Tuesday, January 29, 2013

Tools used: RStudio with ggplot2, plyr, reshape2, and XML packages

The way Super Bowl pools work is this: people who buy in are randomly assigned a pair of numbers on a 10-by-10 grid. At the end of each quarter, the person with the box corresponding to the last digits of the two teams’ scores wins part of the pool.

Since some of us will be participating in a pool like this on Sunday, I thought it’d be useful to look at which numbers are most likely to be profitable.

Q1 Q2


Q3 Final

(R code here).

Sunday, January 27, 2013

Thing practiced: photography

Tools used: Canon Digital Rebel T3i, Canon 40mm pancake lens, and a hat to keep my hair from flying into the shot

Point Reyes National Seashore Point Reyes National Seashore – f/2.8, 1/400 s, ISO 100

Julia Pfeiffer Burns State Park Point Reyes National Seashore
Julia Pfeiffer Burns State Park – f/8, 1/20 s, ISO 100
Point Reyes National Seashore – f/2.8, 1/2000 s, ISO 100

Limekiln State Park Limekiln State Park – f/2.8, 1/20 s, ISO 100

Thursday, January 24, 2013

Thing practiced: programming

Tools used: Sublime Text 2, Inkscape, Google Chrome

screenshot screenshot

I was never much of a gamer, mainly because I really suck at video games. I do remember spending an entire afternoon as a kid playing a Flash game about pancakes, though. For today’s practice I tried to replicate the pancake game, but in HTML5/JavaScript instead of Flash, and with pillows instead of pancakes.

Try it here.

Your goal is to earn points by arranging pillows into piles of the same color. You earn 100 points whenever you make a pile of three same-colored pillows. You can only move the topmost pillow in each stack. To move a pillow, touch the stack it’s in, then touch the stack you’d like to move it to. New pillows are added at an increasing rate as the game level increases, and the game ends when any stack reaches the top of the screen.

Tuesday, January 22, 2013

Thing practiced: nothing

Tools used: OpenSCAD, MakerBot Replicator

No practice – just modified and printed Emmett Lalish’s heart gears.

screwless heart gears screwless heart gears

Sunday, January 20, 2013

Thing practiced: photography

Tools used: Canon Digital Rebel T3i with a Canon 40mm EF f/2.8 pancake lens and a Tiffen circular polarizer

Huntington Library - outside Huntington Library, near main gallery – f/2.8, 1/80 s, ISO 100

Huntington Library - rose garden Huntington Library, rose garden – f/2.8, 1/500 s, ISO 100

Huntington Library - Japanese garden Huntington Library, Japanese garden – f/2.8, 1/80 s, ISO 200

Huntington Library - Japanese garden Huntington Library, Japanese garden – f/2.8, 1/60 s, ISO 320

Huntington Library - bamboo forest Huntington Library - desert garden
Huntington Library, bamboo forest – f/2.8, 1/80 s, ISO 1600
Huntington Library, desert garden – f/2.8, 1/60 s, ISO 200

Huntington Library - lily pond Huntington Library, lily pond – f/2.8, 1/60 s, ISO 1600

Griffith Observatory Griffith Park, walk up to the observatory – f/2.8, 1/25 s, ISO 3200

Griffith Observatory Griffith Observatory, southeast view – f/2.8, 0.3 s, ISO 3200

Griffith Observatory Griffith Observatory, southeast view – f/2.8, 1/25 s, ISO 3200

Thursday, January 17, 2013

Thing practiced: programming

Tools used: Visual Studio 2012 on a Windows machine

(Quick background: The Fourier transform lets us translate any irregular signal into a combination of pure sine waves. The numerical version of this is the discrete Fourier transform (DFT), which is ubiquitous today – it’s responsible for jpegs and mp3s, among other things – thanks to the efficient fast Fourier transform (FFT) family of algorithms.)

Let’s try to implement the FFT in F#. First, here’s the naive DFT algorithm (a direct translation to code of the definition):

let slow_dft xs =
    let N = Array.length xs
    [| for k in 0 .. N - 1 ->
           let mutable x_k = Complex.Zero
           for n = 0 to N - 1 do
               let a = 2.0 * Math.PI / float N *
                       float k * float n
               x_k <- x_k + xs.[n] *
                      Complex((cos a), (sin a))
           x_k |]

This gives an exact answer, but with poor performance – this has O(N2) time complexity. To improve performance, we first reduce problem scope by assuming that the number of input terms is an integral power of 2, letting us use the radix-2 DIT divide-and-conquer algorithm. This splits the DFT into two DFTs – a DFT of the even-numbered and a DFT of the odd-numbered terms:

which gets rearranged as:

This article explains further. Code:

let radix2_fft xs =
    let bit_rev arr =
        let N = Array.length arr
        let mutable j = 0
        let swap (a : _[]) i j =
            let tmp = a.[j]
            a.[j] <- a.[i]
            a.[i] <- tmp
        let rec helper m j =
            let m = m / 2
            let j = j ^^^ m
            if j &&& m = 0 then helper m j else j
        for i = 0 to N - 2 do
            if i < j then swap arr i j
            j <- helper N j
    let radix2_helper (xs : Complex[]) N j m =
        let t = Math.PI * float m / float j
        let a = Complex((cos t), (sin t))
        let mutable i = m
        while i < N do
            let x_i = xs.[i]
            let t = a * xs.[i+j]
            xs.[i] <- x_i + t
            xs.[i+j] <- x_i - t
            i <- i + 2 * j
    let N = Array.length xs
    bit_rev xs
    let mutable j = 1
    while j < N do
        for m = 0 to j - 1 do
            radix2_helper xs N j m
        j <- 2 * j

This is fast – O(n log n) – but only handles signals with lengths of integral powers of 2. To use the radix-2 algorithm for a signal of arbitrary length, we need to increase length to the next smallest power of 2. We can’t zero-pad the signal itself, but if we express the transform as a convolution then pad the convolution we’ll get an exact answer. (My hour’s up, but if you’re interested the convolution algorithm is explained here.)

