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I have a list of 2D points (x1,y1),(x2,y2)......(Xn,Yn) representing a curved segment, is there any formula to determine whether the direction of drawing that segment is clockwise or anti clockwise ?
any help is appreciated
Alternately, you can use a bit of linear algebra. If you have three points a, b, and c, in that order, then do the following:
1) create the vectors u = (b-a) = (b.x-a.x,b.y-a.y) and v = (c-b) ...
2) calculate the cross product uxv = u.x*v.y-u.y*v.x
3) if uxv is -ve then a-b-c is curving in clockwise direction (and vice-versa).
by following a longer curve along in the same manner, you can even detect when as 's'-shaped curve changes from clockwise to anticlockwise, if that is useful.
One possible approach. It should work reasonably well if the sampling of the line represented by your list of points is uniform and smooth enough, and if the line is sufficiently simple.
Subtract the mean to "center" the line.
Convert to polar coordinates to get the angle.
Unwrap the angle, to make sure its increments are meaningful.
Check if total increment is possitive or negative.
I'm assuming you have the data in x and y vectors.
theta = cart2pol(x-mean(x), y-mean(y)); %// steps 1 and 2
theta = unwrap(theta); %// step 3
clockwise = theta(end)<theta(1); %// step 4. Gives 1 if CW, 0 if ACW
This only considers the integrated effect of all points. It doesn't tell you if there are "kinks" or sections with different directions of turn along the way.
A possible improvement would be to replace the average of x and y by some kind of integral. The reason is: if sampling is denser in a region the average will be biased towards that, whereas the integral wouldn't.
Now this is my approach, as mentioned in a comment to the question -
Another approach: draw a line from starting point to ending point. This line is indeed a vector. A CW curve has most of its part on RHS of this line. For CCW, left.
I wrote a sample code to elaborate this idea. Most of the explanation can be found in comments in the code.
clear;clc;close all
%% draw a spiral curve
N = 30;
theta = linspace(0,pi/2,N); % a CCW curve
rho = linspace(1,.5,N);
[x,y] = pol2cart(theta,rho);
clearvars theta rho N
plot(x,y);
hold on
%% find "the vector"
vec(:,:,1) = [x(1), y(1); x(end), y(end)]; % "the vector"
scatter(x(1),y(1), 200,'s','r','fill') % square is the starting point
scatter(x(end),y(end), 200,'^','r','fill') % triangle is the ending point
line(vec(:,1,1), vec(:,2,1), 'LineStyle', '-', 'Color', 'r')
%% find center of mass
com = [mean(x), mean(y)]; % center of mass
vec(:,:,2) = [x(1), y(1); com]; % secondary vector (start -> com)
scatter(com(1), com(2), 200,'d','k','fill') % diamond is the com
line(vec(:,1,2), vec(:,2,2), 'LineStyle', '-', 'Color', 'k')
%% find rotation angle
dif = diff(vec,1,1);
[ang, ~] = cart2pol(reshape(dif(1,1,:),1,[]), reshape(dif(1,2,:),1,[]));
clearvars dif
% now you can tell the answer by the rotation angle
if ( diff(ang)>0 )
disp('CW!')
else
disp('CCW!')
end
One can always tell on which side of the directed line (the vector) a point is, by comparing two vectors, namely, rotating vector [starting point -> center of mass] to the vector [starting point -> ending point], and then comparing the rotation angle to 0. A few seconds of mind-animating can help understand.
I'm searching the way to efficiently find the point on an edge which is the closest point to some other point.
Let's say I know two points which are vertices of the edge. I can calculate the equation of the line that crosses those points.
What is the best way to calculate the point on the edge which is the closest point to some other point in the plane.
I would post an image but I don't have enough reputation points.
Let’s assume the line is defined by the two points (x1,y1), (x2,y2) and the “other point” is (a,b).
The point you’re looking for is (x,y).
You can easily find the equation of the black line. To find the blue line equation use the fact that m1*m2=-1 (m1 and m2 are the slopes of the two lines).
Clearly, the point you’re looking for is the intersection between the two lines.
There are two exceptions to what I was saying:
If x1=x2 then (x,y)=(x1,b).
If y1=y2 then (x,y)=(a,y1).
