Related
Setup
Given some set of nodes within a convex hull, assume the domain contains one or more concave areas:
where blue dots are points, and the black line illustrates the domain. Assume the points are held as a 2D array points of length n, where n is the number of point-pairs.
Let us then triangulate the points, using something like the Delaunay method from scipy.spatial:
As you can see, one may experience the creation of triangles crossing through the domain.
Question
What is a good algorithmic approach to removing any triangles that span outside of the domain? Ideally but not necessarily, where the simplex edges still preserve the domain shape (i.e., no major gaps where triangles are removed).
Since my question is seeming to continue to get a decent amount of activity, I wanted to follow up with the application that I'm currently using.
Assuming that you have your boundary defined, you can use a ray casting algorithm to determine whether or not the polygon is inside the domain.
To do this:
Take the centroid of each polygon as C_i = (x_i,y_i).
Then, imagine a line L = [C_i,(+inf,y_i)]: that is, a line that spans east past the end of your domain.
For each boundary segment s_i in boundary S, check for intersections with L. If yes, add +1 to an internal counter intersection_count; else, add nothing.
After the count of all intersections between L and s_i for i=1..N are calculated:
if intersection_count % 2 == 0:
return True # triangle outside convex hull
else:
return False # triangle inside convex hull
If your boundary is not explicitly defined, I find it helpful to 'map' the shape onto an boolean array and use a neighbor tracing algorithm to define it. Note that this approach assumes a solid domain and you will need to use a more complex algorithm for domains with 'holes' in them.
Here is some Python code that does what you want.
First, building the alpha shape (see my previous answer):
def alpha_shape(points, alpha, only_outer=True):
"""
Compute the alpha shape (concave hull) of a set of points.
:param points: np.array of shape (n,2) points.
:param alpha: alpha value.
:param only_outer: boolean value to specify if we keep only the outer border or also inner edges.
:return: set of (i,j) pairs representing edges of the alpha-shape. (i,j) are the indices in the points array.
"""
assert points.shape[0] > 3, "Need at least four points"
def add_edge(edges, i, j):
"""
Add a line between the i-th and j-th points,
if not in the list already
"""
if (i, j) in edges or (j, i) in edges:
# already added
assert (j, i) in edges, "Can't go twice over same directed edge right?"
if only_outer:
# if both neighboring triangles are in shape, it's not a boundary edge
edges.remove((j, i))
return
edges.add((i, j))
tri = Delaunay(points)
edges = set()
# Loop over triangles:
# ia, ib, ic = indices of corner points of the triangle
for ia, ib, ic in tri.vertices:
pa = points[ia]
pb = points[ib]
pc = points[ic]
# Computing radius of triangle circumcircle
# www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-circumcircle
a = np.sqrt((pa[0] - pb[0]) ** 2 + (pa[1] - pb[1]) ** 2)
b = np.sqrt((pb[0] - pc[0]) ** 2 + (pb[1] - pc[1]) ** 2)
c = np.sqrt((pc[0] - pa[0]) ** 2 + (pc[1] - pa[1]) ** 2)
s = (a + b + c) / 2.0
area = np.sqrt(s * (s - a) * (s - b) * (s - c))
circum_r = a * b * c / (4.0 * area)
if circum_r < alpha:
add_edge(edges, ia, ib)
add_edge(edges, ib, ic)
add_edge(edges, ic, ia)
return edges
To compute the edges of the outer boundary of the alpha shape use the following example call:
edges = alpha_shape(points, alpha=alpha_value, only_outer=True)
Now, after the edges of the outer boundary of the alpha-shape of points have been computed, the following function will determine whether a point (x,y) is inside the outer boundary.
