Sunday, December 16, 2012

Uncountable Infinity and The Diagonalization Method


I took a class on Discrete Mathematics this past semester, and one of the topics we covered very briefly was the concept of countable and uncountable infinity. I didn't end up getting a great grasp on it, however, and the topic I'm writing about today didn't make a lick of sense to me at the time. It was never covered on exams, so I never had to revisit the idea. But, I've been reading through Douglas R. Hofstadter's Gödel, Escher, Bach, and it just so happens that the topic is covered in the book. As it turns out, it's not actually too difficult a thing to understand, so I'm hoping to relay the ideas in this blog post. If you've heard about the concept of multiple infinities but not understood it, hopefully this will shed some light! If you haven't yet heard about the concept of multiple infinities, enjoy the read!

Anyway, on with the show...

Natural Numbers Form an Infinite Set

It is a widely known fact (axiom of ZFC Set Theory) that the set of natural numbers (all non-negative integers) extends on to infinity. That is to say, there is no "largest" natural number. We can order the set of natural numbers in this way:

N_0 & = & 0 \cr
N_1 & = & 1 \cr
N_2 & = & 2 \cr
N_3 & = & 3 \cr
N_4 & = & 4 \cr
N_5 & = & 5 \cr
This ordering comes very naturally to us. The nth term of the set of natural numbers will be, simply, n. Because of this fact, we can tell that this set of natural numbers contains all of them -- if you give me any natural number, it will correspond to the nth number in the set.

There is a similar way of ordering the integers (including negatives) in this fashion. It may not be heavily intuitive at first, because we have to extend in two directions rather than one. But, it is apparent that this ordering of integers will allow us to find the location of any integer in it:

$$ \begin{matrix}
Z_0 & = & 0 \cr
Z_1 & = & 1 \cr
Z_2 & = & -1 \cr
Z_3 & = & 2 \cr
Z_4 & = & -2 \cr
Z_5 & = & 3 \cr
Any integer, positive or negative, is contained in this infinite set. An infinite set that can be ordered in a way such as this is said to be countably infinite.

What About Real Numbers?

For the purposes of this post, we're going to focus on the real numbers between 0 and 1. We can represent them in the following way:

R_0 & = & 0 . & a_0 & a_1 & a_2 & a_3 & a_4 & \cdots \cr
R_1 & = & 0 . & b_0 & b_1 & b_2 & b_3 & b_4 & \cdots \cr
R_2 & = & 0 . & c_0 & c_1 & c_2 & c_3 & c_4 & \cdots \cr
R_3 & = & 0 . & d_0 & d_1 & d_2 & d_3 & d_4 & \cdots \cr
We can see that this set goes on forever, that is, extends infinitely, just as the set of integers and naturals does. However, the set of real numbers is "built" in a different way. Both of these facts are important in what we will observe next.

Cantor's Diagonalization Method

I claim now that I can produce a number that the above set does not contain. To do this, I will be using Georg Cantor's famous Diagonalization Method. Here's how it works.

First, I will grab the $n$th term of each $n$th element in our set of real numbers to compose a new number, like so:

R_0 & = & 0 . & \color{red}{a_0} & a_1 & a_2 & a_3 & a_4 & \cdots \cr
R_1 & = & 0 . & b_0 & \color{red}{b_1} & b_2 & b_3 & b_4 &\cdots \cr
R_2 & = & 0 . & c_0 & c_1 & \color{red}{c_2} & c_3 & c_4 & \cdots \cr
R_3 & = & 0 . & d_0 & d_1 & d_2 & \color{red}{d_3} & d_4 & \cdots \cr
\cdots &&& &&&& \color{red}{\ddots}
So my new number becomes:

0 . & a_0 & b_1 & c_2 & d_3 & \cdots
Now I'm going to perform some transformation on each digit of this number to produce a new digit. A simple way (the simplest way?) to do this would just be to add 1 to each digit. This will produce a new number as follows:

M_0 & = & 0 . & a_0 + 1 & b_1 + 1 & c_2 + 1 & d_3 + 1 & \cdots
We can see $M_0$ cannot be the same as $R_0$ because its first term differs. Same goes for $R_1$ with its second digit, and so on, ad infinitum. Therefore, each element of set of real numbers between 0 and 1 is going to differ from $M_0$ by at least one digit. We can conclude from this observation that our (infinite!) set of real numbers excludes $M_0$. That is to say, our set of real numbers between 0 and 1 isn't actually complete, and cannot actually be complete.

That last part, "cannot be complete," may sound confusing, because, why can't we just add $M_0$ to the set, and call it complete?

Well, let's do it! We'll tack on $M_0$ to the set to produce something like this:

M_0 & = & 0 . & a_0 & b_1 & c_2 & d_3 & e_4 & \cdots \cr
R_0 & = & 0 . & a_0 & a_1 & a_2 & a_3 & a_4 & \cdots \cr
R_1 & = & 0 . & b_0 & b_1 & b_2 & b_3 & b_4 & \cdots \cr
R_2 & = & 0 . & c_0 & c_1 & c_2 & c_3 & c_4 & \cdots \cr
R_3 & = & 0 . & d_0 & d_1 & d_2 & d_3 & d_4 & \cdots \cr
\cdots \cr\
You might foresee what's going to happen next. We can perform diagonalization again:

M_0 & = & 0 . & \color{red}{a_0} & b_1 & c_2 & d_3 & e_4 & f_5 & \cdots \cr
R_0 & = & 0 . & a_0 & \color{red}{a_1} & a_2 & a_3 & a_4 & a_5 & \cdots \cr
R_1 & = & 0 . & b_0 & b_1 & \color{red}{b_2} & b_3 & b_4 & b_5 & \cdots \cr
R_2 & = & 0 . & c_0 & c_1 & c_2 & \color{red}{c_3} & c_4 & c_5 & \cdots \cr
R_3 & = & 0 . & d_0 & d_1 & d_2 & d_3 & \color{red}{d_4} & d_5 & \cdots \cr
\cdots & & & &&&&& \color{red}{\ddots} \cr
$$ produce a new number...

$$ \begin{matrix} 0. & a_0 & a_1 & b_2 & c_3 & d_4 & \cdots \end{matrix} $$
We perform some transformation on its elements (let's add one, again) in order to get a new number, say $M_1$.

$$ \begin{matrix} M_1 & = & 0. & a_0 + 1 & a_1 + 1 & b_2 + 1 & c_3 + 1& d_4 + 1 & \cdots \end{matrix}$$
However, note that $M_1$ must differ from $M_0$ now, as well as every other number in the set, by at least one digit. As such, $M_1$ must not yet exist in the set. We can add $M_1$ to the set, and repeat this process as many times as we want, but we'll always be able to produce a number outside of the set! We call a set with this fascinating property uncountably infinite.

What does this mean? We can see that the set of integers is countably infinite, while the set of real numbers between 0 and 1 is uncountably infinite. We have successfully proven in this blog post that there are actually more numbers between 0 and 1 than there are integers!

Cantor's Diagonalization Method has been used to prove several difficult problems in Mathematics, including the Church-Turing Theorem and Gödel's Incompleteness Theorems. All three of these theorems have had major effects on the nature of Mathematics, so you can see that Cantor's Diagonalization Method can be quite useful!

If you have any questions, or if anything was unclear, please leave a comment!

Until next time,


Thursday, December 6, 2012

Graphs and Topological Sorting in the Functional Paradigm

What is a Graph?

From Wikipedia
In mathematics, a graph is a representation of a set of objects where some pairs of the objects are connected by links. The interconnected objects are represented by mathematical abstractions called vertices, and the links that connect some pairs of vertices are called edges.
Simply put, a graph is just a bunch of points with links between them. A road map is a simple example: roads  being edges, and intersections being vertices.  In fact, Google maps uses graphs for just this purpose! Graphs are widely used in a wide variety of places. Facebook uses graphs to model your friend connections and likes. In fact, the entire internet is just a giant graph; websites act as vertices, with hyperlinks as edges. Graphs are highly useful structures, as they can be used to model many different types of situations, and as such, they will be the focus of this blog post. I am going to discuss one way to represent a graph in the Haskell programming language, and how to functionally solve a common problem using graphs.

Graphs are often represented visually like this:

Graph representing abstract data

This graph links the first six letters of the alphabet in an arbitrary way. This data doesn't really mean anything, but it will serve as a simple foray into the world of graphs, and provides an initial graph to work towards representing in Haskell.

Let's get right to it; here's the data structure we'll be using, along with some convenience methods:

First we define the actual Graph a data type: It's simply a set vertices and edges in the form of 2-tuples (The tuple (a, b) connects vertex a to vertex b), which fits our definition. I've also defined the removeEdge method, which does just what you'd expect. The outbound and inbound methods find the outbound and inbound connections to any point in the graph, respectively. They make use of the polymorphic connections method in order to get this done in a small amount of code. Finally, the Graph module exports the relevant functions at the top of the file.

Now that we've got our framework in order, we can go ahead and build the graph we mentioned above:

We import the Graph module and define a simple Letter data type, then build our Graph from it. The set of vertices are the letters A, B, C, D, E, and F, and the edges are modeled as above.

Now that we know how to build graphs, we can start modeling more important information with them.

Modeling Actual Scenarios using Graphs

Suppose some of the characters from NBC's Parks and Recreation have just finished competing in a dance competition, and we know the following about their rankings:

Leslie beat April.
April beat Ann.
Ron beat April.
Ron beat Ann.
April beat Andy.
Leslie beat Ron.
Andy beat Jerry.
Ron beat Andy.
Ann beat Jerry.
Leslie beat Andy.
Ann beat Andy.

This is a little hard to understand, so why don't we model it as a graph to make it a little more readable? Each person can be represented as a vertex, with outgoing edges representing connections to the people they beat.

A graph of dance competition results

It would be nice to be able to be able to read scenarios like this from a text file containing the important data and parse it into a graph. Let's go ahead and set up a function to do this for us, so we don't have to hard-code each and every graph that we want to use:

Here's our data file, with a list of space-separated connections, one on each line:

And our parsing function:

The graphFromFile function takes a String and returns an IO (Graph String).  The function reads a file, parses it into two important pieces: verts (the set of all unique strings in the file, or, our vertices) and conns (the set of connections between strings in the file). It then builds a Graph from this data, wraps it in the IO monad with return, and gives it back.

Now you might have been wondering from the beginning of this section what the ranking from the dance competition was (maybe you even figured it out on your own!). How do we do this programmatically, using our graph?

Enter Topological Sort

Again, from Wikipedia:
In computer science, a topological sort of a directed graph is a linear ordering of its vertices such that, for every edge uvu comes before v in the ordering.
In our case, this just means that each person must come before all of the people that he or she beat in the competition, in the ordering.

The basic procedure for topological sort is as follows:

L = {} --sorted list
S = Set of vertices with no incoming connections
while S is not empty:
    for each vertex v in S with no incoming connections:
        push v to L
        for each edge e from v to u:
            remove e from graph
            if u has no more incoming connections:
                push u to S

if edges still exist in the graph:
    error: there is at least one cycle in the graph

else return L

If you do not understand this, I urge you to work through topologically sorting a graph on paper first; it's not too tough to understand once you've done it on paper, but can get a little confusing in psuedocode.

The problem with this algorithm is that you see a ton of loops -- control structures that we do not have in Haskell. Therefore, we must rely on recursion, folds, and maps to achieve what we want to do. Here's how it looks:

Our tsort function first finds the elements in the graph with no incoming edges using the function noInbound.  We pass this into a sub-routine tsort' that takes a sorted list l, a list of vertices with no incoming connections (n:s), and a graph g.

We operate on the first element of the set of vertices with no incoming connections n, finding outEdges (the outgoing edges from n), and outNodes (the nodes that n points to). We build a new graph g' with the outEdges removed, and find the nodes in g' with no inbound connections, and add them to s.

We then recursively call tsort' with these new parameters (and prepend our current n to the sorted list), until there are no more nodes to check. At this point, if the edge list in the graph is empty, all is well and we return the list of sorted elements. Otherwise, an error is thrown stating that there is at least one cycle in the graph.

Now that we've got that, we're ready to find out how everyone ranked in the dance competition!

This produces the output:
["Leslie", "Ron", "April", "Ann", "Andy", "Jerry"]

(Of course Jerry lost.)


As you can see, Graphs are very useful data structures. They can be used to model a huge variety of things (see how many more you can come up with, they're everywhere!). Topological sort in particular is a pretty remarkable algorithm, and can be applied in many different situations from the one above. For example, finding paths through college courses with prerequisites. It's even used in UNIX systems to schedule processes according to their dependencies.

Hope you enjoyed the post! Until next time,


Tuesday, October 9, 2012

"Memorizing" the Unit Circle

I, like many others I'm sure, was told that I needed to memorize the unit circle while studying Pre-calculus in High School. This looks like a very daunting task at first glance, and it caused me a lot of frustration the first time I learned it.

I was taught to memorize them straight. I don't know why, because there's a much, much, better way to go about it. I'm going to start with the full unit circle and widdle down "memorization" to the bare minimum. Once you can start to see the reality of what the Unit Circle is, it becomes way easier to interpret.

The Unit Circle
First off, the unit circle is just a circle of radius 1. Knowing this, we can determine that its circumference is $c = 2\pi r = 2\pi 1 = 2\pi$ radians. Points on the unit circle are easy to find knowing that the circumference is just $2\pi$. We just have to figure out what fraction of $2\pi$ our target angle is, and locate that section on the circle.

So what's our goal? We need to be able to extract the trig functions from the unit circle:
$$sin\theta,   cos\theta,   tan\theta,   csc\theta,   sec\theta,   tan\theta$$
at each interval of $\frac{\pi}{6}$, or $30˚$, and each interval of $\frac{\pi}{4}$, or $45˚$.

