y coordinate for a given x cubic bezier

Question

This question is very similar to: Quadratic bezier curve: Y coordinate for a given X?. But this one is cubic...

I'm using the getBezier function to calculate the Y coordinates of a bezier curve. The bezier curve starts always at (0,0) and ends always at (1,1).

I know the X value, so I tried to insert it as percent (I'm a moron). But that didn't work, obviously. Could you provide a solution? It's necessary it's an idiot proof function. Like:

function yFromX (c2x,c2y,c3x,c3y) { //c1 = (0,0) and c4 = (1,1), domainc2 and domainc3 = [0,1]
    //your magic
    return y;
}

Answer 1

Since the problem is so limited (function x(t) is monotonic), we can probably get away with using a pretty cheap method of solution-- binary search.

var bezier = function(x0, y0, x1, y1, x2, y2, x3, y3, t) {
    /* whatever you're using to calculate points on the curve */
    return undefined; //I'll assume this returns array [x, y].
};

//we actually need a target x value to go with the middle control
//points, don't we? ;)
var yFromX = function(xTarget, x1, y1, x2, y2) {
  var xTolerance = 0.0001; //adjust as you please
  var myBezier = function(t) {
    return bezier(0, 0, x1, y1, x2, y2, 1, 1, t);
  };

  //we could do something less stupid, but since the x is monotonic
  //increasing given the problem constraints, we'll do a binary search.

  //establish bounds
  var lower = 0;
  var upper = 1;
  var percent = (upper + lower) / 2;

  //get initial x
  var x = myBezier(percent)[0];

  //loop until completion
  while(Math.abs(xTarget - x) > xTolerance) {
    if(xTarget > x) 
      lower = percent;
    else 
      upper = percent;

    percent = (upper + lower) / 2;
    x = myBezier(percent)[0];
  }
  //we're within tolerance of the desired x value.
  //return the y value.
  return myBezier(percent)[1];
};

This should certainly break on some inputs outside of your constraints.

Answer 2

I used the algorithm from this page and wrote it down in JavaScript. It works for all the cases I have tested so far. (And doesn't use a while loop.)

Call the solveCubicBezier function. Pass in the x values of all the control points and the x value you want to get the y coordinate from. For example:

var results = solveCubicBezier(p0.x, p1.x, p2.x, p3.x, myX);

results is an array containing the 't' values originally passed into the Bezier function. The array can contain 0 to 3 elements, because not all x values have a corresponding y value, and some even have multiple.

function solveQuadraticEquation(a, b, c) {

    var discriminant = b * b - 4 * a * c;

    if (discriminant < 0) {
        return [];

    } else {
        return [
            (-b + Math.sqrt(discriminant)) / (2 * a),
            (-b - Math.sqrt(discriminant)) / (2 * a)
        ];
    }

}

function solveCubicEquation(a, b, c, d) {

    if (!a) return solveQuadraticEquation(b, c, d);

    b /= a;
    c /= a;
    d /= a;

    var p = (3 * c - b * b) / 3;
    var q = (2 * b * b * b - 9 * b * c + 27 * d) / 27;

    if (p === 0) {
        return [ Math.pow(-q, 1 / 3) ];

    } else if (q === 0) {
        return [Math.sqrt(-p), -Math.sqrt(-p)];

    } else {

        var discriminant = Math.pow(q / 2, 2) + Math.pow(p / 3, 3);

        if (discriminant === 0) {
            return [Math.pow(q / 2, 1 / 3) - b / 3];

        } else if (discriminant > 0) {
            return [Math.pow(-(q / 2) + Math.sqrt(discriminant), 1 / 3) - Math.pow((q / 2) + Math.sqrt(discriminant), 1 / 3) - b / 3];

        } else {

            var r = Math.sqrt( Math.pow(-(p/3), 3) );
            var phi = Math.acos(-(q / (2 * Math.sqrt(Math.pow(-(p / 3), 3)))));

            var s = 2 * Math.pow(r, 1/3);

            return [
                s * Math.cos(phi / 3) - b / 3,
                s * Math.cos((phi + 2 * Math.PI) / 3) - b / 3,
                s * Math.cos((phi + 4 * Math.PI) / 3) - b / 3
            ];

        }

    }
}

function roundToDecimal(num, dec) {
    return Math.round(num * Math.pow(10, dec)) / Math.pow(10, dec);
}

function solveCubicBezier(p0, p1, p2, p3, x) {

    p0 -= x;
    p1 -= x;
    p2 -= x;
    p3 -= x;

    var a = p3 - 3 * p2 + 3 * p1 - p0;
    var b = 3 * p2 - 6 * p1 + 3 * p0;
    var c = 3 * p1 - 3 * p0;
    var d = p0;

    var roots = solveCubicEquation(
        p3 - 3 * p2 + 3 * p1 - p0,
        3 * p2 - 6 * p1 + 3 * p0,
        3 * p1 - 3 * p0,
        p0
    );

    var result = [];
    var root;
    for (var i = 0; i < roots.length; i++) {
        root = roundToDecimal(roots[i], 15);
        if (root >= 0 && root <= 1) result.push(root);
    }

    return result;

}