How to test if a line segment intersects an axis-aligned rectange in 2D?
How to test if a line segment intersects an axis-aligned rectange in 2D? The segment is defined with its two ends: p1, p2. The rectangle is defined with top-left and bottom-right points.
Answer
The original poster wanted to DETECT an intersection between a line segment and a polygon. There was no need to LOCATE the intersection, if there is one. If that's how you meant it, you can do less work than Liang-Barsky or Cohen-Sutherland:
Let the segment endpoints be p1=(x1 y1) and p2=(x2 y2).
Let the rectangle's corners be (xBL yBL) and (xTR yTR).
Then all you have to do is
A. Check if all four corners of the rectangle are on the same side of the line. The implicit equation for a line through p1 and p2 is:
F(x y) = (y2-y1)*x + (x1-x2)*y + (x2*y1-x1*y2)
If F(x y) = 0, (x y) is ON the line.
If F(x y) > 0, (x y) is "above" the line.
If F(x y) < 0, (x y) is "below" the line.
Substitute all four corners into F(x y). If they're all negative or all positive, there is no intersection. If some are positive and some negative, go to step B.
B. Project the endpoint onto the x axis, and check if the segment's shadow intersects the polygon's shadow. Repeat on the y axis:
If (x1 > xTR and x2 > xTR), no intersection (line is to right of rectangle).
If (x1 < xBL and x2 < xBL), no intersection (line is to left of rectangle).
If (y1 > yTR and y2 > yTR), no intersection (line is above rectangle).
If (y1 < yBL and y2 < yBL), no intersection (line is below rectangle).
else, there is an intersection. Do Cohen-Sutherland or whatever code was mentioned in the other answers to your question.
You can, of course, do B first, then A.
Alejo