Are two rectangles overlapping each other?
10,890
Solution 1
I think it should be
boolean isOverLap = (r1x1 < r2x2) && (r1x2 > r2x1) && (r1y1 < r2y2) && (r1y2 > r2y1);
(easier to read without all the negations)
You can see this as the opposite of the four cases that each on their own guarantee that an overlap cannot happen:
boolean isNonOverLap = (r1x1 >= r2x2) || (r1x2 <= r2x1) || (r1y1 >= r2y2) || (r1y2 <= r2y1);
Solution 2
Here is my solution, I tested it as well.
package small_Progs;
class Point {
int x, y;
Point(int x, int y) {
this.x = x;
this.y = y;
}
}
class Rectangle {
Point lt, lb, rt, rb;
Rectangle(Point lt, Point lb, Point rt, Point rb) {
this.lt = lt;
this.lb = lb;
this.rt = rt;
this.rb = rb;
}
}
public class OverlappingRectagles {
public static void main(String arg[]) {
Point lt1 = new Point(3, 8);
Point lb1 = new Point(3, 5);
Point rt1 = new Point(6, 8);
Point rb1 = new Point(6, 5);
Point lt2 = new Point(5, 6);
Point lb2 = new Point(5, 3);
Point rt2 = new Point(9, 6);
Point rb2 = new Point(9, 3);
Point lt3 = new Point(3, 7);
Point lb3 = new Point(3, 6);
Point rt3 = new Point(5, 7);
Point rb3 = new Point(5, 6);
Point lt4 = new Point(1, 2);
Point lb4 = new Point(1, 1);
Point rt4 = new Point(2, 2);
Point rb4 = new Point(2, 1);
Rectangle r1 = new Rectangle(lt1, lb1, rt1, rb1);
Rectangle r2 = new Rectangle(lt2, lb2, rt2, rb2);
Rectangle r3 = new Rectangle(lt3, lb3, rt3, rb3);
Rectangle r4 = new Rectangle(lt4, lb4, rt4, rb4);
OverlappingRectagles obj = new OverlappingRectagles();
obj.isOverLapping(r1, r2);
obj.isOverLapping(r1, r3);
obj.isOverLapping(r1, r4);
}
private void isOverLapping(Rectangle rect1, Rectangle rect2) {
Point l1 = rect1.lt;
Point l2 = rect2.lt;
Point r1 = rect1.rb;
Point r2 = rect2.rb;
if (l1.y < l2.y || l2.y < r1.y) {
System.out.println("Not Overlapping");
} else if (l1.x > r2.x || l2.x > r1.x) {
System.out.println("Not Overlapping");
} else {
if ((l1.y > r2.y && l2.y > r1.y) || (l2.y > r1.y && r2.y > r2.y)) {
System.out.println("Overlapping");
} else {
System.out.println("Not Overlapping");
}
}
}
}
Author by
George
Updated on June 04, 2022Comments
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George about 2 years
I studied this question, and followed the answer to implement my own version in Java. I think it is close... but still incorrect. Could you please give me some suggestion on the error?
The full source code can be found here:
// Determine if it is inside boolean isInside = ((r1x1 >= r2x1) && (r1x2 >= r2x2) && (r1y1 >= r2y1) && (r1y2 <= r2y2)); // Determine if it is overlap boolean isOverLap = (!(r1x1 >= r2x2) && !(r1x2 <= r2x2) && !(r1y2 >= r2y1) && !(r1y1 <= r2y2)); // Determine if it is NOT overlap boolean isNotOverLap = ((r1x1 >= r2x2) || (r1x2 <= r2x2) || (r1y2 >= r2y1) || (r1y1 <= r2y2));
According to the textbook I am studying, this is supposed to be:
r2 overlap r1
. But my program outputr2 does not overlap r1
.Enter the r1's center x, y coordinates, width and height 1 2 3 5.5 Enter the r2's center x, y coordinates, width and height 3 4 4.5 5 Rectangle 1: (-0.50, 4.75), (2.50, -0.75) Rectangle 2: (0.75, 6.50), (5.25, 1.50) r2 does not overlap r1
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arshajii almost 11 yearsWhat's wrong with
boolean isNotOverLap = !isOverLap
? :P -
Marko Topolnik almost 11 yearsIs this a riddle? What error?
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Hovercraft Full Of Eels almost 11 yearsTitle edited:
"Is you is, or is you ain't, my constituency?"
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George almost 11 yearsThank for pointing it out, but I was asking for
isOverLap
andisNotOverLap
...