def is_inside(x, y, points, edges, eps=1.0e-10):
intersection_counter = 0
for i, j in edges:
assert abs((points[i,1]-y)*(points[j,1]-y)) > eps, 'Need to handle these end cases separately'
y_in_edge_domain = ((points[i,1]-y)*(points[j,1]-y) < 0)
if y_in_edge_domain:
upper_ind, lower_ind = (i,j) if (points[i,1]-y) > 0 else (j,i)
upper_x = points[upper_ind, 0]
upper_y = points[upper_ind, 1]
lower_x = points[lower_ind, 0]
lower_y = points[lower_ind, 1]
# is_left_turn predicate is evaluated with: sign(cross_product(upper-lower, p-lower))
cross_prod = (upper_x - lower_x)*(y-lower_y) - (upper_y - lower_y)*(x-lower_x)
assert abs(cross_prod) > eps, 'Need to handle these end cases separately'
point_is_left_of_segment = (cross_prod > 0.0)
if point_is_left_of_segment:
intersection_counter = intersection_counter + 1
return (intersection_counter % 2) != 0
On the input shown in the above figure (taken from my previous answer) the call is_inside(1.5, 0.0, points, edges) will return True, whereas is_inside(1.5, 3.0, points, edges) will return False.
Note that the is_inside function above does not handle degenerate cases. I added two assertions to detect such cases (you can define any epsilon value that fits your application). In many applications this is sufficient, but if not and you encounter these end cases, they need to be handled separately.
See, for example, here on robustness and precision issues when implementing geometric algorithms.
One of Classic DT algorithms generates first a bounding triangle, then adds all new triangles sorted by x, then prunes out all triangles having a vertex in the supertriangle.
At least from the provided image one can derive the heuristics of pruning out also some triangles having all vertices on the concave hull. Without a proof, the triangles to be pruned out have a negative area when their vertices are sorted in the same order as the concave hull is defined.
This may need the concave hull to be inserted as well, and to be pruned out.
Since my question is seeming to continue to get a decent amount of activity, I wanted to follow up with the application that I'm currently using.
Assuming that you have your boundary defined, you can use a ray casting algorithm to determine whether or not the polygon is inside the domain.
To do this:
Take the centroid of each polygon as C_i = (x_i,y_i).
Then, imagine a line L = [C_i,(+inf,y_i)]: that is, a line that spans east past the end of your domain.
For each boundary segment s_i in boundary S, check for intersections with L. If yes, add +1 to an internal counter intersection_count; else, add nothing.
After the count of all intersections between L and s_i for i=1..N are calculated:
if intersection_count % 2 == 0:
return True # triangle outside convex hull
else:
return False # triangle inside convex hull
If your boundary is not explicitly defined, I find it helpful to 'map' the shape onto an boolean array and use a neighbor tracing algorithm to define it. Note that this approach assumes a solid domain and you will need to use a more complex algorithm for domains with 'holes' in them.
You can try a constrained delaunay algorithm for example with sloan algoritm or cgal library.
[1] A Brute-Force Constrained Delaunay Triangulation?
A simple but elegant way is to loop over the triangels and check wether they are within our domain or not. The shapely package could do the trick for you.
for more on this please check the following post: https://gis.stackexchange.com/a/352442
Note that triangulation in shapely is also implemented, even for MultiPoin objects.
I used it, the performance was amazing and the code was only like five lines.
Compute the triangles centroid an check if it's inside the polygon using this algorithm.
Hi sorry for the confusing title.
I'm trying to make a race track using points. I want to draw 3 rectangles which form my roads. However I don't want these rectangles to overlap, I want to leave an empty space between them to place my corners (triangles) meaning they only intersect at a single point. Since the roads have a common width I know the width of the rectangles.
I know the coordinates of the points A, B and C and therefore their length and the angles between them. From this I think I can say that the angles of the yellow triangle are the same as those of the outer triangle. From there I can work out the lengths of the sides of the blue triangles. However I don't know how to find the coordinates of the points of the blue triangles or the length of the sides of the yellow triangle and therefore the rectangles.
This is an X-Y problem (asking us how to accomplish X because you think it would help you solve a problem Y better solved another way), but luckily you gave us Y so I can just answer that.
What you should do is find the lines that are the edges of the roads, figure out where they intersect, and proceed to calculate everything else from that.
First, given 2 points P and Q, we can write down the line between them in parameterized form as f(t) = P + t(Q - P). Note that Q - P = v is the vector representing the direction of the line.
Second, given a vector v = (x_v, y_v) the vector (y_v, -x_v) is at right angles to it. Divide by its length sqrt(x_v**2 + y_v**2) and you have a unit vector at right angles to the first. Project P and Q a distance d along this vector, and you've got 2 points on a parallel line at distance d from your original line.
