The Inner Product as a Decision Rule

The standard inner product of two vectors has some nice geometric properties. Given two vectors x, y \in \mathbb{R}^n, where by x_i I mean the i-th coordinate of x, the standard inner product (which I will interchangeably call the dot product) is defined by the formula

\displaystyle \langle x, y \rangle = x_1 y_1 + \dots + x_n y_n

This formula, simple as it is, produces a lot of interesting geometry. An important such property, one which is discussed in machine learning circles more than pure math, is that it is a very convenient decision rule.

In particular, say we’re in the Euclidean plane, and we have a line L passing through the origin, with w being a unit vector perpendicular to L (“the normal” to the line).

decision-rule-1

If you take any vector x, then the dot product \langle x, w \rangle is positive if x is on the same side of L as w, and negative otherwise. The dot product is zero if and only if x is exactly on the line L, including when x is the zero vector.

decision-rule-2

Left: the dot product of w and x is positive, meaning they are on the same side of w. Right: The dot product is negative, and they are on opposite sides.

Here is an interactive demonstration of this property. Click the image below to go to the demo, and you can drag the vector arrowheads and see the decision rule change.

decision-rule

Click above to go to the demo

The code for this demo is available in a github repository.

It’s always curious, at first, that multiplying and summing produces such geometry. Why should this seemingly trivial arithmetic do anything useful at all?

The core fact that makes it work, however, is that the dot product tells you how one vector projects onto another. When I say “projecting” a vector x onto another vector w, I mean you take only the components of x that point in the direction of w. The demo shows what the result looks like using the red (or green) vector.

In two dimensions this is easy to see, as you can draw the triangle which has x as the hypotenuse, with w spanning one of the two legs of the triangle as follows:

decision-rule-3.png

If we call a the (vector) leg of the triangle parallel to w, while b is the dotted line (as a vector, parallel to L), then as vectors x = a + b. The projection of x onto w is just a.

Another way to think of this is that the projection is x, modified by removing any part of x that is perpendicular to w. Using some colorful language: you put your hands on either side of x and w, and then you squish x onto w along the line perpendicular to w (i.e., along b).

And if w is a unit vector, then the length of a—that is, the length of the projection of x onto w—is exactly the inner product product \langle x, w \rangle.

Moreover, if the angle between x and w is larger than 90 degrees, the projected vector will point in the opposite direction of w, so it’s really a “signed” length.

decision-rule-4

Left: the projection points in the same direction as w. Right: the projection points in the opposite direction.

And this is precisely why the decision rule works. This 90-degree boundary is the line perpendicular to w.

More technically said: Let x, y \in \mathbb{R}^n be two vectors, and \langle x,y \rangle their dot product. Define by \| y \| the length of y, specifically \sqrt{\langle y, y \rangle}. Define by \text{proj}_{y}(x) by first letting y' = \frac{y}{\| y \|}, and then let \text{proj}_{y}(x) = \langle x,y' \rangle y'. In words, you scale y to a unit vector y', use the result to compute the inner product, and then scale y so that it’s length is \langle x, y' \rangle. Then

Theorem: Geometrically, \text{proj}_y(x) is the projection of x onto the line spanned by y.

This theorem is true for any n-dimensional vector space, since if you have two vectors you can simply apply the reasoning for 2-dimensions to the 2-dimensional plane containing x and y. In that case, the decision boundary for a positive/negative output is the entire n-1 dimensional hyperplane perpendicular to y (the projected vector).

In fact, the usual formula for the angle between two vectors, i.e. the formula \langle x, y \rangle = \|x \| \cdot \| y \| \cos \theta, is a restatement of the projection theorem in terms of trigonometry. The \langle x, y' \rangle part of the projection formula (how much you scale the output) is equal to \| x \| \cos \theta. At the end of this post we have a proof of the cosine-angle formula above.

Part of why this decision rule property is so important is that this is a linear function, and linear functions can be optimized relatively easily. When I say that, I specifically mean that there are many known algorithms for optimizing linear functions, which don’t have obscene runtime or space requirements. This is a big reason why mathematicians and statisticians start the mathematical modeling process with linear functions. They’re inherently simpler.

In fact, there are many techniques in machine learning—a prominent one is the so-called Kernel Trick—that exist solely to take data that is not inherently linear in nature (cannot be fruitfully analyzed by linear methods) and transform it into a dataset that is. Using the Kernel Trick as an example to foreshadow some future posts on Support Vector Machines, the idea is to take data which cannot be separated by a line, and transform it (usually by adding new coordinates) so that it can. Then the decision rule, computed in the larger space, is just a dot product. Irene Papakonstantinou neatly demonstrates this with paper folding and scissors. The tradeoff is that the size of the ambient space increases, and it might increase so much that it makes computation intractable. Luckily, the Kernel Trick avoids this by remembering where the data came from, so that one can take advantage of the smaller space to compute what would be the inner product in the larger space.

Next time we’ll see how this decision rule shows up in an optimization problem: finding the “best” hyperplane that separates an input set of red and blue points into monochromatic regions (provided that is possible). Finding this separator is core subroutine of the Support Vector Machine technique, and therein lie interesting algorithms. After we see the core SVM algorithm, we’ll see how the Kernel Trick fits into the method to allow nonlinear decision boundaries.


Proof of the cosine angle formula

Theorem: The inner product \langle v, w \rangle is equal to \| v \| \| w \| \cos(\theta), where \theta is the angle between the two vectors.

Note that this angle is computed in the 2-dimensional subspace spanned by v, w, viewed as a typical flat plane, and this is a 2-dimensional plane regardless of the dimension of v, w.

Proof. If either v or w is zero, then both sides of the equation are zero and the theorem is trivial, so we may assume both are nonzero. Label a triangle with sides v,w and the third side v-w. Now the length of each side is \| v \|, \| w\|, and \| v-w \|, respectively. Assume for the moment that \theta is not 0 or 180 degrees, so that this triangle is not degenerate.

The law of cosines allows us to write

\displaystyle \| v - w \|^2 = \| v \|^2 + \| w \|^2 - 2 \| v \| \| w \| \cos(\theta)

Moreover, The left hand side is the inner product of v-w with itself, i.e. \| v - w \|^2 = \langle v-w , v-w \rangle. We’ll expand \langle v-w, v-w \rangle using two facts. The first is trivial from the formula, that inner product is symmetric: \langle v,w \rangle = \langle w, v \rangle. Second is that the inner product is linear in each input. In particular for the first input: \langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle and \langle cx, z \rangle = c \langle x, z \rangle. The same holds for the second input by symmetry of the two inputs. Hence we can split up \langle v-w, v-w \rangle as follows.

\displaystyle \begin{aligned} \langle v-w, v-w \rangle &= \langle v, v-w \rangle - \langle w, v-w \rangle \\ &= \langle v, v \rangle - \langle v, w \rangle - \langle w, v \rangle +  \langle w, w \rangle \\ &= \| v \|^2 - 2 \langle v, w \rangle + \| w \|^2 \\ \end{aligned}

Combining our two offset equations, we can subtract \| v \|^2 + \| w \|^2 from each side and get

\displaystyle -2 \|v \| \|w \| \cos(\theta) = -2 \langle v, w \rangle,

Which, after dividing by -2, proves the theorem if \theta \not \in \{0, 180 \}.

Now if \theta = 0 or 180 degrees, the vectors are parallel, so we can write one as a scalar multiple of the other. Say w = cv for c \in \mathbb{R}. In that case, \langle v, cv \rangle = c \| v \| \| v \|. Now \| w \| = | c | \| v \|, since a norm is a length and is hence non-negative (but c can be negative). Indeed, if v, w are parallel but pointing in opposite directions, then c < 0, so \cos(\theta) = -1, and c \| v \| = - \| w \|. Otherwise c > 0 and \cos(\theta) = 1. This allows us to write c \| v \| \| v \| = \| w \| \| v \| \cos(\theta), and this completes the final case of the theorem.

\square

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Hamming’s Code

Or how to detect and correct errors

Last time we made a quick tour through the main theorems of Claude Shannon, which essentially solved the following two problems about communicating over a digital channel.

  1. What is the best encoding for information when you are guaranteed that your communication channel is error free?
  2. Are there any encoding schemes that can recover from random noise introduced during transmission?

The answers to these questions were purely mathematical theorems, of course. But the interesting shortcoming of Shannon’s accomplishment was that his solution for the noisy coding problem (2) was nonconstructive. The question remains: can we actually come up with efficiently computable encoding schemes? The answer is yes! Marcel Golay was the first to discover such a code in 1949 (just a year after Shannon’s landmark paper), and Golay’s construction was published on a single page! We’re not going to define Golay’s code in this post, but we will mention its interesting status in coding theory later. The next year Richard Hamming discovered another simpler and larger family of codes, and went on to do some of the major founding work in coding theory. For his efforts he won a Turing Award and played a major part in bringing about the modern digital age. So we’ll start with Hamming’s codes.

We will assume some basic linear algebra knowledge, as detailed our first linear algebra primer. We will also use some basic facts about polynomials and finite fields, though the lazy reader can just imagine everything as binary \{ 0,1 \} and still grok the important stuff.

hamming-3

Richard Hamming, inventor of Hamming codes. [image source]

What is a code?

The formal definition of a code is simple: a code C is just a subset of \{ 0,1 \}^n for some n. Elements of C are called codewords.

This is deceptively simple, but here’s the intuition. Say we know we want to send messages of length k, so that our messages are in \{ 0,1 \}^k. Then we’re really viewing a code C as the image of some encoding function \textup{Enc}: \{ 0,1 \}^k \to \{ 0,1 \}^n. We can define C by just describing what the set is, or we can define it by describing the encoding function. Either way, we will make sure that \textup{Enc} is an injective function, so that no two messages get sent to the same codeword. Then |C| = 2^k, and we can call k = \log |C| the message length of C even if we don’t have an explicit encoding function.

Moreover, while in this post we’ll always work with \{ 0,1 \}, the alphabet of your encoded messages could be an arbitrary set \Sigma. So then a code C would be a subset of tuples in \Sigma^n, and we would call q = |\Sigma|.

So we have these parameters n, k, q, and we need one more. This is the minimum distance of a code, which we’ll denote by d. This is defined to be the minimum Hamming distance between all distinct pairs of codewords, where by Hamming distance I just mean the number of coordinates that two tuples differ in. Recalling the remarks we made last time about Shannon’s nonconstructive proof, when we decode an encoded message y (possibly with noisy bits) we look for the (unencoded) message x whose encoding \textup{Enc}(x) is as close to y as possible. This will only work in the worst case if all pairs of codewords are sufficiently far apart. Hence we track the minimum distance of a code.

So coding theorists turn this mess of parameters into notation.

Definition: A code C is called an (n, k, d)_q-code if

  • C \subset \Sigma^n for some alphabet \Sigma,
  • k = \log |C|,
  • C has minimum distance d, and
  • the alphabet \Sigma has size q.

The basic goals of coding theory are:

  1. For which values of these four parameters do codes exist?
  2. Fixing any three parameters, how can we optimize the other one?

In this post we’ll see how simple linear-algebraic constructions can give optima for one of these problems, optimizing k for d=3, and we’ll state a characterization theorem for optimizing k for a general d. Next time we’ll continue with a second construction that optimizes a different bound called the Singleton bound.

Linear codes and the Hamming code

A code is called linear if it can be identified with a linear subspace of some finite-dimensional vector space. In this post all of our vector spaces will be \{ 0,1 \}^n, that is tuples of bits under addition mod 2. But you can do the same constructions with any finite scalar field \mathbb{F}_q for a prime power q, i.e. have your vector space be \mathbb{F}_q^n. We’ll go back and forth between describing a binary code q=2 over \{ 0,1 \} and a code in $\mathbb{F}_q^n$. So to say a code is linear means:

  • The zero vector is a codeword.
  • The sum of any two codewords is a codeword.
  • Any scalar multiple of a codeword is a codeword.

Linear codes are the simplest kinds of codes, but already they give a rich variety of things to study. The benefit of linear codes is that you can describe them in a lot of different and useful ways besides just describing the encoding function. We’ll use two that we define here. The idea is simple: you can describe everything about a linear subspace by giving a basis for the space.

Definition: generator matrix of a (n,k,d)_q-code C is a k \times n matrix G whose rows form a basis for C.

There are a lot of equivalent generator matrices for a linear code (we’ll come back to this later), but the main benefit is that having a generator matrix allows one to encode messages x \in \{0,1 \}^k by left multiplication xG. Intuitively, we can think of the bits of x as describing the coefficients of the chosen linear combination of the rows of G, which uniquely describes an element of the subspace. Note that because a k-dimensional subspace of \{ 0,1 \}^n has 2^k elements, we’re not abusing notation by calling k = \log |C| both the message length and the dimension.

For the second description of C, we’ll remind the reader that every linear subspace C has a unique orthogonal complement C^\perp, which is the subspace of vectors that are orthogonal to vectors in C.

Definition: Let H^T be a generator matrix for C^\perp. Then H is called a parity check matrix.

Note H has the basis for C^\perp as columns. This means it has dimensions n \times (n-k). Moreover, it has the property that x \in C if and only if the left multiplication xH = 0. Having zero dot product with all columns of H characterizes membership in C.

The benefit of having a parity check matrix is that you can do efficient error detection: just compute yH on your received message y, and if it’s nonzero there was an error! What if there were so many errors, and just the right errors that y coincided with a different codeword than it started? Then you’re screwed. In other words, the parity check matrix is only guarantee to detect errors if you have fewer errors than the minimum distance of your code.

So that raises an obvious question: if you give me the generator matrix of a linear code can I compute its minimum distance? It turns out that this problem is NP-hard in general. In fact, you can show that this is equivalent to finding the smallest linearly dependent set of rows of the parity check matrix, and it is easier to see why such a problem might be hard. But if you construct your codes cleverly enough you can compute their distance properties with ease.

Before we do that, one more definition and a simple proposition about linear codes. The Hamming weight of a vector x, denoted wt(x), is the number of nonzero entries in x.

Proposition: The minimum distance of a linear code C is the minimum Hamming weight over all nonzero vectors x \in C.

Proof. Consider a nonzero x \in C. On one hand, the zero vector is a codeword and wt(x) is by definition the Hamming distance between x and zero, so it is an upper bound on the minimum distance. In fact, it’s also a lower bound: if x,y are two nonzero codewords, then x-y is also a codeword and wt(x-y) is the Hamming distance between x and y.

\square

So now we can define our first code, the Hamming code. It will be a (n, k, 3)_2-code. The construction is quite simple. We have fixed d=3, q=2, and we will also fix l = n-k. One can think of this as fixing n and maximizing k, but it will only work for n of a special form.

We’ll construct the Hamming code by describing a parity-check matrix H. In fact, we’re going to see what conditions the minimum distance d=3 imposes on H, and find out those conditions are actually sufficient to get d=3. We’ll start with 2. If we want to ensure d \geq 2, then you need it to be the case that no nonzero vector of Hamming weight 1 is a code word. Indeed, if e_i is a vector with all zeros except a one in position i, then e_i H = h_i is the i-th row of H. We need e_i H \neq 0, so this imposes the condition that no row of H can be zero. It’s easy to see that this is sufficient for d \geq 2.

Likewise for d \geq 3, given a vector y = e_i + e_j for some positions i \neq j, then yH = h_i + h_j may not be zero. But because our sums are mod 2, saying that h_i + h_j \neq 0 is the same as saying h_i \neq h_j. Again it’s an if and only if. So we have the two conditions.

