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.
I would like to check the intersection of two squares which are not axis-aligned. I know how to do it for axis-aligned squares. Can I extend the same idea?
Basically I want to distribute squares by gaussian distribution on the x-y plane in +ve quadrant say, but two squares should not be intersecting so I have to shift the original center of the square. Suggestions are welcome.
Separating axis theorem (link2) is suitable for effective checking of convex polygon intersection. For squares prerequisite (normals and so on) calculation becomes especially simple.
After almost 4-5 hours of thinking, I got one click. Just sharing if anyone needs this. PseudoCode
Input: Square1 and Square2
Bool(Sqaure1,square2)
for vertex v0 to v3 of sqaure 1
angle = 0;
for vertex P0 to P3 of square 2 (In anticlockwise of clockwise way)
angle = angle + angle(Pj,vi,P((j+1)%4));
if(abs(angle-360) == 0)
return true;
return false;
IDEA: point of one square inside or on the line of another square will have angle sum of 360 with all points if two squares intersects.So you can call function twice by swapping arguments and if anyone returns true then answer is yes.(Call twice or check if angle sum is 0 or 360 exactly)
If I am right, you can proceed as follows:
rotate both Sa and Sb so that Sa becomes axis-aligned;
check overlap of Sa' with the bounding box of Sb';
rotate both Sa and Sb so that Sb becomes axis-aligned;
check overlap of the bounding box of Sa" with Sb".
If there is a configuration with no overlap, then the squares are disjoint.
In analytical terms, the no-overlap conditions will have a form like
C1 + C2 (|cos(t1+t2)| + |sin(t1+t2)|) < 2D Min(|cos(t1)|, (|sin(t1)|)
where C1, C2 are the sides, D the distance between centers and t1, t2 are the angles between sides and the center line, and similar by exchange of 1/2.
I am looking for an algorithm that, given two rectangles that overlap partially or totally, finds the ordered list of vertexes that defines the polygon representing the sum of the two rectangles.
To be more specific:
I have as input two ordered list of points, representing the two rectangles
I know how to find the vertexes of the resulting polygon, which is formed by the vertexes of each rectangle which are outside the other rectangle, plus the intersection points between each edge of one rectangle with each edge of the other
I don't currently know how to order into an array the points, obtained as explained above, so that the element j and j+1 of the array represents the two vertexes of the same edge (this is what I mean by ordered list of vertexes).
Thanks in advance for any help
UPDATE :
I found a way to sort vertexes to obtain a polygon, as follows:
compute the vertexes centroid ( coords average )
sort the vertexes by the angle formed between the segment from the centroid and the vertex and any reference line passing by the centroid (e.g. the X axis ).
However, although I consistently obtain a polygon enclosing the two rectangles, without holes or intersecting edges, it is not always the polygon I want ( sometimes it includes extra area not belonging to one of the input rectangles ).
So I'm going back to the solution pointed in one of the comments, which is also described here:
polygon union without holes
Once you have the 4 vertices, you merely find the farther point using the distance formula (since it seems we can't make the assumption of a collinear or unrotated beginning rects)
So if you have points a = (xA, yA), b, c, d and you know these 4 points make a rectangle
float dist(Point a, Point b){
float dx = a.x - b.x;
float dy = a.y - b.y;
return Math.sqrt(dx * dx + dy * dy);
}
//somewhere else, where u need it
//put point A into index 0
Point curFarthest = b;
float distance = dist(a, b);
if (dist(a, c) > distance){
curFarther = c;
distance = dist(a, c);
} else if (dist(a, d) > distance){
curFarther = d;
curFarthest = dist(a, d);
}
//store curFarthest into index 2
// store the rest (exculding points a and curFarthest)
// into index 1 and 3 in no particular order
I am working on the same problem but I use a different approach(work still in progress).
Find the intersection points.
Distance of each point(vertices) with its neighboring connected points.
Using Dinics Algorithm find the Maxmimum flow.
Note: there will be a few special cases. But then again my problems revolves around polygons having 1 common point(vertice).
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);
}