找到两点系列的交集？

Question

A line is here defined as a series of 2D node points. Now, I have two such lines A and B .

A=[(0, 0), (1, 1), (2.1, 3), (4,7)]
B=[(2, 0), (2, 6)]

When one draws them on the paper, one can easily see the two lines intersect at a point that is NOT a node member of either A or B .

However, both A and B indeed cross this point . That is, the point indeed lies both on A and B , just not collide with the node points.

I now wish to find the intersection point.

(a gentle reminder once again: the intersection is on A and B , but it may not be a node)

What I come up with now is to use a polynomial to fit each point series. In this way, I can solve for the intersection with the equations. However, it seems to be a quite stupid way to myself.

Is there any smart way to do so?

I am speaking Python, but any generic answers are also very much welcomed.

Answer 1

for i in range(0, len(A), 2):
    line1 = A[i:i+2]
    for j in range(0, len(B), 2):
        line1 = A[j:j+2]
        point_of_intersect = intersection(line1, line2)
        if point_of_intersect:
            print point_of_intersect

where the function intersection is defined according to this Wikipedia entry.

Answer 2

Having 2 points X(x1, x2), Y(y1, y2) , you can determine the line equation such as:

(x-x1)/(x2-x1) = (y-y1)/(y2-y1)

Do this for line A and line B.

You will obtain for line

A: y = m1*x+n1
B: y = m2*x+n2

Now, you just need to find the value of y and x which respects both equations above.

Answer 3

If your lines have very few segments, then testing all possible segment pairs is practical, but it doesn't scale very well.

The most common solution to this problem is the Bentley-Ottman algorithm. It's easy to search the internet for pseudo-code or even libraries which implement it; it's a simple sweep-line algorithm whose execution time is O( (n + k) log n) where n is the number of line segments (the sum of the segments in both lines) and k is the number of intersections. If you only want to know if there are any intersections, you can stop at the first intersection found, at which point the algorithm reduces to O(n log n) .

This assumes that you know that A and B do not contain internal crossings. A naive implementation of Bentley-Ottman (where the two sets of line segments are just processed indiscriminately) will report all crossings, including self-crossings.