Surfaces" by Gerhard H Bendels, Ruwen Schnabel and Reinhard Klein. One of the criterions mentioned is an Angle Criterion in which we need to determine angles between consecutive points in a KD Tree(Radially Nearest Neighbors to a given point).
See Image:
Angle between points
I am using Open3D to extract a KD Tree but I believe it is giving me an unsorted list of points rather than a list of consecutive points.
See Image:
List of Nearest neighbors
The point below '______' is the point of interest and rest are it's neighbors. Now my question is,
How do I know which point is next to which point?
And if that's not possible to know, How can I find the angles as shown in the first image.
I just need the angles to find the boundary probability for each point, so an answer would really help me progress.
Thanks
What I've Tried so far..
I have tried generating vectors out of all the points and calculated the angles using dot product. But that seems wrong because I believe I may be calculating dot products between first and third point.
I am dealing with a reverse-engineering problem regarding road geometry and estimation of design conditions.
Suppose you have a set of points obtained from the measurement of positions of a road. This road has straight sections as well as curve sections. Straight sections are, of course, represented by lines, and curves are represented by circles of unknown center and radius. There are, as well, transition sections, which may be clothoids / Euler spirals or any other usual track transition curve. A representation of the track may look like this:
We know in advance that the road / track was designed taking this transition + circle + transition principle into account for every curve, yet we only have the measurement points, and the goal is to find the parameters describing every curve on the track, this is, the transition parameters as well as the circle's center and radius.
I have written some code using a nonlinear optimization algorithm, where a user can select start and end points and fit a circle that to the arc section between them, as it shows in next figure:
However, I don't find a suitable way to take the transition into account. After giving it some thought I came to think that this s because, given a set of discrete points -with their measurement error- representing a full curve, it is not entirely clear where to consider it "begins" and where it "ends" and, moreover, it is less clear where to consider the transition, the proper circle and the exit transition "begin" and "end".
Is there any work on this subject which I may have missed? is there a proper way to fit the whole transition + curve + transition structure into the set of points?
As far as I know, there's no method to fit a sequence clothoid1-circle-clothoid2 into a given set of points.
Basic facts are that two points define a straight, and three points define a unique circle.
The clothoid is far more complex, because you need: The parameter A, the final radius Rf, an initial point px,py, the radius Ri at that point, and the tangent T (angle with X-axis) at that point.
These are 5 data you may use to find the solution.
Due to clothoid coords are calculated by expanded Fresnel integrals (see https://math.stackexchange.com/a/3359006/688039 a little explanation), and then apply a translation & rotation, there's no an easy way to fit this spiral into a set of given points.
When I've had to deal with your issue, what I've done is:
Calculate the radius for triplets of consecutive points: p1p2p3, p2p3p4, p3p4p5, etc
Observe the sequence of radius. Similar values mean a circle, increasing/decreasing values mean a clothoid; Big values would mean a straight.
For each basic element (line, circle) find the most probably characteristics (angles, vertices, radius) by hand or by some regression method. Many times the common sense is the best.
For a spiral you may start with aproximated values, taken from the adjacent elements. These values may very well be the initial angle and point, and the initial and final radius. Then you need to iterate, playing with Fresnel and 'space change' until you find a "good" parameter A. Then repeat with small differences in the other values, those you took from adjacents.
Make the changes you consider as good. For example, many values (A, radius) use to be integers, without decimals, just because it was easier for the designer to type.
If you can make a small applet to do these steps then it's enough. Using a typical roads software helps, but doesn't avoid you the iteration process.
If the points are dense compared to the effective radii of curvature, estimate the local curvature by least square fitting of a circle on a small number of points, taking into account that the curvature is most of the time zero.
You will obtain a plot with constant values and ramps that connect them. You can use an estimate of the slope at the inflection points to figure out the transition points.
I would like to find two vertices of two meshes (1 vertex per mesh) that define the closest distance between them. Or the two triangles would be fine I guess.
However I'm not sure how to search for this in CGAL's documentation, I'm sure that this is doable with some existing tool (probably based on a 3d distance field and/or AABBs). Could I please get a hint (keywords/link) on what to look for?
