Determine if two rectangles overlap each other?

Question

I am trying to write a C++ program that takes the following inputs from the user to construct rectangles (between 2 and 5): height, width, x-pos, y-pos. All of these rectang

Answer 1

Java code to figure out if Rectangles are contacting or overlapping each other

...

for ( int i = 0; i < n; i++ ) {
    for ( int j = 0; j < n; j++ ) {
        if ( i != j ) {
            Rectangle rectangle1 = rectangles.get(i);
            Rectangle rectangle2 = rectangles.get(j);

            int l1 = rectangle1.l; //left
            int r1 = rectangle1.r; //right
            int b1 = rectangle1.b; //bottom
            int t1 = rectangle1.t; //top

            int l2 = rectangle2.l;
            int r2 = rectangle2.r;
            int b2 = rectangle2.b;
            int t2 = rectangle2.t;

            boolean topOnBottom = t2 == b1;
            boolean bottomOnTop = b2 == t1;
            boolean topOrBottomContact = topOnBottom || bottomOnTop;

            boolean rightOnLeft = r2 == l1;
            boolean leftOnRight = l2 == r1;
            boolean rightOrLeftContact = leftOnRight || rightOnLeft;

            boolean leftPoll = l2 <= l1 && r2 >= l1;
            boolean rightPoll = l2 <= r1 && r2 >= r1;
            boolean leftRightInside = l2 >= l1 && r2 <= r1;
            boolean leftRightPossiblePlaces = leftPoll || rightPoll || leftRightInside;

            boolean bottomPoll = t2 >= b1 && b2 <= b1;
            boolean topPoll = b2 <= b1 && t2 >= b1;
            boolean topBottomInside = b2 >= b1 && t2 <= t1;
            boolean topBottomPossiblePlaces = bottomPoll || topPoll || topBottomInside;


            boolean topInBetween = t2 > b1 && t2 < t1;
            boolean bottomInBetween = b2 > b1 && b2 < t1;
            boolean topBottomInBetween = topInBetween || bottomInBetween;

            boolean leftInBetween = l2 > l1 && l2 < r1;
            boolean rightInBetween = r2 > l1 && r2 < r1;
            boolean leftRightInBetween = leftInBetween || rightInBetween;

            if ( (topOrBottomContact && leftRightPossiblePlaces) || (rightOrLeftContact && topBottomPossiblePlaces) ) {
                path[i][j] = true;
            }
        }
    }
}

...

Answer 2

In the question, you link to the maths for when rectangles are at arbitrary angles of rotation. If I understand the bit about angles in the question however, I interpret that all rectangles are perpendicular to one another.

A general knowing the area of overlap formula is:

Using the example:

   1   2   3   4   5   6

1  +---+---+
   |       |   
2  +   A   +---+---+
   |       | B     |
3  +       +   +---+---+
   |       |   |   |   |
4  +---+---+---+---+   +
               |       |
5              +   C   +
               |       |
6              +---+---+

1) collect all the x coordinates (both left and right) into a list, then sort it and remove duplicates

1 3 4 5 6

2) collect all the y coordinates (both top and bottom) into a list, then sort it and remove duplicates

1 2 3 4 6

3) create a 2D array by number of gaps between the unique x coordinates * number of gaps between the unique y coordinates.

4 * 4

4) paint all the rectangles into this grid, incrementing the count of each cell it occurs over:

   1   3   4   5   6

1  +---+
   | 1 | 0   0   0
2  +---+---+---+
   | 1 | 1 | 1 | 0
3  +---+---+---+---+
   | 1 | 1 | 2 | 1 |
4  +---+---+---+---+
     0   0 | 1 | 1 |
6          +---+---+

5) As you paint the rectangles, its easy to intercept the overlaps.

Answer 3

A and B be two rectangle. C be their covering rectangle.

four points of A be (xAleft,yAtop),(xAleft,yAbottom),(xAright,yAtop),(xAright,yAbottom)
four points of A be (xBleft,yBtop),(xBleft,yBbottom),(xBright,yBtop),(xBright,yBbottom)

A.width = abs(xAleft-xAright);
A.height = abs(yAleft-yAright);
B.width = abs(xBleft-xBright);
B.height = abs(yBleft-yBright);

C.width = max(xAleft,xAright,xBleft,xBright)-min(xAleft,xAright,xBleft,xBright);
C.height = max(yAtop,yAbottom,yBtop,yBbottom)-min(yAtop,yAbottom,yBtop,yBbottom);

A and B does not overlap if
(C.width >= A.width + B.width )
OR
(C.height >= A.height + B.height) 

It takes care all possible cases.