  • No row of H may be zero.
  • All rows of H must be distinct.

That is, any parity check matrix with those two properties defines a distance 3 linear code. The only question that remains is how large can n  be if the vectors have length n-k = l? That’s just the number of distinct nonzero binary strings of length l, which is 2^l - 1. Picking any way to arrange these strings as the rows of a matrix (say, in lexicographic order) gives you a good parity check matrix.

Theorem: For every l > 0, there is a (2^l - 1, 2^l - l - 1, 3)_2-code called the Hamming code.

Since the Hamming code has distance 3, we can always detect if at most a single error occurs. Moreover, we can correct a single error using the Hamming code. If x \in C and wt(e) = 1 is an error bit in position i, then the incoming message would be y = x + e. Now compute yH = xH + eH = 0 + eH = h_i and flip bit i of y. That is, whichever row of H you get tells you the index of the error, so you can flip the corresponding bit and correct it. If you order the rows lexicographically like we said, then h_i = i as a binary number. Very slick.

Before we move on, we should note one interesting feature of linear codes.

Definition: A code is called systematic if it can be realized by an encoding function that appends some number n-k “check bits” to the end of each message.

The interesting feature is that all linear codes are systematic. The reason is as follows. The generator matrix G of a linear code has as rows a basis for the code as a linear subspace. We can perform Gaussian elimination on G and get a new generator matrix that looks like [I \mid A] where I is the identity matrix of the appropriate size and A is some junk. The point is that encoding using this generator matrix leaves the message unchanged, and adds a bunch of bits to the end that are determined by A. It’s a different encoding function on \{ 0,1\}^k, but it has the same image in \{ 0,1 \}^n, i.e. the code is unchanged. Gaussian elimination just performed a change of basis.

If you work out the parameters of the Hamming code, you’ll see that it is a systematic code which adds \Theta(\log n) check bits to a message, and we’re able to correct a single error in this code. An obvious question is whether this is necessary? Could we get away with adding fewer check bits? The answer is no, and a simple “information theoretic” argument shows this. A single index out of n requires \log n bits to describe, and being able to correct a single error is like identifying a unique index. Without logarithmically many bits, you just don’t have enough information.

The Hamming bound and perfect codes

One nice fact about Hamming codes is that they optimize a natural problem: the problem of maximizing d given a fixed choice of n, k, and q. To get this let’s define V_n(r) denote the volume of a ball of radius r in the space \mathbb{F}_2^n. I.e., if you fix any string (doesn’t matter which) x, V_n(r) is the size of the set \{ y : d(x,y) \leq r \}, where d(x,y) is the hamming distance.

There is a theorem called the Hamming bound, which describes a limit to how much you can pack disjoint balls of radius r inside \mathbb{F}_2^n.

Theorem: If an (n,k,d)_2-code exists, then

\displaystyle 2^k V_n \left ( \left \lfloor \frac{d-1}{2} \right \rfloor \right ) \leq 2^n

Proof. The proof is quite simple. To say a code C has distance d means that for every string x \in C there is no other string y within Hamming distance d of x. In other words, the balls centered around both x,y of radius r = \lfloor (d-1)/2 \rfloor are disjoint. The extra difference of one is for odd d, e.g. when d=3 you need balls of radius 1 to guarantee no overlap. Now |C| = 2^k, so the total number of strings covered by all these balls is the left-hand side of the expression. But there are at most 2^n strings in \mathbb{F}_2^n, establishing the desired inequality.

\square

Now a code is called perfect if it actually meets the Hamming bound exactly. As you probably guessed, the Hamming codes are perfect codes. It’s not hard to prove this, and I’m leaving it as an exercise to the reader.

The obvious follow-up question is whether there are any other perfect codes. The answer is yes, some of which are nonlinear. But some of them are “trivial.” For example, when d=1 you can just use the identity encoding to get the code C = \mathbb{F}_2^n. You can also just have a code which consists of a single codeword. There are also some codes that encode by repeating the message multiple times. These are called “repetition codes,” and all three of these examples are called trivial (as a definition). Now there are some nontrivial and nonlinear perfect codes I won’t describe here, but here is the nice characterization theorem.

Theorem [van Lint ’71, Tietavainen ‘73]: Let C be a nontrivial perfect (n,d,k)_q code. Then the parameters must either be that of a Hamming code, or one of the two:

  • A (23, 12, 7)_2-code
  • A (11, 6, 5)_3-code

The last two examples are known as the binary and ternary Golay codes, respectively, which are also linear. In other words, every possible set of parameters for a perfect code can be realized as one of these three linear codes.

So this theorem was a big deal in coding theory. The Hamming and Golay codes were both discovered within a year of each other, in 1949 and 1950, but the nonexistence of other perfect linear codes was open for twenty more years. This wrapped up a very neat package.

Next time we’ll discuss the Singleton bound, which optimizes for a different quantity and is incomparable with perfect codes. We’ll define the Reed-Solomon and show they optimize this bound as well. These codes are particularly famous for being the error correcting codes used in DVDs. We’ll then discuss the algorithmic issues surrounding decoding, and more recent connections to complexity theory.

Until then!

Posts in this series:

Linear Programming and the Simplex Algorithm

In the last post in this series we saw some simple examples of linear programs, derived the concept of a dual linear program, and saw the duality theorem and the complementary slackness conditions which give a rough sketch of the stopping criterion for an algorithm. This time we’ll go ahead and write this algorithm for solving linear programs, and next time we’ll apply the algorithm to an industry-strength version of the nutrition problem we saw last time. The algorithm we’ll implement is called the simplex algorithm. It was the first algorithm for solving linear programs, invented in the 1940’s by George Dantzig, and it’s still the leading practical algorithm, and it was a key part of a Nobel Prize. It’s by far one of the most important algorithms ever devised.

As usual, we’ll post all of the code written in the making of this post on this blog’s Github page.

Slack variables and equality constraints

The simplex algorithm can solve any kind of linear program, but it only accepts a special form of the program as input. So first we have to do some manipulations. Recall that the primal form of a linear program was the following minimization problem.

\min \left \langle c, x \right \rangle \\ \textup{s.t. } Ax \geq b, x \geq 0

where the brackets mean “dot product.” And its dual is

\max \left \langle y, b \right \rangle \\ \textup{s.t. } A^Ty \leq c, y \geq 0

The linear program can actually have more complicated constraints than just the ones above. In general, one might want to have “greater than” and “less than” constraints in the same problem. It turns out that this isn’t any harder, and moreover the simplex algorithm only uses equality constraints, and with some finicky algebra we can turn any set of inequality or equality constraints into a set of equality constraints.

We’ll call our goal the “standard form,” which is as follows:

\max \left \langle c, x \right \rangle \\ \textup{s.t. } Ax = b, x \geq 0

It seems impossible to get the usual minimization/maximization problem into standard form until you realize there’s nothing stopping you from adding more variables to the problem. That is, say we’re given a constraint like:

\displaystyle x_7 + x_3 \leq 10,

we can add a new variable \xi, called a slack variable, so that we get an equality:

\displaystyle x_7 + x_3 + \xi = 10

And now we can just impose that \xi \geq 0. The idea is that \xi represents how much “slack” there is in the inequality, and you can always choose it to make the condition an equality. So if the equality holds and the variables are nonnegative, then the x_i will still satisfy their original inequality. For “greater than” constraints, we can do the same thing but subtract a nonnegative variable. Finally, if we have a minimization problem “\min z” we can convert it to \max -z.

So, to combine all of this together, if we have the following linear program with each kind of constraint,

Screen Shot 2014-10-05 at 12.06.19 AM

We can add new variables \xi_1, \xi_2, and write it as

Screen Shot 2014-10-05 at 12.06.41 AM

By defining the vector variable x = (x_1, x_2, x_3, \xi_1, \xi_2) and c = (-1,-1,-1,0,0) and A to have -1, 0, 1 as appropriately for the new variables, we see that the system is written in standard form.

