I have two 2D rotated rectangles, defined as an (center x,center y, height, width) and an angle of rotation (0-360°). How would I calculate the area of intersection of these two rotated rectangles.
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Project the vertex to the axis and check those points intersection. – Netwave Jun 28 '17 at 08:49
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6Possible duplicate of [Area of Intersection of Two Rotated Rectangles](https://stackoverflow.com/questions/11670028/area-of-intersection-of-two-rotated-rectangles) – custom_user Jun 28 '17 at 08:58
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1This is less a software engineering problem than a mathematical problem – Philipp Jul 13 '17 at 09:54
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5Its a mathematical problem for a mathematician. As a software engineer I come across this type of problem in collision detection, ocr, computer vision and other systems for which I need functional code rather than mathematical theory – Bug Jul 13 '17 at 11:59
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Can we assume that two of the sides will not be collinear? – Jared Goguen Jul 21 '17 at 20:15
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There is a nice expalanation of a polygon area computation here: https://www.youtube.com/watch?v=0KjG8Pg6LGk. Together with links and hints from the linked duplicate question (areas: intersection = polygon1 + polygon2 - aggregated polygon) the missing part is to convert your data to (x,y) coordinates of vertices. i.e. some easy sine and cosine calculations. – VPfB Jul 22 '17 at 19:00
3 Answers
Such tasks are solved using computational geometry packages, e.g. Shapely:
import shapely.geometry
import shapely.affinity
class RotatedRect:
def __init__(self, cx, cy, w, h, angle):
self.cx = cx
self.cy = cy
self.w = w
self.h = h
self.angle = angle
def get_contour(self):
w = self.w
h = self.h
c = shapely.geometry.box(-w/2.0, -h/2.0, w/2.0, h/2.0)
rc = shapely.affinity.rotate(c, self.angle)
return shapely.affinity.translate(rc, self.cx, self.cy)
def intersection(self, other):
return self.get_contour().intersection(other.get_contour())
r1 = RotatedRect(10, 15, 15, 10, 30)
r2 = RotatedRect(15, 15, 20, 10, 0)
from matplotlib import pyplot
from descartes import PolygonPatch
fig = pyplot.figure(1, figsize=(10, 4))
ax = fig.add_subplot(121)
ax.set_xlim(0, 30)
ax.set_ylim(0, 30)
ax.add_patch(PolygonPatch(r1.get_contour(), fc='#990000', alpha=0.7))
ax.add_patch(PolygonPatch(r2.get_contour(), fc='#000099', alpha=0.7))
ax.add_patch(PolygonPatch(r1.intersection(r2), fc='#009900', alpha=1))
pyplot.show()

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1@Bug once you have the shape of the intersection `r1.intersection(r2)` (which is a `Shapely` object) you only need to access its `area` attribute: `r1.intersection(r2).area`. – Leon Jul 18 '17 at 10:57
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I have installed shapely through pip install shapely, it shows Requirement already satisfied: shapely in /usr/local/lib/python2.7/dist-packages, but when I run the code, it always tell us ImportError: No module named affinity, any clue? – user824624 Jul 25 '17 at 04:25
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@user824624 Given the scarce information you've provided about your problem I have no clue as to what may have gone wrong. I think you better post a separate question for that problem and provide enough details. – Leon Jul 25 '17 at 17:58
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Here is a solution that does not use any libraries outside of Python's standard library.
Determining the area of the intersection of two rectangles can be divided in two subproblems:
- Finding the intersection polygon, if any;
- Determine the area of the intersection polygon.
Both problems are relatively easy when you work with the
vertices (corners) of the rectangles. So first you have to determine
these vertices. Assuming the coordinate origin is in the center
of the rectangle, the vertices are,
starting from the lower left in a counter-clockwise direction:
(-w/2, -h/2)
, (w/2, -h/2)
, (w/2, h/2)
, and (-w/2, h/2)
.
Rotating this over the angle a
, and translating them
to the proper position of the rectangle's center, these become:
(cx + (-w/2)cos(a) - (-h/2)sin(a), cy + (-w/2)sin(a) + (-h/2)cos(a))
, and similar for the other corner points.
