GameDev.netPixel Coordinates to Hexagonal Coordinates

Pixel Coordinates to Hexagonal Coordinates
by Amit Patel
24 May 1996
I'll post the routines I use to calculate which hex the mouse is in. First, I should explain the hexagon size and layout.
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
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