The following Python function finds the point (if you don’t know Python just think of it as a psudo-code):
def get_closest_point( x1,y1, x2,y2, a,b ):
if x1==x2: return (x1,b)
if y1==y2: return (a,y1)
m1 = (y2-y1)/(x2-x1)
m2 = -1/m1
x = (m1*x1-m2*a+b-y1) / (m1-m2)
y = m2*(x-a)+b
return (x,y)
You have three zones to consider. The "perpendicular" approach is for the zone in the middle:
For the other two zones the distance is the distance to the nearest segment endpoint.
The equation for the segment is:
y[x] = m x + b
Where
m -> -((Ay - By)/(-Ax + By)),
b -> -((-Ax By + Ay By)/(Ax - By))
And the perpendiculars have slope -1/m
The equations for the perpendicular passing thru A is:
y[x] = (-Ax + By)/(Ay - By) x + (Ax^2 + Ay^2 - Ax By - Ay By)/(Ay - By)
And the perpendicular passing thru B is the same exchanging the A's and B's in the equation above.
So you can know in which region lies your point introducing its x coordinate in the above equations and then comparing the y coordinate of the point with the result of y[x]
Edit
How to find in which region lies your point?
Let's suppose Ax ≤ Bx (if it's the other way, just change the point labels in the following formulae)
We will call your point {x0,y0}
1) Calculate
f[x0] = (-Ax + By)/(Ay - By) x0 + (Ax^2 + Ay^2 - Ax By - Ay By)/(Ay - By)
and compare with y0.
If y0 > f[x0], then your point lies in the green field in the figure above and the nearest point is A.
2) Else, Calculate
g[x0] = (-Bx + Ay)/(By - Ay) x0 + (Bx^2 + By^2 - Bx Ay - By Ay)/(By - Ay)
and compare with y0.
If y0 < g[x0], then your point lies in the yellow field in the figure above and the nearest point is B.
3) Else, you are in the "perpendicular light blue zone", and any of the other answer tell you how to calculate the nearest point and distance (I am not going to plagiarize :))
HTH!
I can describe what you want to do in geometric terms, but I don't have the algorithm at hand. Will that help?
Anyway, you want to draw a line which contains the stray point and is perpendicular to the edge. I think the slopes are a negative inverse relation between perpendicular lines, if that helps.
Then you want to find the intersection of the two lines.
Let's stick with the 2D case to save typing. It's been a while, so please forgive any elementary mistakes in my algebra.
The line forming the edge between the two points (x1, y1), (x2, y2) is represented as a function
y = mx + b
(You get to figure out m and b yourself, but it's elementary)
What you want to do is minimize the distance from your point (p1, p2) to a point on this line, i.e.
(p1-x)^2 + (p2-y)^2 (equation I)
subject to the equation
y = mx + b (equation II)
Substitute equation II into equation I and solve for x. You'll get two solutions; pick the one which gives the smaller value in equation I.
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);
}
Given a point (pX, pY) and a circle with a known center (cX,cY) and radius (r), what is the shortest amount of code you can come up with to find the point on the circle closest to (pX, pY) ?
I've got some code kind of working but it involves converting the circle to an equation of the form (x - cX)^2 + (y - cY)^2 = r^2 (where r is radius) and using the equation of the line from point (pX, pY) to (cX, cY) to create a quadratic equation to be solved.
Once I iron out the bugs it'll do, but it seems such an inelegant solution.
where P is the point, C is the center, and R is the radius, in a suitable "mathy" language:
V = (P - C); Answer = C + V / |V| * R;
where |V| is length of V.
OK, OK
double vX = pX - cX;
double vY = pY - cY;
double magV = sqrt(vX*vX + vY*vY);
double aX = cX + vX / magV * R;
double aY = cY + vY / magV * R;
easy to extend to >2 dimensions.
i would make a line from the center to the point, and calc where that graph crosses the circle oO i think not so difficult
Solve it mathematically first, then translate into code. Remember that the shortest line between a point and the edge of a circle will also pass through its center (as stated by #litb).
The shortest distance point lies at the intersection of circumference and line passing through the center and the input point. Also center, input and output points lie on a straight line
let the center be (xc, yc) and shortest point from input (xi, yi) be (x,y) then
sqrt((xc-x)^2 + (yc-y)^2) = r
since center, input and output points lie on a straight line, slope calculated between
any of two of these points should be same.
(yc-yi)/(xc-xi) = (y-yc)/(x-xc)
4.solving equations 2&3 should give us the shortest point.
