Given a point's coordinates, how can I determine if it is within an arbitrary shape?
The shape is defined by an array of points, I do not know where the shape is 'closed', the part I really need help is to work out where the shape is closed.
Here's an image to illustrate what I mean a little better:
Easiest way to do it is cast a ray from that point and count how many times it crosses the boundary. If it is odd, the point is inside, even the point is outside.
Wiki: http://en.wikipedia.org/wiki/Point_in_polygon
Note that this only works for manifold shapes.
If you want to determine whether or not a point P is in an arbitrary shape, I would simply run a flood fill starting at P. If your flood fill leaves a pre-determined bounding box, you are outside the shape. Otherwise if your flood fill terminates, then you're within the shape :)
I believe this algorithm is O(N^2) where N is the number of points, since the maximum area is proportional to N^2.
Wikipedia: Flood Fill
Actually, if you are given an array of points, you can check the closeness of the shape as follows:
Consider pairs of points P[i] and P[i+1] given in the array - they form some segment of the border of your shape. What you need to check is if there exist two such segments that intersect, which can be checked in O(N^2) time (just by checking all possible pairs of such segments). If there exists an intersection, that means that your shape is closed.
Note: you must be attentive not to forget to check the segment P[0],P[n-1] either (i.e. first and last points in the array).
Related
I am reviewing some MATLAB code that is publicly available at the following location:
https://github.com/mattools/matGeom/blob/master/matGeom/geom2d/orientedBox.m
This is an implementation of the rotating calipers algorithm on the convex hull of a set of points in order to compute an oriented bounding box. My review was to understand intuitively how the algorithm works however I seek clarification on certain lines within the file which I am confused on.
On line 44: hull = bsxfun(#minus, hull, center);. This appears to translate all the points within the convex hull set so the calculated centroid is at (0,0). Is there any particular reason why this is performed? My only guess would be that it allows straightforward rotational transforms later on in the code, as rotating about the real origin would cause significant problems.
On line 71 and 74: indA2 = mod(indA, nV) + 1; , indB2 = mod(indB, nV) + 1;. Is this a trick in order to prevent the access index going out of bounds? My guess is to prevent out of bounds access, it will roll the index over upon reaching the end.
On line 125: y2 = - x * sit + y * cot;. This is the correct transformation as the code behaves properly, but I am not sure why this is actually used and different from the other rotational transforms done later and also prior (with the calls to rotateVector). My best guess is that I am simply not visualizing what rotation needs to be done in my head correctly.
Side note: The external function calls vectorAngle, rotateVector, createLine, and distancePointLine can all be found under the same repository, in files named after the function name (as per MATLAB standard). They are relatively uninteresting and do what you would expect aside from the fact that there is normalization of vector angles going on.
I'm the author of the above piece of code, so I can give some explanations about it:
First of all, the algorithm is indeed a rotating caliper algorithm. In the current implementation, only the width of the algorithm is tested (I did not check the west and est vertice). Actually, it seems the two results correspond most of the time.
Line 44 -> the goal of translating to origin was to improve numerical accuracy. When a polygon is located far away from the origin, coordinates may be large, and close together. Many computation involve products of coordinates. By translating the polygon around the origin, the coordinates are smaller, and the precision of the resulting products are expected to be improved. Well, to be honest, I did not evidenced this effect directly, this is more a careful way of coding than a fix…
Line 71-74! Yes. The idea is to find the index of the next vertex along the polygon. If the current vertex is the last vertex of the polygon, then the next vertex index should be 1. The use of modulo rescale between 0 and N-1. The two lines ensure correct iteration.
Line 125: There are several transformations involved. Using the rotateVector() function, one simply computes the minimal with for a given edge. On line 125, one rotate the points (of the convex hull) to align with the “best” direction (the one that minimizes the width). The last change of coordinates (lines 132->140) is due to the fact that the center of the oriented box is different from the centroid of the polygon. Then we add a shift, which is corrected by the rotation.
I did not really look at the code, this is an explanation of how the rotating calipers work.