This is the kind of tedious transformation we can automate with a program. Assuming there are n variables, the input consists of the vector c of length n, and three matrix-vector pairs (A, b) representing the three kinds of constraints. It’s a bit annoying to describe, but the essential idea is that we compute a rectangular “identity” matrix whose diagonal entries are \pm 1, and then join this with the original constraint matrix row-wise. The reader can see the full implementation in the Github repository for this post, though we won’t use this particular functionality in the algorithm that follows.

There are some other additional things we could do: for example there might be some variables that are completely unrestricted. What you do in this case is take an unrestricted variable z and replace it by the difference of two unrestricted variables z' - z''.  For simplicity we’ll ignore this, but it would be a fruitful exercise for the reader to augment the function to account for these.

What happened to the slackness conditions?

The “standard form” of our linear program raises an obvious question: how can the complementary slackness conditions make sense if everything is an equality? It turns out that one can redo all the work one did for linear programs of the form we gave last time (minimize w.r.t. greater-than constraints) for programs in the new “standard form” above. We even get the same complementary slackness conditions! If you want to, you can do this entire routine quite a bit faster if you invoke the power of Lagrangians. We won’t do that here, but the tool shows up as a way to work with primal-dual conversions in many other parts of mathematics, so it’s a good buzzword to keep in mind.

In our case, the only difference with the complementary slackness conditions is that one of the two is trivial: \left \langle y^*, Ax^* - b \right \rangle = 0. This is because if our candidate solution x^* is feasible, then it will have to satisfy Ax = b already. The other one, that \left \langle x^*, A^Ty^* - c \right \rangle = 0, is the only one we need to worry about.

Again, the complementary slackness conditions give us inspiration here. Recall that, informally, they say that when a variable is used at all, it is used as much as it can be to fulfill its constraint (the corresponding dual constraint is tight). So a solution will correspond to a choice of some variables which are either used or not, and a choice of nonzero variables will correspond to a solution. We even saw this happen in the last post when we observed that broccoli trumps oranges. If we can get a good handle on how to navigate the set of these solutions, then we’ll have a nifty algorithm.

Let’s make this official and lay out our assumptions.

Extreme points and basic solutions

Remember that the graphical way to solve a linear program is to look at the line (or hyperplane) given by \langle c, x \rangle = q and keep increasing q (or decreasing it, if you are minimizing) until the very last moment when this line touches the region of feasible solutions. Also recall that the “feasible region” is just the set of all solutions to Ax = b, that is the solutions that satisfy the constraints. We imagined this picture:

The constraints define a convex area of "feasible solutions." Image source: Wikipedia.

The constraints define a convex area of “feasible solutions.” Image source: Wikipedia.

With this geometric intuition it’s clear that there will always be an optimal solution on a vertex of the feasible region. These points are called extreme points of the feasible region. But because we will almost never work in the plane again (even introducing slack variables makes us relatively high dimensional!) we want an algebraic characterization of these extreme points.

If you have a little bit of practice with convex sets the correct definition is very natural. Recall that a set X is convex if for any two points x, y \in X every point on the line segment between x and y is also in X. An algebraic way to say this (thinking of these points now as vectors) is that every point \delta x + (1-\delta) y \in X when 0 \leq \delta \leq 1. Now an extreme point is just a point that isn’t on the inside of any such line, i.e. can’t be written this way for 0 < \delta < 1. For example,

A convex set with extremal points in red. Image credit Wikipedia.

A convex set with extremal points in red. Image credit Wikipedia.

Another way to say this is that if z is an extreme point then whenever z can be written as \delta x + (1-\delta) y for some 0 < \delta < 1, then actually x=y=z. Now since our constraints are all linear (and there are a finite number of them) they won’t define a convex set with weird curves like the one above. This means that there are a finite number of extreme points that just correspond to the intersections of some of the constraints. So there are at most 2^n possibilities.

Indeed we want a characterization of extreme points that’s specific to linear programs in standard form, “\max \langle c, x \rangle \textup{ s.t. } Ax=b, x \geq 0.” And here is one.

Definition: Let A be an m \times n matrix with n \geq m. A solution x to Ax=b is called basic if at most m of its entries are nonzero.

The reason we call it “basic” is because, under some mild assumptions we describe below, a basic solution corresponds to a vector space basis of \mathbb{R}^m. Which basis? The one given by the m columns of A used in the basic solution. We don’t need to talk about bases like this, though, so in the event of a headache just think of the basis as a set B \subset \{ 1, 2, \dots, n \} of size m corresponding to the nonzero entries of the basic solution.

Indeed, what we’re doing here is looking at the matrix A_B formed by taking the columns of A whose indices are in B, and the vector x_B in the same way, and looking at the equation A_Bx_B = b. If all the parts of x that we removed were zero then this will hold if and only if Ax=b. One might worry that A_B is not invertible, so we’ll go ahead and assume it is. In fact, we’ll assume that every set of m columns of A forms a basis and that the rows of A are also linearly independent. This isn’t without loss of generality because if some rows or columns are not linearly independent, we can remove the offending constraints and variables without changing the set of solutions (this is why it’s so nice to work with the standard form).

Moreover, we’ll assume that every basic solution has exactly m nonzero variables. A basic solution which doesn’t satisfy this assumption is called degenerate, and they’ll essentially be special corner cases in the simplex algorithm. Finally, we call a basic solution feasible if (in addition to satisfying Ax=b) it satisfies x \geq 0. Now that we’ve made all these assumptions it’s easy to see that choosing m nonzero variables uniquely determines a basic feasible solution. Again calling the sub-matrix A_B for a basis B, it’s just x_B = A_B^{-1}b. Now to finish our characterization, we just have to show that under the same assumptions basic feasible solutions are exactly the extremal points of the feasible region.

Proposition: A vector x is a basic feasible solution if and only if it’s an extreme point of the set \{ x : Ax = b, x \geq 0 \}.

Proof. For one direction, suppose you have a basic feasible solution x, and say we write it as x = \delta y + (1-\delta) z for some 0 < \delta < 1. We want to show that this implies y = z. Since all of these points are in the feasible region, all of their coordinates are nonnegative. So whenever a coordinate x_i = 0 it must be that both y_i = z_i = 0. Since x has exactly n-m zero entries, it must be that y, z both have at least n-m zero entries, and hence y,z are both basic. By our non-degeneracy assumption they both then have exactly m nonzero entries. Let B be the set of the nonzero indices of x. Because Ay = Az = b, we have A(y-z) = 0. Now y-z has all of its nonzero entries in B, and because the columns of A_B are linearly independent, the fact that A_B(y-z) = 0 implies y-z = 0.

In the other direction, suppose  that you have some extreme point x which is feasible but not basic. In other words, there are more than m nonzero entries of x, and we’ll call the indices J = \{ j_1, \dots, j_t \} where t > m. The columns of A_J are linearly dependent (since they’re t vectors in \mathbb{R}^m), and so let \sum_{i=1}^t z_{j_i} A_{j_i} be a nontrivial linear combination of the columns of A. Add zeros to make the z_{j_i} into a length n vector z, so that Az = 0. Now

A(x + \varepsilon z) = A(x - \varepsilon z) = Ax = b

And if we pick \varepsilon sufficiently small x \pm \varepsilon z will still be nonnegative, because the only entries we’re changing of x are the strictly positive ones. Then x = \delta (x + \varepsilon z) + (1 - \delta) \varepsilon z for \delta = 1/2, but this is very embarrassing for x who was supposed to be an extreme point. \square

Now that we know extreme points are the same as basic feasible solutions, we need to show that any linear program that has some solution has a basic feasible solution. This is clear geometrically: any time you have an optimum it has to either lie on a line or at a vertex, and if it lies on a line then you can slide it to a vertex without changing its value. Nevertheless, it is a useful exercise to go through the algebra.