Trig functions, multiply by r, and add pX or pY as appropriate.
Treat the centre of the circular as your origin, convert the co-ordinates of (pX, pY) to polar co-ordinates, (theta, r') replace r' with the original circle's r and convert back to cartesian co-ordinates (and adjust for the origin).
You asked for the shortest code, so here it is. In four lines it can be done, although there is still a quadratic.
I've considered the point to be outside the circle.
I've not considered what happens if the point is directly above or below the circle center, that is cX=pX.
m=(cY-pY)/(cX-pX); //slope
b=cY-m*cX; //or Py-m*Px. Now you have a line in the form y=m*x+b
X=( (2mcY)*((-2*m*cY)^2-4*(cY^2+cX^2-b^2-2*b*cY-r^2)*(-1-m^2))^(1/2) )/(2*(cY^2+cX^2-b^2-2*bc*Y-r^2));
Y=mX+b;
1] Get an equation for a line connecting the point and the circle center.
2] Move along the line a distance of one radius from the center to find the point on the circle. That is: radius=a^2+b^2 which is: r=((cY-Y)+(cX-X))^(1/2)
3] Solve quadratically. X=quadratic_solver(r=((cY-Y)+(cX-X))^(1/2),X) which if you substitute in Y=m*X+b you get that hell above.
4] X and Y are your results on the circle.
I am rather certain I have made an error somewhere, please leave a comment if anyone finds something. Of course it is degenerate, one answer is furthest from your point and the other is closest.
Easy way to think about it in terms of a picture, and easy to turn into code: Take the vector (pX - cX, pY - cY) from the center to the point. Divide by its length sqrt(blah blah blah), multiply by radius. Add this to (cX, cY).
Here is a simple method I use in unity... for the math kn00bs amongst us.
Its dependent on the transform orientation but it works nicely. I am doing a postion.z = 0 but just fatten the axis of the 2d circle you are not using.
//Find closest point on circle
Vector3 closestPoint = transform.InverseTransformPoint(m_testPosition.position);
closestPoint.z = 0;
closestPoint = closestPoint.normalized * m_radius;
Gizmos.color = Color.yellow;
Gizmos.DrawWireSphere(transform.TransformPoint(closestPoint), 0.01f);
I'm looking for an algorithm to detect if two rectangles intersect (one at an arbitrary angle, the other with only vertical/horizontal lines).
Testing if a corner of one is in the other ALMOST works. It fails if the rectangles form a cross-like shape.
It seems like a good idea to avoid using slopes of the lines, which would require special cases for vertical lines.
The standard method would be to do the separating axis test (do a google search on that).
In short:
Two objects don't intersect if you can find a line that separates the two objects. e.g. the objects / all points of an object are on different sides of the line.
The fun thing is, that it's sufficient to just check all edges of the two rectangles. If the rectangles don't overlap one of the edges will be the separating axis.
In 2D you can do this without using slopes. An edge is simply defined as the difference between two vertices, e.g.
edge = v(n) - v(n-1)
You can get a perpendicular to this by rotating it by 90°. In 2D this is easy as:
rotated.x = -unrotated.y
rotated.y = unrotated.x
So no trigonometry or slopes involved. Normalizing the vector to unit-length is not required either.
If you want to test if a point is on one or another side of the line you can just use the dot-product. the sign will tell you which side you're on:
// rotated: your rotated edge
// v(n-1) any point from the edge.
// testpoint: the point you want to find out which side it's on.
side = sign (rotated.x * (testpoint.x - v(n-1).x) +
rotated.y * (testpoint.y - v(n-1).y);
Now test all points of rectangle A against the edges of rectangle B and vice versa. If you find a separating edge the objects don't intersect (providing all other points in B are on the other side of the edge being tested for - see drawing below). If you find no separating edge either the rectangles are intersecting or one rectangle is contained in the other.
The test works with any convex polygons btw..
Amendment: To identify a separating edge, it is not enough to test all points of one rectangle against each edge of the other. The candidate-edge E (below) would as such be identified as a separating edge, as all points in A are in the same half-plane of E. However, it isn't a separating edge because the vertices Vb1 and Vb2 of B are also in that half-plane. It would only have been a separating edge if that had not been the case
http://www.iassess.com/collision.png
Basically look at the following picture:
If the two boxes collide, the lines A and B will overlap.
