In a road network graph, the nodes are represented by their coordinates (x,y) and the edges have a weight equal to the euclidean distance between the two connected nodes.
In an A* search algorithm executed over a road network graph, is the heuristic defined as the euclidean distance consistent (h(m)<=h(n)+d(m,n) for any edge (m,n))?
Yes, the Euclidean distance heuristic is consistent. This property is known as the triangle inequality
It also works on the surface of the Earth, if you use great-circle distances, even though that geometry is non-Euclidean.
Be careful to do it correctly, though -- calculating with latitude and longitude as if they were cartesian coordinates can sometimes produce the wrong answer.
Related
I am looking for an algorithm to calculate the following:
I have:
A 3D triangle mesh. The triangles do not necessarily lie in one plane. The angle between the norm vectors of two neighbouring triangles is less then 90 degrees.
Two points. The two points lie either on an edge of the triangle mesh or inside a triangle of the mesh.
I need to calculate the polyline which represents the shortest path between the two points on the mesh.
What is the simplest and/or most effective strategy to do this?
As it stands, your problem is not well defined; there can be many solutions depending on the direction used to "project" the line segment onto the mesh.
Once you have chosen the direction of projection, flatten the mesh onto a plane perpendicular to the direction of projection. At this point, your mesh is a collection of 2d edges (line segments); just determine the intersection (if any) of each edge with your target line segment.
Edit:
The updated question is now well defined. Since my answer to the original question (above) has been marked as accepted, presumably that means the information given in the comments below are actually what was really being "accepted" for the update question. I'll summarize:
A google search of "shortest distance on 3d mesh" turns up some relevant information, like Shortest Path Approximation on Triangulated Meshes
Also, see: https://stackoverflow.com/a/10389377/294949 -- danh
Since your start/end points potentially lie anywhere on the mesh (not restricted to vertices) I guess you are searching for the geodesic shortest path (not Dikstra shortest path following edges). A nice algorithm is implemented in geometry-central: http://geometry-central.net/surface/algorithms/flip_geodesics/
The algorithm is described in the paper "You Can Find Geodesic Paths in Triangle Meshes by Just Flipping Edges".
A standard approach to this task of finding the shortest path polyline (or geodesic) on the surface of triangular mesh between two given points consists of two steps:
Find a path approximation between two points
Iteratively adjust it to make it locally shortest everywhere.
The first step (path approximation) can be computed, for example, using
Dijkstra algorithm, which considers only paths along mesh edges (no crossing of mesh triangles),
Or some variations of Dijkstra algorithm, better suited for near planar surfaces like A*-search,
Or it can be Fast marching method, which can find also the paths crossing the triangles, however not guaranteed to produce a geodesic line.
The next step is iterative adjustment of the approximate path till it becomes truly locally shortest. Some recent articles are really promising here, like Just Flipping Edges. But it requires to construct a path network from the original mesh before operating, so it can be really expensive if your task is just to find one shortest path on the mesh.
A more classical way is to consider every piece of current path approximation between it enters two consecutive vertices, and unfold the strip of crossed triangles on plane. Then find the shortest path in the planar strip, which is a task that can be solved exactly in a linear time, for example by Shortest Paths in Polygons method by Wolfgang Mulzer. The crossing of this line with the edges will give the shortest path on the mesh in between two vertices.
Then for every vertex on the path approximation, two walks around this vertex are evaluated using same unfolding in hope a path around will be shorter then the path exactly via the vertex. The last steps are repeated till convergence.
Below is an example of geodesic path on a mesh with 2 million triangles:
Left - from the application MeshInspector: the time of iterative adjustment is highlighted in green and it is only a small fraction of total time.
Right - the picture from Just Flipping Edges article, where total time is not given, but it is presumably even higher due to the necessity to construct path network from the mesh.
I need to find a largest inscribed circle of a convex polygon, I've searched many sites and I get that this can be done by using Delaunay triangulation. I found a thread in CGAL discussion with an algorithm using CGAL:
You can compute this easily with CGAL:
First, compute the Delaunay triangulation of the points.
