Select pixels in a hexagon around a center point

Question

I would like to implement a simple function into my code to get an list of pixel coordinates which are in a (hypothetical) hexagon of a certain size around a center point (x,y or also linear RGBA byte array, but I can convert later).

Maybe there's a simple solution I have not thought about. Could you think of a neat way to implement this?

I don't have a solution yet. I'm just grabbing pixels in a circular fashion, but would like to have a hexagon instead. — user3530347, Apr 16 '14 at 00:17

score 0 · Answer 1 · answered Apr 14 '14 at 04:14

All you really need is a list of pixels in one quadrant of the hexagon. Then you can simply "reflect" the x and y coordinates to get the full hexagon (subject to screen bounds of course).

First, I would make the assertion that I want one of my hexagon sides to be horizontally aligned. I would also make the assumption that by "size of my hexagon" I mean the length (let's call is L) of the vertical line from the center of my hexagon to the bottom side (which is horizontally aligned). Then, I would do algebra and trig on a hexagon of this alignment with size L, assuming that the origin (0,0) is my center point.

I know, then that all of the points within the bounds [0,0,L/sqrt(3),L] (that is [x-offset, y-offset, width, height]) are definitely within my hexagon. So add all these points to my list.

List<Point> pointsInHexagonQuadrant = new List<Point>();
for (int i = 0; i < L/Math.Sqrt(3); i++)  //I'm ignoring any casting, you may have to fix.
{
    for (int j = 0; j <= L; j++)
    {
         pointsInHexagonQuadrant.Add(new Point(i,j));
    }
}

I know by trig and algebra that the right-most point of my hexagon is at (2*L/sqrt(3),0) and from L/sqrt(3) to 2*L/sqrt(3) the equation of the hexagon's sloped side is y=sqrt(3)*x-2*L. I want all the points whose y coordinate is less than that.

for(int i = L/Math.Sqrt(3); i <= 2*L/Math.Sqrt(3); i++)
{
     for (int j = 0; j <= Math.Sqrt(3)*i-2*L; j++)
     {
          pointsInHexagonQuadrant.Add(new Point(i,j));
     }
}

Add this point you have one quadrant of the hexagon, like this:

(0,0)          (2L/sqrt(3),0)
   ---------------
   |            /
   |           /
   |          /
   |         /
   |        /
   |-------/
(0,L)      (L/sqrt(3),L)

To get the full hexagon, you "reflect" across the x and y axes...

List<Point> pointsInMyHexagon = new List<Point>();
foreach (Point p in pointsInHexagonQuadrant)
{
    pointsInMyHexagon.Add(new Point(p.X,p.Y));
    pointsInMyHexagon.Add(new Point(-p.X,p.Y));
    pointsInMyHexagon.Add(new Point(p.X,-p.Y));
    pointsInMyHexagon.Add(new Point(-p.X,-p.Y));
}

Now offset the hexagon to put the center back on your (x,y) point.

foreach (Point p in pointsInMyHexagon)
{
    p.Offset(myCenterPoint.X, myCenterPoint.Y);
}

It might be crude but the concept should work.

Select pixels in a hexagon around a center point

1 Answers1