Note that this will have to be done on both the X and the Y axis, and both need to overlap for the rectangles to collide.
There is a good article in gamasutra.com which answers the question (the picture is from the article).
I did similar algorithm 5 years ago and I have to find my code snippet to post it here later
Amendment: The Separating Axis Theorem states that two convex shapes do not overlap if a separating axis exists (i.e. one where the projections as shown do not overlap). So "A separating axis exists" => "No overlap". This is not a bi-implication so you cannot conclude the converse.
In Cocoa you could easily detect whether the selectedArea rect intersects your rotated NSView's frame rect.
You don't even need to calculate polygons, normals an such. Just add these methods to your NSView subclass.
For instance, the user selects an area on the NSView's superview, then you call the method DoesThisRectSelectMe passing the selectedArea rect. The API convertRect: will do that job. The same trick works when you click on the NSView to select it. In that case simply override the hitTest method as below. The API convertPoint: will do that job ;-)
- (BOOL)DoesThisRectSelectMe:(NSRect)selectedArea
{
NSRect localArea = [self convertRect:selectedArea fromView:self.superview];
return NSIntersectsRect(localArea, self.bounds);
}
- (NSView *)hitTest:(NSPoint)aPoint
{
NSPoint localPoint = [self convertPoint:aPoint fromView:self.superview];
return NSPointInRect(localPoint, self.bounds) ? self : nil;
}
m_pGladiator's answer is right and I prefer to it.
Separating axis test is simplest and standard method to detect rectangle overlap. A line for which the projection intervals do not overlap we call a separating axis. Nils Pipenbrinck's solution is too general. It use dot product to check whether one shape is totally on the one side of the edge of the other. This solution is actually could induce to n-edge convex polygons. However, it is not optmized for two rectangles.
the critical point of m_pGladiator's answer is that we should check two rectangles' projection on both axises (x and y). If two projections are overlapped, then we could say these two rectangles are overlapped. So the comments above to m_pGladiator's answer are wrong.
for the simple situation, if two rectangles are not rotated,
we present a rectangle with structure:
struct Rect {
x, // the center in x axis
y, // the center in y axis
width,
height
}
we name rectangle A, B with rectA, rectB.
if Math.abs(rectA.x - rectB.x) < (Math.abs(rectA.width + rectB.width) / 2)
&& (Math.abs(rectA.y - rectB.y) < (Math.abs(rectA.height + rectB.height) / 2))
then
// A and B collide
end if
if any one of the two rectangles are rotated,
It may needs some efforts to determine the projection of them on x and y axises. Define struct RotatedRect as following:
struct RotatedRect : Rect {
double angle; // the rotating angle oriented to its center
}
the difference is how the width' is now a little different:
widthA' for rectA: Math.sqrt(rectA.width*rectA.width + rectA.height*rectA.height) * Math.cos(rectA.angle)
widthB' for rectB: Math.sqrt(rectB.width*rectB.width + rectB.height*rectB.height) * Math.cos(rectB.angle)
if Math.abs(rectA.x - rectB.x) < (Math.abs(widthA' + widthB') / 2)
&& (Math.abs(rectA.y - rectB.y) < (Math.abs(heightA' + heightB') / 2))
then
// A and B collide
end if
Could refer to a GDC(Game Development Conference 2007) PPT www.realtimecollisiondetection.net/pubs/GDC07_Ericson_Physics_Tutorial_SAT.ppt
The accepted answer about the separating axis test was very illuminating but I still felt it was not trivial to apply. I will share the pseudo-code I thought, "optimizing" first with the bounding circle test (see this other answer), in case it might help other people. I considered two rectangles A and B of the same size (but it is straightforward to consider the general situation).
1 Bounding circle test:
function isRectangleACollidingWithRectangleB:
if d > 2 * R:
return False
...
Computationally is much faster than the separating axis test. You only need to consider the separating axis test in the situation that both circles collide.
2 Separating axis test
The main idea is:
Consider one rectangle. Cycle along its vertices V(i).
Calculate the vector Si+1: V(i+1) - V(i).
Calculate the vector Ni using Si+1: Ni = (-Si+1.y, Si+1.x). This vector is the blue from the image. The sign of the dot product between the vectors from V(i) to the other vertices and Ni will define the separating axis (magenta dashed line).
Calculate the vector Si-1: V(i-1) - V(i). The sign of the dot product between Si-1 and Ni will define the location of the first rectangle with respect to the separating axis. In the example of the picture, they go in different directions, so the sign will be negative.