To do this, we need to know a little bit about the way that the coordinate system on the unit circle works. At any given point on the unit circle, we can make a triangle starting from the center of the circle to said point that rests on the x axis.  The hypotenuse of this triangle will always be $1$, simply by the definition of the unit circle. We know that $sin = \frac{opposite}{hypotenuse}$ and $cos = \frac{adjacent}{hypotenuse}$ for any angle on a triangle. We have noted that our $hypotenuse$ will always be $1$, so the $sin$ of any angle on the unit circle will simply be the $\frac{opposite}{1}$, or the length of the opposite side, otherwise known as our $y$ coordinate. Similarly, $cos = \frac{adjacent}{1}$, or the length of the adjacent side: our $x$ coordinate.

Knowing this, the cases of $0\pi,   \frac{\pi}{2},   \pi, $ and $\frac{3\pi}{2}$ become trivial -- All we need to do is point out which coordinate is $0$, and which one is $1$, and we have both $cos$ and $sin$.

So, what about the rest of them? These are the only three fractions that I like to keep in mind when dealing with the unit circle angles. Commit these to memory: 
$$\frac{1}{2},   \frac{\sqrt2}{2},   \frac{\sqrt3}{2}$$

These (and their negations) will be our $cos$ and $sin$ values for any "special" point on the unit circle, and as such, they are all we need to remember (the rest of the trig functions can be derived from these values).

Let's first just deal with the first quadrant angles, $\frac{\pi}{6},   \frac{\pi}{4},   \frac{\pi}{3}$. The first thing that I want to do is order the fractions above from smallest to largest. I've actually written them in the correct order; so I won't bother writing them out again. Now, let's think about this for a second. In the first quadrant, out of our three points, where is $sin$ (our $y$ coordinate) the smallest? We start at $0$, and our $y$ value gets closer and closer to $1$ as the angle increases. This means that the smallest angle of our three ($\frac{\pi}{6}$) gets the smallest $sin$ value. Following this convention, we can say that our middle angle ($\frac{\pi}{4}$) gets the middle $sin$ value ($\frac{\sqrt2}{2}$), and our largest angle ($\frac{\pi}{3}$) gets the largest $sin$ value ($\frac{\sqrt3}{2}$). 

But wait! Didn't we say that our $cos$ value is simply the $y$ coordinate? Yes, and as such, this idea can be applied to the $cos$ values as well. $cos$ ($x$) is decreasing as we move from $0$ to $2\pi$ on our unit circle, though, so as angles increase, $cos$ decreases. Apply the logic from before, and we have:
 $$cos \frac{\pi}{6} = \frac{\sqrt3}{2},   cos \frac{\pi}{4} = \frac{\sqrt2}{2},   cos \frac{\pi}{3} = \frac{1}{2}$$

Knowing this, you know the rest of the values on the unit circle. Why? Because we can see where the $x$ or $y$ coordinates are negative, and it is obvious which one is larger than the other at any point. For example, take $\frac{5\pi}{6}$. Here, we can see that the $x$ coordinate is negative, but the $y$ coordinate is positive. So, we have a positive $sin$ and a negative $cos$. The length of the $y$ side is clearly shorter than the length of the $x$ side, so we can see that $sin \frac{5\pi}{6} = \frac{1}{2}$ and $cos \frac{5\pi}{6} = -\frac{\sqrt3}{2}$. Note: At intervals of $\frac{\pi}{4}$, we will always have the same ratio for both $sin$ and $cos$: $\frac{sqrt2}{2}$. It's just a matter of getting the signs correct!

So, what about the rest of the trig functions? Well, let's start with $tan$. We know that $tan\theta = \frac{opposite}{adjacent}$, right? Well, our $adjacent$ side on the unit circle is just $cos$. Our $opposite$ side is just $sin$. Therefore, on the unit circle, $tan\theta = \frac{sin\theta}{cos\theta}$. Since we can get $sin$ and $cos$, we can just divide them out and get our $tan$ value. Simple.

$sec\theta$,   $csc\theta$,  and $cot\theta$ then become trivial. All we need to do is take the reciprocals of their counterparts ($cos\theta$, $sec\theta$, and $tan\theta$ respectively). $\frac{1}{2}$ becomes $2$, $\frac{\sqrt2}{2}$ becomes $\sqrt2$, and so on.

And that's all of the trig functions for the "essential points" on the unit circle. All that is truly necessary is the recalling of three simple fractions and a few key concepts, and the whole thing becomes easy.

Hope this helps!


