1
$\begingroup$

How to find diagonals of any convex quadrilateral?

I have only Points which are corners P1(x1,y1), P2(x2,y2), P3(x3,y3), P4(x4,y4). How can I distinguish diagonals from sides?

I couldn't find an answer anywhere on the web.

  • 0
    For a general convex polygon, such an algorithm is called a "convex hull". With that search term you will find a lot of information on the Web. For four vertices one can do it simply by checking the angles from any one corner to the other three. It is worth the effort to think it through yourself. – 2017-02-04
  • 0
    I didn't think this way. Thanks a lot for comment it clears a lot to me. – 2017-02-04
  • 0
    @JackD'Aurizio: The Question claims that the quadrilateral having the four vertices is convex, so if the convex hull is a triangle, we would have a contradiction. – 2017-02-05
  • 0
    @hardmath: my bad, I didn't notice the term *convex*. – 2017-02-05

1 Answers 1

1

For a well-known implementation in 2D of the general polygon orientation, see:

[2D Convex Hulls and Extreme Points]

CGAL provides implementations of several classical algorithms for computing the counterclockwise sequence of extreme points for a set of points in two dimensions (i.e., the counterclockwise sequence of points on the convex hull).

A tailored approach to the convex hull/cyclic orientation of a quadrilateral might be the following.

Pick an interior point $Q=(\overline{x},\overline{y})$ of the convex quadrilateral by averaging the four corner points $P_1=(x_1,y_1),P_2=(x_2,y_2),P_3=(x_3,y_3),P_4=(x_4,y_4)$. Sort the vectors from $Q$ to each $P_i$ by angle of elevation/depression. The endpoints of diagonals will occupy alternating entries of this sorted list.

It is possible to avoid explicit use of trigonometry in this sorting by pigeonholing the vectors $P_i - Q$ into quadrants (based on signs of the coordinate components) and sorting within a quadrant based on "steepness" of the slopes of the vectors.

Another computational approach would be to solve the segment intersection problem for $\overline{P_1 P_2}$ and $\overline{P_3 P_4}$. If they intersect (in a point belonging to both line segments), then these are the diagonals we seek. If they don't intersect, then these are a pair of sides. Unfortunately we may need to repeat the test (once) if the latter occurs, in order to determine which pairs of points form the diagonals.

A minimal amount of computation?

Given three points $A=(a_x,a_y),B=(b_x,b_y),C=(c_x,c_y)$, we can determine if the triangular loop $A\to B\to C\to A$ goes counterclockwise (or not) from the sign of the $3\times 3$ determinant:

$$ \left| \begin{array}{ccc} a_x & a_y & 1 \\ b_x & b_y & 1 \\ c_x & c_y & 1 \end{array} \right| = (b_x-a_x)(c_y-a_y)-(c_x-a_x)(b_y-a_y) $$

This determinant is positive if the loop goes counterclockwise, negative if it goes clockwise, and zero exactly when the three points are co-linear.

If $\overline{P_1P_2}$ is a diagonal, then the loops $P_1P_2P_3$ and $P_1P_2P_4$ will have opposite orientations, but if $\overline{P_1P_2}$ is a side, then they will have the same orientation.

Since $P_1P_2P_3$ gives the same orientation if we consider it as $P_2P_3P_1$, it suffices to check at most $3$ triangular loop orientations. If loops $P_1P_2P_3$ and $P_1P_2P_4$ have the same orientation, we only need to check the orientation of $P_2P_3P_4$. If it has opposite orientation to $P_1P_2P_3$, then $\overline{P_2P_3}$ is a diagonal; otherwise, if the orientation is again the same, then $\overline{P_2P_4}$ is a diagonal.