A simple way to determine the intersection polygon is the following: you start with one rectangle as the candidate intersection polygon. Then you apply the process of sequential cutting (as described here. In short: you take each edges of the second rectangle in turn, and remove all parts from the candidate intersection polygon that are on the "outer" half plane defined by the edge (extended in both directions). Doing this for all edges leaves the candidate intersection polygon with only the parts that are inside the second rectangle or on its boundary.
The area of the resulting polygon (defined by a series of vertices) can be calculated from the coordinates of the vertices. You sum the cross products of the vertices of each edge (again in counter-clockwise order), and divide that by two. See e.g. www.mathopenref.com/coordpolygonarea.html
Enough theory and explanation. Here is the code:
from math import pi, cos, sin
class Vector:
def __init__(self, x, y):
self.x = x
self.y = y
def __add__(self, v):
if not isinstance(v, Vector):
return NotImplemented
return Vector(self.x + v.x, self.y + v.y)
def __sub__(self, v):
if not isinstance(v, Vector):
return NotImplemented
return Vector(self.x - v.x, self.y - v.y)
def cross(self, v):
if not isinstance(v, Vector):
return NotImplemented
return self.x*v.y - self.y*v.x
class Line:
# ax + by + c = 0
def __init__(self, v1, v2):
self.a = v2.y - v1.y
self.b = v1.x - v2.x
self.c = v2.cross(v1)
def __call__(self, p):
return self.a*p.x + self.b*p.y + self.c
def intersection(self, other):
# See e.g. https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection#Using_homogeneous_coordinates
if not isinstance(other, Line):
return NotImplemented
w = self.a*other.b - self.b*other.a
return Vector(
(self.b*other.c - self.c*other.b)/w,
(self.c*other.a - self.a*other.c)/w
)
def rectangle_vertices(cx, cy, w, h, r):
angle = pi*r/180
dx = w/2
dy = h/2
dxcos = dx*cos(angle)
dxsin = dx*sin(angle)
dycos = dy*cos(angle)
dysin = dy*sin(angle)
return (
Vector(cx, cy) + Vector(-dxcos - -dysin, -dxsin + -dycos),
Vector(cx, cy) + Vector( dxcos - -dysin, dxsin + -dycos),
Vector(cx, cy) + Vector( dxcos - dysin, dxsin + dycos),
Vector(cx, cy) + Vector(-dxcos - dysin, -dxsin + dycos)
)
def intersection_area(r1, r2):
# r1 and r2 are in (center, width, height, rotation) representation
# First convert these into a sequence of vertices
rect1 = rectangle_vertices(*r1)
rect2 = rectangle_vertices(*r2)
# Use the vertices of the first rectangle as
# starting vertices of the intersection polygon.
intersection = rect1
# Loop over the edges of the second rectangle
for p, q in zip(rect2, rect2[1:] + rect2[:1]):
if len(intersection) <= 2:
break # No intersection
line = Line(p, q)
# Any point p with line(p) <= 0 is on the "inside" (or on the boundary),
# any point p with line(p) > 0 is on the "outside".
# Loop over the edges of the intersection polygon,
# and determine which part is inside and which is outside.
new_intersection = []
line_values = [line(t) for t in intersection]
for s, t, s_value, t_value in zip(
intersection, intersection[1:] + intersection[:1],
line_values, line_values[1:] + line_values[:1]):
if s_value <= 0:
new_intersection.append(s)
if s_value * t_value < 0:
