Can you say what this graphics is?

Hi, it is Dr. NISHIO Hirokazu, a mathematical artist!

Today's question is:


Hint: There are no randomness. There are no magic numbers in my code (except for the paper's width, height and an 2)

Here is a close-up of the graphics:

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I struggled to show the interesting mathematical interesting structures and got this:



It contains some magic numbers (for folding, spacing etc.) but more easy to understand than the first 'pure' graphics.

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P.S. (2010-01-07)

Ivanhoe said

Without a high-resolution version of the complete (or at least the upper left corner) of the artwork it seems hard to guess.

Oh, it is very good question. The upper left corner is the most important part of the artwork. It helps you to solve the quiz.

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Ivanhoe said...

I would say that this figure the last bit or the sum parity of some 2D function, am I close ?

It is reciprocal numbers in binary notation. 1/2 = 0.1, 1/3 = 0.01010101..., 1/4 = 0.01, 1/5 = 0.00110011, etc. I omit preceding 0s, and make 0 as black, 1 as white. Left-most column corresponts to 1/2.

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)

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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.