I have two rectangles, the red rectangle (can move) and the blue rectangle. Both have: x, y, width, height.
How can I say in a programming language such as Java when there is a collision between the blue and the red rectangle?
One of the simpler forms of collision detection is between two rectangles that are axis aligned — meaning no rotation. The algorithm works by ensuring there is no gap between any of the 4 sides of the rectangles. Any gap means a collision does not exist.
Rectangle rect1 = new Rectangle(100, 100, 200, 240); Rectangle rect2 = new Rectangle(120, 80, 80, 120); Rectangle intersection = rect1. intersection(rect2); To use java. awt.
Since Java doesn't have an intersect function (really!?) you can do collision detection by simply comparying the X and Y, Width and Height values of the bounding boxes (rectangle) for each of the objects that could potentially collide.
Two rectangles do not overlap if one of the following conditions is true. 1) One rectangle is above top edge of other rectangle. 2) One rectangle is on left side of left edge of other rectangle. We need to check above cases to find out if given rectangles overlap or not.
if (RectA.X1 < RectB.X2 && RectA.X2 > RectB.X1 &&
RectA.Y1 < RectB.Y2 && RectA.Y2 > RectB.Y1)
Say you have Rect A, and Rect B. Proof is by contradiction. Any one of four conditions guarantees that no overlap can exist:
Cond1. If A's left edge is to the right of the B's right edge, - then A is Totally to right Of B
Cond2. If A's right edge is to the left of the B's left edge, - then A is Totally to left Of B
Cond3. If A's top edge is below B's bottom edge, - then A is Totally below B
Cond4. If A's bottom edge is above B's top edge, - then A is Totally above B
So condition for Non-Overlap is
Cond1 Or Cond2 Or Cond3 Or Cond4
Therefore, a sufficient condition for Overlap is the opposite (De Morgan)
Not Cond1 And Not Cond2 And Not Cond3 And Not Cond4 This is equivalent to:
A's Left Edge to left of B's right edge, and
A's right edge to right of B's left edge, and
A's top above B's bottom, and
A's bottom below B's Top
Note 1: It is fairly obvious this same principle can be extended to any number of dimensions. Note 2: It should also be fairly obvious to count overlaps of just one pixel, change the < and/or the > on that boundary to a <= or a >=.
If you are having a hard time visualizing why it works, I made an example page at silentmatt.com/intersection.html where you can drag rectangles around and see the comparisons.
In java , to detect if two when two rectangles collide, you can use intersects()
method
Sample Code:
Rectangle r1 = new Rectangle(x1,y1,x2,y2);
Rectangle r2 = new Rectangle(x1,y1,x2,y2);
if(r1.intersects(r2))
{
//what to happen when collision occurs goes here
}
bool isIntersect(
int Ax, int Ay, int Aw, int Ah,
int Bx, int By, int Bw, int Bh)
{
return
Bx + Bw > Ax &&
By + Bh > Ay &&
Ax + Aw > Bx &&
Ay + Ah > By;
}
You have to check both the intersection along the x-axis and along the y-axis. If any of them is missing, there is no collision between rectangles.
Code for 1-D:
boolean overlaps(double point1, double length1, double point2, double length2)
{
double highestStartPoint = Math.max(point1, point2);
double lowestEndPoint = Math.min(point1 + length1, point2 + length2);
return highestStartPoint < lowestEndPoint;
}
You have to call it for both x and y:
boolean collision(double x1, double x2, double y1, double y2, double width1, double width2, double height1, double height2)
{
return overlaps(x1, width1, x2, width2) && overlaps (y1, height1, y2, height2);
}
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