I've been pointed to the Optimal Distances CGAL package, but it's not exactly what I want, since it outputs the distance and the coordinates, so finding the vertex ID is an additional computational overhead.
I've already implemented a collision detection with CGAL to find the triangle-triangle intersection in a triangle-soup, using AABB-trees. I guess that I should be somehow close to this, although now a simple soup with all me object-triangles wouldn't do the job.
The solution found was this:
CGAL's Optimal Distances package can give an approximation of the closest distance between the convex hulls of two meshes, without explicitly computing the hulls. As a result one gets the shortest distance between these hulls, and the coordinates of the 2 points that lie on them and define this distance.
Then these coordinates can be used as a search-query in kd-trees that contains the original vertices of the meshes in order to find the closest vertices.
In case one mesh is non-convex, the hull that CGAL is using is very approximate, so convex decomposition might be necessary. In such a case one would have to check distances for each convex part and then take the shortest distance.
The above would result in something like this:
enter link description here
A naive approach is to find, for each edge in the polygon, the point on that edge closest to the given point, and then take the one that's closest. Is there a faster algorithm? My goal is to implement a 2D Super Mario Galaxy-style platformer.
Apparently this can be done with Voronoi regions, as in this video: http://www.youtube.com/watch?v=Ldh2YKobuWo
However, I can't find any Voronoi algorithms that deal with edges as well as points. Ideas?
Calculate the point-line distance for each of the edges, then pick the shortest one. There is no shortcut. This site has a good explanation and even implementations in various languages.
However, finding "the point on that edge closest to the given point" is a computationally unnecessary intermediate result.
If the polygon is convex, then the overhead of the voronoi calculation far exceeds that of the naive approach.
If this is run many times, and each time the point changes slightly, you only need to check 3 segments (think about it: as you move around, assuming many checks, then the closest edge will only change to an adjacent edge)
I would like to deconstruct the following polygon shown in blue removing all of the points from the polygon that cause concavity.
Currently, what I have been attempting to do is:
Take each point out of the polygon
Test the point to see if it falls within the polygon created by the rest of the set
If true remove the point
If false keep the point
This works in most cases, but in the previous case the points at (2,3) and (2,4) will not both be removed. In both cases either one of the points will be removed, but the other will not depending on the order which the array is passed in.
What I am wondering is this:
Is there some way to test to see if the polygon I am dealing with happens to have one of these cases (IE: 3 points of failure in a row?)
or
Is there simply a more effective way of creating convex polygons?
Thank you.
I think perhaps you're looking for the convex hull?
The first algorithm that springs to mind is QuickHull. Initially, take the leftmost and the rightmost points, l and r. They must be on the hull.
Construct a first guess at the hull that's two outward faces, one from l to r and one from r to l. So you have a polygon with zero volume.
Divide all remaining points into those in front of lr and those in front of rl.
From then on, while any face has any points in front of it:
find the furthest point from the face
delete this edge and replace it with two edges, one from the original start point to the furthest point and one from the furthest point to the original end point
of all the points that were in front of the old face, put those in front of the first of the new faces you've added into its in front set, put those in front of the second into its in front set and don't retain any reference to those now inside
At the end you'll have the convex hull.
Why not simply compute the convex hull of the points?
This is a well studied problem with a number of algorithms in books and online. A method of "sweeping angles" is particularly common, eg.
http://courses.csail.mit.edu/6.854/06/scribe/s25-rasmu-sweepline.pdf
What you are looking for is known as "convex hull" finding. Look here at wikipedia for algorithms for this problem. The "gift wrapping" algorithm is easy to implement. When yout found the hull, just remove all points that are not part of the hull.
Be aware that the Convex Hull is already implemented in some languages/environments.
Example in Mathematica:
<< ComputationalGeometry`;
data2D = {{4.4, 14}, {6.7, 15.25}, {6.9,12.8}, {2.1, 11.1}, {9.5, 14.9},
{13.2, 11.9}, {10.3, 12.3}, {6.8, 9.5}, {3.3, 7.7}, {0.6, 5.1},
{5.3, 2.4}, {8.45, 4.7}, {11.5,9.6}, {13.8, 7.3}, {12.9, 3.1},
{11, 1.1}};
PlanarGraphPlot[data2D, ConvexHull[data2D]]
Output: