Give me a reason to hang on
Give the strength to continue
In this land full of pain

How can I continue
to walk among the lifeless…the cowards and the ill-bred
while knowing that the noble ones are dying…or dead

In this reign of fear, you bend…or disappear
Ozymandias now has pathetic pretender of silly sneer and cold command
His colossal wrecks stretch everywhere, further spreading despair

The servile and hungry rise in thunderous applause to Works of their new god
The smell of decay stifling, but the fouls reckon it ambrosia
Cowardice begetting dystopia

Mathematical recreation is fun. I came across a nice Mind your Decisions video and thought it would be cool to use Julia to find that special number. Have a look at the video here before we continue

Julia can be amazingly expressive. In line 14 are getting all the prime factors of all the numbers from 1 to 100 using the map function. We are using the type BigInt since will be generating pretty large numbers. The standard Int will simply overflow. We then use reduce to get all the common factors for the numbers from 1 to 100.

If your are slight confused lets take a bit slower. First we generate a one dimensional array of BigInts from 1 to 100.

BigInt.(1:100)

This makes use of Julia‘s very powerful dot factorization syntax. We then map that array into another array of Primes.Factorization{BigInt} through the factor function in the Primes package. The Primes.Factorization type is subtype of Associative. I learned that from reading the implementation.

The cool thing about Associative types that the can be very conveniently accessed. Lets have a closer look.

a=factor(24)println(a)

yields

Primes.Factorization(2=>3,3=>1)

That is

You can then do the following

print(a[2])

yielding

3

This means that the number 24 has a 2 raised to the third power as a factor.

Or, be more adventurous and try

print(a[5])

which will print a

0

This makes perfect sense as and 1 is always a factor of any number. We will always get 0 exponent for any number that is not a prime factor of a given number. The Primes.jl package authors have done great use of Julia‘s very powerful type system.

To solve this problem we need to look at all the prime factors for all the numbers from 1 to 100 find the highest exponent of the of each of those prime factors. To do that we implement the getBigestFactor function when does that for any two prime factorizations. We plug that into reduce and et voilà!

Finally we multiply them to get our special number at line 16

n=69720375229712477164533808935312303556800

Are we done yet?

As the saying goes “there is more than one way to skin a cat”. Some are more elegant than others.
Here is a more elegant and computationally more efficient way.

Here we just get the prime numbers in the range from 1 to n (line 3). Using those primes we then get the highest exponents that will not yield a number outside the range (line 4). Finally, we package everything as a Primes.Factorization (line 5).

Recently, I watched a cool mind your decsions video on an inscribed circle and rectangle puzzle. In the video they showed a diagram that was not scale. I wanted to get a sense of how these differently shaped areas will match.

There was a cool ratio between the outer and inner circle radii that is expressed as