Cycle for all vertices j of the second square and calculate the vector Sij = V(j) - V(i).
If for any vertex V(j), the sign of the dot product of the vector Sij with Ni is the same as with the dot product of the vector Si-1 with Ni, this means both vertices V(i) and V(j) are on the same side of the magenta dashed line and, thus, vertex V(i) does not have a separating axis. So we can just skip vertex V(i) and repeat for the next vertex V(i+1). But first we update Si-1 = - Si+1. When we reach the last vertex (i = 4), if we have not found a separating axis, we repeat for the other rectangle. And if we still do not find a separating axis, this implies there is no separating axis and both rectangles collide.
If for a given vertex V(i) and all vertices V(j), the sign of the dot product of the vector Sij with Ni is different than with the vector Si-1 with Ni (as occurs in the image), this means we have found the separating axis and the rectangles do not collide.
In pseudo-code:
function isRectangleACollidingWithRectangleB:
...
#Consider first rectangle A:
Si-1 = Vertex_A[4] - Vertex_A[1]
for i in Vertex_A:
Si+1 = Vertex_A[i+1] - Vertex_A[i]
Ni = [- Si+1.y, Si+1.x ]
sgn_i = sign( dot_product(Si-1, Ni) ) #sgn_i is the sign of rectangle A with respect the separating axis
for j in Vertex_B:
sij = Vertex_B[j] - Vertex_A[i]
sgn_j = sign( dot_product(sij, Ni) ) #sgnj is the sign of vertex j of square B with respect the separating axis
if sgn_i * sgn_j > 0: #i.e., we have the same sign
break #Vertex i does not define separating axis
else:
if j == 4: #we have reached the last vertex so vertex i defines the separating axis
return False
Si-1 = - Si+1
#Repeat for rectangle B
...
#If we do not find any separating axis
return True
You can find the code in Python here.
Note:
In this other answer they also suggest for optimization to try before the separating axis test whether the vertices of one rectangle are inside the other as a sufficient condition for colliding. However, in my trials I found this intermediate step to actually be less efficient.
Check to see if any of the lines from one rectangle intersect any of the lines from the other. Naive line segment intersection is easy to code up.
If you need more speed, there are advanced algorithms for line segment intersection (sweep-line). See http://en.wikipedia.org/wiki/Line_segment_intersection
One solution is to use something called a No Fit Polygon. This polygon is calculated from the two polygons (conceptually by sliding one around the other) and it defines the area for which the polygons overlap given their relative offset. Once you have this NFP then you simply have to do an inclusion test with a point given by the relative offset of the two polygons. This inclusion test is quick and easy but you do have to create the NFP first.
Have a search for No Fit Polygon on the web and see if you can find an algorithm for convex polygons (it gets MUCH more complex if you have concave polygons). If you can't find anything then email me at howard dot J dot may gmail dot com
Here is what I think will take care of all possible cases.
Do the following tests.
Check any of the vertices of rectangle 1 reside inside rectangle 2 and vice versa. Anytime you find a vertex that resides inside the other rectangle you can conclude that they intersect and stop the search. THis will take care of one rectangle residing completely inside the other.
If the above test is inconclusive find the intersecting points of each line of 1 rectangle with each line of the other rectangle. Once a point of intersection is found check if it resides between inside the imaginary rectangle created by the corresponding 4 points. When ever such a point is found conclude that they intersect and stop the search.
If the above 2 tests return false then these 2 rectangles do not overlap.
If you're using Java, all implementations of the Shape interface have an intersects method that take a rectangle.
Well, the brute force method is to walk the edges of the horizontal rectangle and check each point along the edge to see if it falls on or in the other rectangle.
The mathematical answer is to form equations describing each edge of both rectangles. Now you can simply find if any of the four lines from rectangle A intersect any of the lines of rectangle B, which should be a simple (fast) linear equation solver.
-Adam
You could find the intersection of each side of the angled rectangle with each side of the axis-aligned one. Do this by finding the equation of the infinite line on which each side lies (i.e. v1 + t(v2-v1) and v'1 + t'(v'2-v'1) basically), finding the point at which the lines meet by solving for t when those two equations are equal (if they're parallel, you can test for that) and then testing whether that point lies on the line segment between the two vertices, i.e. is it true that 0 <= t <= 1 and 0 <= t' <= 1.