Answer 4

This is from exercise 3.28 from the book Introduction to Java Programming- Comprehensive Edition. The code tests whether the two rectangles are indenticle, whether one is inside the other and whether one is outside the other. If none of these condition are met then the two overlap.

**3.28 (Geometry: two rectangles) Write a program that prompts the user to enter the center x-, y-coordinates, width, and height of two rectangles and determines whether the second rectangle is inside the first or overlaps with the first, as shown in Figure 3.9. Test your program to cover all cases. Here are the sample runs:

Enter r1's center x-, y-coordinates, width, and height: 2.5 4 2.5 43 Enter r2's center x-, y-coordinates, width, and height: 1.5 5 0.5 3 r2 is inside r1

Enter r1's center x-, y-coordinates, width, and height: 1 2 3 5.5 Enter r2's center x-, y-coordinates, width, and height: 3 4 4.5 5 r2 overlaps r1

Enter r1's center x-, y-coordinates, width, and height: 1 2 3 3 Enter r2's center x-, y-coordinates, width, and height: 40 45 3 2 r2 does not overlap r1

import java.util.Scanner;

public class ProgrammingEx3_28 {
public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    System.out
            .print("Enter r1's center x-, y-coordinates, width, and height:");
    double x1 = input.nextDouble();
    double y1 = input.nextDouble();
    double w1 = input.nextDouble();
    double h1 = input.nextDouble();
    w1 = w1 / 2;
    h1 = h1 / 2;
    System.out
            .print("Enter r2's center x-, y-coordinates, width, and height:");
    double x2 = input.nextDouble();
    double y2 = input.nextDouble();
    double w2 = input.nextDouble();
    double h2 = input.nextDouble();
    w2 = w2 / 2;
    h2 = h2 / 2;

    // Calculating range of r1 and r2
    double x1max = x1 + w1;
    double y1max = y1 + h1;
    double x1min = x1 - w1;
    double y1min = y1 - h1;
    double x2max = x2 + w2;
    double y2max = y2 + h2;
    double x2min = x2 - w2;
    double y2min = y2 - h2;

    if (x1max == x2max && x1min == x2min && y1max == y2max
            && y1min == y2min) {
        // Check if the two are identicle
        System.out.print("r1 and r2 are indentical");

    } else if (x1max <= x2max && x1min >= x2min && y1max <= y2max
            && y1min >= y2min) {
        // Check if r1 is in r2
        System.out.print("r1 is inside r2");
    } else if (x2max <= x1max && x2min >= x1min && y2max <= y1max
            && y2min >= y1min) {
        // Check if r2 is in r1
        System.out.print("r2 is inside r1");
    } else if (x1max < x2min || x1min > x2max || y1max < y2min
            || y2min > y1max) {
        // Check if the two overlap
        System.out.print("r2 does not overlaps r1");
    } else {
        System.out.print("r2 overlaps r1");
    }

}
}

Answer 5

Ask yourself the opposite question: How can I determine if two rectangles do not intersect at all? Obviously, a rectangle A completely to the left of rectangle B does not intersect. Also if A is completely to the right. And similarly if A is completely above B or completely below B. In any other case A and B intersect.

What follows may have bugs, but I am pretty confident about the algorithm:

struct Rectangle { int x; int y; int width; int height; };

bool is_left_of(Rectangle const & a, Rectangle const & b) {
   if (a.x + a.width <= b.x) return true;
   return false;
}
bool is_right_of(Rectangle const & a, Rectangle const & b) {
   return is_left_of(b, a);
}

bool not_intersect( Rectangle const & a, Rectangle const & b) {
   if (is_left_of(a, b)) return true;
   if (is_right_of(a, b)) return true;
   // Do the same for top/bottom...
 }

bool intersect(Rectangle const & a, Rectangle const & b) {
  return !not_intersect(a, b);
}

Answer 6

I have implemented a C# version, it is easily converted to C++.

public bool Intersects ( Rectangle rect )
{
  float ulx = Math.Max ( x, rect.x );
  float uly = Math.Max ( y, rect.y );
  float lrx = Math.Min ( x + width, rect.x + rect.width );
  float lry = Math.Min ( y + height, rect.y + rect.height );

  return ulx <= lrx && uly <= lry;
}