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Challenges computer science Programming Python

The mathematics behind the solution for Challenge No. 5

If you take a look at the various solutions people proposed for the last challenge of generating a specific permutation, you’ll see that they are very similar. Most of them are based on some form of div-mod usage. The reason this is so, is because all of these solutions are using the Factorial Base.

What does that mean?
Note that we usually encounter div-mods when we want to find the representation of a number in a certain base. That should already pique your interest. Now consider that a base’s digits need not have the same weight. For example, consider how we count the number of seconds since the start of the week:

seconds of the last minute, A (at most 60-1)
minutes of the last hour, B (at most 60-1)
hours of the last day, C (at most (24-1)
days of the last week, D (at most 7-1)

So given A, B, C, D, we would say that the number of seconds is:
A + 60*B + 24*C + 7*D. This certainly looks like a base transformation. To go back, we would use divmod.

The factorial base is just the same, with the numbers n, n-1, … 1. Note that in the factorial base, you can only represent a finite number of numbers – n!. This should not be surprising – this is what we set out to do in the first place!
The thing that I found really amazing about this is that all the people to whom I posed this challenge came up with almost the same “way” of solving it.

Other interesting curiosities regarding bases can be found in Knuth’s book, “The Art of Computer Programming”, volume 2, Section 4.1.

Categories
Fractals Math Programming Python

Fractals in 10 minutes no. 3 – The Dragon

When first I looked through the pages of the book “Hacker’s Delight”, I found myself looking at the chapter about bases. There I learned a very curious fact – with the digits of 0,1 and the base of -2, you can represent any integer. Right afterwards I learned something even more interesting – with the digits of 0,1 and the base of 1-i, you can represent and number of the form a+bi where a and b are integers. Having nothing to do with this curious fact, I let the subject go.
Some time later, I was reading through Knuth’s “Art of Computer Programming”, and found that with the base of (1-i)^-1, and digits of 0,1 you can generate the dragon fractal!

The dragon fractal

Generating the fractal is quite simple actually:

def create_dragon_set(n):
    """calculate the dragon set, according to Knuth"""
    s = set([0.0+0.0j])
    for i in range(n):
        new_power = (1.0-1.0j)**(-i)
        s |= set(x+new_power for x in s)
    return s

(By the way, can you do it better?)

The annoying part is converting the complex numbers to drawable integer points. After doing so, I used PIL to draw the jpeg.
Here’s a link to the code.