Theorem. Whenever a linear program is feasible and bounded, it has a basic feasible solution.

Proof. Let x be an optimal solution to the LP. If x has at most m nonzero entries then it’s a basic solution and by the non-degeneracy assumption it must have exactly m nonzero entries. In this case there’s nothing to do, so suppose that x has r > m nonzero entries. It can’t be a basic feasible solution, and hence is not an extreme point of the set of feasible solutions (as proved by the last theorem). So write it as x = \delta y + (1-\delta) z for some feasible y \neq z and 0 < \delta < 1.

The only thing we know about x is it’s optimal. Let c be the cost vector, and the optimality says that \langle c,x \rangle \geq \langle c,y \rangle, and \langle c,x \rangle \geq \langle c,z \rangle. We claim that in fact these are equal, that y, z are both optimal as well. Indeed, say y were not optimal, then

\displaystyle \langle c, y \rangle < \langle c,x \rangle = \delta \langle c,y \rangle + (1-\delta) \langle c,z \rangle

Which can be rearranged to show that \langle c,y \rangle < \langle c, z \rangle. Unfortunately for x, this implies that it was not optimal all along:

\displaystyle \langle c,x \rangle < \delta \langle c, z \rangle + (1-\delta) \langle c,z \rangle = \langle c,z \rangle

An identical argument works to show z is optimal, too. Now we claim we can use y,z to get a new solution that has fewer than r nonzero entries. Once we show this we’re done: inductively repeat the argument with the smaller solution until we get down to exactly m nonzero variables. As before we know that y,z must have at least as many zeros as x. If they have more zeros we’re done. And if they have exactly as many zeros we can do the following trick. Write w = \gamma y + (1- \gamma)z for a \gamma \in \mathbb{R} we’ll choose later. Note that no matter the \gamma, w is optimal. Rewriting w = z + \gamma (y-z), we just have to pick a \gamma that ensures one of the nonzero coefficients of z is zeroed out while maintaining nonnegativity. Indeed, we can just look at the index i which minimizes z_i / (y-z)_i and use \delta = - z_i / (y-z)_i. \square.

So we have an immediate (and inefficient) combinatorial algorithm: enumerate all subsets of size m, compute the corresponding basic feasible solution x_B = A_B^{-1}b, and see which gives the biggest objective value. The problem is that, even if we knew the value of m, this would take time n^m, and it’s not uncommon for m to be in the tens or hundreds (and if we don’t know m the trivial search is exponential).

So we have to be smarter, and this is where the simplex tableau comes in.

The simplex tableau

Now say you have any basis B and any feasible solution x. For now x might not be a basic solution, and even if it is, its basis of nonzero entries might not be the same as B. We can decompose the equation Ax = b into the basis part and the non basis part:

A_Bx_B + A_{B'} x_{B'} = b

and solving the equation for x_B gives

x_B = A^{-1}_B(b - A_{B'} x_{B'})

It may look like we’re making a wicked abuse of notation here, but both A_Bx_B and A_{B'}x_{B'} are vectors of length m so the dimensions actually do work out. Now our feasible solution x has to satisfy Ax = b, and the entries of x are all nonnegative, so it must be that x_B \geq 0 and x_{B'} \geq 0, and by the equality above A^{-1}_B (b - A_{B'}x_{B'}) \geq 0 as well. Now let’s write the maximization objective \langle c, x \rangle by expanding it first in terms of the x_B, x_{B'}, and then expanding x_B.

\displaystyle \begin{aligned} \langle c, x \rangle & = \langle c_B, x_B \rangle + \langle c_{B'}, x_{B'} \rangle \\  & = \langle c_B, A^{-1}_B(b - A_{B'}x_{B'}) \rangle + \langle c_{B'}, x_{B'} \rangle \\  & = \langle c_B, A^{-1}_Bb \rangle + \langle c_{B'} - (A^{-1}_B A_{B'})^T c_B, x_{B'} \rangle \end{aligned}

If we want to maximize the objective, we can just maximize this last line. There are two cases. In the first, the vector c_{B'} - (A^{-1}_B A_{B'})^T c_B \leq 0 and A_B^{-1}b \geq 0. In the above equation, this tells us that making any component of x_{B'} bigger will decrease the overall objective. In other words, \langle c, x \rangle \leq \langle c_B, A_B^{-1}b \rangle. Picking x = A_B^{-1}b (with zeros in the non basis part) meets this bound and hence must be optimal. In other words, no matter what basis B we’ve chosen (i.e., no matter the candidate basic feasible solution), if the two conditions hold then we’re done.

Now the crux of the algorithm is the second case: if the conditions aren’t met, we can pick a positive index of c_{B'} - (A_B^{-1}A_{B'})^Tc_B and increase the corresponding value of x_{B'} to increase the objective value. As we do this, other variables in the solution will change as well (by decreasing), and we have to stop when one of them hits zero. In doing so, this changes the basis by removing one index and adding another. In reality, we’ll figure out how much to increase ahead of time, and the change will correspond to a single elementary row-operation in a matrix.

Indeed, the matrix we’ll use to represent all of this data is called a tableau in the literature. The columns of the tableau will correspond to variables, and the rows to constraints. The last row of the tableau will maintain a candidate solution y to the dual problem. Here’s a rough picture to keep the different parts clear while we go through the details.

tableau

But to make it work we do a slick trick, which is to “left-multiply everything” by A_B^{-1}. In particular, if we have an LP given by c, A, b, then for any basis it’s equivalent to the LP given by c, A_B^{-1}A, A_{B}^{-1} b (just multiply your solution to the new program by A_B to get a solution to the old one). And so the actual tableau will be of this form.

tableau-symbols

When we say it’s in this form, it’s really only true up to rearranging columns. This is because the chosen basis will always be represented by an identity matrix (as it is to start with), so to find the basis you can find the embedded identity sub-matrix. In fact, the beginning of the simplex algorithm will have the initial basis sitting in the last few columns of the tableau.

Let’s look a little bit closer at the last row. The first portion is zero because A_B^{-1}A_B is the identity. But furthermore with this A_B^{-1} trick the dual LP involves A_B^{-1} everywhere there’s a variable. In particular, joining all but the last column of the last row of the tableau, we have the vector c - A_B^T(A_B^{-1})^T c, and setting y = A_B^{-1}c_B we get a candidate solution for the dual. What makes the trick even slicker is that A_B^{-1}b is already the candidate solution x_B, since (A_B^{-1}A)_B^{-1} is the identity. So we’re implicitly keeping track of two solutions here, one for the primal LP, given by the last column of the tableau, and one for the dual, contained in the last row of the tableau.

I told you the last row was the dual solution, so why all the other crap there? This is the final slick in the trick: the last row further encodes the complementary slackness conditions. Now that we recognize the dual candidate sitting there, the complementary slackness conditions simply ask for the last row to be non-positive (this is just another way of saying what we said at the beginning of this section!). You should check this, but it gives us a stopping criterion: if the last row is non-positive then stop and output the last column.

The simplex algorithm

Now (finally!) we can describe and implement the simplex algorithm in its full glory. Recall that our informal setup has been:

  1. Find an initial basic feasible solution, and set up the corresponding tableau.
  2. Find a positive index of the last row, and increase the corresponding variable (adding it to the basis) just enough to make another variable from the basis zero (removing it from the basis).
  3. Repeat step 2 until the last row is nonpositive.
  4. Output the last column.

This is almost correct, except for some details about how increasing the corresponding variables works. What we’ll really do is represent the basis variables as pivots (ones in the tableau) and then the first 1 in each row will be the variable whose value is given by the entry in the last column of that row. So, for example, the last entry in the first row may be the optimal value for x_5, if the fifth column is the first entry in row 1 to have a 1.

