Find area of overlapping rectangles in Python / OpenCV with a raw list of points

Question

I have the coordinates x1 , y1 , x2 , y2 of a rectangle and a list of other coordinates for other rectangles.

I want to compare the value of the one I already have over with the others to see if they overlap more than 50% of the original rectangle.

I checked other resources but still, I could get my head around it:

• https://codereview.stackexchange.com/questions/151309/check-if-two-rectangles-overlap
• https://codereview.stackexchange.com/questions/31352/overlapping-rectangles

Answer 1

Let's first simplify your problem to a single dimension:

You have an interval A = [a0, a1] and want to find out how much does another interval B = [b0, b1] intersects with it. I will represent A with = and B with - .

There are 6 possible scenarios:

• A contains B , intersection = b1 - b0

   a0 b0 b1 a1 ============== ------ 

• B contains A , intersection = a1 - a0

   b0 a0 a1 b1 ====== -------------- 

• B intersects A from the left, intersection = b1 - a0

   b0 a0 b1 a1 ========== ---------- 

• B intersects A from the right, intersection = a1 - b0

   a0 b0 a1 b1 ========== ---------- 

• B is to the left of A , intersection = 0

   b0 b1 a0 a1 ====== ------ 

• B is to the right of A , intersection = 0

   a0 a1 b0 b1 ====== ------ 

Based on that, we can define a function that, given two intervals A = [a0, a1] and B = [b0, b1] , returns how much they intersect:

def calculateIntersection(a0, a1, b0, b1):
    if a0 >= b0 and a1 <= b1: # Contained
        intersection = a1 - a0
    elif a0 < b0 and a1 > b1: # Contains
        intersection = b1 - b0
    elif a0 < b0 and a1 > b0: # Intersects right
        intersection = a1 - b0
    elif a1 > b1 and a0 < b1: # Intersects left
        intersection = b1 - a0
    else: # No intersection (either side)
        intersection = 0

    return intersection

That's pretty much everything you need. To find out how much do two rectangles intersect, you just need to execute this function on both the X and the Y axis and multiply those amounts to get the intersecting area:

# The rectangle against which you are going to test the rest and its area:
X0, Y0, X1, Y1, = [0, 0, 10, 10]
AREA = float((X1 - X0) * (Y1 - Y0))

# Rectangles to check
rectangles = [[15, 0, 20, 10], [0, 15, 10, 20], [0, 0, 5, 5], [0, 0, 5, 10], [0, 5, 10, 100], [0, 0, 100, 100]]

# Intersecting rectangles:
intersecting = []

for x0, y0, x1, y1 in rectangles:       
    width = calculateIntersection(x0, x1, X0, X1)        
    height = calculateIntersection(y0, y1, Y0, Y1)        
    area = width * height
    percent = area / AREA

    if (percent >= 0.5):
        intersecting.append([x0, y0, x1, y1])

The result will be:

• [15, 0, 20, 10] does not intersect in the X-axis, so width = 0 .
• [0, 15, 10, 20] does not intersect in the Y-axis, so height = 0 .
• [0, 0, 5, 5] intersects only 25% .
• [0, 0, 5, 10] intersects 50% and will be added to intersecting .
• [0, 5, 10, 100] intersects 50% and will be added to intersecting .
• [0, 0, 100, 100] intersects 100% and will be added to intersecting .

Answer 2

The overlap area is the product of the overlap width and the overlap height.

To find the width overlap of the rectangles XY and xy , take the rightmost of the left edges and the leftmost of the right edges. Numerically, max(X1, x1) and min(X2, x2) . Thus the overlap width is min(X2, x2) - max(X1, x1) . If this is negative, there is no overlap at all. Repeat for the height.

Thus, the area of overlap can be expressed with a single formula:

max(0, min(X2, x2) - max(X1, x1)) . max(0, min(Y2, y2) - max(Y1, y1))

[Total cost: six comparisons, two subtractions, one product (plus one comparison to the target area).]

As a micro-optimization, you can conclude "no" immediately if

min(X2, x2) - max(X1, x1) < 50% (X2 - X1)

[With some probability, the cost is reduced to three comparisons and one subtraction.]