Random mesh in Adobe Illustrator

Question

Is there any way to create this kind of random mesh in Adobe Illustrator besides doing that manually? Not interested in these bold dots, just the mesh.

Answer 1

Quickly adapted form this answer. Be sure to have round joints and non filled as your color setting or you get random stuff:

#target illustrator

var delunlay = getDelunlay();
var points = []

//~ var sel =  activeDocument.selection;
//~ for(var i=0; i < sel.length; i++){
//~     points.push(sel[i].position);
//~ }

for (var i = 0;i <40;i++){
    points.push([getRandomInt(100,400),getRandomInt(-200,-50)]);
}

var tris = delunlay.triangulate(points);


for(var i = 0; i < tris.length;i+=3 ){
    var obj = activeDocument.pathItems.add();
    obj.setEntirePath([points[tris[i]], points[tris[i+1]],points[tris[i+2]]]);
    obj.closed = true;
}


function getRandomInt(min, max) {
  min = Math.ceil(min);
  max = Math.floor(max);
  return Math.floor(Math.random() * (max - min)) + min; //The maximum is exclusive and the minimum is inclusive
}

// https://github.com/ironwallaby/delaunay/blob/master/delaunay.js
// slightly modified to work here 
function getDelunlay() {
  "use strict";

  var EPSILON = 1.0 / 1048576.0;

  function supertriangle(vertices) {
    var xmin = Number.POSITIVE_INFINITY,
        ymin = Number.POSITIVE_INFINITY,
        xmax = Number.NEGATIVE_INFINITY,
        ymax = Number.NEGATIVE_INFINITY,
        i, dx, dy, dmax, xmid, ymid;

    for(i = vertices.length; i--; ) {
      if(vertices[i][0] < xmin) xmin = vertices[i][0];
      if(vertices[i][0] > xmax) xmax = vertices[i][0];
      if(vertices[i][1] < ymin) ymin = vertices[i][1];
      if(vertices[i][1] > ymax) ymax = vertices[i][1];
    }

    dx = xmax - xmin;
    dy = ymax - ymin;
    dmax = Math.max(dx, dy);
    xmid = xmin + dx * 0.5;
    ymid = ymin + dy * 0.5;

    return [
      [xmid - 20 * dmax, ymid -      dmax],
      [xmid            , ymid + 20 * dmax],
      [xmid + 20 * dmax, ymid -      dmax]
    ];
  }

  function circumcircle(vertices, i, j, k) {
    var x1 = vertices[i][0],
        y1 = vertices[i][1],
        x2 = vertices[j][0],
        y2 = vertices[j][1],
        x3 = vertices[k][0],
        y3 = vertices[k][1],
        fabsy1y2 = Math.abs(y1 - y2),
        fabsy2y3 = Math.abs(y2 - y3),
        xc, yc, m1, m2, mx1, mx2, my1, my2, dx, dy;

    /* Check for coincident points */
    if(fabsy1y2 < EPSILON && fabsy2y3 < EPSILON)
      throw new Error("Eek! Coincident points!");

    if(fabsy1y2 < EPSILON) {
      m2  = -((x3 - x2) / (y3 - y2));
      mx2 = (x2 + x3) / 2.0;
      my2 = (y2 + y3) / 2.0;
      xc  = (x2 + x1) / 2.0;
      yc  = m2 * (xc - mx2) + my2;
    }

    else if(fabsy2y3 < EPSILON) {
      m1  = -((x2 - x1) / (y2 - y1));
      mx1 = (x1 + x2) / 2.0;
      my1 = (y1 + y2) / 2.0;
      xc  = (x3 + x2) / 2.0;
      yc  = m1 * (xc - mx1) + my1;
    }

    else {
      m1  = -((x2 - x1) / (y2 - y1));
      m2  = -((x3 - x2) / (y3 - y2));
      mx1 = (x1 + x2) / 2.0;
      mx2 = (x2 + x3) / 2.0;
      my1 = (y1 + y2) / 2.0;
      my2 = (y2 + y3) / 2.0;
      xc  = (m1 * mx1 - m2 * mx2 + my2 - my1) / (m1 - m2);
      yc  = (fabsy1y2 > fabsy2y3) ?
        m1 * (xc - mx1) + my1 :
        m2 * (xc - mx2) + my2;
    }

    dx = x2 - xc;
    dy = y2 - yc;
    return {i: i, j: j, k: k, x: xc, y: yc, r: dx * dx + dy * dy};
  }

  function dedup(edges) {
    var i, j, a, b, m, n;

    for(j = edges.length; j; ) {
      b = edges[--j];
      a = edges[--j];

      for(i = j; i; ) {
        n = edges[--i];
        m = edges[--i];

        if((a === m && b === n) || (a === n && b === m)) {
          edges.splice(j, 2);
          edges.splice(i, 2);
          break;
        }
      }
    }
  }

  var Delaunay = {
    triangulate: function(vertices, key) {
      var n = vertices.length,
          i, j, indices, st, open, closed, edges, dx, dy, a, b, c;

