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.