However, this doesn't cover the case when one rectangle completely covers the other. That you can cover by testing whether all four points of either rectangle lie inside the other rectangle.
This is what I would do, for the 3D version of this problem:
Model the 2 rectangles as planes described by equation P1 and P2, then write P1=P2 and derive from that the line of intersection equation, which won't exist if the planes are parallel (no intersection), or are in the same plane, in which case you get 0=0. In that case you will need to employ a 2D rectangle intersection algorithm.
Then I would see if that line, which is in the plane of both rectangles, passes through both rectangles. If it does, then you have an intersection of 2 rectangles, otherwise you don't (or shouldn't, I might have missed a corner case in my head).
To find if a line passes through a rectangle in the same plane, I would find the 2 points of intersection of the line and the sides of the rectangle (modelling them using line equations), and then make sure the points of intersections are with in range.
That is the mathematical descriptions, unfortunately I have no code to do the above.
Another way to do the test which is slightly faster than using the separating axis test, is to use the winding numbers algorithm (on quadrants only - not angle-summation which is horrifically slow) on each vertex of either rectangle (arbitrarily chosen). If any of the vertices have a non-zero winding number, the two rectangles overlap.
This algorithm is somewhat more long-winded than the separating axis test, but is faster because it only require a half-plane test if edges are crossing two quadrants (as opposed to up to 32 tests using the separating axis method)
The algorithm has the further advantage that it can be used to test overlap of any polygon (convex or concave). As far as I know, the algorithm only works in 2D space.
Either I am missing something else why make this so complicated?
if (x1,y1) and (X1,Y1) are corners of the rectangles, then to find intersection do:
xIntersect = false;
yIntersect = false;
if (!(Math.min(x1, x2, x3, x4) > Math.max(X1, X2, X3, X4) || Math.max(x1, x2, x3, x4) < Math.min(X1, X2, X3, X4))) xIntersect = true;
if (!(Math.min(y1, y2, y3, y4) > Math.max(Y1, Y2, Y3, Y4) || Math.max(y1, y2, y3, y4) < Math.min(Y1, Y2, Y3, Y4))) yIntersect = true;
if (xIntersect && yIntersect) {alert("Intersect");}
I implemented it like this:
bool rectCollision(const CGRect &boundsA, const Matrix3x3 &mB, const CGRect &boundsB)
{
float Axmin = boundsA.origin.x;
float Axmax = Axmin + boundsA.size.width;
float Aymin = boundsA.origin.y;
float Aymax = Aymin + boundsA.size.height;
float Bxmin = boundsB.origin.x;
float Bxmax = Bxmin + boundsB.size.width;
float Bymin = boundsB.origin.y;
float Bymax = Bymin + boundsB.size.height;
// find location of B corners in A space
float B0x = mB(0,0) * Bxmin + mB(0,1) * Bymin + mB(0,2);
float B0y = mB(1,0) * Bxmin + mB(1,1) * Bymin + mB(1,2);
float B1x = mB(0,0) * Bxmax + mB(0,1) * Bymin + mB(0,2);
float B1y = mB(1,0) * Bxmax + mB(1,1) * Bymin + mB(1,2);
float B2x = mB(0,0) * Bxmin + mB(0,1) * Bymax + mB(0,2);
float B2y = mB(1,0) * Bxmin + mB(1,1) * Bymax + mB(1,2);
float B3x = mB(0,0) * Bxmax + mB(0,1) * Bymax + mB(0,2);
float B3y = mB(1,0) * Bxmax + mB(1,1) * Bymax + mB(1,2);
if(B0x<Axmin && B1x<Axmin && B2x<Axmin && B3x<Axmin)
return false;
if(B0x>Axmax && B1x>Axmax && B2x>Axmax && B3x>Axmax)
return false;
if(B0y<Aymin && B1y<Aymin && B2y<Aymin && B3y<Aymin)
