Algorithm for creating cells by spiral on the hexagonal field

Question

Help to find an algorithm for creating cells by spiral on the hexagonal field.

Look at the image:

Let's imagine an dimensionless 2d array. The X axis is the blue line, Y is horizontal, spiral is red.

I need to add cells from the central point x0y0 to point N by spiral

Tell me the way to solve the problem, please. Thank you!

Answer 1

I'd suggest changing the cells numbering sligtly, so that X remains the same when you go down and right (or up and left). Then simple algorithm like the following should work:

  int x=0, y=0;   
  add(x, y); // add the first cell
  int N=1 
  for( int N=1; <some condition>; ++N ) {
    for(int i=0; i<N; ++i) add(++x, y);  // move right
    for(int i=0; i<N-1; ++i) add(x, ++y); // move down right. Note N-1
    for(int i=0; i<N; ++i) add(--x, ++y); // move down left
    for(int i=0; i<N; ++i) add(--x, y); // move left
    for(int i=0; i<N; ++i) add(x, --y); // move up left
    for(int i=0; i<N; ++i) add(++x, --y); // move up right
  }

This generates the points as follows:

After a transformation we get:

Answer 2

(the circles have a diameter of 1)

Here's a function to get position i:

  void getHexPosition( int i, ref double x, ref double y )
  {
     if ( i == 0 ) { x = y = 0; return; }

     int layer = (int) Math.Round( Math.Sqrt( i/3.0 ) );

     int firstIdxInLayer = 3*layer*(layer-1) + 1;
     int side = (i - firstIdxInLayer) / layer; // note: this is integer division
     int idx  = (i - firstIdxInLayer) % layer;                  
     x =  layer * Math.Cos( (side - 1) * Math.PI/3 ) + (idx + 1) * Math.Cos( (side + 1) * Math.PI/3 );
     y = -layer * Math.Sin( (side - 1) * Math.PI/3 ) - (idx + 1) * Math.Sin( (side + 1) * Math.PI/3 );
  }

Scaling the result by Math.Sqrt(.75) gives

If you're interested in the skewed coordinates like in shura's answer:

  int[] h = { 1, 1, 0, -1, -1, 0, 1, 1, 0 };
  void getHexSkewedPosition( int i, ref int hx, ref int hy )
  {
     if ( i == 0 ) { hx = hy = 0; return; }

     int layer = (int) Math.Round( Math.Sqrt( i/3.0 ) );

     int firstIdxInLayer = 3*layer*(layer-1) + 1;
     int side = (i - firstIdxInLayer) / layer;
     int idx  = (i - firstIdxInLayer) % layer;

     hx = layer*h[side+0] + (idx+1) * h[side+2];
     hy = layer*h[side+1] + (idx+1) * h[side+3];
  }

  void getHexPosition( int i, ref double hx, ref double hy )
  {
     int x = 0, y = 0;
     getHexSkewedPosition( i, ref x, ref y );
     hx = x - y * .5;
     hy = y * Math.Sqrt( .75 );
  }

Answer 3

Imagine you had a normal grid with squares instead of hexagons, create the spiral using that grid, then draw it by shifting say, every odd y to the left by m pixels, that'll give you that effect.

Answer 4

You can pick hexes one at a time by using an appropriate score function to select the best of the six not-yet-selected adjacent hexes to the hex selected the previous round. I think a score function that works is to pick the closest to (0,0) (forces selecting hexes in one "shell" at a time), breaking ties by choosing the closest to (1,0) (forces a consistent spiral direction in the new shell). Distance in the hex grid can be computed using the following function:

double grid_distance(int dx, int dy) {
  double real_dx = dx + y/2.0;
  double real_dy = dy * sqrt(3)/2.0;
  return sqrt(real_dx * real_dx + real_dy * real_dy);
}

Answer 5

YOu could do it by simulating directions. If your directions are "0 points up", then increment by 1 as you go clock-wise, the following should do:

Pick a centre cell.
Pick the second cell (ideally in direction 0).
Set direction to 2.

While you have more cells to mark:
  if the cell in (direction+1)%6 is free:
    set direction = (direction+1)%6
  mark current cell as used
  go to cell in direction

Answer 6

I loved @shura's way of approaching the problem, but couldn't get that exact algorithm to work. Also, I'm using a 2x1 hexagon spacing (where x cells are spaced 2 apart, and every other x item is hidden).

Here's what I got working (though in JavaScript):

    //Hexagon spiral algorithm, modified from 
    for(var n=1; n<=rings; ++n) {
        x+=2; add(x, y);
        for(var i=0; i<n-1; ++i) add(++x,++y); // move down right. Note N-1
        for(var i=0; i<n; ++i) add(--x,++y); // move down left
        for(var i=0; i<n; ++i) { x-=2; add(x, y); } // move left
        for(var i=0; i<n; ++i) add(--x,--y); // move up left
        for(var i=0; i<n; ++i) add(++x, --y); // move up right
        for(var i=0; i<n; ++i) { x+=2; add(x, y); }  // move right
    }