I've read this article and try to use it with my project whitch is a point to several polygons shortest distance.
https://medium.com/analytics-vidhya/calculating-distances-from-points-to-polygon-borders-in-python-a-paris-example-3b597e1ea291
enter image description here
but it says"In order to keep our algorithm lean, let’s not account for these specific cases and always calculate the distance to the middle point of our lines. Especially in accurately defined polygons (on a small space), the differences are negligible."
So now it won't work if I want to know the shortest distance,and it'll sometimes got the wrong vertex(I'll figure out later) is there any solutions to know the polygon and the points shortest distance?
btw I'm not in English so sorry for the grammer...
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.
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.
For example the light source is coming from 1,3,-5 and object is at 4,-2,-1.
Algebraic formula is going to give the answer as 3. [1,3-5].[4,-2,-1]
= 1*4 + 3*-2 + -5*-1 = 3
But what does this 3 means? How do I know if my object is shaded with this number 3? Or is there more to it? I did look around and unable to find anything conclusive. Would be great if someone could give some insight. Thank you.
Judging from answers, pondering if I am understanding my question wrong. I was trying to get my head around the following question:
For a point on a convex surface, with the normal n=(n1,n2,n3)and light
direction l = (l1,l2,l3), determine if the point can be seen by light
source.
Using a dot product between two points makes no sense. Essentially, a dot product gives a measure of how similar two vectors are. When applied to points, the value will be related to the similarity of the direction to the points from the origin, as well as their distance from it. That metric doesn't make much sense, as you found out with that '3'.
To determine the amount of illumination, you want to be using a dot product between the normalized vector of the direction from the surface to the light and the surface normal. The result will be a value from -1 to 1, which you can interpret as an illumination factor for simple gouraud shading. In pseudocode:
illumination = max(0, dot(normalize(lightPosition - positionOnSurface), surfaceNormal))
Determining if a light hits an object is an entirely different problem called occlusion, and not really something you express in as mathematical formula. It's about testing what objects are in the path from the light to your target object.
The dot product can tell you on what side of a line a point is. The triangle is formed by three lines. If you are on the same side of all three lines then you are inside the triangle. You can use three dot products to test for each of the three sides. See slide 23 on this link http://comp575.web.unc.edu/files/2010/09/06raytracing1.pdf.
To detect if a point is in a polygon, you project a line from the point, to infinity, and see how many of polygon's vertices it intersects with... simple enough. My problem is that if the ray intersects the polygon on one of the points, then it is counted as intersecting two segments, and considered outside the polygon. I altered my function to make it only count one of the segments when the ray intersects a point of the polygon, but there are cases where a line could intersect the point while still being outside as well. Take this image as an example:
If you assume the point in the top left is "infinity", and cast a ray to either of the other points, both intersect at a point of the polygon, and would count as intersecting the same number of vertices even though one is inside, and one is outside.
Is there a way to compensate for that, or do I just have to assume that those fringe cases won't pop up?
If the ray crosses a side exactly on a vertex, only count that side if the other vertex is above the ray. That will fix your corner case.
For example in the picture you posted, the lower ray crosses two sides of the square at the top-left vertex, but one side is above the ray and the other below, so that contributes 1 and the target point is found to be inside. The upper ray crosses two sides at the top-right vertex, both sides are below the ray, so they contribute 0 to the count and the target point is found to be outside.
Update:
I remembered reading an article which describes a technique for dealing with singular cases in general. Please read my other answer if interested.
While my first answer should do the trick for this simple problem, I can't help but mention that there exist general techniques for dealing with these kinds of special cases.
This article describes a technique for dealing with these kinds of issues in general. And one of the first examples they provide happens to be the algorithm you ask about!
The idea is to apply Automatic differentiation aka Dual numbers to compute symbolic perturbations.
By the way the same technique can also be used to avoid handling 0/0 as a special case in programs!
Here is the blog post I originally learned this from, it gives some great background to the technique, and the author blogs a lot about automatic differentiation (AD).
Despite appearances AD is a very practical technique especially in languages with good support for operator overloading (eg: C++, Haskell, Python ...) and I have used it in "real life" (industrial applications in C++).
Send ray in another direction.
If you try n+1 different directions (n is number of polygon points) one of them surely will not pass through any vertex.
This will simplify the code compared to consideration of corner cases.
Worst case becomes O(n)*CheckComplexity(n) which is likely O(n^2). If it's not acceptable, you can just sort all vertices by direction from the point to them and select middle of some interval. This will give O(n*log n).
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)