As we describe the algorithm, we’ll illustrate it running on a simple example. In doing this we’ll see what all the different parts of the tableau correspond to from the previous section in each step of the algorithm.

example

Spoiler alert: the optimum is x_1 = 2, x_2 = 1 and the value of the max is 8.

So let’s be more programmatically formal about this. The main routine is essentially pseudocode, and the difficulty is in implementing the helper functions

def simplex(c, A, b):
   tableau = initialTableau(c, A, b)

   while canImprove(tableau):
      pivot = findPivotIndex(tableau)
      pivotAbout(tableau, pivot)

   return primalSolution(tableau), objectiveValue(tableau)

Let’s start with the initial tableau. We’ll assume the user’s inputs already include the slack variables. In particular, our example data before adding slack is

c = [3, 2]
A = [[1, 2], [1, -1]]
b = [4, 1]

And after adding slack:

c = [3, 2, 0, 0]
A = [[1,  2,  1,  0],
     [1, -1,  0,  1]]
b = [4, 1]

Now to set up the initial tableau we need an initial feasible solution in mind. The reader is recommended to work this part out with a pencil, since it’s much easier to write down than it is to explain. Since we introduced slack variables, our initial feasible solution (basis) B can just be (0,0,1,1). And so x_B is just the slack variables, c_B is the zero vector, and A_B is the 2×2 identity matrix. Now A_B^{-1}A_{B'} = A_{B'}, which is just the original two columns of A we started with, and A_B^{-1}b = b. For the last row, c_B is zero so the part under A_B^{-1}A_B is the zero vector. The part under A_B^{-1}A_{B'} is just c_{B'} = (3,2).

Rather than move columns around every time the basis B changes, we’ll keep the tableau columns in order of (x_1, \dots, x_n, \xi_1, \dots, \xi_m). In other words, for our example the initial tableau should look like this.

[[ 1,  2,  1,  0,  4],
 [ 1, -1,  0,  1,  1],
 [ 3,  2,  0,  0,  0]]

So implementing initialTableau is just a matter of putting the data in the right place.

def initialTableau(c, A, b):
   tableau = [row[:] + [x] for row, x in zip(A, b)]
   tableau.append(c[:] + [0])
   return tableau

As an aside: in the event that we don’t start with the trivial basic feasible solution of “trivially use the slack variables,” we’d have to do a lot more work in this function. Next, the primalSolution() and objectiveValue() functions are simple, because they just extract the encoded information out from the tableau (some helper functions are omitted for brevity).

def primalSolution(tableau):
   # the pivot columns denote which variables are used
   columns = transpose(tableau)
   indices = [j for j, col in enumerate(columns[:-1]) if isPivotCol(col)]
   return list(zip(indices, columns[-1]))

def objectiveValue(tableau):
   return -(tableau[-1][-1])

Similarly, the canImprove() function just checks if there’s a nonnegative entry in the last row

def canImprove(tableau):
   lastRow = tableau[-1]
   return any(x &gt; 0 for x in lastRow[:-1])

Let’s run the first loop of our simplex algorithm. The first step is checking to see if anything can be improved (in our example it can). Then we have to find a pivot entry in the tableau. This part includes some edge-case checking, but if the edge cases aren’t a problem then the strategy is simple: find a positive entry corresponding to some entry j of B', and then pick an appropriate entry in that column to use as the pivot. Pivoting increases the value of x_j (from zero) to whatever is the largest we can make it without making some other variables become negative. As we’ve said before, we’ll stop increasing x_j when some other variable hits zero, and we can compute which will be the first to do so by looking at the current values of x_B = A_B^{-1}b (in the last column of the tableau), and seeing how pivoting will affect them. If you stare at it for long enough, it becomes clear that the first variable to hit zero will be the entry x_i of the basis for which x_i / A_{i,j} is minimal (and A_{i,j} has to be positve). This is because, in order to maintain the linear equalities, every entry of x_B will be decreased by that value during a pivot, and we can’t let any of the variables become negative.

All of this results in the following function, where we have left out the degeneracy/unboundedness checks.

def findPivotIndex(tableau):
   # pick first nonzero index of the last row
   column = [i for i,x in enumerate(tableau[-1][:-1]) if x &gt; 0][0]
   quotients = [(i, r[-1] / r[column]) for i,r in enumerate(tableau[:-1]) if r[column] &gt; 0]

   # pick row index minimizing the quotient
   row = min(quotients, key=lambda x: x[1])[0]
   return row, column

For our example, the minimizer is the (1,0) entry (second row, first column). Pivoting is just doing the usual elementary row operations (we covered this in a primer a while back on row-reduction). The pivot function we use here is no different, and in particular mutates the list in place.

def pivotAbout(tableau, pivot):
   i,j = pivot

   pivotDenom = tableau[i][j]
   tableau[i] = [x / pivotDenom for x in tableau[i]]

   for k,row in enumerate(tableau):
      if k != i:
         pivotRowMultiple = [y * tableau[k][j] for y in tableau[i]]
         tableau[k] = [x - y for x,y in zip(tableau[k], pivotRowMultiple)]

And in our example pivoting around the chosen entry gives the new tableau.

[[ 0.,  3.,  1., -1.,  3.],
 [ 1., -1.,  0.,  1.,  1.],
 [ 0.,  5.,  0., -3., -3.]]

In particular, B is now (1,0,1,0), since our pivot removed the second slack variable \xi_2 from the basis. Currently our solution has x_1 = 1, \xi_1 = 3. Notice how the identity submatrix is still sitting in there, the columns are just swapped around.

There’s still a positive entry in the bottom row, so let’s continue. The next pivot is (0,1), and pivoting around that entry gives the following tableau:

[[ 0.        ,  1.        ,  0.33333333, -0.33333333,  1.        ],
 [ 1.        ,  0.        ,  0.33333333,  0.66666667,  2.        ],
 [ 0.        ,  0.        , -1.66666667, -1.33333333, -8.        ]]

And because all of the entries in the bottom row are negative, we’re done. We read off the solution as we described, so that the first variable is 2 and the second is 1, and the objective value is the opposite of the bottom right entry, 8.

To see all of the source code, including the edge-case-checking we left out of this post, see the Github repository for this post.

Obvious questions and sad answers

An obvious question is: what is the runtime of the simplex algorithm? Is it polynomial in the size of the tableau? Is it even guaranteed to stop at some point? The surprising truth is that nobody knows the answer to all of these questions! Originally (in the 1940’s) the simplex algorithm actually had an exponential runtime in the worst case, though this was not known until 1972. And indeed, to this day while some variations are known to terminate, no variation is known to have polynomial runtime in the worst case. Some of the choices we made in our implementation (for example, picking the first column with a positive entry in the bottom row) have the potential to cycle, i.e., variables leave and enter the basis without changing the objective at all. Doing something like picking a random positive column, or picking the column which will increase the objective value by the largest amount are alternatives. Unfortunately, every single pivot-picking rule is known to give rise to exponential-time simplex algorithms in the worst case (in fact, this was discovered as recently as 2011!). So it remains open whether there is a variant of the simplex method that runs in guaranteed polynomial time.

But then, in a stunning turn of events, Leonid Khachiyan proved in the 70’s that in fact linear programs can always be solved in polynomial time, via a completely different algorithm called the ellipsoid method. Following that was a method called the interior point method, which is significantly more efficient. Both of these algorithms generalize to problems that are harder than linear programming as well, so we will probably cover them in the distant future of this blog.

Despite the celebratory nature of these two results, people still use the simplex algorithm for industrial applications of linear programming. The reason is that it’s much faster in practice, and much simpler to implement and experiment with.

The next obvious question has to do with the poignant observation that whole numbers are great. That is, you often want the solution to your problem to involve integers, and not real numbers. But adding the constraint that the variables in a linear program need to be integer valued (even just 0-1 valued!) is NP-complete. This problem is called integer linear programming, or just integer programming (IP). So we can’t hope to solve IP, and rightly so: the reader can verify easily that boolean satisfiability instances can be written as linear programs where each clause corresponds to a constraint.

