calculating collisions between 2 objects with known size and position

Question

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Determine if two rectangles overlap each other?

Considering I have 2 squares for which I know the x and y positions and I also know the size, what would be the formula to use if I wanted to see if the objects collide with eachother.

if(   ((shapeA->getX() - shapeA->getSize()) > (player->getX() - player->getSize())
    && (shapeA->getX() + shapeA->getSize()) < (player->getX() + player->getSize()))
    && (shapeA->getY() - shapeA->getSize()  > player->getY() - player->getSize()
    && (shapeA->getY() + shapeA->getSize()) < (player->getY() + player->getSize()))
                )

This works, but it works strange (not all the time). I must be missing something

You should use the debugger (or just add helpful print statements) to identify the values of all variables involved, and figure out why it doesn't work for the failing cases. — Oliver Charlesworth, May 31 '12 at 10:07
find the equation of all edges of the squares, and check if two edges are intersecting or not.. — ardiyu07, May 31 '12 at 10:12

score 4 · Accepted Answer · answered May 31 '12 at 10:20

It's very easy to check whether a rectangle intersects or touches another rectangle. Have a look at the following picture:

Rectangular intersection

As you can see, two rectangles intersect if the intersections between ([x,x+a] and [X,X+A]) and ([y,y+b] and [Y,Y+B]) both aren't empty.

struct Rectangle{
    bool intersects(const Rectangle&);

    unsigned int a; //!< width of the rectangle
    unsigned int b; //!< height of the rectangle
    unsigned int x; //!< x position
    unsigned int y; //!< y position
};

bool Rectangle::intersects(const Rectangle& oRectangle){        
    return  (x < oRectangle.x + oRectangle.a) && // [x,x+a], [X,X+A] intersection
            (oRectangle.x < x + a)            && // [x,x+a], [X,X+A] intersection
            (y < oRectangle.y + oRectangle.b) && // [y,y+b], [Y,Y+B] intersection
            (oRectangle.y < y + b);              // [y,y+b], [Y,Y+B] intersection
}

So your code should be

if(((shapeA->getX() + shapeA->getSize()) > (player->getX()) // x intersection
    && (shapeA->getX() < (player->getX() + player->getSize())) // x intersection
    && (shapeA->getY() < player->getY() + player->getSize()  // y intersection
    && (shapeA->getY() + shapeA->getSize()) > player->getY())  // y intersection
)

rwst · Answer 2 · 2012-06-02T15:46:33.760

2

Assuming that getX/Y gives the bottom left corner of the square,

shapeMinX = shapeA; shapeMaxX = shapeB;
if (shapeA()->getX() > shapeB()->getX())
  swap (shapeMinX, shapeMaxX);
shapeMinY = shapeA; shapeMaxY = shapeB;
if (shapeA()->getY() > shapeB()->getY())
  swap (shapeMinY, shapeMaxY);

collision = (shapeMinX->getX()+shapeMinX->size() >= shapeMaxX()->getX) || (shapeMinX->getY()+shapeMinY->size() >= shapeMaxY()->getY);

edited Jun 02 '12 at 15:46

answered May 31 '12 at 10:14

rwst

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This answer lacks explanation (+I doubt you can do this with one `or`, but I could be mistaken, I won't know until you explain). – KillianDS May 31 '12 at 10:17
You probably mean `swap(shapeMinX, shapeMaxX)` and not `switch(shapeMinX, shapeMaxX)`. – RedX May 31 '12 at 10:20

score 2 · Answer 3 · answered May 31 '12 at 10:29

You do wrong tests, try this:

int left_bound_A=  shapeA->getX()-shapeA->getSize();
int right_bound_A= shapeA->getX()+shapeA->getSize();
int top_bound_A= shapeA->getY()-shapeA->getSize();
int bottom_bound_A= shapeA->getY()+shapeA->getSize();

int left_bound_B=  shapeB->getX()-shapeB->getSize();
int right_bound_B= shapeB->getX()+shapeB->getSize();
int top_bound_B= shapeB->getY()-shapeB->getSize();
int bottom_bound_B= shapeB->getY()+shapeB->getSize();

if( left_bound_A < right_bound_B &&
    right_bound_A > left_bound_B &&
    top_bound_A > bottom_bound_B &&
    bottom_bound_A < top_bound_B ) colide(shapeA,shapeB);

The general way is to test for shape intersection. If you implement a Box or Rectangle class, the code simplify to:

 Box colision= intersect( shapeA->getBoundBox(), shapeB->getBoundBox() );
 if( colision.have_positive_area() )
     colide(shapeA,shapeB,colision);

Why i don't get that popup that thread have new answers and I need to reload? Fuuuuu... — Arpegius, May 31 '12 at 10:31

score 0 · Answer 4 · edited May 23 '17 at 12:06

0

Yes, the way to check if two rectangles is simple. Just as a suggestion, if you then want to compute all the possible intersection between the rectangles in a list of rectangles, preorder them by increasing x of their border and then start the comparison exploiting this relation. This question may be of help

Fast hiding of intersecting rectangles can be of interest

edited May 23 '17 at 12:06

Community

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answered May 31 '12 at 11:01

linello

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calculating collisions between 2 objects with known size and position

4 Answers4