How can I check if two segments intersect?

Solution 1:

User @i_4_got points to this page with a very efficent solution in Python. I reproduce it here for convenience (since it would have made me happy to have it here):

def ccw(A,B,C):
    return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)

# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
    return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)

Solution 2:

The equation of a line is:

f(x) = A*x + b = y

For a segment, it is exactly the same, except that x is included on an interval I.

If you have two segments, defined as follow:

Segment1 = {(X1, Y1), (X2, Y2)}
Segment2 = {(X3, Y3), (X4, Y4)}

The abcisse Xa of the potential point of intersection (Xa,Ya) must be contained in both interval I1 and I2, defined as follow :

I1 = [min(X1,X2), max(X1,X2)]
I2 = [min(X3,X4), max(X3,X4)]

And we could say that Xa is included into :

Ia = [max( min(X1,X2), min(X3,X4) ),
      min( max(X1,X2), max(X3,X4) )]

Now, we need to check that this interval Ia exists :

if (max(X1,X2) < min(X3,X4)):
    return False  # There is no mutual abcisses

So, we have two line formula, and a mutual interval. Your line formulas are:

f1(x) = A1*x + b1 = y
f2(x) = A2*x + b2 = y

As we got two points by segment, we are able to determine A1, A2, b1 and b2:

A1 = (Y1-Y2)/(X1-X2)  # Pay attention to not dividing by zero
A2 = (Y3-Y4)/(X3-X4)  # Pay attention to not dividing by zero
b1 = Y1-A1*X1 = Y2-A1*X2
b2 = Y3-A2*X3 = Y4-A2*X4

If the segments are parallel, then A1 == A2 :

if (A1 == A2):
    return False  # Parallel segments

A point (Xa,Ya) standing on both line must verify both formulas f1 and f2:

Ya = A1 * Xa + b1
Ya = A2 * Xa + b2
A1 * Xa + b1 = A2 * Xa + b2
Xa = (b2 - b1) / (A1 - A2)   # Once again, pay attention to not dividing by zero

The last thing to do is check that Xa is included into Ia:

if ( (Xa < max( min(X1,X2), min(X3,X4) )) or
     (Xa > min( max(X1,X2), max(X3,X4) )) ):
    return False  # intersection is out of bound
else:
    return True

In addition to this, you may check at startup that two of the four provided points are not equals to avoid all that testing.