Wednesday, August 15, 2012

Haskell: Implementing a Fractional Data Type and Investigating the Expansion of the Square Root of 2

A couple of weeks ago, I completed Project Euler #57: Investigating the Expansion of the Square Root of 2. I found the problem really interesting, since it required me to write up my own fractional operations (addition, division, etc.) Today, I decided that I wanted to take my minimally defined fractional library and make it a more full-featured one. First, I'll first walk through the building of the data type and implementing fractional operations, and then I'll get to my solution of the Project Euler problem.

First thing's first. A Fraction is defined as a numerator and a denominator. My fractions consist of only Integers, and here's how the data type is defined:

I wanted fractions to show up in the form x / y, so that is what the instance of Show I defined does.

So, now that we have a data type, what can we do with it? Here's a list of common functions that I think are necessary in the definition of a fraction: Addition, Subtraction, Multiplication, Division, Exponentiation, Simplification, Negation, getting the Numerator, and getting the Denominator.

Let's start with the simple ones:

Those are are numerator and denominator functions. We simply take the first or second argument of the Frac in order to get the parameter that we need, and disregard the other one (using _) since it's unused.

Simplifying is also simple enough. We just need to divide the numerator and the denominator by the greatest common divisor of the two. Since Haskell gives us a GCD function, this is pretty darn simple. Here's how it's implemented:

Easy! We just make a new fraction out of the divided values. The function `quot` is basically just integer division that truncates the result towards 0. The second half of that is unimportant in this instance, since we're dividing by the GCD of the numbers and the result will always be an integer value.

Okay, so we have a couple of functions. Great! But, what about implementing addition, subtraction, etc? Well, basically what we want to do here is define a couple of instances of numeric types for our Fractional data type.

The first instance we need to derive is Num, which has a few operations that we want:

The three "common" operators (+, -, *) are defined here, which means that we can now evaluate expressions such as Frac 1 2 * Frac 1 2. Cool, right? Those three operations are fairly self-explanatory, and the code (hopefully) isn't too tough to follow. There are also three other function definitions here, that maybe aren't quite as clear. The function negate simply turns a negative fraction positive, or a positive fraction negative. The function abs takes the absolute value of the fraction. This is fairly simple; we just use a function that maps abs (Integer absolute value) over the numerator and denominator of the fraction. The last is signum, which looks probably the most complicated, but all it actually does is tells us whether the Fraction is less than, greater than, or equal to 0 (returning -1, 1, and 0, respectively).

Cool, so since we got all of those functions out of Num, where can we find the rest? We're missing /, so we'll make our Fraction an instance of Fractional. Seems appropriate, right?

Cool, we get division out of that, which is simple enough! We just take the reciprocal of the second fraction, and multiply the two. This may look a little funky, so I'll explain that last.

The other two functions defined here are recip and fromRational. recip simply flips the numerator and denominator in a Fraction, and this is easily handled with pattern matching. fromRational takes a rational number (which is provided the numerator and denominator functions) and turns it into a Fraction. Knowing what the numerator and denominator functions are, this function is incredibly trivial.

Okay, so about that division function. Our division appears to take only one argument, but it actually takes two. (*) f f' is just syntactic sugar for f * f'. We want to compose the function f* with the function recip f', so we use the function (*), apply it to f, and then apply that to the function recip, which is then called on the second argument of the function. This kind of function is hard to read for a lot of people, but it definitely comes more naturally with Haskell practice.

Alright! We've got plenty of functions at our disposal now, so what's next? Well, we want to be able to compare Fractions, so let's go ahead and make it an instance of Eq and Ord, which allow us to check equivalency and order, respectively.

There's a lot of functions that are similar in structure to our / function from earlier, so understanding what is happening with those will make these much easier to understand.

Starting with Eq, we have two functions, /= (not equals) and ==. == simply checks to see if the simplified version of f and the simplified version of f' have the same numerators and denominators. /= basically just returns the opposite of == by calling not on the result.

Ord isn't too tough either. First we have compare, which returns a comparator (LT, GT, or EQ) based on the relationship between two Fractions. The inequality functions are all designed around this function. The max and min functions are simple, and designed around our inequality functions.

So, what's left? I decided I wanted to experiment, so I decided to also make Fraction an instance of Monoid, and give our Fraction a simpler constructor. Here's the rest!

The instance of Monoid defines a couple of functions: mempty, mappend, and mconcat. mempty is the minimal value for the Monoid. I chose 0 (which gets automatically boxed into a  Fraction). mappend is a combiner function, and I thought that addition of fractions fit this bill well enough, so I simply gave it an alias. mconcat concatenates a list of fractions with some function, and in our case, sums the entire list.