      /* Bail if there aren't enough vertices to form any triangles. */
      if(n < 3)
        return [];

      /* Slice out the actual vertices from the passed objects. (Duplicate the
       * array even if we don't, though, since we need to make a supertriangle
       * later on!) */
      vertices = vertices.slice(0);

      if(key)
        for(i = n; i--; )
          vertices[i] = vertices[i][key];

      /* Make an array of indices into the vertex array, sorted by the
       * vertices' x-position. Force stable sorting by comparing indices if
       * the x-positions are equal. */
      indices = new Array(n);

      for(i = n; i--; )
        indices[i] = i;

      indices.sort(function(i, j) {
        var diff = vertices[j][0] - vertices[i][0];
        return diff !== 0 ? diff : i - j;
      });

      /* Next, find the vertices of the supertriangle (which contains all other
       * triangles), and append them onto the end of a (copy of) the vertex
       * array. */
      st = supertriangle(vertices);
      vertices.push(st[0], st[1], st[2]);

      /* Initialize the open list (containing the supertriangle and nothing
       * else) and the closed list (which is empty since we havn't processed
       * any triangles yet). */
      open   = [circumcircle(vertices, n + 0, n + 1, n + 2)];
      closed = [];
      edges  = [];

      /* Incrementally add each vertex to the mesh. */
      for(i = indices.length; i--; edges.length = 0) {
        c = indices[i];

        /* For each open triangle, check to see if the current point is
         * inside it's circumcircle. If it is, remove the triangle and add
         * it's edges to an edge list. */
        for(j = open.length; j--; ) {
          /* If this point is to the right of this triangle's circumcircle,
           * then this triangle should never get checked again. Remove it
           * from the open list, add it to the closed list, and skip. */
          dx = vertices[c][0] - open[j].x;
          if(dx > 0.0 && dx * dx > open[j].r) {
            closed.push(open[j]);
            open.splice(j, 1);
            continue;
          }

          /* If we're outside the circumcircle, skip this triangle. */
          dy = vertices[c][1] - open[j].y;
          if(dx * dx + dy * dy - open[j].r > EPSILON)
            continue;

          /* Remove the triangle and add it's edges to the edge list. */
          edges.push(
            open[j].i, open[j].j,
            open[j].j, open[j].k,
            open[j].k, open[j].i
          );
          open.splice(j, 1);
        }

        /* Remove any doubled edges. */
        dedup(edges);

        /* Add a new triangle for each edge. */
        for(j = edges.length; j; ) {
          b = edges[--j];
          a = edges[--j];
          open.push(circumcircle(vertices, a, b, c));
        }
      }

      /* Copy any remaining open triangles to the closed list, and then
       * remove any triangles that share a vertex with the supertriangle,
       * building a list of triplets that represent triangles. */
      for(i = open.length; i--; )
        closed.push(open[i]);
      open.length = 0;

      for(i = closed.length; i--; )
        if(closed[i].i < n && closed[i].j < n && closed[i].k < n)
          open.push(closed[i].i, closed[i].j, closed[i].k);

      /* Yay, we're done! */
      return open;
    },
    contains: function(tri, p) {
      /* Bounding box test first, for quick rejections. */
      if((p[0] < tri[0][0] && p[0] < tri[1][0] && p[0] < tri[2][0]) ||
         (p[0] > tri[0][0] && p[0] > tri[1][0] && p[0] > tri[2][0]) ||
         (p[1] < tri[0][1] && p[1] < tri[1][1] && p[1] < tri[2][1]) ||
         (p[1] > tri[0][1] && p[1] > tri[1][1] && p[1] > tri[2][1]))
        return null;

      var a = tri[1][0] - tri[0][0],
          b = tri[2][0] - tri[0][0],
          c = tri[1][1] - tri[0][1],
          d = tri[2][1] - tri[0][1],
          i = a * d - b * c;

      /* Degenerate tri. */
      if(i === 0.0)
        return null;

      var u = (d * (p[0] - tri[0][0]) - b * (p[1] - tri[0][1])) / i,
          v = (a * (p[1] - tri[0][1]) - c * (p[0] - tri[0][0])) / i;

      /* If we're outside the tri, fail. */
      if(u < 0.0 || v < 0.0 || (u + v) > 1.0)
        return null;

      return [u, v];
    }
  };
  return Delaunay;
}

Running this results in different image each time but heres an example:

You could use some more structured random starting point if you wish this is just fast to do. And hey you asked for random.