# Points are on opposite sides.
# Add the intersection of the lines to new_intersection.
intersection_point = line.intersection(Line(s, t))
new_intersection.append(intersection_point)
intersection = new_intersection
# Calculate area
if len(intersection) <= 2:
return 0
return 0.5 * sum(p.x*q.y - p.y*q.x for p, q in
zip(intersection, intersection[1:] + intersection[:1]))
if __name__ == '__main__':
r1 = (10, 15, 15, 10, 30)
r2 = (15, 15, 20, 10, 0)
print(intersection_area(r1, r2))

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1No wonder you are a compatriot of [the](http://www.win.tue.nl/~mdberg/) [three](http://www.cs.uu.nl/staff/marc.html) [Marks](http://www.cs.uu.nl/staff/markov.html) coauthoring the [Computational Geometry book](http://www.cs.uu.nl/geobook/). I have a feeling that you may know some of them in person. – Leon Jul 23 '17 at 18:49
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BTW, you should note in your answer that it works for computing the intersection of *any* polygon with *any convex* polygon. – Leon Jul 23 '17 at 19:05
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Thank you for this answer, its very clear and useful! Is there a mathematical name for the operation happening in `Line.__call__()`. It seems like its related to this? https://math.stackexchange.com/questions/274712/calculate-on-which-side-of-a-straight-line-is-a-given-point-located – dgoldman May 03 '20 at 15:19
intersection, pnt = contourIntersection(rect1, rect2)
After looking at the possible duplicate page for this problem I couldn't find a completed answer for python so here is my solution using masking. This function will work with complex shapes on any angle, not just rectangles
You pass in the 2 contours of your rotated rectangles as parameters and it returns 'None' if no intersection occurs or an image of the intersected area and the left/top position of that image in relation to the original image the contours were taken from
Uses python, cv2 and numpy
import cv2
import math
import numpy as np
def contourIntersection(con1, con2, showContours=False):
# skip if no bounding rect intersection
leftmost1 = tuple(con1[con1[:, :, 0].argmin()][0])
topmost1 = tuple(con1[con1[:, :, 1].argmin()][0])
leftmost2 = tuple(con2[con2[:, :, 0].argmin()][0])
topmost2 = tuple(con2[con2[:, :, 1].argmin()][0])
rightmost1 = tuple(con1[con1[:, :, 0].argmax()][0])
bottommost1 = tuple(con1[con1[:, :, 1].argmax()][0])
rightmost2 = tuple(con2[con2[:, :, 0].argmax()][0])
bottommost2 = tuple(con2[con2[:, :, 1].argmax()][0])
if rightmost1[0] < leftmost2[0] or rightmost2[0] < leftmost1[0] or bottommost1[1] < topmost2[1] or bottommost2[1] < topmost1[1]:
return None, None
# reset top / left to 0
left = leftmost1[0] if leftmost1[0] < leftmost2[0] else leftmost2[0]
top = topmost1[1] if topmost1[1] < topmost2[1] else topmost2[1]
newCon1 = []
for pnt in con1:
newLeft = pnt[0][0] - left
newTop = pnt[0][1] - top
newCon1.append([newLeft, newTop])
# next
con1_new = np.array([newCon1], dtype=np.int32)
newCon2 = []
for pnt in con2:
newLeft = pnt[0][0] - left
newTop = pnt[0][1] - top
newCon2.append([newLeft, newTop])
# next
con2_new = np.array([newCon2], dtype=np.int32)
# width / height
right1 = rightmost1[0] - left
bottom1 = bottommost1[1] - top
right2 = rightmost2[0] - left
bottom2 = bottommost2[1] - top
width = right1 if right1 > right2 else right2
height = bottom1 if bottom1 > bottom2 else bottom2
# create images
img1 = np.zeros([height, width], np.uint8)
cv2.drawContours(img1, con1_new, -1, (255, 255, 255), -1)
img2 = np.zeros([height, width], np.uint8)
cv2.drawContours(img2, con2_new, -1, (255, 255, 255), -1)
# mask images together using AND
imgIntersection = cv2.bitwise_and(img1, img2)
if showContours:
img1[img1 > 254] = 128
img2[img2 > 254] = 100
imgAll = cv2.bitwise_or(img1, img2)
cv2.imshow('Merged Images', imgAll)
# end if
if not imgIntersection.sum():
return None, None
# trim
while not imgIntersection[0].sum():