A fundamental property is that the tightest bounding box is such that one of its sides overlaps an edge of the hull. So what you do is essentially
try every edge in turn;
for a given edge, seen as being horizontal, south, find the farthest vertices north, west and east;
evaluate the area or the perimeter of the rectangle that they define;
remember the best area.
It is important to note that when you switch from an edge to the next, the N/W/E vertices can only move forward, and are readily found by finding the next decrease of the relevant coordinate. This is how the total processing time is linear in the number of edges (the search for the initial N/E/W vertices takes 3(N-3) comparisons, then the updates take 3(N-1)+Nn+Nw+Ne comparisons, where Nn, Nw, Ne are the number of moves from a vertex to the next; obviously Nn+Nw+Ne = 3N in total).
The modulos are there to implement the cyclic indexing of the edges and vertices.
As you may have already noticed, in the newer versions of matlab the
boundary function (which computes the boundary for a set of 2d or 3d points) has been improved.
Now it is possible to give the function a parameter called 'shrink factor'. If the shrink factor is 0, then the boundary traced is the traditional convex hull. The boundary is more shrinked when the shrink parameter is bigger. The default value for the shrink factor is 0.5, in case you don't specify any value.
So, I understand its use and what it does (actually I've already used the function in a project), but I don't know how it works. What are the geometrical principles of this shrink factor?
Thanks!
Found your question while loooking for the answer myself. Hope you've solved it by now. I've figured it out and in case someone else finds this question, here's my understanding of the boundary() function.
The boundary function is an implementation of alpha shapes. Using alpha shapes, a set of points can be assigned a polygon by using a set of circles of a specific radius:
imagine an arbitrary shape drawn around the points and proceed to remove as much of this shape as possible using circles of a specific radius. Continue as long as possible, without enclosing any points. A small radius will mean more "material" can be removed, a larger radius means less "removal", i.e. a small radius creates a close cropped shape whereas an infinite radius recreates a convex hull of the set. The points determined to be edge points are then conencted with straight edges. This can create hollow areas inside the point set.
See e.g. http://doc.cgal.org/latest/Alpha_shapes_2/index.html
MATLAB has an alphashape() function which calculates alphashapes with all possible alpha radii giving different shapes. This is used in the boundary function.
boundary() workflow:
(1) Create alphashape
(2) Find critical alpha radius, needed to create a single region for alpha shape
(3) Extract all alphavalues that create unique shapes above this critical value
(4) Use the shrink factor, S, to select a single alpha value to use.
Example: with S=0.25, use alpha radius with index (1-.25)*numel(alphavalues>=alpha_crit). This creates an alpha shape
using the 75th smallest alpha radius giving rise to a single region
(for S=0.25).
If S=1 (max shrink), gives the lowest alpha-radius that gives a single
region for the alpha-shape.
If S=0 (no shrink), gives the maximum alpha-radius that gives a unique
shape. (Incraesing alpha radius further has no effect).
(5) set the threshold for filling in holes in the alphashape to be the same as the alphashape's area, i.e. fill in all holes
(6) Return the indices of the original point cloud correspocing to the vertices of this alphashape.
The relevant section of the boundary.m file (lines 79-86)
Acrit = shp.criticalAlpha('one-region'); %alpha-radius required for single region
spec = shp.alphaSpectrum();%all alphavalues
idx = find(spec==Acrit);
subspec = spec(1:idx);%alphavalues up to criticalAlpha
subspec = flipud(subspec);%reverse order
idx = max(ceil((1-S)*numel(subspec)),1); %find index from shrink factor
alphaval = subspec(idx);
shp.Alpha = alphaval; %set alpha value of alpha shape
shp.HoleThreshold = areavol; % remove holes in interior
Hope this is clear enough and useful to someone.
I use MATLAB R2014b
YiraDati's answer provides great details.
You can also type "open boundary" in command windows, and then all procedures are written in boundary function. And all subfunctions shown in boundary function are accessible using matlab documentation, like area(), criticalAlpha(), alphaSpectrum(), etc...