There are two such parallel lines. Given a point on the line and a point off of the line, the sign of the dot product of your normal vector with the vector between those two lines tells you whether you've found the parallel line on the same side as the other, or on the opposite side.
You just need to figure out where they intersect. But figuring out where lines P1 + t*v1 and P2 + s*v2 intersect can be done by setting up 2 equations in 2 variables and solving that. Which calculation you can carry out.
And now you have sufficient information to calculate the edges of the roads, which edges are inside, and every intersection in your diagram. Which lets you figure out anything else that you need.
Slightly different approach with a bit of trigonometry:
Define vectors
b = B - A
c = C - A
uB = Normalized(b)
uC = Normalized(c)
angle
Alpha = atan2(CrossProduct(b, c), DotProduct(b,c))
HalfA = Alpha / 2
HalfW = Width / 2
uB_Perp = (-uB.Y, ub.X) //unit vector, perpendicular to b
//now calculate points:
P1 = A + HalfW * (uB * ctg(HalfA) + uB_Perp) //outer blue triangle vertice
P2 = A + HalfW * (uB * ctg(HalfA) - uB_Perp) //inner blue triangle vertice, lies on bisector
(I did not consider extra case of too large width)
Given a rectangle defined by 4 points is there an algorithm to shift a point so that it lies inside the rectangle moving it the shortest distance.
I should have clarified the target rectangle has an arbitrary rotation. Also I am not so concerned with the translation vector I simply want to know which point within a arbitrary rectangle is closest to said point
The shortest distance to move the point to the rectangle will either move the point to one of the corners or one of the sides (assuming the point isn't already inside the rectangle which you could test for like a bounding box). So what you want to do is find the distance from the point to each of the segments representing the sides. This is a point-segment distance and there are lots of places to find examples online.
Here's a clean example that only returns distance, but I've highlighted the three geometric cases of 1. Nearest to first point on segment 2. Nearest to second point on segment 3. Nearest to the side.
float dist_Point_to_Segment( Point P, Segment S)
{
Vector v = S.P1 - S.P0;
Vector w = P - S.P0;
double c1 = dot(w,v); // Point is closest to first point on segment
if ( c1 <= 0 )
return distance(P, S.P0);
double c2 = dot(v,v); // Point is closest to second point on segment
if ( c2 <= c1 )
return distance(P, S.P1);
double b = c1 / c2; // Point is closest to a side
Point Pb = S.P0 + b * v;
return distance(P, Pb);
}
So you just have to do the above for each of the four sides of the rectangle. You'll want to identify which case (1, 2, or 3) gave you the closest point. That will give you the translation vector to the rectangle.
I should mention, there are optimizations you can make if you want to speed up your implementation. For example, if you test two adjacent sides and you find that the closest point for both is the same vertex, then you can stop there because that must be the closest point on the rectangle to the point.
Assuming that the polygon does not self-intersect, what would be the most efficient way to do this? The polygon has N vertices.
I know that it can be calculated with the coordinates but is there another general way?
The signed area, A(T), of the triangle T = ((x1, y1), (x2, y2), (x3, y3)) is defined to be 1/2 times the determinant of the following matrix:
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|
The determinant is -y1*x2 + x1*y2 + y1*x3 - y2*x3 - x1*y3 + x2*y3.
Given a polygon (convex or concave) defined by the vertices p[0], p[1], ..., p[N - 1], you can compute the area of the polygon as follows.
area = 0
for i in [0, N - 2]:
area += A((0, 0), p[i], p[i + 1])
area += A((0, 0), p[N - 1], p[0])
area = abs(area)
Using the expression for the determinant above, you can compute A((0, 0), p, q) efficiently as 0.5 * (-p.y*q.x + p.x*q.y). A further improvement is to do the multiplication by 0.5 only once:
area = 0
for i in [0, N - 2]:
area += -p[i].y * p[i+1].x + p[i].x * p[i+1].y
area += -p[N-1].y * p[0].x + p[N-1].x * p[0].y
area = 0.5 * abs(area)
This is a linear time algorithm, and it is trivial to parallelize. Note also that it is an exact algorithm when the coordinates of your vertices are all integer-valued.