Our type constructor (%) takes two integers and boxes them up into a Fraction, which is then simplified. Easy enough.

One final note on all of this. You may have noticed that exponentiation (^) isn't implemented here. But, it actually is! Turns out, any data type that has a Num  instance defined (like our Fraction) can use the ^ operator. So things like (1 % 2)^2 actually get evaluated properly. (In this case, to 1 % 4).

All right! Let's use this library to solve  Project Euler #57! Knowing how to use the Fraction library, this should be relatively easy to follow. Here we go:

Fairly simple. We directly implement the function and run it on each iteration, building a list as we go. We then filter our list so that the only entries left are the ones that have more digits in the numerator than the denominator. Lastly, we evaluate the length of that list, which gives us the answer.

Pretty neat, huh? Here's the full Fraction.hs source file:

Until next time!


Sunday, August 5, 2012

My first real code golf program!

I've always been interested in Code golf, which essentially boils down to creating the shortest possible program that completes some task.

I finally got around to completing one of these challenges, and though I wasn't really that close to winning, it was fun to mess around with it it was one of the shorter Haskell solutions!

So, the challenge I attempted was Reddit's Tiny Code Challenge #4, asking for the shortest program that finds the longest identical subsequence in a string.

Directly from the challenge page:
For example:
aaabbbbccc aabccdcdd -> aab, bcc
adeaabbbbccc aabccdedcdd -> aab, bcc
abcdef ghijk -> nothing
abcdef fghijk -> f
So, here was my solution in Haskell:

Let's break this down into steps, since it probably looks like a bunch of mumbo-jumbo!

And there you have it! 193 characters of Haskell code got us a pretty cool program.

Here's a little bit of output: 
>tc04.exe "aaabbbbccc" "aabccdcdd"
>tc04.exe "abcdef" "ghijk"
>tc04.exe "abcdef" "fghijk"
>tc04.exe "fhqwhgads" "where are all the dragons?"
Cool! :)


Tuesday, July 17, 2012

Introducing fhck, a brainf*ck interpreter written in Haskell!

If you haven't heard of brainf*ck, you should totally check it out, because it's probably the coolest programming language of them all!

Also, probably one of the simplest...

That's why I decided to write an interpreter for it!

This is the product of about 3 days of work, give or take, but the sessions were long and I put a lot into it.

If you're a Haskeller reading this, please feel free to critique it (I'm sure it could be improved in many places)!

If you're not, please feel free to check it out anyway! I encourage everyone to play around with brainfuck a little bit because it's actually a  lot of fun.

If you're not familiar with the language, a quick look through brainfuck's Wikipedia page should help you to understand what's going on in brainfuck (it's not too complex).

With the interpreter, you can write a file, say thisisbf.b, containing the classic "Hello, world!" brainfuck program directly from the wikipedia page:

and pass it to the interpreter:

on Windows:
$ build\windows\fhck.exe thisisbrainfuck.b
or Linux (or presumably Mac OSX, though I haven't tried):
$ build/linux/fhck thisisbrainfuck.b
...and it should print "Hello, world!" to your console!

Neat, right? (I think so!)

Download link/source is located here: fhck 1.0


Wednesday, July 11, 2012

99 Haskell Problems #13: Run-length Encoding

I've been working on The Ninety-Nine Haskell Problems lately, and I've come across a particular one that seems to be practical, and was a lot of fun to implement!

The 99 Problems start out with a lot of list-based exercises, and eventually you're asked to build up to a point where you can implement a function that performs Run-length encoding on a String. Basically what this means is, if we have a string, say "aabbcccd", we'd group together the adjacent, identical characters, count them, and then output the characters along with their counts in a new list.

Thus, the previous example "aabbcccd" would output something like this: a2b2c3d

A couple of the 99 Problems ask you to implement this indirectly, meaning, actually group the identical characters into sub-lists and count the length of those lists. This is fairly trivial, making use of Haskell's built-in function group, which takes a list and separates it in exactly the manner we would need in order to get sub-lists of adjacent identical values.

The real fun came in Problem 13, where it asks for a direct implementation of the run-length encoding algorithm, meaning, we're no longer allowed to split up the groups in the way that we were before. Additionally, the problem asks that we use a data wrapper on each ending value so we are able to discern Single values (example in above: d) vs Multiple values (ex: a2).

Here's what I came up with:

Let's take a look at this.

