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.)
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.
- Amit
This article was originally written by Alizerin |
Copyright Open Source Institute, 2006 
