Wednesday, May 7, 2014

Point inside a triangle or NOT

Problem

Given three corner points of a triangle, and one more point P. Write a function to check whether P lies within the triangle or not.

Example

For example, consider the following program, the function should return true for P(10, 15) and false for P’(30, 15)
              B(10,30)
                / \
               /   \
              /     \
             /   P   \      P'
            /         \
     A(0,0) ----------- C(20,0) 


Solution

Method 1 - Area of triangles made by point and other 2 vertices of triangle
Let the coordinates of three corners be (x1, y1), (x2, y2) and (x3, y3). And coordinates of the given point P be (x, y)
  1. Calculate area of the given triangle, i.e., area of the triangle ABC in the above diagram. Area A = [ x1(y2 - y3) + x2(y3 - y1) + x3(y1-y2)]/2
  2. Area A1 = Calculate area of the triangle PAB (using the same formula)
  3. Area A2 = Calculate area of the triangle PBC
  4. Area A3 = Calculate area of the triangle PAC
  5. If P lies inside the triangle, then A1 + A2 + A3 must be equal to A.
 Here is the code
float area(int x1, int y1, int x2, int y2, int x3, int y3)
{
   return Math.abs((x1*(y2-y3) + x2*(y3-y1)+ x3*(y1-y2))/2.0);
}

boolean isInside(int x1, int y1, int x2, int y2, int x3, int y3, int x, int y)
{   
   // Calculate area of triangle ABC 
   float A = area (x1, y1, x2, y2, x3, y3);
 
   // Calculate area of triangle PBC   
   float A1 = area (x, y, x2, y2, x3, y3);
 
   // Calculate area of triangle PAC   
   float A2 = area (x1, y1, x, y, x3, y3);
 
   // Calculate area of triangle PAB    
   float A3 = area (x1, y1, x2, y2, x, y);
   
   // Check if sum of A1, A2 and A3 is same as A 
   return (A == A1 + A2 + A3);
}

But can this differentiate whether the point lies on the edges, I don't thing so. It can only tell whether it is in triangle or not.

Method 2 - Using Barycentric coordinates
Solve the following equation system:
p = p0 + (p1 - p0) * s + (p2 - p0) * t
The point p is inside the triangle if 0 <= s <= 1 and 0 <= t <= 1 and s + t <= 1.
s,t and 1 - s - t are called the barycentric coordinates of the point p.

Now, the barycentric coordinates, generally called alpha, beta, and gamma, are calculated as follows:
float alpha = ((p2.y - p3.y)*(p.x - p3.x) + (p3.x - p2.x)*(p.y - p3.y)) /
        ((p2.y - p3.y)*(p1.x - p3.x) + (p3.x - p2.x)*(p1.y - p3.y));
float beta = ((p3.y - p1.y)*(p.x - p3.x) + (p1.x - p3.x)*(p.y - p3.y)) /
       ((p2.y - p3.y)*(p1.x - p3.x) + (p3.x - p2.x)*(p1.y - p3.y));
float gamma = 1.0f - alpha - beta;
If all of alpha, beta, and gamma are greater than 0, then the point p lies within the triangle made of points p1, p2, and p3.
The explanation behind this is that a point inside a triangle can be described using the points of the triangle, and three coefficients (one for each point, in the range [0,1]):
p = (alpha)*p1 + (beta)*p2 + (gamma)*p3
Rearranging this function gives you the formula to compute barycentric coordinates, but I feel like the steps to do so might be beyond the scope of the question. They are provided on the Wikipedia page that I linked up top.

References

Reactions:

0 comments:

Post a Comment