return false;
if(B0y>Aymax && B1y>Aymax && B2y>Aymax && B3y>Aymax)
return false;
float det = mB(0,0)*mB(1,1) - mB(0,1)*mB(1,0);
float dx = mB(1,2)*mB(0,1) - mB(0,2)*mB(1,1);
float dy = mB(0,2)*mB(1,0) - mB(1,2)*mB(0,0);
// find location of A corners in B space
float A0x = (mB(1,1) * Axmin - mB(0,1) * Aymin + dx)/det;
float A0y = (-mB(1,0) * Axmin + mB(0,0) * Aymin + dy)/det;
float A1x = (mB(1,1) * Axmax - mB(0,1) * Aymin + dx)/det;
float A1y = (-mB(1,0) * Axmax + mB(0,0) * Aymin + dy)/det;
float A2x = (mB(1,1) * Axmin - mB(0,1) * Aymax + dx)/det;
float A2y = (-mB(1,0) * Axmin + mB(0,0) * Aymax + dy)/det;
float A3x = (mB(1,1) * Axmax - mB(0,1) * Aymax + dx)/det;
float A3y = (-mB(1,0) * Axmax + mB(0,0) * Aymax + dy)/det;
if(A0x<Bxmin && A1x<Bxmin && A2x<Bxmin && A3x<Bxmin)
return false;
if(A0x>Bxmax && A1x>Bxmax && A2x>Bxmax && A3x>Bxmax)
return false;
if(A0y<Bymin && A1y<Bymin && A2y<Bymin && A3y<Bymin)
return false;
if(A0y>Bymax && A1y>Bymax && A2y>Bymax && A3y>Bymax)
return false;
return true;
}
The matrix mB is any affine transform matrix that converts points in the B space to points in the A space. This includes simple rotation and translation, rotation plus scaling, and full affine warps, but not perspective warps.
It may not be as optimal as possible. Speed was not a huge concern. However it seems to work ok for me.
Here is a matlab implementation of the accepted answer:
function olap_flag = ol(A,B,sub)
%A and B should be 4 x 2 matrices containing the xy coordinates of the corners in clockwise order
if nargin == 2
olap_flag = ol(A,B,1) && ol(B,A,1);
return;
end
urdl = diff(A([1:4 1],:));
s = sum(urdl .* A, 2);
sdiff = B * urdl' - repmat(s,[1 4]);
olap_flag = ~any(max(sdiff)<0);
This is the conventional method, go line by line and check whether the lines are intersecting. This is the code in MATLAB.
C1 = [0, 0]; % Centre of rectangle 1 (x,y)
C2 = [1, 1]; % Centre of rectangle 2 (x,y)
W1 = 5; W2 = 3; % Widths of rectangles 1 and 2
H1 = 2; H2 = 3; % Heights of rectangles 1 and 2
% Define the corner points of the rectangles using the above
R1 = [C1(1) + [W1; W1; -W1; -W1]/2, C1(2) + [H1; -H1; -H1; H1]/2];
R2 = [C2(1) + [W2; W2; -W2; -W2]/2, C2(2) + [H2; -H2; -H2; H2]/2];
R1 = [R1 ; R1(1,:)] ;
R2 = [R2 ; R2(1,:)] ;
plot(R1(:,1),R1(:,2),'r')
hold on
plot(R2(:,1),R2(:,2),'b')
%% lines of Rectangles
L1 = [R1(1:end-1,:) R1(2:end,:)] ;
L2 = [R2(1:end-1,:) R2(2:end,:)] ;
%% GEt intersection points
P = zeros(2,[]) ;
count = 0 ;
for i = 1:4
line1 = reshape(L1(i,:),2,2) ;
for j = 1:4
line2 = reshape(L2(j,:),2,2) ;
point = InterX(line1,line2) ;
if ~isempty(point)
count = count+1 ;
P(:,count) = point ;
end
end
end
%%
if ~isempty(P)
fprintf('Given rectangles intersect at %d points:\n',size(P,2))
plot(P(1,:),P(2,:),'*k')
end
the function InterX can be downloaded from: https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections?focused=5165138&tab=function
I have a simplier method of my own, if we have 2 rectangles:
R1 = (min_x1, max_x1, min_y1, max_y1)
R2 = (min_x2, max_x2, min_y2, max_y2)
They overlap if and only if:
Overlap = (max_x1 > min_x2) and (max_x2 > min_x1) and (max_y1 > min_y2) and (max_y2 > min_y1)
You can do it for 3D boxes too, actually it works for any number of dimensions.
Enough has been said in other answers, so I'll just add pseudocode one-liner:
!(a.left > b.right || b.left > a.right || a.top > b.bottom || b.top > a.bottom);
Check if the center of mass of all the vertices of both rectangles lies within one of the rectangles.