The question is
a.write a function which finds the circle with the minimal area s.t it bounds a given list of points (use fminsearch and give appropriate plot).
b.If you managed do the same for sphere (find one with minimal volume)
What I've tried so far:
%%Main function
function minarea= mincircle(points)
maxx=max(points(1,:));
maxy=max(points(2m:));
radius=max(maxx,maxy);
minarea=fminsearch(#(x) circle(x,r,c),[0,0])
end
%%This function is supposed to give equalation of circle
function eq=circle(x,r,c)
eq=(x(1)-c(1)).^2+(x(2)-c(2)).^2 %=r?
% and here I don't know how to insert r:
end`
For better understanding I'll attach a sketch.
In these terms I want to find the area of the circle whose center is in O
Note: I don't believe that the circle you drew is the smallest possible bounding circle. It should be a little smaller, up and to the right, and should touch at least two points on its perimeter.
Approaching the problem
We have a set of points, and we want to draw a circle that encompasses all of them. The problem is that you need three bits of information to define a circle: the X and Y coordinates of the circle's center, and the circle's radius. So the problem doesn't seem straightforward.
However, there is a related problem that is much easier to solve. Suppose the circle's center is fixed. From that point, we make a circle grow concentrically outwards so that it becomes bigger and bigger. At some point, the circle will encompass one of the points in our set. As it gets bigger, it will encompass a second point, and a third, until all the points in our set fall within our circle. Clearly, as soon as the last point in the set falls within our circle, we have the smallest possible circle that encompasses all the points, given that we started by fixing the center point of the circle.
Moreover, we can determine what the radius of this circle is. It is simply the maximum distance from any point in the set to the center of the circle, since we stop when the last point is touched by the perimeter of our expanding circle.
The next problem is to determine What is the best starting point to place the center of our circle? Clearly if the starting point is far away from all the points in our set, then the radius must be very large to even encompass one point in the set. Intuitively, it must be "in the middle" of our points somewhere. But where, exactly?
Using fminsearch
My suggestion is that you want to find the point P(x, y) that minimises how large you have to grow the circle to encompass all the points in the set. And we're in luck that we can use fminsearch to find P.
According to the fminsearch documentation, the function you pass in must be a function of one parameter (which may be an array), and it must return a scalar. The idea is that you want the output of your function to be as small as possible, and you want to find out what inputs will make that possible.
In our case, we want to write a function that outputs the size of our circle, given the center of the circle as input. That way, fminsearch will find the center of the smallest possible circle that will still encompass all the points. I'm going to write a function that outputs the radius required to encompass all the points given a center point P.
pointsX = [..]; % X-coordinates of points in the set
pointsY = [..]; % Y-coordinates of points in the set
function r = radiusFromPoint(P)
px = P(1);
py = P(2);
distanceSquared = (pointsX - px).^2 + (pointsY - py).^2;
r = sqrt(max(distanceSquared));
end
Then we want to use fminsearch to find the point that gives us the smallest radius. I've just naively used the origin (0, 0) as my starting estimate, but you may have a better idea (like using the first point in the set)
P0 = [0, 0]; % starting estimate
[P, radiusMin] = fminsearch(#radiusFromPoint, P0);
The circle is defined by its center at P and radius of radiusMin.
And I'll leave it to you to plot the output and generalize to the 3D case!
Actually, while you may need it to complete your homework assignment (I assume that is what this is) you don't really need to use an optimizer at all. The minboundcircle code posted with my minimal bounding tools does it without use of an optimizer. (There is also a minboundsphere tool.)
Regardless, you might find a few tricks in there that will be useful. At the very least, learn how to reduce the size of the problem (and so the speed of solution) by use of a convex hull. After all, it is only the points on the convex hull that can determine a minimal bounding circle. All other points are simply a waste of CPU time.
is it possible to find the center of the big black spot(the area which with the set?)
I've tried to loop through all points which are in the set, sum their locationד and eventually divided by the num of points which are in the set.
it didn't work as expected because the the set isn't formed, e.g its not a perfect sphere or a square therefore the center always changes. is there another way of finding the center?
thanks!