So, first I defined a new data type called Count that can be either a Single that contains a character, or a Multiple that contains a character and it's count (as an Int)

What encodeDirect does is actually parses a list of Chars (which, in Haskell, can be represented as a String, like "aabbcccd") into a list that looks something like this:

[Multiple 2 a, Multiple 2 b, Multiple 3 c, Single d]

The procedure is as follows:

First, we use encode every character in our input set by using a foldr and an encode function in order to get a list that contains 2-tuples with our character as the second value, and its count as the first. (   [(2, 'a'), (2, 'b') ... ]   )

The encode function might look a little cryptic, but broken down, it makes sense. First, if we have an empty resultant list of Counts, we need to put the first character we encounter in the resultant list. Once we have a resultant list (no matter how small), we are able to encode the other values by following some logic:
1. If the first character value in the resultant list matches what we are currently trying encode, we simply need to step up the count on that entry.
2. Otherwise, we need to add the new character to the front of the list with a count of 1.

When the foldr is complete, we have a nice list that looks just like how I described previously.

Now, to get everything wrapped up into a Count declaration, we simply map the toCount function over the list we just generated. The toCount function takes one of our 2-tuple values and creates a Single value out of it if its count is equivalent to 1, otherwise it packs it into a Multiple value.

Once this is mapped, we have our resultant list!

encodeDirect "aabbcccd" = [Multiple 2 a, Multiple 2 b, Multiple 3 c, Single d] 

PS:I picked up a bit of new syntax from a video I watched last night, located here:  Sokoban Live Coding. I learned a lot from it. If you're interested in Haskell, check it out.

Until next time!


Tuesday, April 3, 2012

Functional Sorting in Haskell

Sorting is a fairly major topic of study when it comes to programming, and there are tons of ways to do it. I don't know how interesting these algorithms are to other people, but they have always been an interest of mine. I think they're neat because they illustrate that there can be several ways to tackle even the simplest of tasks (which, in my opinion, is one of the things that makes programming so fun).

Since I'm always looking for new things to try, I've decided to take a few of the common sorting algorithms and implement them on my own in Haskell. For those of you who aren't familiar with it, here's a brief overview.

Most programming languages (Java, C++, Python, Ruby, Javascript, etc) are defined as imperative, which basically means that programs are executed line-by-line. Many of the aforementioned languages are also defined as object-oriented, meaning that virtually everything is a part of an object. I won't get into the details about OOP here, since we're talking about Haskell, but you can read about it if you'd like.  Haskell differs from these languages by conforming to a paradigm in which everything is defined as a function. Objects are not present here, and iterative steps don't quite work in the same way --  nothing is being executed line-by-line -- all instructions are defined in functions. This is a rather mind-blowing concept, but it's extremely cool once you can get a mental grasp on it.

Back to the point of the blog post. I'll be covering how to implement three sorting algorithms in a functional manner: Bubble Sort, Insertion Sort, and finally, Quicksort. If you don't have a background in Haskell, this might all come off as programming jargon, but if you're looking for an interesting new look at these algorithms you've seen a billion times implemented imperatively, or are interested in picking up Haskell as a hobby or something, I'd recommend reading on.

Bubble Sort:
The bubble sort algorithm is as follows:
1. Take an array's first two elements (we'll call the first one x, and the second one y)
2. If x is less than y, leave those elements alone. If not, swap y with x, so that the array now reads
[y, x, (the rest of the array)]
3. Repeat this for each element in the array, swapping as it goes along.
4. Check to see if the list is sorted. If not, rerun the algorithm on the result of the last iteration of the function.
5. When list is sorted, return the result.

Here it is in Haskell:

Rather long, but effective. Bubble Sort isn't the most efficient sorting algorithm, especially since it has to do all of that checking if it's already sorted. I'm not a big fan of this one.

Insertion Sort:
Insertion sort is pretty simple to follow. We start with an array, as always, and then follow these steps:
1. Pop the first element of the array off of the array, and populate a new array with it.
2. For each element after the first in the array, insert it at it's proper sorted location in the array.
3. Return the final list of sorted numbers.

Let's see what this looks like in Haskell:

How elegant is that? This could take many lines in an imperative language, but we make use of Haskell's recursive foldr method to populate our sorted list, and the appropriately named function insert taken from the Data.List package helps us to populate it in the exact manner we're looking for.

The Quicksort algorithm can be pretty intimidating, especially at first glance. This is because it makes use of recursion to accomplish it's sorting, which is a particularly tough concept to fully understand when it comes to programming. The instructions for Quicksort look a little something like this:
1. Choose any point in the array as a "pivot" point.
2. Create two new arrays, one populated with the numbers in the array less than the pivot, and another with the numbers greater than the pivot.
3. Recursively call Quicksort on both of the lists, until these lists (inevitably) turn up empty, at which point the function returns an empty list.
4. At this point, the recursion unwinds itself, and we concatenate the recursively sorted arrays together. (I would recommend reading up on this algorithm if the above instructions do not make sense; I am not the best at explaining things)

Sounds complicated, right? Here's a Haskell implementation:

What? It only takes five lines to implement Quicksort? This is why Haskell is so cool to me. Compare this with something like, say, a Java implementation, and you'll see what I mean. It's just so simple! We're simply defining quickSort as a quickSorted list of smaller numbers concatenated with a singleton list containing a pivot, concatenated with a quickSorted list of larger numbers. Easy to read, though the algorithm may seem very complicated.