This brings up a very interesting theoretical issue: if we take an integer program and just remove the integrality constraints, and solve the resulting linear program, how far away are the two solutions? If they’re close, then we can hope to give a good approximation to the integer program by solving the linear program and somehow turning the resulting solution back into an integer solution. In fact this is a very popular technique called LP-rounding. We’ll also likely cover that on this blog at some point.

Oh there’s so much to do and so little time! Until next time.

Learning to Love Complex Numbers

This post is intended for people with a little bit of programming experience and no prior mathematical background.

So let’s talk about numbers.

Numbers are curious things. On one hand, they represent one of the most natural things known to humans, which is quantity. It’s so natural to humans that even newborn babies are in tune with the difference between quantities of objects between 1 and 3, in that they notice when quantity changes much more vividly than other features like color or shape.

But our familiarity with quantity doesn’t change the fact that numbers themselves (as an idea) are a human invention. And they’re not like most human inventions, the kinds where you have to tinker with gears or circuits to get a machine that makes your cappuccino. No, these are mathematical inventions. These inventions exist only in our minds.

Numbers didn’t always exist. A long time ago, back when the Greeks philosophers were doing their philosophizing, negative numbers didn’t exist! In fact, it wasn’t until 1200 AD that the number zero was first considered in Europe. Zero, along with negative numbers and fractions and square roots and all the rest, were invented primarily to help people solve more problems than they could with the numbers they had available. That is, numbers were invented primarily as a way for people to describe their ideas in a useful way. People simply  wondered “is there a number whose square gives you 2?” And after a while they just decided there was and called it \sqrt{2} because they didn’t have a better name for it. 

But with these new solutions came a host of new problems. You see, although I said mathematical inventions only exist in our minds, once they’re invented they gain a life of their own. You start to notice patterns in your mathematical objects and you have to figure out why they do the things they do. And numbers are a perfectly good example of this: once I notice that I can multiply a number by itself, I can ask how often these “perfect squares” occur. That is, what’s the pattern in the numbers 1^2, 2^2, 3^2, 4^2, \dots? If you think about it for a while, you’ll find that square numbers have a very special relationship with odd numbers.

Other times, however, the things you invent turn out to make no sense at all, and you can prove they never existed in the first place! It’s an odd state of affairs, but we’re going to approach the subject of complex numbers from this mindset. We’re going to come up with a simple idea, the idea that negative numbers can be perfect squares, and explore the world of patterns it opens up. Along the way we’ll do a little bit of programming to help explore, give some simple proofs to solidify our intuition, and by the end we’ll see how these ideas can cause wonderful patterns like this one:

mandelbrot

The number i

Let’s bring the story back around to squares. One fact we all remember about numbers is that squaring a number gives you something non-negative. 7^2 = 49, (-2)^2 = 4, 0^2 = 0, and so on. But it certainly doesn’t have to be this way. What if we got sick of that stupid fact and decided to invent a new number whose square was negative? Which negative, you ask? Well it doesn’t really matter, because I can always stretch it larger or smaller so that it’s square is -1.

Let’s see how: if you say that your made-up number x makes x^2 = -7, then I can just use \frac{x}{\sqrt{7}} to get a number whose square is -1. If you’re going to invent a number that’s supposed to interact with our usual numbers, then you have to be allowed to add, subtract, and multiply x with regular old real numbers, and the usual properties would have to still work. So it would have to be true that (x / \sqrt{7})^2 = x^2 / \sqrt{7}^2 = -7/7 = -1.

So because it makes no difference (this is what mathematicians mean by, “without loss of generality”) we can assume that the number we’re inventing will have a square of negative one. Just to line up with history, let’s call the new number i. So there it is: i exists and i^2 = -1. And now that we are “asserting” that i plays nicely with real numbers, we get these natural rules for adding and subtracting and multiplying and dividing. For example

  • 1 + i is a new number, which we’ll just call 1+i. And if we added two of these together, (1+ i) + (1+i), we can combine the real parts and the i parts to get 2 + 2i. Same goes for subtraction. In general a complex number looks like a + bi, because as we’ll see in the other points you can simplify every simple arithmetic expression down to just one “real number” part and one “real number times i” part.
  • We can multiply 3 \cdot i, and we’ll just call it 3i, and we require that multiplication distributes across addition (that the FOIL rule works). So that, for example, (2 - i)(1 + 3i) = (2 + 6i - i - 3i^2) = (2 + 3) + (6i - i) = (5 + 5i).
  • Dividing is a significantly more annoying. Say we want to figure out what 1 / (1+i) is (in fact, it’s not even obvious that this should look like a regular number! But it does). The 1 / a notation just means we’re looking for a number which, when we multiply by the denominator a, we get back to 1. So we’re looking to find out when (a + bi)(1 + i) = 1 + 0i where a and b are variables we’re trying to solve for. If we multiply it out we get (a-b) + (a + b)i = 1 + 0i, and since the real part and the i part have to match up, we know that a - b = 1 and a + b = 0. If we solve these two equations, we find that a = 1/2, b = -1/2 works great. If we want to figure out something like (2 + 3i) / (1 - i), we just find out what 1 / (1- i) is first, and then multiply the result by (2+3i).

So that was tedious and extremely boring, and we imagine you didn’t even read it (that’s okay, it really is boring!). All we’re doing is establishing ground rules for the game, so if you come across some arithmetic that doesn’t make sense, you can refer back to this list to see what’s going on. And once again, for the purpose of this post, we’re asserting that all these laws hold. Maybe some laws follow from others, but as long as we don’t come up with any nasty self-contradictions we’ll be fine.

And now we turn to the real questions: is i the only square root of -1? Does i itself have a square root? If it didn’t, we’d be back to where we started, with some numbers (the non-i numbers) having square roots while others don’t. And so we’d feel the need to make all the i numbers happy by making up more numbers to be their square roots, and then worrying what if these new numbers don’t have square roots and…gah!

I’ll just let you in on the secret to save us from this crisis. It turns out that i does have a square root in terms of other i numbers, but in order to find it we’ll need to understand i from a different angle, and that angle turns out to be geometry.

Geometry? How is geometry going to help me understand numbers!?

It’s a valid question and part of why complex numbers are so fascinating. And I don’t mean geometry like triangles and circles and parallel lines (though there will be much talk of angles), I mean transformations in the sense that we’ll be “stretching,” “squishing,” and “rotating” numbers. Maybe another time I can tell you why for me “geometry” means stretching and rotating; it’s a long but very fun story.

The clever insight is that you can represent complex numbers as geometric objects in the first place. To do it, you just think of a + bi as a pair of numbers (a,b), (the pair of real part and i part), and then plot that point on a plane. For us, the x-axis will be the “real” axis, and the y-axis will be the i-axis. So the number (3 - 4i) is plotted 3 units in the positive x direction and 4 units in the negative y direction. Like this:

single-complex-number

The “j” instead of “i” is not a typo, but a disappointing fact about the programming language we used to make this image. We’ll talk more about why later.

We draw it as an arrow for a good reason. Stretching, squishing, rotating, and reflecting will all be applied to the arrow, keeping its tail fixed at the center of the axes. Sometimes the arrow is called a “vector,” but we won’t use that word because here it’s synonymous with “complex number.”

So let’s get started squishing stuff.

Stretching, Squishing, Rotating

Before we continue I should clear up some names. We call a number that has an i in it a complex number, and we call the part without the i the real part (like 2 in 2-i) and the part with i the complex part.

Python is going to be a great asset for us in exploring complex numbers, so let’s jump right into it. It turns out that Python natively supports complex numbers, and I wrote a program for drawing complex numbers. I used it to make the plot above. The program depends on a library I hate called matplotlib, and so the point of the program is to shield you from as much pain as possible and focus on complex numbers. You can use the program by downloading it from this blog’s Github page, along with everything else I made in writing this post. All you need to know how to do is call a function, and I’ve done a bit of window dressing removal to simplify things (I really hate matplotlib).