Look for centers of mandelbrot set hyperbolic components
http://fraktal.republika.pl/eigensolve.html
That shape is a perfect cardioid (no other similar shapes in the Mandelbrot set are perfect cardioids, they are somewhat distorted).
Check Wikipedia, you'll find the equation for calculating the cardioid itself: http://en.wikipedia.org/wiki/Cardioid
Note, however that at the borders of the Mandelbrot map (where the black and non-black area meets) deciding whether a point belongs to the map or not, heavily depends on how many times you iterate z = z^2 + c. If you iterate it 50 times, the main cardioid will be smaller than if you iterate it 500 times because with 500 iteration more points at the border will go to infinity.
It's usually popular to work with polygons with their vertices sorted CW or CCW in vectors(2*1 or 1*2 matrices). However, how to state polygons with holes in vectors?
I'm going to apply various process on these polygons, so I want a way of representing with which I could work easily or efficiently.(i.e how to state that kind of polygons in my program in order to ease my algorithms?)
polygons are 2D and I'm programming in MATLAB.
EDIT 1 : I'm going to calculate visibility graph of these polygons(with or without holes).
As others have mentioned, a polygon with holes can be represented as an exterior boundary, plus zero or more interior boundaries, all of which are mutually nonoverlapping*. If you use nonzero winding number to determine inside/outside, be sure to specify your interior boundaries in the opposite direction as the exterior boundaries (counterclockwise for exterior and clockwise for interior, or vice-versa) so that the contour integrals are zero inside the holes.
FYI, tis kind of definition/representation has been formalized in the OpenGIS Simple Features Specification (PDF).
As far as representation:
I'd probably have a cell array of K Nx2 matrices, where the first element in the cell array is the exterior boundary, and the remaining elements (if any) in the cell array are the interior boundaries. I would use a cell array because there may not be the same number of points on each boundary.
*nonoverlapping = except at individual points, e.g. a diamond inside a square:
You can break a polygon with a hole in it into two shapes without a hole. When you're doing contour integration in a complex plane, you can create a "cut" from one edge of the polygon that brings you to the edge of the hole; integrate around one side of the hole and back; then traverse around the other side for the second polygon. You end up with two path integrals along each cut that cancel each other out.
"visibility graph" - is this for a radiation view factor calculation with shading? Or a ray-tracing graphics algorithm?
A polygon, plus a list of polygonal holes. Just be sure the various polygons don't intersect.
What do you plan to do with this thing?
It sounds like each hole is just a polygon inside the polygon itself. Perhaps you could store a vector like you describe for the outer polygon, then a vector of more polygon vectors for the holes.
Presumably you'll want to have a tree structure if you want this to be as generic as possible (i.e. polygons with polygonal holes that have polygons inside them with holes inside that, ...). Matlab isn't really great at representing tree structures efficiently, but here's one idea...
Have a struct-array of polygons.
Each polygon is a struct with two fields, 'corners', and 'children'.
The 'corners' field contains a matrix of (x,y) coordinates of the corners, accessed as "data{polyIdx}.corners(:,cornerIdx)".
The 'children' field is a struct-array of polygons.
Here's an example of some code to make a triangle with bogus children that are holes (they aren't really valid though because they will likely overlap:
polygon = struct;
npoints = 3;
polygon.corners = rand(2,npoints);
polygon.children = struct;
nchildren = 5;
for c=1:nchildren
polygon.children(c).corners = rand(2,npoints);
polygon.children(c).children = struct;
end
You could continue to recursively define children that alternate between creating holes and filling them.
What exactly do you mean under "a visibility graph" ?
Two "full" poligons, two states possible, either +1 or -1.
If you're representing a hole, you've got one with state +1 and one with state -1, which represents a hole, resulting in state 0.
If you've got overlapping polygons, you'll end up with resultant state >1. Then you can calculate the borders of a new polygon.
If you've got two polygons with holes that intersect, then first calculate the state of a new polygon which consists of outer borders of the two old ones, then deal with holes.
Anyways, ... I think you get the general principle.
Have no idea how to do it in matlab, I used it only marginally so far, and even that for very simple things.