<p>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. </p>

<p>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?</p>

<p>Thank you very much :)</p>

CC BY-SA 4.0
2020-02-24T12:41:01.897Z
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).
lightxbulb
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2020-02-24T12:48:21.890Z
Thank you very much, are you aware of something for squares as well?
Andreas
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2020-02-24T13:06:11.070Z
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".
lightxbulb
CC BY-SA 4.0
2020-02-24T13:34:07.867Z
You could make 2 triangles from the quad - sure. You could do a number of other things though, hence why it is ill-posed.
lightxbulb
CC BY-SA 4.0
2020-02-24T13:39:56.990Z
Another thing to note us that such a decomposition will make the problem piecewise-linear, which is not the same as linear.
lightxbulb
CC BY-SA 4.0
2020-02-24T13:58:22.587Z
Thank you very much for your replies, I guess this is one of the cases where I would have had to have been in the lecture to be not be tripped up by it :)
Andreas
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2020-02-24T14:26:38.710Z
The question is stupid as I mentioned, I may be jumping to conclusions, but if such questions are commonplace in the slides I would have skipped too and read from an actual resource.
lightxbulb
CC BY-SA 4.0
2020-02-25T08:56:22.180Z
I don't understand why we are assuming there is a problem. Once you accept that the linear mapping is fixed by defining 3 (non colinear) XY locations, with associated UV values, it's done.
Simon F
<texture><interpolation>
Is linear interpolation in a 2D texture possible (not quadratic like bilinear interpolation)?
Andreas
# Is linear interpolation in a 2D texture possible (not quadratic like bilinear interpolation)?

(Asked by: username_0 on 2020-02-24)

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 :)

**Tags:** texture, interpolation

> **Comments:**
> **username_1**: 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).  
> **username_0**: Thank you very much, are you aware of something for squares as well?  
> **username_1**: 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".  
> **username_1**: You could make 2 triangles from the quad - sure. You could do a number of other things though, hence why it is ill-posed.  
> **username_1**: Another thing to note us that such a decomposition will make the problem piecewise-linear, which is not the same as linear.  
> **username_0**: Thank you very much for your replies, I guess this is one of the cases where I would have had to have been in the lecture to be not be tripped up by it :)  
> **username_1**: The question is stupid as I mentioned, I may be jumping to conclusions, but if such questions are commonplace in the slides I would have skipped too and read from an actual resource.  
> **username_2**: I don't understand why we are assuming there is a problem. Once you accept that the linear mapping is fixed by defining 3 (non colinear) XY locations, with associated UV values, it's done.  

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## 0 answers