Hope you enjoyed the read!


Wednesday, March 21, 2012

Hey! [Insert obligatory first post statement here]

I've been messing around with Unity3d for the past week or so, and I have to say, it's very fun and simple to use.

That said, one of the first things that I thought to do after watching a bunch of tutorials and reading into various sample scripts was to implement an intuitive third person camera interface. I've seen several tutorials regarding how to move around using the WASD keys and the like, but not much regarding how to piece together a nice third person camera system. So, I decided to go ahead and make one. 

Please keep in mind that I am a beginnerThere may be code issues here when things start getting more complicated, but this does work fine with a character on a flat plane (and should work for inclines mostly). This covers camera panning and zooming, as well as character movement with respect to the camera.
The first thing that we need to do is handle panning around the player character. I created a sphere object (as our player character), gave it the Rigidbody property, and created a basic floor out of a cube for  our player to run around on. I gave the scene a Point Light object to see everything better, and then started the scripting.

Luckily, Unity3d has a standard asset script that gives us basic mouse-based camera control. The script is called MouseOrbit.js, and can be found by searching through the Standard Assets or at Standard Assets > Scripts > Camera Scripts > MouseOrbit. This script basically takes a target as a parameter (in our case, the sphere), and circles around that target based on mouse controls. 

I thought that this should be tweaked in a couple of ways, so here's what I changed about it:

1. Only rotate around the target object when the right mouse is held down ( this seems to be how most third-person RPGs do things these days ).

2. Implement a zoom function using the mouse wheel!

I won't get into the details of how I did this, I'll just post my modified MouseOrbit.js file. It should be rather easy to follow:

With that script attached to our Main Camera object, we should be able to perform all of our cool new functions. Zooming can be done by scrolling in and out, and circular panning around our sphere (assuming that the sphere is attached to the script as the target) will only happen if the Right Mouse button is held down.

Now that that is working, the obvious next step would be to be able to move our character around, so that's exactly what I decided to do!

We can create a new javascript (I named mine MoveCharacter.js) to attach to our character to perform movement relative to the camera.

Moving our object with respect to the camera is a bit tricky, as you'll see. Initially I thought this would be as simple as adding to Z to move forward, subtracting from Z to move backward, and so on. However, this will only move our character in one direction, regardless of the camera's position. This means that, even if our camera is facing west, our character  will still only move north when we press the 'w' key. This is clearly a problem (but we'll fix it)!

We need to perform a little trigonometry to get our sphere moving in the proper direction. I had a little bit of help from the community over at Stack Overflow with this one. As it turns out, moving forward in space can be done like so:

 First, we take the x and z coordinates of our object with the Sin and Cos values of our current rotation and place them into a new vector (Sin(rotation), 0, Cos(rotation) ). This is our displacement vector, which adjusts for the rotation of our object.

Second, we take this displacement vector and perform our movement functions on it -- In our case, we want to multiply it by our speed value and adjust it for slower computers using our delta time

Lastly, we take this new displacement vector, adjusted for rotation and applied speed, and we add it to our position in space.

The resultant code for moving forward using the 'w' key can be implemented inside of our update function as follows:

Note: the xDisplacement, yDisplacement, and displacement variables must be defined before we can use them!

Moving backwards is basically the same, except we want to subtract our displacement instead of adding it. The code for moving backwards is as simple as mapping our input key to "s" instead of "w" and replacing line 7 in the above code with:

transform.position -= displacement;

Moving left and right, however, is a little bit more tricky, but it's nothing unmanageable. Basically all that we need to do is add 90 degrees to our angle, and follow the same basic instructions as before. If you look at a Cartesian plane and assess what we're doing so far, this makes perfect sense. What we're essentially doing is pretending that our object is actually facing directly to the right of its actual orientation, and moving forward and backward based on that rotation. This moves our character left and right. 

That's basically everything that this script does (there are different speeds for moving in different directions, but that's easy to grasp in the code), so I'll just go ahead and post the full MoveCharacter.js script for you here:

Attaching this script to your main character object (whatever it may be) will allow movement based upon its rotation. 

Hope you enjoyed this! If you have any questions or suggestions or anything, just post a comment!