Then, iterate on all the finite faces of the triangulation.
For each finite face f
compute its circumcenter c
locate c in the triangulation (to speed up things, you can give one
vertex of f as starting hint for the point location)
if the face returned by locate(c,hint) is finite, then the circumcenter
c lies in the convex hull of the points, so, f is a candidate
if f is such a candidate face, compute its squared circumradius
keep only the face with minimum squared circumradius
The CGAL manual (chapter 2D triangulation, together with a few things
from the kernel) shows every basic function to do this.
I was a bit confused with the last part of this algorithm. When I read it what I understand from it is that the minimum circumradius of the triangulation face is the radius for the largest inscibed circle. But from examples of polygon with Delaunay triangulation, it seems that even the smallest circumcircle sometimes cannot fit inside the polygon, so how can this has the same radius as the largest inscribed circle?
Maximum inscribed circle in polygons.
The classical computational-geometry solution to the maximum inscribed circle problem for polygons is to use the generalized Voronoi diagram of the polygon's faces resp. the medial axis of the polygon. This approach works in a more general setting like polygons with holes, see this stackoverflow answer to a similar question.
Convex input.
The convexity of your input polygon, however, gives the problem more structure, which I would like to comment on. Consider the following convex input polygon (black), the Voronoi diagram (blue), and the maximum inscribed circle (green) centered on a Voronoi node.
The classical Voronoi-based solution is to (i) compute the Voronoi diagram and (ii) take the Voronoi node with largest clearance (i.e., distance to its defining faces).
The Voronoi diagram of a polygon with holes (i.e., the set of vertices and edges) can be computed in O(n log n) time, c.f. Fortune's algorithm (1986). Later Chin et alii (1999) gave an O(n) algorithm for the medial axis of a simple polygon.
For convex polygons, however, a time-optimal algorithm for Voronoi diagram that runs in O(n) time was already known in 1989 due to Aggarwal et alii. This algorithm follows basically the following idea: Consider the grey offset curves moving inwards at unit speed. If you project this movement into three-space where the z-axis is time you get a unit-slop roof over the polygon:
This roof model could also be characterized as follows: Put a half-space on each polygon edge at 45° slope with polygon (such that they contain the polygon) and intersect them all. So if you can quickly compute the intersect of half-spaces then you can also quickly compute Voronoi diagrams of convex polygons. Actually, for the maximum inscribed circle problem we do not need to go back to the Voronoi diagram but take the one peak of the roof, which marks the center of the maximum inscribed circle.
Now the half-spaces are dualized to points, and then the intersection of half-spaces corresponds the convex hull of its dual points. Aggarwal et al. now found an O(n) algorithm for the convex hull of points that stem from this setting.
A summary of this construction that leads to a Voronoi diagram algorithm for convex polyhedra in any dimension can be found in a blog article of mine.
Simple & fast implementation. A simpler algorithm to compute the Voronoi diagram is motivated by straight skeletons. For convex polygons the Voronoi diagram and the straight skeleton are the same.
The algorithm behind the straight-skeleton implementation Stalgo basically simulates the evolution of the wavefront structure (the grey offset curves). For convex polygons this reduces to finding the sequence of edges that collapse.
So a simple O(n log n) algorithm could look like this:
Construct a circular list of the polygon edges. Compute the collapse time of each edge during wavefront propagation, and insert this event into a priority queue.
Until the queue is empty: Take out the next edge-collapse event: Remove the edge from the circular structure and update the collapse times of the neighboring edges of the removed edge.
Actually, you can simplify the above algorithm further: You do not need to update edge collapses in the priority queue but simply insert new ones: Since the new collapse time of edges are strictly lower, you always get the right event first and dismiss the others and the queue is not growing larger than 2n. Hence, you do not compromise the O(n log n) time complexity.