Link to Wikipedia article on this algorithm
The best way to approach this problem that I can think of is to consider the polygon as several triangles, find their areas separately, and sum them for the total area. All polygons, regular, or irregular, are essentially just a bunch of triangle (cut a quadrilateral diagonally to make two triangles, a pentagon in two cuts from one corner to the two most opposite ones, and the pattern continues on). This is quite simple to put to code.
A general algorithm for this can be coded as follows:
function polygonArea(Xcoords, Ycoords) {
numPoints = len(Xcoords)
area = 0; // Accumulates area in the loop
j = numPoints-1; // The last vertex is the 'previous' one to the first
for (i=0; i<numPoints; i++)
{ area = area + (Xcoords[j]+Xcoords[i]) * (Ycoords[j]-Ycoords[i]);
j = i; //j is previous vertex to i
}
return area/2;
}
Xcoords and Ycoords are arrays, where Xcoords stores the X coordinates, and Ycoords the Y coordinates.
The algorithm iteratively constructs the triangles from previous vertices.
I modified this from the algorithm provided Here by Math Open Ref
It should be relatively painless to adapt this to whatever form you are storing your coordinates in, and whatever language you are using for your project.
The "Tear one ear at a time" algorithm works, provided the triangle you remove does not contain "holes" (other vertices of the polygon).
That is, you need to choose the green triangle below, not the red one:
However, it is always possible to do so (Can't prove it mathematically right now, but you'l have to trust me). You just need to walk the polygon's vertices and perform some inclusion tests until you find a suitable triple.
Source: I once implemented a triangulation of arbitrary, non-intersecting polygons based on what I read in Computational Geometry in C by Joseph O'Rourke.
Take 3 consecutive points from the polygon.
Calculate the area of the resulting triangle.
Remove the middle of the 3 points from the polygon.
Do a test to see if the removed point is inside the remaining polygon or not. If it's inside subtract the triangle area from the total, otherwise add it.
Repeat until the polygon consists of a single triangle, and add that triangle's area to the total.
Edit: to solve the problem given by #NicolasMiari simply make two passes, on the first pass only process the vertices that are inside the remainder polygon, on the second pass process the remainder.
Actually this is a classic problem as SO user Victor put it (in another SO question regarding which tasks to ask during an interview).
I couldn't do it in an hour (sigh) so what is the algorithm that calculates the number of integer points within a triangle?
EDIT: Assume that the vertices are at integer coordinates. (otherwise it becomes a problem of finding all points within the triangle and then subtracting all the floating points to be left with only the integer points; a less elegant problem).
Assuming the vertices are at integer coordinates, you can get the answer by constructing a rectangle around the triangle as explained in Kyle Schultz's An Investigation of Pick's Theorem.
For a j x k rectangle, the number of interior points is
I = (j – 1)(k – 1).
For the 5 x 3 rectangle below, there are 8 interior points.
(source: uga.edu)
For triangles with a vertical leg (j) and a horizontal leg (k) the number of interior points is given by
I = ((j – 1)(k – 1) - h) / 2
where h is the number of points interior to the rectangle that are coincident to the hypotenuse of the triangles (not the length).
(source: uga.edu)
For triangles with a vertical side or a horizontal side, the number of interior points (I) is given by
(source: uga.edu)
where j, k, h1, h2, and b are marked in the following diagram
(source: uga.edu)
Finally, the case of triangles with no vertical or horizontal sides can be split into two sub-cases, one where the area surrounding the triangle forms three triangles, and one where the surrounding area forms three triangles and a rectangle (see the diagrams below).
The number of interior points (I) in the first sub-case is given by
(source: uga.edu)
where all the variables are marked in the following diagram
(source: uga.edu)
The number of interior points (I) in the second sub-case is given by
(source: uga.edu)
where all the variables are marked in the following diagram
(source: uga.edu)
Pick's theorem (http://en.wikipedia.org/wiki/Pick%27s_theorem) states that the surface of a simple polygon placed on integer points is given by:
A = i + b/2 - 1
Here A is the surface of the triangle, i is the number of interior points and b is the number of boundary points. The number of boundary points b can be calculated easily by summing the greatest common divisor of the slopes of each line:
b = gcd(abs(p0x - p1x), abs(p0y - p1y))
+ gcd(abs(p1x - p2x), abs(p1y - p2y))
+ gcd(abs(p2x - p0x), abs(p2y - p0y))
The surface can also be calculated. For a formula which calculates the surface see https://stackoverflow.com/a/14382692/2491535 . Combining these known values i can be calculated by:
i = A + 1 - b/2
My knee-jerk reaction would be to brute-force it:
Find the maximum and minimum extent of the triangle in the x and y directions.
