How to calculate Bezier curves' bounding box

Hi, it is NISHIO Hirokazu. Today I wanted to make a lot of Bezier curves same size. And I got it.

Here is what I want to draw:


To calculate bounding box of cubic Bezier seems easy, especially you know its parametric form. See Bézier curve - Wikipedia, the free encyclopedia

However, there are some pitfall. The derivative of Bezier equation is usually quadratic equation but not always. Solutions of the derivative may out of range, etc.

I publish following source code under MIT License. Feel free to use it.

def calc_box(start, curves):
    P0 = start
    bounds = [[P0[0]], [P0[1]]]

    for c in curves:
        P1, P2, P3 = (
            (c[0], c[1]), 
            (c[2], c[3]), 
            (c[4], c[5]))

        bounds[0].append(P3[0])
        bounds[1].append(P3[1])

        for i in [0, 1]:
            f = lambda t: (
                (1-t)**3 * P0[i] 
                + 3 * (1-t)**2 * t * P1[i] 
                + 3 * (1-t) * t**2 * P2[i]
                + t**3 * P3[i])

            b = 6 * P0[i] - 12 * P1[i] + 6 * P2[i]
            a = -3 * P0[i] + 9 * P1[i] - 9 * P2[i] + 3 * P3[i]
            c = 3 * P1[i] - 3 * P0[i]

            if a == 0:
                if b == 0:
                    continue
                t = -c / b
                if 0 < t < 1: 
                    bounds[i].append(f(t))
                continue

            b2ac = b ** 2 - 4 * c * a
            if b2ac < 0: 
                continue
            t1 = (-b + sqrt(b2ac))/(2 * a)
            if 0 < t1 < 1: bounds[i].append(f(t1))
            t2 = (-b - sqrt(b2ac))/(2 * a)
            if 0 < t2 < 1: bounds[i].append(f(t2))

        P0 = P3

    x = min(bounds[0])
    w = max(bounds[0]) - x
    y = min(bounds[1])
    h = max(bounds[1]) - y
    return (x, y, w, h)

---

Blaze Boy asked me about the structure, thank you. for each c in curves is 6-tuples: P1, P2, P3 = ((c[0], c[1]), (c[2], c[3]), (c[4], c[5])) P0 and P3 are terminal points of each bezier curves.