Monday, January 14, 2013

Thing practiced: data visualization

Tools used: RStudio, ggplot2, and Inkscape

For today’s practice I tried to recreate the graphics in Nathan Yau’s unemployment map from 2009 using R and ggplot2.

Code for the solution is here.

The Bureau of Labor Statistics makes state-, region-, county-, and city-level unemployment data available in flat files via FTP here. I used mapping data from the R maps library to draw the regions, but the county names there don’t match the BLS county names exactly – I removed references to “County”, “Parish”, and “City” in the county names to solve some of the mismatch problems, but a few counties are still missing. Also, I didn’t bother plotting and insetting Alaska or Hawaii.

The result (this would’ve been more interesting if I’d waited until the 2012 data was released, but still): unemployment choropleth

Thursday, January 10, 2013

Thing practiced: hacking stuff together

Tools used: Jekyll (blog engine), RDiscount (Markdown parser), Pygments (code highlighter), MathJax (LaTeX renderer), Sublime Text 2 (for building templates and writing posts)

The idea behind Jekyll – content-centric, lightweight blogging – suits me, and so far I love it. Boy is this some ugly site, though. Visual design next.

Wednesday, January 9, 2013

Thing practiced: math

Tools used: pen, paper, and MonoDevelop


In the equation
where x, y, and n are positive integers, what is the least value of n where the number of distinct solutions (i.e. solutions where x, y does not recur as y, x) exceeds 1000?


Since x, y, and n are positive integers, we know that x and y must each be greater than n. Thus, we can rewrite (1) as
where a and b are again positive integers.

Doing some algebra, we get:

This means that each distinct pair of divisors of n2 represents a distinct solution to (1). Thus, if we can find the number of divisors that n2 has, we’ll know that half that number (well, actually divisors(n2) + 1 / 2 because one of the divisor pairs – n * n = n2 – won’t be doubled) is the number of solutions to (1).

Now we just need an algorithm to find the number of divisors of a given integer m. An easy way to do this would be to enumerate each of the numbers up to sqrt(m) and count each number that divides m evenly as a divisor. Since we’re going to use this algorithm many times, though, this brute force approach won’t work.

Instead, we’ll use the divisor function. This tells us that we can obtain the number of divisors of m from its prime factorization. Specifically, the number of divisors of m is the product over all distinct prime factors of the power of the prime factor, plus 1. (That is, if we can write
then we have .) Think of it this way: each distinct divisor of m is some distinct combination of m’s prime factors, and each prime factor can appear 0, 1, …, or si times, so multiplying through for each of the prime factors we get (7).)

Now we just need an algorithm to factor a number into primes. We can use a (pseudo-)sieve of Eratosthenes to generate a seed list of potential primes:

// This is not optimized, and not the real sieve
// -- see Melissa E. O'Neill, The Genuine Sieve
// of Eratosthenes -- because it doesn't matter
// here and my brain hurts.

// Returns a list of primes up to a given maximum.
let primes max =
    let rec sieve list =
        match list with
        | [] -> []
        | hd :: rest ->
            hd :: (sieve <| List.filter (fun x -> x % hd <> 0) rest)
    sieve [ 2 .. max ]

We use the list of primes to obtain the prime factorization:

// Returns the prime factorization of a number
// as a list of tuples of (prime, exponent). This
// only looks for prime factors up to a given
// maximum. Again this isn't optimized, because
// it doesn't matter for this size of problem.
// Definitely won't work for #110 though.
let primeFactors m max =
    let rec primeFactorsHelper (m, factorsSoFar) primesToTest =
        let singleFactorHelper m p factorsSoFar =
            let rec exponentHelper m p e =
                if m % p = 0 then exponentHelper (m/p) p (e+1)
                else (m, e)
            let (rem, exp) = exponentHelper m p 0
            rem, (p, exp) :: factorsSoFar
        match primesToTest with
        | [] -> factorsSoFar
        | hd :: rest when hd > m -> factorsSoFar
        | hd :: rest when m % hd = 0 ->
            primeFactorsHelper (singleFactorHelper m hd factorsSoFar) rest
        | hd :: rest -> primeFactorsHelper (m, factorsSoFar) rest
    primeFactorsHelper (m, []) (primes max)

Factoring n2 still takes some time, though, as n2 grows much more quickly than n. Luckily, we can factor n instead. From (6) we can see that so with the same combinatorics that gave us (7) we have

Now, finally, we can find the solution using (9):

// Returns the number of solutions for a given n.
// This still only uses prime factors up to a
// given maximum.
let solutions m maxPrime =
    primeFactors m maxPrime
    |> List.fold (fun acc (p, e) -> acc * ((2*e) + 1)) 1

And obtain the solution using an (empirically-determined-to-be) appropriate max prime seed:

// Returns the solution, finally, I think.
let problem108answer =
    { 1 .. System.Int32.MaxValue }
    |> Seq.find (fun n -> (((solutions n 17) + 1) / 2) > 1000)

This gives us a final answer of 180,180.

Meant to have a crack at #110 today also, but this problem took an embarassingly long time to solve.

Monday, January 7, 2013

Thing practiced: drawing

Tools used: index card, pencil, eraser, and two pens

Keeping an eye on the gay marriage debate in France…

girls in Paris