Loop over all combinations of integer points within those extents.
For each set of points, use one of the standard tests (Same side or Barycentric techniques, for example) to see if the point lies within the triangle. Since this sort of computation is a component of algorithms for detecting intersections between rays/line segments and triangles, you can also check this link for more info.
This is called the "Point in the Triangle" test.
Here is an article with several solutions to this problem: Point in the Triangle Test.
A common way to check if a point is in a triangle is to find the vectors connecting the point to each of the triangle's three vertices and sum the angles between those vectors. If the sum of the angles is 2*pi (360-degrees) then the point is inside the triangle, otherwise it is not.
Ok I will propose one algorithm, it won't be brilliant, but it will work.
First, we will need a point in triangle test. I propose to use the "Barycentric Technique" as explained in this excellent post:
http://www.blackpawn.com/texts/pointinpoly/default.html
Now to the algorithm:
let (x1,y1) (x2,y2) (x3,y3) be the triangle vertices
let ymin = floor(min(y1,y2,y3)) ymax = ceiling(max(y1,y2,y3)) xmin = floor(min(x1,x2,x3)) ymax = ceiling(max(x1,x2,3))
iterating from xmin to xmax and ymin to ymax you can enumerate all the integer points in the rectangular region that contains the triangle
using the point in triangle test you can test for each point in the enumeration to see if it's on the triangle.
It's simple, I think it can be programmed in less than half hour.
I only have half an answer for a non-brute-force method. If the vertices were integer, you could reduce it to figuring out how to find how many integer points the edges intersect. With that number and the area of the triangle (Heron's formula), you can use Pick's theorem to find the number of interior integer points.
Edit: for the other half, finding the integer points that intersect the edge, I suspect that it's the greatest common denominator between the x and y difference between the points minus one, or if the distance minus one if one of the x or y differences is zero.
Here's another method, not necessarily the best, but sure to impress any interviewer.
First, call the point with the lowest X co-ord 'L', the point with the highest X co-ord 'R', and the remaining point 'M' (Left, Right, and Middle).
Then, set up two instances of Bresenham's line algorithm. Parameterize one instance to draw from L to R, and the second to draw from L to M. Run the algorithms simultaneously for X = X[L] to X[M]. But instead of drawing any lines or turning on any pixels, count the pixels between the lines.
After stepping from X[L] to X[M], change the parameters of the second Bresenham to draw from M to R, then continue to run the algorithms simultaneously for X = X[M] to X[R].
This is very similar to the solution proposed by Erwin Smout 7 hours ago, but using Bresenham instead of a line-slope formula.
I think that in order to count the columns of pixels, you will need to determine whether M lies above or below the line LR, and of course special cases will arise when two points have the same X or Y co-ordinate. But by the time this comes up, your interviewer will be suitably awed and you can move on to the next question.
Quick n'dirty pseudocode:
-- Declare triangle
p1 2DPoint = (x1, y1);
p2 2DPoint = (x2, y2);
p3 2DPoint = (x3, y3);
triangle [2DPoint] := [p1, p2, p3];
-- Bounding box
xmin float = min(triangle[][0]);
xmax float = max(triangle[][0]);
ymin float = min(triangle[][1]);
ymax float = max(triangle[][1]);
result [[float]];
-- Points in bounding box might be inside the triangle
for x in xmin .. xmax {
for y in ymin .. ymax {
if a line starting in (x, y) and going in any direction crosses one, and only one, of the lines between the points in the triangle, or hits exactly one of the corners of the triangle {
result[result.count] = (x, y);
}
}
}
I have this idea -
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle. Let 'count' be the number of integer points forming the triangle.
If we need the points on the triangle edges then using Euclidean Distance formula http://en.wikipedia.org/wiki/Euclidean_distance, the length of all three sides can be ascertained.
