找出点是否位于矩形内

Question

如何找到一个点位于给定4个点的2D矩形内？

Answer 1

Transform the point to a coordinate frame aligned with the rectangle, then the problem becomes axis-aligned and trivial.

If the rectangle consists of the following 4 points:

a  b
c  d

Then get the "x-axis" and "y-axis" of the rectangle as:

x = Normalize(d-c)
y = Normalize(a-c)

Then construct a rotation matrix using x and y as columns:

r = [ x | y ]

If you're using 3-d coordinates, we need az axis:

z = CrossProduct(x, y)
r = [ x | y | z ]

Your transform matrix from world coordinates to your rectangle's axis-aligned coordinates becomes:

T = [ r^T | -r^T * c ]
    [ 0^T |     1    ]

Here we've chosen the lower-left corner c to be the local origin. "r^T" is r transposed. "0^T" is either a 2-d or 3-d row-vector filled with zeros. 1 is just a one. Note that this is just the inverse of the simpler rectangle-to-world transform, which is

T^-1 = [ r   | c ]
       [ 0^T | 1 ]

We can use T to transform the point to axis-aligned coordinates. Remember to pad p with a trailing 1, since T is a homogeneous matrix.

tp = T * p;  // Don't forget to pad p with a trailing 1 before multiplying.

// Checks that p isn't below or to the left of the rectangle.
for ( int d = 0; d < num_dimensions; ++d ) {
  if ( tp[d] < 0.0 ) {
    return false;
  }
}

// Checks that p isn't to the right of the rectangle
double width = Length(d-c);
if ( tp[0] > width ) {
  return false;
}

// Checks that p isn't above the rectangle.
double height = Length(a-c);
if ( tp[1] > height ) {
  return false;
}

// p must be inside or on the rectangle.
return true

If you're using 3d coordinates, note that the above disregards the local z value of transformed point tp. Even if p is out of the plane of the rectangle, the above behaves as if it's been projected to the rectangle surface. If you want to check for coplanarity, just do the following beforehand:

if ( fabs(tp[2]) > some_small_positive_number ) {
   return false;  // point is out of the rectangle's plane.
}

Answer 2

I think this might answer your question

• full disclosure - I went to Drexel for my grad dregree

Answer 3

To make it OpenGl specific:

I suppose your 2D rectangle is in screen coordinates!

first:

 gluProject (bli, bla, blorp, ...);

to get from 3d to screen coordinates.

Then: Noah's suggestion.

Only shoot yourself if your point in question is already 2D ;)

Answer 4

For a non-axis-aligned rectangle, use the same algorithm as for general polygons: the point-in-polygon test:

Imagine a ray pointing rightward from the test point. Test whether each line in the polygon crosses the ray. If an even number of lines crosses the ray, then the point is outside the polygon. If an odd number of lines crosses the ray, then the point is inside the polygon.

In the case of a rectangle, between zero and two lines will cross the ray.

If a line touches the ray but does not cross it, the result is ambiguous. Therefore, in your calculations, imagine that the ray is an infinitely small amount ɛ higher than its y coordinate, so that it is impossible for a line to touch the ray without crossing it.

Given your test point (x,y) and line (x1,y1,x2,y2), testing whether a line crosses the ray is pretty simple. Assume, without loss of generality, that y1 < y2. Then

if y < y2 and y >= y1:
    let x0 = x1 + (y-y1)/(y2-y1) * (x2-x1) // crossing point (x0,y)
    if x0 > x:
        crossing_detected++

http://en.wikipedia.org/wiki/Point_in_polygon

Answer 5

It's easy to test if a point lies in a triangle, so you can split your rectangle in two triangles and test these. See eg http://www.blackpawn.com/texts/pointinpoly/default.html

Answer 6

Generic test for point in quadrilateral is sufficient. The quad is defined as an ordered series of points. Handles both clockwise and counter-clockwise winding:

typedef struct {float x; float y} vec2;
bool pointIsInQuad(const vec2 point, const vec2 quad[4])
{
    bool sides[4];
    for (int i = 0; i < 4; i++) {
        sides[i] = ((point.x - quad[i].x)*(quad[(i + 1)%4].y - quad[i].y) - (point.y - quad[i].y)*(quad[(i + 1)%4].x - quad[i].x)) > 0.0f;
    }
    return ((sides[0] == sides[1]) && (sides[0] == sides[2]) && (sides[0] == sides[3]));
}