Convex hull of 4 points

Question

I would like an algorithm to calculate the convex hull of 4 2D points. I have looked at the algorithms for the generalized problem, but I wonder if there is a simple solution for 4 points.

Answer 1

Take three of the points, and determine whether their triangle is clockwise or counterclockwise::

triangle_ABC= (A.y-B.y)*C.x + (B.x-A.x)*C.y + (A.x*B.y-B.x*A.y)

For a right-handed coordinate system, this value will be positive if ABC is counterclockwise, negative for clockwise, and zero if they are collinear. But, the following will work just as well for a left-handed coordinate system, as the orientation is relative.

Compute comparable values for three triangles containing the fourth point:

triangle_ABD= (A.y-B.y)*D.x + (B.x-A.x)*D.y + (A.x*B.y-B.x*A.y)
triangle_BCD= (B.y-C.y)*D.x + (C.x-B.x)*D.y + (B.x*C.y-C.x*B.y)
triangle_CAD= (C.y-A.y)*D.x + (A.x-C.x)*D.y + (C.x*A.y-A.x*C.y)

If all three of {ABD,BCD,CAD} have the same sign as ABC, then D is inside ABC, and the hull is triangle ABC.

If two of {ABD,BCD,CAD} have the same sign as ABC, and one has the opposite sign, then all four points are extremal, and the hull is quadrilateral ABCD.

If one of {ABD,BCD,CAD} has the same sign as ABC, and two have the opposite sign, then the convex hull is the triangle with the same sign; the remaining point is inside it.

If any of the triangle values are zero, the three points are collinear and the middle point is not extremal. If all four points are collinear, all four values should be zero, and the hull will be either a line or a point. Beware of numerical robustness problems in these cases!

For those cases where ABC is positive:

ABC  ABD  BCD  CAD  hull
------------------------
 +    +    +    +   ABC
 +    +    +    -   ABCD
 +    +    -    +   ABDC
 +    +    -    -   ABD
 +    -    +    +   ADBC
 +    -    +    -   BCD
 +    -    -    +   CAD
 +    -    -    -   [should not happen]