For the maximum inscribed circle problem you can simplify the above algorithm even further: The Voronoi node (resp. straight skeleton node) you look for is due to the collapse of the final triangle at the end of the loop over the priority queue.
This algorithm should be quick in practice and only a few lines of code.
The last step can mean to select the minimum face of the triangle. Then rinse and repeat.
I have a set of rectangles which I need to cluster together, based on euclidean distance between them.The situation is explained in the attached image. .
One possible approach is to take the center of each rectangle and cluster the center points using K means (distance function would be euclidean distance in XY plane). However, I would like to know if there is any other approach to this problem, which does not approximate a rectangle by it's central point, but also takes the actual shape of the rectangle into consideration.
Have a look at algorithms such as DBSCAN and OPTICS that can be used with arbitrary data types as long as you can define a distance between them (such as the minimum rectangle-to-rectangle distance).
K-means is probably not so good, as it is designed for point data with squared euclidean distance (= sum of squares, within-cluster-variance).
One way to formulate this problem is to look at each rectangle i, and each pair of rectangles (i,j) having distance d(i,j), and then forming a distance matrix from those. This distance measure d could be distance between rectangle centers or something more fancy like distance between closest points on rectangles.
Then, apply a clustering algorithm that takes a distance matrix as input, where you define your distance matrix D as the matrix where element (i,j) is d(i,j).
Related: Clustering with a distance matrix
Anony-Mousse's answer has some nice suggestions for algorithms you could use to cluster given the distance matrix.
We used Spectral Clustering with left_x, right_x, top_y, bottom_y coordinates as features with pretty good results.
The problem is to find the closest features between two 2d concave polygons. The features can be vertex, edge. So result can be any combination of features. Is there any simple solution with complexity better than O(m*n)? where m, n - number of edges of polygons respectively. The polygons are coplanar.
An algorithm in O(n.log(m)) seems to exists, please see this paper, and this question.
An optimization of mine you could try: (not tested)
If your polygons are most of the time far apart enough, you can build the two convex hull and fall back on the easiest problem of finding the Hausdorff distance between two convex polygons (solution in O(n+m)). If the distance is 0, you have to fall back to the O(m.log(n)) case, but it's worth it if you are most of the time in the "convex hull" case with a positive distance.
Post-Scriptum. I just realized that in order of the postulate to work, you also need to check that the closest features from the convex hulls belongs to the original concave polygon. If not, it's easy to find a counter-example (imagine a polygon in shape of the letter C with another round just nearby: CO).
The updated postulate is then: The Hausdorff distance d between two concave polygons is the Hausdorff distance between their convex hulls, if d > 0, and both closest features are part of the original polygons.
The proof of this is left as an exercice to the reader.
Is there an algorithm that gives me coordinates of vertices in a graph, when I give him a weighted graph and the edge weights between vertices points to the distance between vertices?
Something like:
public _ArrayOfCoordinatesForVertices_ **super_hyper_algorithm**(weighted_graph){
return _foo_;
}
This is in general not possible: Imagine a graph with 3 nodes n1, n2, and n3.
now consider the following distances:
n1-n2: 4
n1-n3: 1
n2-n3: 1
(This violates the triangle inquality).
What you are referring to is called Multidimensional scaling (MDS) and you should find plenty of implementations now you know how to search for it.
Like others said, to some extent, it is impossible to draw a perfect graph without violating some of your constraints (the distances between points). MDS algorithms are specifically targeted at minimizing such violations.
If the graph is drawn in the Euclidean Space you can't do that, because, as pointed out in this answer you could violate the Triangle Inequality.
Usually, you can visually represent edges' weights by using different colour (i.e. by mapping weights to a colour-map), or by using different thickness of the edges (i.e. by mapping weights to a thickness scale).
OK i have found a library for python and it creates a graph image for me :) and i can give weights for edges like attribute: Weight of edge. In dot, the heavier the weight, the shorter, straighter and more vertical the edge is.