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Honeycomb hexagonal grid?

I'm trying to draw a hexagonal grid in a honeycomb shape. So far I'm able to draw it in a rectangular shape but I don't know how to convert my for loop to make a honeycomb shape instead.

This is what I currently have

<html>

<body>
    <canvas width='1080' height='720' id='hexmap'></canvas>
</body>
<script>

    window.addEventListener('DOMContentLoaded', (event) => {

        var canvas = document.getElementById('hexmap');

    var hexHeight,
        hexRadius,
        hexRectangleHeight,
        hexRectangleWidth,
        hexagonAngle = 0.523598776, // 30 degrees in radians
        sideLength = 36,
        boardWidth = 10,
        boardHeight = 10;

    hexHeight = Math.sin(hexagonAngle) * sideLength;
    hexRadius = Math.cos(hexagonAngle) * sideLength;
    hexRectangleHeight = sideLength + 2 * hexHeight;
    hexRectangleWidth = 2 * hexRadius;
    var ctx = canvas.getContext('2d');

        ctx.fillStyle = "#000000";
        ctx.strokeStyle = "#CCCCCC";
        ctx.lineWidth = 1;

    drawBoard(ctx, boardWidth, boardHeight);
    
    function drawBoard(canvasContext, width, height) {

        var i,j;
        //this loop generates a rectangular hexagon grid
        for(i = 0; i < width; ++i) {
            for(j = 0; j < height; ++j) {
                drawHexagon(
                    ctx, 
                    i * hexRectangleWidth + ((j % 2) * hexRadius), 
                    j * (sideLength + hexHeight), 
                    false
                );
            }
        }
    }

    function drawHexagon(canvasContext, x, y, fill) {           
        var fill = fill || false;

        canvasContext.beginPath();
        canvasContext.moveTo(x + hexRadius, y);
        canvasContext.lineTo(x + hexRectangleWidth, y + hexHeight);
        canvasContext.lineTo(x + hexRectangleWidth, y + hexHeight + sideLength);
        canvasContext.lineTo(x + hexRadius, y + hexRectangleHeight);
        canvasContext.lineTo(x, y + sideLength + hexHeight);
        canvasContext.lineTo(x, y + hexHeight);
        canvasContext.closePath();

        if(fill) {
            canvasContext.fill();
        } else {
            canvasContext.stroke();
        }
    }

})
</script>

</html>

Which results in this shape

What I'd like to achieve though is a shape like this

I was able to do it using like 13 separate for loops, shifting the hexagon over manually each time but it wasn't very practical nor automated.

about 4 years ago · Juan Pablo Isaza

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Accepted Answer · 2022-06-30T14:37:28.534Z

If we set some conditions, we can derive an algorithm quite easily. Let the conditions be:

width & height have to be equal
width & height have to be odd numbers

Now let's look at your shape, which meets the condition as it's width & height is 13. A closer look reveals that we have 7 hexagons in the first row, 8 in the second, 9 in the third and so on up to 13 hexagons at row 7. Afterwards the number of hexagons decreases by one per row until reaching the last row 13.

So the number of hexagons per row can be expressed as:

hexagons = width - (Math.abs(Math.floor(width / 2) - i));

Where i is the row.

Likewise the horizontal starting position of each row decrements by half a hexagon's width until reaching the center.

xStart = (width - 3) % 4 == 0 ? Math.ceil((width - hexagons) / 2) : Math.floor((width - hexagons) / 2);

Now all that's left to do is modifying your for-loop to start at xStart up to xStart+hexagons.

for (j = xStart; j < xStart+hexagons; j++)

Here's a complete example:

var canvas = document.getElementById('hexmap');

var hexHeight,
  hexRadius,
  hexRectangleHeight,
  hexRectangleWidth,
  hexagonAngle = 0.523598776, // 30 degrees in radians
  sideLength = 9,
  boardWidth = 13,
  boardHeight = 13;

hexHeight = Math.sin(hexagonAngle) * sideLength;
hexRadius = Math.cos(hexagonAngle) * sideLength;
hexRectangleHeight = sideLength + 2 * hexHeight;
hexRectangleWidth = 2 * hexRadius;
var ctx = canvas.getContext('2d');

ctx.fillStyle = "#000000";
ctx.strokeStyle = "#CCCCCC";
ctx.lineWidth = 1;

drawBoard(ctx, boardWidth, boardHeight);

function drawBoard(canvasContext, width, height) {
  var i, j, hexagons, xStart;
  //this loop generates a rectangular hexagon grid
  for (i = 0; i < height; i++) {
    hexagons = width - (Math.abs(Math.floor(width / 2) - i));
    xStart = (width - 3) % 4 == 0 ? Math.ceil((width - hexagons) / 2) : Math.floor((width - hexagons) / 2);

    for (j = xStart; j < xStart + hexagons; j++) {
      drawHexagon(
        ctx,
        j * hexRectangleWidth + ((i % 2) * hexRadius),
        i * (sideLength + hexHeight),
        false
      );
    }
  }
}

function drawHexagon(canvasContext, x, y, fill) {
  var fill = fill || false;

  canvasContext.beginPath();
  canvasContext.moveTo(x + hexRadius, y);
  canvasContext.lineTo(x + hexRectangleWidth, y + hexHeight);
  canvasContext.lineTo(x + hexRectangleWidth, y + hexHeight + sideLength);
  canvasContext.lineTo(x + hexRadius, y + hexRectangleHeight);
  canvasContext.lineTo(x, y + sideLength + hexHeight);
  canvasContext.lineTo(x, y + hexHeight);
  canvasContext.closePath();

  if (fill) {
    canvasContext.fill();
  } else {
    canvasContext.stroke();
  }
}

document.getElementById("slider").oninput = (e) => {
  ctx.clearRect(0, 0, canvas.width, canvas.height)
  drawBoard(ctx, e.target.value, e.target.value);
}

<input type="range" min="3" max="27" value="13" step="2" id="slider"><br>
<canvas width='400' height='300' id='hexmap'></canvas>

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203

Honeycomb hexagonal grid?

1 Respuestas

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Encuentra empleos remotos

Andres GPT