What would be a clean way to define a function $F : R^2 \rightarrow R^2$ that maps coordinates from one quadrilateral domain to another like the below image? (Assuming we know the coordinates of the 4 points of each quadrilateral)

Does it matter if one of them is non-convex or is it possible to get a single nice function that handles all cases? If non-convex cases complicate the mapping, what would be a nice function assuming that both are convex?

  • $\begingroup$ If both are convex, map the corners of square A to the corners of square B. Every point in A is a convex combination of the vertices, so you map it to the corresponding combination of the vertices of B. If A is not convex, but B is convex, you can choose 3 of the vertices of A so that the triangle they determine covers all of A. Map these three vertices to 3 vertices of B and proceed as above. Some convex combinations of the 3 vertices don't lie in A, but you don't map them. The image of the map will not be all of B. I A isn't convex, I don't have a good idea. $\endgroup$
    – saulspatz
    Commented Feb 8, 2018 at 22:33
  • $\begingroup$ It depends on what other properties you want the map to have. Two common choices are a planar perspective transformation (homography) and bilinear interpolation. $\endgroup$
    – amd
    Commented Feb 8, 2018 at 23:22
  • $\begingroup$ @saulspatz: That doesn't fix the result uniquely, because each internal point has a one-dimensional range of convex combinations, which will not all correspond to the same point in B. $\endgroup$ Commented Feb 9, 2018 at 2:06

1 Answer 1


This can be done using range-split operation, $A,B \in (\mathbb{R},\mathbb{R})$, $x \in [0..1]$:

  • RangeSplit($A$,$B$,$x$) = $(1.0-x) \cdot A+x\cdot B$.

If your original rectangle coordinates are $(x,y), x \in [0..w], y \in [0..h]$, and corners of new quad is $(A..B),(C..D)$, then the final solution would be:

  • $(x_0,y_0) = (x/w,y/h)$
  • $(x_r,y_r)$ = RangeSplit(RangeSplit($A$,$B$,$x_0$),RangeSplit($C$,$D$,$x_0$),$y_0$)
  • $\begingroup$ this isn't really an answer to "Function to map 2D coordinates from one quadrilateral to another". Instead it's an answer "Function to map 2D coordinates from one axis aligned rectangle to another" right? $\endgroup$
    – gman
    Commented Feb 23, 2019 at 15:04

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