I'm writing a program that will take in four $(x, y, z)$ points that form a quadrilateral, and produce two sets of $(x, y, z)$ points that form the two largest triangles that comprise the quadrilateral.

I'm having some difficulty in coming up with a way to determine which points make up the diagonals of the quad. Obviously if I can figure out two points that are guaranteed to make up a diagonal, the two triangles will be comprised of those two points plus one of the other remaining points.

So, given four points:

  • $A = (x_A, y_A, z_A)$
  • $B = (x_B, y_B, z_B)$
  • $C = (x_C, y_C, z_C)$
  • $D = (x_D, y_D, z_D)$

How can I determine which points form a diagonal of the quadrilateral?

  • $\begingroup$ Is your quadrilateral guaranteed to be convex? $\endgroup$ – Wouter Sep 25 '18 at 7:04
  • $\begingroup$ Yes, the quadrilateral will be convex. $\endgroup$ – ImpromptuMusic Sep 25 '18 at 7:05
  • $\begingroup$ Is it also planar? $\endgroup$ – Wouter Sep 25 '18 at 7:26
  • $\begingroup$ Yes, the quadrilaterals will all be planar, not skewed. $\endgroup$ – ImpromptuMusic Sep 25 '18 at 8:13
  • $\begingroup$ Find areas of all four possible triangles. In one combination if sum of triangles $ ABC,BCD$ equals that of $BAD,ADC$ then $BC,AD$ are the required diagonals. Else try other combination. $\endgroup$ – Narasimham Oct 4 '18 at 18:58

I think this approach is the most straightforward for planar, convex, 3D quadrilaterals:

enter image description here

Suppose that you have four points $(A,B,C,D)$ and that you want to check if AC is really a diagonal of the quadrilateral. If it is, the angles $\angle CAB$ and $\angle CAD$ will have different orientations (cunterclockwise/clockwise).

This means that vectors $\vec{c}_1$ and $\vec{c}_2$ defined as cross products:



...will be both perpendicular to the plane of the quadrilateral $ABCD$ but with opposite directions. This means that their scalar product must be negative:

$$s = \vec{c}_1 \cdot \vec{c}_2 < 0$$

So the algorithm goes as follows:

$$\vec AC=(x_C-y_A)\vec{i}+(y_C-y_A)\vec{j}+(z_C-z_A)\vec{k}=p_1\vec{i}+q_1\vec{j}+r_1\vec{k}$$

$$\vec AB=(x_B-x_A)\vec{i}+(y_B-y_A)\vec{j}+(z_B-z_A)\vec{k}=p_2\vec{i}+q_2\vec{j}+r_2\vec{k}$$

$$\vec AD=(x_D-x_A)\vec{i}+(y_D-y_A)\vec{j}+(z_D-z_A)\vec{k}=p_3\vec{i}+q_3\vec{j}+r_3\vec{k}$$




If $s<0$, $AC$ is a diagonal of the quadrilateral.

If not, you should try only once more, swapping places for $C$ and $B$ (or $C$ and $D$). If you fail again, $AD$ (or $AB$) has to be a diagonal (you don't have to check that).

EDIT: The method can be transformed into software code fairly easy. All computations are straightforward and not too expensive. And there is only one 'if' condition along the way.

  • 1
    $\begingroup$ Note that you needn't necessarily compute all of $s$: the sign of any one (non-zero) term $c_{1\star}c_{2\star}$ is all you need in order to determine whether the vectors point in opposite directions. Also note that you needn't actually compute each $c_{1\star}$ as an exact value: all you need to know is whether it's positive/negative/zero. That said, introducing the necessary logic handling on a CPU may be less performant than simply throwing the full vector calculations at a GPU. $\endgroup$ – Blue Sep 25 '18 at 14:28
  • $\begingroup$ @Blue: You are right, there is a lot of room for optimization, the part which I skipped. Thanks for spotting that. I do like to optimize the software, but I always try to make it work first :) $\endgroup$ – Oldboy Sep 25 '18 at 14:41
  • $\begingroup$ "Make it work, then make it pretty." :) $\endgroup$ – Blue Sep 25 '18 at 14:43
  • $\begingroup$ @Blue 1) Make it work 2) Make it work correctly 3) Make it work fast. In software industry steps 2) and 3) are optional. $\endgroup$ – Oldboy Sep 25 '18 at 14:53
  • $\begingroup$ Fantastic! Thanks for the help! $\endgroup$ – ImpromptuMusic Sep 26 '18 at 6:57

If the quarilateral is convex and planar, then this will work.

Given 4 points, there are 3 possibilities that might be the diagonals

  • AB,CD
  • AC,BD
  • AD,BC

Suppose AB,CD are the diagonals, then the intersection of AB and CD lies between A and B, and also between C and D. You can test this for all three possibilities, it will be true for just one.


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