Perspective correct interpolation

Reading this very useful series about the graphics pipeline I realized I had never thought about what is the proper way to interpolate between vertex attributes (outputted by the vertex, domain or geometry shader to the pixel shader) using the vertices of the triangle after perspective divison. I spent some time thinking on that, and although it’s not hard to figure out and the result is described in the link above, I thought I describe one possible way to get the solution.

Let’s assume we have a triangle before perspective divison with vertices $(p_0, w_0)$ , $(p_1, w_1)$ , $(p_2,w_2)$ with some corresponding vertex attributes $c_0$ , $c_1$ , $c_2$ (e.g. texture coordinates, color, normal). After perspective divison we have $p'_0=\frac{p_0}{w_0}$ , $p'_1=\frac{p_1}{w_1}$ , $p'_2=\frac{p_2}{w_2}$ . Let’s have $p'=(1-t'_1-t'_2)p'_0+t'_1p'_1+t'_2p'_2$ . The question is, how should we compute the proper vertex attribute $c$ corresponding to $p'$ ?

Let’s think backward: assume we have $p=(1-t_1-t_2)p_0+t_1p_1+t_2p_2$ . Then, of course, the vertex attribute corresponding to $p$ is simply $c= (1-t_1-t_2)c_0+t_1c_1+t_2c_2$ . Now, what is the point $p'$ we get from $p$ after perspective divison? Using the notation $w = (1-t_1-t_2)w_0+t_1w_1+t_2w_2$ we have

$p' =\frac{p}{w} \\ \quad = \frac{(1-t_1-t_2)p_0+t_1p_1+t_2p_2 }{w} \\ \quad =\frac{(1-t_1-t_2)w_0}{w}\frac{p_0}{w_0}+ \frac{t_1w_1}{w}\frac{p_1}{w_1} + \frac{t_2w_2}{w}\frac{p_2}{w_2} \\ \quad = \frac{(1-t_1-t_2)w_0}{w}p'_0+ \frac{t_1w_1}{w}p'_1 + \frac{t_2w_2}{w}p'_2$

Well, it’s easy to check that the sum of the coefficients in the above interpolation between $p'_0$ , $p'_1$ and $p'_2$ is $1$ . So by setting $t_1'= \frac{t_1w_1}{w}$ and $t_2'= \frac{t_2w_2}{w}$ , we can write $p'= (1-t'_1-t'_2)p'_0+t'_1p'_1+t'_2p'_2$ .

And basically, we are done: for a point $p'= (1-t'_1-t'_2)p'_0+t'_1p'_1+t'_2p'_2$ , the corresponding vertex attribute is $c= (1-t_1-t_2)c_0+t_1c_1+t_2c_2$ , where $t_1$ and $t_2$ can be computed from the two equations $t_1'= \frac{t_1w_1}{w}$ and $t_2'= \frac{t_2w_2}{w}$ , where $w = (1-t_1-t_2)w_0+t_1w_1+t_2w_2$ .

Of course, we can solve for $t_1$ and $t_2$ explicitly (it’s just a system a linear equations with two variable), and we get

$t_1 = \frac{t'_1w^{-1}_1}{\tilde{w}}$ , and $t_2 = \frac{t'_2w^{-1}_2}{\tilde{w}}$ ,

where $\tilde{w} = (1-t'_0-t'_1)w^{-1}_0 + t'_1w^{-1}_1 +t'_2 w^{-1}_2$ . We can go further by substituting $t_1$ and $t_2$ into the equation for $c$

$c= (1-t_1-t_2)c_0+t_1c_1+t_2c_2 \\ (1- \frac{t'_1w^{-1}_1}{\tilde{w}} - \frac{t'_2w^{-1}_2}{\tilde{w}} )c_0 + \frac{t'_1w^{-1}_1}{\tilde{w}} c_1 + \frac{t'_2w^{-1}_2}{\tilde{w}} c_2 \\ = ((\tilde{w} - t'_1w^{-1}_1 - t'_2w^{-1}_2)c_0 + t'_1w^{-1}_1 c_1 + t'_1w^{-1}_2 c_2)/\tilde{w} \\= ((1-t'_1-t'_2)\frac{c_0}{w_0} + t'_1\frac{c_1}{w_1} + t'_2\frac{c_2}{w_2})/\tilde{w}$

Therefore, to do the interpolation in practice, we have two main choices:

For given $t'_1$ and $t'_2$ values (that is, basically, for a given pixel inside the triangle) we calculate $t_1$ and $t_2$ , and use them to calculate all vertex attributes by a simple linear interpolation.
For all vertex attributes $c_i$ , we compute the corresponding $\frac{c_i}{w_i}$ values, and use them to calculate the vertex attributes directly from given $t'_1$ and $t'_2$ .