I am currently studying for a computer graphics exam and on the slides about bilinear interpolation it is said that it is in reality quadratic interpolation.

An (on the slides) unanswered question then asks what interpolation on a 2D texture would be linear, but I can't really think of a way that is not cheating like going nearest neigbour in one dimension. Did I miss something or is it a trick question?

Thank you very much :)

  • $\begingroup$ Barycentric interpolation is linear, but it works for triangles only: $c = (1 - \xi - \eta)c_1 + \xi c_2 + \eta c_3$. For quadrilaterals it us not unique (there is redundancy). $\endgroup$
    – lightxbulb
    Commented Feb 24, 2020 at 12:41
  • $\begingroup$ Thank you very much, are you aware of something for squares as well? $\endgroup$
    – Andreas
    Commented Feb 24, 2020 at 12:48
  • $\begingroup$ There are generalized barycentric coordinates, but those are not exactly linear. A linear interpolation on a 2d texture would be either the interpolation in a column or a row - so actually 1D interpolation. Honestly that question in your slides is just ill-posed and stupid - it's equivalent to your lecturer saying "guess what I am thinking". $\endgroup$
    – lightxbulb
    Commented Feb 24, 2020 at 13:06
  • $\begingroup$ Couldn’t you use independent barycentric interpolation on the lower left and upper right triangles of each texel, effectively turning the texture map into a triangular grid? $\endgroup$
    – user2500
    Commented Feb 24, 2020 at 13:18
  • $\begingroup$ You could make 2 triangles from the quad - sure. You could do a number of other things though, hence why it is ill-posed. $\endgroup$
    – lightxbulb
    Commented Feb 24, 2020 at 13:34


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