imgIntersection = np.delete(imgIntersection, (0), axis=0)
top += 1
while not imgIntersection[-1].sum():
imgIntersection = np.delete(imgIntersection, (-1), axis=0)
while not imgIntersection[:, 0].sum():
imgIntersection = np.delete(imgIntersection, (0), axis=1)
left += 1
while not imgIntersection[:, -1].sum():
imgIntersection = np.delete(imgIntersection, (-1), axis=1)
return imgIntersection, (left, top)
# end function
To complete the answer so you can use the above function with the values of CenterX, CenterY, Width, Height and Angle of 2 rotated rectangles I have added the below functions. Simple change the Rect1 and Rect2 properties at the bottom of the code to your own
def pixelsBetweenPoints(xy1, xy2):
X = abs(xy1[0] - xy2[0])
Y = abs(xy1[1] - xy2[1])
return int(math.sqrt((X ** 2) + (Y ** 2)))
# end function
def rotatePoint(angle, centerPoint, dist):
xRatio = math.cos(math.radians(angle))
yRatio = math.sin(math.radians(angle))
xPotted = int(centerPoint[0] + (dist * xRatio))
yPlotted = int(centerPoint[1] + (dist * yRatio))
newPoint = [xPotted, yPlotted]
return newPoint
# end function
def angleBetweenPoints(pnt1, pnt2):
A_B = pixelsBetweenPoints(pnt1, pnt2)
pnt3 = (pnt1[0] + A_B, pnt1[1])
C = pixelsBetweenPoints(pnt2, pnt3)
angle = math.degrees(math.acos((A_B * A_B + A_B * A_B - C * C) / (2.0 * A_B * A_B)))
# reverse if above horizon
if pnt2[1] < pnt1[1]:
angle = angle * -1
# end if
return angle
# end function
def rotateRectContour(xCenter, yCenter, height, width, angle):
# calc positions
top = int(yCenter - (height / 2))
left = int(xCenter - (width / 2))
right = left + width
rightTop = (right, top)
centerPoint = (xCenter, yCenter)
# new right / top point
rectAngle = angleBetweenPoints(centerPoint, rightTop)
angleRightTop = angle + rectAngle
angleRightBottom = angle + 180 - rectAngle
angleLeftBottom = angle + 180 + rectAngle
angleLeftTop = angle - rectAngle
distance = pixelsBetweenPoints(centerPoint, rightTop)
rightTop_new = rotatePoint(angleRightTop, centerPoint, distance)
rightBottom_new = rotatePoint(angleRightBottom, centerPoint, distance)
leftBottom_new = rotatePoint(angleLeftBottom, centerPoint, distance)
leftTop_new = rotatePoint(angleLeftTop, centerPoint, distance)
contourList = [[leftTop_new], [rightTop_new], [rightBottom_new], [leftBottom_new]]
contour = np.array(contourList, dtype=np.int32)
return contour
# end function
# rect1
xCenter_1 = 40
yCenter_1 = 20
height_1 = 200
width_1 = 80
angle_1 = 45
rect1 = rotateRectContour(xCenter_1, yCenter_1, height_1, width_1, angle_1)
# rect2
xCenter_2 = 80
yCenter_2 = 25
height_2 = 180
width_2 = 50
angle_2 = 123
rect2 = rotateRectContour(xCenter_2, yCenter_2, height_2, width_2, angle_2)
intersection, pnt = contourIntersection(rect1, rect2, True)
if intersection is None:
print('No intersection')
else:
print('Area of intersection = ' + str(int(intersection.sum() / 255)))
cv2.imshow('Intersection', intersection)
# end if
cv2.waitKey(0)

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The drawback of this solution is that it computes the result via [rasterisation](https://en.wikipedia.org/wiki/Rasterisation), which introduces inaccuracy, is slow and the result may not be directly usable (if we need it as a polygon rather than as an image). See [my answer](https://stackoverflow.com/a/45141648/6394138) for a better option. – Leon Jul 17 '17 at 10:24
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Was attempting to calc a value rather than a shape. Also was thinking the contours may be pulled from an image so returned a result which allows the mapping to the individual intersecting pixels. Only single line to turn image into polygon, didnt want to clutter code – Bug Jul 18 '17 at 09:52
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After running time tests on above code using time.time() with all output to screen turned off it ran at < minimum time able to be recorded by time.time(). Also avoids unnecessary processing if no possibility of intersection occurs – Bug Jul 18 '17 at 10:48