Update: See Amit's guide to hex grids (2013) for a more visual and comprehensive explanation.

From: amitp@Xenon.Stanford.EDU (Amit Patel)
Newsgroups: rec.games.programmer
Subject: Re: Hexagon playfield query
Date: 24 May 1996 18:22:11 GMT
Organization: Computer Science Department, Stanford University.
      

I'll post the routines I use to calculate which hex the mouse is in. First, I should explain the hexagon size and layout.

    ___     ___
   /   \___/   \
   \___/   \___/
   /   \___/   \
   \___/   \___/
      

Each hexagon is 28 x 24 pixels, but since the columns overlap, the distance from the center of one hex to the center of the next column's hex is 21.

My coordinate system is offset-grid with no gaps. The lower left is (1,1); as you go up, the N coordinate increases. (I call them (M,N) instead of (X,Y) to distinguish between the hex and square coordinates.) Every other column is pushed up half a hexagon height.

First, this is the approach based on a rec.games.programmer article saved on my web pages. It is based on the view that hexagons are a projection of three dimensional cubes onto a plane. (See that web page for an explanation.)

// Note:  HexCoord is a struct that just stores hex coordinates
HexCoord PointToHex( int xp, int yp )
{
    // NOTE:  HexCoord(0,0)'s x() and y() just define the origin
    //        for the coordinate system; replace with your own
    //        constants.  (HexCoord(0,0) is the origin in the hex
    //        coordinate system, but it may be offset in the x/y
    //        system; that's why I subtract.)
    double x = 1.0 * ( xp - HexCoord(0,0).x() ) / HexXSpacing;
    double y = 1.0 * ( yp - HexCoord(0,0).y() ) / HexYSpacing;
    double z = -0.5 * x - y;
           y = -0.5 * x + y;
    int ix = floor(x+0.5);
    int iy = floor(y+0.5);
    int iz = floor(z+0.5);
    int s = ix+iy+iz;
    if( s )
    {
        double abs_dx = fabs(ix-x);
        double abs_dy = fabs(iy-y);
        double abs_dz = fabs(iz-z);
        if( abs_dx >= abs_dy && abs_dx >= abs_dz )
            ix -= s;
        else if( abs_dy >= abs_dx && abs_dy >= abs_dz )
            iy -= s;
        else
            iz -= s;
    }
    return HexCoord( ix, ( iy - iz + (1-ix%2) ) / 2 );
}
      

Now, here's another approach that I'm now using. It's not as general, but it's faster.

HexCoord PointToHex( int xp, int yp )
{
    // NOTE:  First we subtract the origin of the coordinate
    //        system; replace with your own values
    xp -= X_ORIGIN;
    yp -= Y_ORIGIN;
    int row = 1 + yp / 12;
    int col = 1 + xp / 21;
    int diagonal[2][12] = {
        {7,6,6,5,4,4,3,3,2,1,1,0},
        {0,1,1,2,3,3,4,4,5,6,6,7}
    };

    if( diagonal[(row+col)%2][yp%12] >= xp%21 )
        col--;
    return HexCoord( col, (row-(col%2))/2 );
}
      

In this approach, I first figure out which "half row" the (x,y) lies in, and put that in `row'. Each hexagon occupies two half rows, but every other column chooses different half rows to start with.

Then I figure out which column I'm in, approximately, and put that in `col'. (Each approximate column is 21 pixels wide.)

    |  ____|
    | /    \
    |/     |\
    |\     |/|
    | \____/ 28 = hex width
    |      |
    0      21
      

The vertical lines marks the approximate column boundary. The half row and the column number tells me whether I need to look at the / diagonal or the \ diagonal.

I then look at the pixel locations of the diagonal. I can use the y coordinate (modulo the half row height) as an index into the diagonal. If the x coordinate (modulo the column width) is less than the diagonal value, then I need to move the coordinate to the left.

September 1999: Jason W. Goodwin writes to me with a fix for negative coordinates (which I don't use in my code, but may affect my readers):

From: Jason W Goodwin
To: amitp@CS.Stanford.EDU

I'm using one of the algorithms listed on your game programming
information web page (BTW, I haven't gone through it thoroughly
yet, but it looks like a great resource) for transforming screen
coordinates into hexagonal coordinates.

Specifically the first one listed at:
http://www-cs-students.stanford.edu/~amitp/Articles/GridToHex.html

Have you noticed a problem with it giving the wrong coordinates
once it should start going into the negatives?

Specifically, it gives N+1 when N < 0 and also errs whenever
M is < 0 and odd.

I made the following small change to the last line, and it works
fine for me now (I've only tested this against -100 <= m <= 100,
-100 <= n <= 100, so there might be an error at 101, I don't know)

//   Note that for ix = -1, -3, -5, ... : 1 - (ix % 2) = 2.
// Original Line:
//   return HexCoord( ix, ( iy - iz + (1-ix%2) ) / 2 );

// Added Lines:
  if ((s = iy - iz) < 0)
    iy = s - 1 + ((ix+1) & 1); // this should be !(ix&1), but I
  else                         // haven't checked it, and it might
    iy = s + 1 - (ix & 1);     // be compiler dependant.

  return HexCoord( ix, iy / 2 );

-- 
Jason Goodwin
"O Theos mou! Echo ten labrida en te mou kephale!"