Two Rectangles intersection

后端 未结 7 2022
北海茫月
北海茫月 2020-11-27 16:39

I have two rectangles characterized by 4 values each :

Left position X, top position Y, width W and height H:

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7条回答
  • 2020-11-27 16:51

    Best example..

    /**
     * Check if two rectangles collide
     * x_1, y_1, width_1, and height_1 define the boundaries of the first rectangle
     * x_2, y_2, width_2, and height_2 define the boundaries of the second rectangle
     */
    boolean rectangle_collision(float x_1, float y_1, float width_1, float height_1, float x_2, float y_2, float width_2, float height_2)
    {
      return !(x_1 > x_2+width_2 || x_1+width_1 < x_2 || y_1 > y_2+height_2 || y_1+height_1 < y_2);
    }
    

    and also one other way see this link ... and code it your self..

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  • 2020-11-27 17:01

    Using a coordinate system where (0, 0) is the left, top corner.

    I thought of it in terms of a vertical and horizontal sliding windows and come up with this:

    (B.Bottom > A.Top && B.Top < A.Bottom) && (B.Right > A.Left && B.Left < A.Right)

    Which is what you get if you apply DeMorgan’s Law to the following:

    Not (B.Bottom < A.Top || B.Top > A.Bottom || B.Right < A.Left || B.Left > A.Right)

    1. B is above A
    2. B is below A
    3. B is left of A
    4. B is right of A
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  • 2020-11-27 17:01

    If the rectangles' coordinates of the lower left corner and upper right corner are :
    (r1x1, r1y1), (r1x2, r1y2) for rect1 and
    (r2x1, r2y1), (r2x2, r2y2) for rect2
    (Python like code below)

        intersect = False
        for x in [r1x1, r1x2]:
            if (r2x1<=x<=r2x2):
                for y in [r1y1, r1y2]:
                    if (r2y1<=y<=r2y2):
                        intersect = True
                        return intersect
                    else:
                        for Y in [r2y1, r2y2]:
                            if (r1y1<=Y<=r1y2):
                                intersect = True
                                return intersect
            else:  
                for X in [r2x1, r2x2]:
                    if (r1x1<=X<=r1x2):
                        for y in [r2y1, r2y2]:
                            if (r1y1<=y<=r1y2):
                                intersect = True
                                return intersect
                            else:
                                for Y in [r1y1, r1y2]:
                                    if (r2y1<=Y<=r2y2):
                                        intersect = True
                                        return intersect
        return intersect
    
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  • 2020-11-27 17:04

    Should the two rectangles have the same dimensions you can do:

    if (abs (x1 - x2) < w && abs (y1 - y2) < h) {
        // overlaps
    }
    
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  • 2020-11-27 17:05

    I just tried with a c program and wrote below.

    #include<stdio.h>
    
    int check(int i,int j,int i1,int j1, int a, int b,int a1,int b1){
        return (\
        (((i>a) && (i<a1)) && ((j>b)&&(j<b1))) ||\ 
        (((a>i) && (a<i1)) && ((b>j)&&(b<j1))) ||\ 
        (((i1>a) && (i1<a1)) && ((j1>b)&&(j1<b1))) ||\ 
        (((a1>i) && (a1<i1)) && ((b1>j)&&(b1<j1)))\
        );  
    }
    int main(){
        printf("intersection test:(0,0,100,100),(10,0,1000,1000) :is %s\n",check(0,0,100,100,10,0,1000,1000)?"intersecting":"Not intersecting");
        printf("intersection test:(0,0,100,100),(101,101,1000,1000) :is %s\n",check(0,0,100,100,101,101,1000,1000)?"intersecting":"Not intersecting");
        return 0;
    }
    
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  • 2020-11-27 17:05

    if( X1<=X2+W2 && X2<=X1+W1 && Y1>=Y2-H2 && Y2>=Y1+H1 ) Intersect

    In the question Y is the top position..

    Note: This solution works only if rectangle is aligned with X / Y Axes.

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