How do I find beginning/end vertices of branches on an L-Systems tree? (p5.js)

Question

I'm trying to get the x, y coordinates of branch endpoints on a simple L-Systems tree. The idea is to create a p5.Vector(x, y) and push it to an array.

Right now, I'm able to draw ellipses marking the desired points by setting their origin to (0, -len), but I have a problem. When I try to push (0, -len) as a new p5.Vector(x, y) to an array, every single point has an x coordinate of 0, albeit with the correct y coordinate.

I know that it has something to do with translating the coordinate back to (width/2, height), but I'm just not able to figure out the correct calculation. I've even tried tan(angle) * (y1 - y2) but it's not quite right. TIA!

var axiom = 'F';
var sentence = axiom;
var len = 300;

var count = 0;
var flowerArr = [];

var rules = [];
rules[0] = {
    a: 'F',
    b: 'G[+F][-F]GF'
};

rules[1] = {
    a: 'G',
    b: 'GG'
};


function setup() {
    createCanvas(window.innerWidth, window.innerHeight);
    stroke(10);
    smooth();
    turtle();
}


function turtle() {
    background(255);
    strokeWeight(1);
    angle = radians(Math.random() * (25 - 15) + 15);
    resetMatrix();
    translate(width / 2, height);

    for (var i = 0; i < sentence.length; i++) {
        var current = sentence.charAt(i);
        var randomSeed = 2;
        if (current == 'F' || current == 'G') {
            ellipse(0, -len, 5);
            line(0, 0, 0, -len);
            translate(0, -len);
        } else if (current == '+') {
            let positiveRotation = angle * Math.random() * randomSeed;
            rotate(positiveRotation);
        } else if (current == '-') {
            let negativeRotation = -angle * Math.random() * randomSeed;
            rotate(negativeRotation);
        } else if (current == '[') {
            push();
        } else if (current == ']') {
            pop();
            count++;
        }
    }
    if (i >= sentence.length) {
        finished = true;
        console.log("done", count);
    }
}

function generateStems(iterations) {
    for (i = iterations - 1; i > 0 ; i--) {
        branch();
    }
}

function branch() {
    len *= Math.random() * (.52 - .45) + .45;
    var nextSentence = '';
    for (var i = 0; i < sentence.length; i++) {
        var current = sentence.charAt(i);
        var found = false;
        for (var j = 0; j < rules.length; j++) {
            if (current == rules[j].a) {
                found = true;
                nextSentence += rules[j].b;
                break;
            }
        }
        if (!found) {
            nextSentence += current;
        }
    }
    sentence = nextSentence;
    turtle();
}

function draw() {
    generateStems(4);
    noLoop();
}

George Profenza · Accepted Answer

As far as I know, at the moment, p5.js support for vector/matrix operations and coordinate space conversion isn't quite there yet.

In theory you could manually keep track of every single transformation (translate/rotate) and manually compute it to get the transformed positions, howeve, in practice this may be error prone and cumbersome.

In Processing you could rely on PMatrix's mult(PVector) method to transform a point from one coordinate system to another, but not in p5.js at the moment. Functions like screenX()/screenY() simplify this even further.

Here's a basic example (note the usage of P3D):

PVector v1 = new PVector();
float len = 100;

void setup(){
  size(300,300,P3D);
  noFill();
  strokeWeight(3);
}

void draw(){
  background(255);
  // isolate coordinate system
  pushMatrix();
  // apply a set of transformations
  translate(width / 2, height);
  translate(0,-len);
  rotate(radians(45));
  // draw a blue rectangle from the corner to illustrate this transformed position
  stroke(0,0,192);
  rect(0,0,30,30);
  // further transform
  translate(90,0);
  // draw a rect rectangle
  stroke(192,0,0);
  rect(0,0,30,30);
  // use screenX/screenY to calculate the transformed coordinates
  v1.set(screenX(0,0,0),screenY(0,0,0)); 
  popMatrix();

  // simply draw a (green) circle on top at the same transformed coordinates, without being in that local coordinate system 
  stroke(0,192,0);
  ellipse(v1.x, v1.y, 30, 30);
}

At the moment, for practical reasons, if computing the transformed locations is a must, I would recommend porting your code to Processing.

Update Based on your comment it is easier to use the L-System to introduce a new rule for the flower.

Let's say * represents a flower, you could modify your rule to include it for example as the last instruction: b: 'G[+F][-F]GF' becomes b: 'G[+F][-F]GF*'

then it's just a matter of handling that symbol as you traverse the current sentence:

var axiom = 'F';
var sentence = axiom;
var len = 300;

var count = 0;
var flowerArr = [];

var rules = [];
rules[0] = {
    a: 'F',
    b: 'G[+F][-F]GF*'
};

rules[1] = {
    a: 'G',
    b: 'GG'
};


function setup() {
    createCanvas(630, 630);
    stroke(10);
    noFill();
    smooth();
    turtle();
}


function turtle() {
    background(255);
    strokeWeight(1);
    angle = radians(Math.random() * (25 - 15) + 15);
    resetMatrix();
    translate(width / 2, height);
    for (var i = 0; i < sentence.length; i++) {
        var current = sentence.charAt(i);
        var randomSeed = 2;
        if (current == 'F' || current == 'G') {
            ellipse(0, -len, 5);
            line(0, 0, 0, -len);
            translate(0, -len);
        // flower rule
        } else if (current == '*') {
            flower(6,len * 0.618033);
        } else if (current == '+') {
            let positiveRotation = angle * Math.random() * randomSeed;
            rotate(positiveRotation);
        } else if (current == '-') {
            let negativeRotation = -angle * Math.random() * randomSeed;
            rotate(negativeRotation);
        } else if (current == '[') {
            push();
        } else if (current == ']') {
            pop();
            count++;
        }
    }
    if (i >= sentence.length) {
        finished = true;
        // console.log("done", count);
    }
}

function flower(sides, sideLength){
  beginShape();
  let angleIncrement = TWO_PI / sides;
  for(let i = 0 ; i <= sides; i++){
    vertex(cos(angleIncrement * i) * sideLength,
           sin(angleIncrement * i) * sideLength);
  }
  endShape();
}

function generateStems(iterations) {
    for (i = iterations - 1; i > 0 ; i--) {
        branch();
    }
}

function branch() {
    len *= Math.random() * (.52 - .45) + .45;
    var nextSentence = '';
    for (var i = 0; i < sentence.length; i++) {
        var current = sentence.charAt(i);
        var found = false;
        for (var j = 0; j < rules.length; j++) {
          
            if (current == rules[j].a) {
              
                found = true;
                nextSentence += rules[j].b;
                break;
            }
        }
        if (!found) {
            nextSentence += current;
        }
    }
    sentence = nextSentence;
    turtle();
}

function draw() {
    generateStems(5);
    noLoop();
}

As further explorations here are a couple of fun JS implementations of L-system to play with:

How do I find beginning/end vertices of branches on an L-Systems tree? (p5.js)

Answers (1)

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