The sum of length of all three sides - 3, would give that count.
To find the number of points inside the triangle we need to use a triangle fill algorithm and instead of doing the actual rendering i.e. executing drawpixel(x,y), just go through the loops and keep updating the count as we loop though.
A triangle fill algorithm from
Fundamentals of Computer Graphics by
Peter Shirley,Michael Ashikhmin
should help. Its referred here http://www.gidforums.com/t-20838.html
cheers
I'd go like this :
Take the uppermost point of the triangle (the one with the highest Y coordinate). There are two "slopes" starting at that point. It's not the general solution, but for easy visualisation, think of one of both "going to the left" (decreasing x coordinates) and the other one "going to the right".
From those two slopes and any given Y coordinate less than the highest point, you should be able to compute the number of integer points that appear within the bounds set by the slopes. Iterating over decreasing Y coordinates, add all those number of points together.
Stop when your decreasing Y coordinates reach the second-highest point of the triangle.
You have now counted all points "above the second-highest point", and you are now left with the problem of "counting all the points within some (much smaller !!!) triangle, of which you know that its upper side parallels the X-axis.
Repeat the same procedure, but now with taking the "leftmost point" instead of the "uppermost", and with proceedding "by increasing x", instead of by "decreasing y".
After that, you are left with the problem of counting all the integer points within a, once again much smaller, triangle, of which you know that its upper side parallels the X-axis, and its left side parallels the Y-axis.
Keep repeating (recurring), until you count no points in the triangle you're left with.
(Have I now made your homework for you ?)
(wierd) pseudo-code for a bit-better-than-brute-force (it should have O(n))
i hope you understand what i mean
n=0
p1,p2,p3 = order points by xcoordinate(p1,p2,p3)
for int i between p1.x and p2.x do
a = (intersection point of the line p1-p2 and the line with x==i).y
b = (intersection point of the line p1-p3 and the line with x==i).y
n += number of integers between floats (a, b)
end
for i between p2.x+1 and p3.x do
a = (intersection point of the line p2-p3 and the line with x==i).y
b = (intersection point of the line p1-p3 and the line with x==i).y
n += number of integers between floats (a, b)
end
this algorithm is rather easy to extend for vertices of type float (only needs some round at the "for i.." part, with a special case for p2.x being integer (there, rounded down=rounded up))
and there are some opportunities for optimization in a real implementation
Here is a Python implementation of #Prabhala's solution:
from collections import namedtuple
from fractions import gcd
def get_points(vertices):
Point = namedtuple('Point', 'x,y')
vertices = [Point(x, y) for x, y in vertices]
a, b, c = vertices
triangle_area = abs((a.x - b.x) * (a.y + b.y) + (b.x - c.x) * (b.y + c.y) + (c.x - a.x) * (c.y + a.y))
triangle_area /= 2
triangle_area += 1
interior = abs(gcd(a.x - b.x, a.y - b.y)) + abs(gcd(b.x - c.x, b.y - c.y)) + abs(gcd(c.x - a.x, c.y - a.y))
interior /= 2
return triangle_area - interior
Usage:
print(get_points([(-1, -1), (1, 0), (0, 1)])) # 1
print(get_points([[2, 3], [6, 9], [10, 160]])) # 289
I found a quite useful link which clearly explains the solution to this problem. I am weak in coordinate geometry so I used this solution and coded it in Java which works (at least for the test cases I tried..)
Link
public int points(int[][] vertices){
int interiorPoints = 0;
double triangleArea = 0;
int x1 = vertices[0][0], x2 = vertices[1][0], x3 = vertices[2][0];
int y1 = vertices[0][1], y2 = vertices[1][1], y3 = vertices[2][1];
triangleArea = Math.abs(((x1-x2)*(y1+y2))
+ ((x2-x3)*(y2+y3))
+ ((x3-x1)*(y3+y1)));
triangleArea /=2;
triangleArea++;
interiorPoints = Math.abs(gcd(x1-x2,y1-y2))
+ Math.abs(gcd(x2-x3, y2-y3))
+ Math.abs(gcd(x3-x1, y3-y1));
interiorPoints /=2;
return (int)(triangleArea - interiorPoints);
}