D3 Examples - Prim’s Algorithm III

Prim’s Algorithm III

A spanning tree of the canvas is generated using Prim’s algorithm and then flooded with color. Hue encodes Manhattan distance from the root of the tree. (This is not an optimal visual encoding, but it suffices and is pretty.)

Spanning trees can also be used to generate mazes. See a maze generated with Prim’s algorithm flooded with color, and compare color floods of spanning trees generated with Prim’s algorithm to random traversal, randomized depth-first traversal and Wilson’s algorithm.

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<!DOCTYPE html>
<meta charset="utf-8">
<canvas width="960" height="500"></canvas>
<script src="//d3js.org/d3.v3.min.js"></script>
<script>
​
var canvas = d3.select("canvas"),
    context = canvas.node().getContext("2d"),
    width = canvas.property("width"),
    height = canvas.property("height");
​
var worker = new Worker("generate-prims.js");
worker.postMessage({width: width, height: height});
worker.addEventListener("message", function(event) {
  worker.terminate();
​
  var N = 1 << 0,
      S = 1 << 1,
      W = 1 << 2,
      E = 1 << 3;
​
  var cells = event.data,
      distance = 0,
      visited = new Array(width * height),
      frontier = [(height - 1) * width],
      image = context.createImageData(width, height);
​
  function flood() {
    var frontier1 = [],
        i0,
        n0 = frontier.length,
        i1,
        color = d3.hsl((distance += .5) % 360, 1, .5).rgb();
​
    for (var i = 0; i < n0; ++i) {
      i0 = frontier[i] << 2;
      image.data[i0 + 0] = color.r;
      image.data[i0 + 1] = color.g;
      image.data[i0 + 2] = color.b;
      image.data[i0 + 3] = 255;
    }
​
    for (var i = 0; i < n0; ++i) {
      i0 = frontier[i];
      if (cells[i0] & E && !visited[i1 = i0 + 1]) visited[i1] = true, frontier1.push(i1);
      if (cells[i0] & W && !visited[i1 = i0 - 1]) visited[i1] = true, frontier1.push(i1);
      if (cells[i0] & S && !visited[i1 = i0 + width]) visited[i1] = true, frontier1.push(i1);
      if (cells[i0] & N && !visited[i1 = i0 - width]) visited[i1] = true, frontier1.push(i1);
    }
​
    frontier = frontier1;
    return !frontier1.length;
  }
​
  d3.timer(function() {
    for (var i = 0, done; i < 3 && !(done = flood()); ++i);
    context.putImageData(image, 0, 0);
    return done;
  });
});
​
</script>
​
    