Question

Given the triangle with vertices (a,b,c):

        c

      /   \

    /       \

  /           \

 a  -  -  -  -  b

Which is then subdivided into four triangles by halving each of the edges:

              c

            /    \

          /        \

     ca /            \ bc
           _   _   _
      /\              /\

    /    \          /    \

  /        \      /        \

a  -  -  -  - ab  -  -   -   -b

Which results in four triangles (a, ab, ca), (b, bc, ab), (c, ca, bc), (ab, bc, ca).

Now given a point p. How do I determine in which triangle p lies, given that p is within the outer triangle (a, b, c)?

Currently I intend to use ab as the origin. Check whether it is to the left of right of the line “ca – ab” using the perp of “ca – ab” and checking the sign against the dot product of “ab – a” and the perp vector and the vector “p – ab”. If it is the same or the dot product is zero then it must be in (a, ab, ca)… Continue with this procedure with the other outer triangles (b, ba, ab) & (c, ca, ba). In the end if it didn’t match with these it must be contained within the inner triangle (ab, bc, ca).

Is there a better way to do it?

EDIT

Here is a little more info of the intended application of the algorithm:

I’m using this as a subdivision mask to generate a fine mesh over which I intend to interpolate. Each of the triangles will be subdivided similarly up to a specified depth. I want to determine the triangle (at the maximum depth) within which the point p lies. With this I can evaluate a function at the point p using interpolation over the triangle. There is a class of triangles which is right-angled and they do comprise a significant portion, but they’re much easier to work with and this algorithm isn’t intended for them.

Answer 1

• If the point is above ca/bc (i.e. in the top grey triangle) it’s easy.

• If the point is left of ca (i.e. in the left grey triangle) it’s easy.

• If the point is right of bc (i.e. in the right grey triangle) it’s easy.

• If the point is in the middle, all you have to do is determine if the point is above or below the black V.

  You can do that by calulcating the y value of the line for the x value of the point and compare the result to the y value of the point.

  if (y' > (y * x') / x)
  {
      // center triangle
  }
  else
  {
      // right triangle
  }
  

Is this the most efficient method? I don’t know.