Here’s the function header:

# plotComplexNumbers : [complex] -> None
# display a plot of the given list of complex numbers
def plotComplexNumbers(numbers):
   ...

Before we show some examples of how to use it, we have to understand how to use complex numbers in Python. It’s pretty simple, except that Python was written by people who hate math, and so they decided the complex number would be represented by j instead of i (people who hate math are sometimes called “engineers,” and they use j out of spite. Not really, though).

So in Python it’s just like any other computation. For example:

>>> (1 + 1j)*(4 - 2j) == (6+2j)
True
>>> 1 / (1+1j)
(0.5-0.5j)

And so calling the plotting function with a given list of complex numbers is as simple as importing the module and calling the function

from plotcomplex import plot
plot.plotComplexNumbers([(-1+1j), (1+2j), (-1.5 - 0.5j), (.6 - 1.8j)])

Here’s the result

example-complex-plot

So let’s use plots like this one to explore what “multiplication by i” does to a complex number. It might not seem exciting at first, but I promise there’s a neat punchline.

Even without plotting it’s pretty easy to tell what multiplying by i does to some numbers. It takes 1 to i, moves i to i^2 = -1, it takes -1 to -i, and -i to -i \cdot i = 1.

What’s the pattern in these? well if we plot all these numbers, they’re all at right angles in counter-clockwise order. So this might suggest that multiplication by i does some kind of rotation. Is that always the case? Well lets try it with some other more complicated numbers. Click the plots below to enlarge.

Well, it looks close but it’s hard to tell. Some of the axes are squished and stretched, so it might be that our images don’t accurately represent the numbers (the real world can be such a pain). Well when visual techniques fail, we can attempt to prove it.

Clearly multiplying by i does some kind of rotation, maybe with other stuff too, and it shouldn’t be so hard to see that multiplying by i does the same thing no matter which number you use (okay, the skeptical readers will say that’s totally hard to see, but we’ll prove it super rigorously in a minute). So if we take any number and multiply it by i once, then twice, then three times, then four, and if we only get back to where we started at four multiplications, then each rotation had to be a quarter turn.

Indeed,

\displaystyle (a + bi) i^4 = (ai - b) i^3 = (-a - bi) i^2 = (-ai + b) i = a + bi

This still isn’t all that convincing, and we want to be 100% sure we’re right. What we really need is a way to arithmetically compute the angle between two complex numbers in their plotted forms. What we’ll do is find a way to measure the angle of one complex number with the x-axis, and then by subtraction we can get angles between arbitrary points. For example, in the figure below \theta = \theta_1 - \theta_2.

angle-example

One way to do this is with trigonometry: the geometric drawing of a + bi is the hypotenuse of a right triangle with the x-axis.

triangle-example

And so if r is the length of the arrow, then by the definition of sine and cosine, \cos(\theta) = a/r, \sin(\theta) = b/r. If we have r, \theta, and r > 0, we can solve for a unique a and b, so instead of representing a complex number in terms of the pair of numbers (a,b), we can represent it with the pair of numbers (r, \theta). And the conversion between the two is just

a + bi = r \cos(\theta) + (r \sin(\theta)) i

The (r, \theta) representation is called the polar representation, while the (a,b) representation is called the rectangular representation or the Cartesian representation. Converting between polar and Cartesian coordinates fills the pages of many awful pre-calculus textbooks (despite the fact that complex numbers don’t exist in classical calculus). Luckily for us Python has built-in functions to convert between the two representations for us.

>>> import cmath
>>> cmath.polar(1 + 1j)
(1.4142135623730951, 0.7853981633974483)
>>> z = cmath.polar(1 + 1j)
>>> cmath.rect(z[0], z[1])
(1.0000000000000002+1j)

It’s a little bit inaccurate on the rounding, but it’s fine for our purposes.

So how do we compute the angle between two complex numbers? Just convert each to the polar form, and subtract the second coordinates. So if we get back to our true goal, to figure out what multiplication by i does, we can just do everything in polar form. Here’s a program that computes the angle between two complex numbers.

def angleBetween(z, w):
   zPolar, wPolar = cmath.polar(z), cmath.polar(w)
   return wPolar[1] - zPolar[1]

print(angleBetween(1 + 1j, (1 + 1j) * 1j))
print(angleBetween(2 - 3j, (2 - 3j) * 1j))
print(angleBetween(-0.5 + 7j, (-0.5 + 7j) * 1j))

Running it gives

1.5707963267948966
1.5707963267948966
-4.71238898038469

Note that the decimal form of \pi/2 is 1.57079…, and that the negative angle is equivalent to \pi/2 if you add a full turn of 2\pi to it. So programmatically we can see that for every input we try multiplying by i rotates 90 degrees.

But we still haven’t proved it works. So let’s do that now. To say what the angle is between r \cos (\theta) + ri \sin (\theta) and i \cdot [r \cos (\theta) + ri \sin(\theta)] = -r \sin (\theta) + ri \cos(\theta), we need to transform the second number into the usual polar form (where the i is on the sine part and not the cosine part). But we know, or I’m telling you now, this nice fact about sine and cosine:

\displaystyle \sin(\theta + \pi/2) = cos(\theta)
\displaystyle \cos(\theta + \pi / 2) = -\sin(\theta)

This fact is maybe awkward to write out algebraically, but it’s just saying that if you shift the whole sine curve a little bit you get the cosine curve, and if you keep shifting it you get the opposite of the sine curve (and if you kept shifting it even more you’d eventually get back to the sine curve; they’re called periodic for this reason).

So immediately we can rewrite the second number as r \cos(\theta + \pi/2) + i r \sin (\theta + \pi/2). The angle is the same as the original angle plus a right angle of \pi/2. Neat!

Applying this same idea to (a + bi) \cdot (c + di), it’s not much harder to prove that multiplying two complex numbers in general multiplies their lengths and adds their angles. So if a complex number z has its magnitude r smaller than 1, multiplying by z squishes and rotates whatever is being multiplied. And if the magnitude is greater than 1, it stretches and rotates. So we have a super simple geometric understanding of how arithmetic with complex numbers works. And as we’re about to see, all this stretching and rotating results in some really weird (and beautifully mysterious!) mathematics and programs.

But before we do that we still have one question to address, the question that started this whole geometric train of thought: does i have a square root? Indeed, I’m just looking for a number such that, when I square its length and double its angle, I get i = \cos(\pi/2) + i \sin(\pi/2). Indeed, the angle we want is \pi/4, and the length we want is r = 1, which means \sqrt{i} = \cos(\pi/4) + i \sin(\pi/4). Sweet! There is another root if you play with the signs, see if you can figure it out.

In fact it’s a very deeper and more beautiful theorem (“theorem” means “really important fact”) called the fundamental theorem of algebra. And essentially it says that the complex numbers are complete. That is, we can always find square roots, cube roots, or anything roots of numbers involving i. It actually says a lot more, but it’s easier to appreciate the rest of it after you do more math than we’re going to do in this post.

On to pretty patterns!

The Fractal

So here’s a little experiment. Since every point in the plane is the end of some arrow representing a complex number, we can imagine transforming the entire complex plane by transforming each number by the same rule. The most interesting simple rule we can think of: squaring! So though it might strain your capacity for imagination, try to visualize the idea like this. Squaring a complex number is the same as squaring it’s length and doubling its angle. So imagine: any numbers whose arrows are longer than 1 will grow much bigger, arrows shorter than 1 will shrink, and arrows of length exactly one will stay the same length (arrows close to length 1 will grow/shrink much more slowly than those far away from 1). And complex numbers with small positive angles will increase their angle, but only a bit, while larger angles will grow faster.

Here’s an animation made by Douglas Arnold showing what happens to the set of complex numbers a + bi with 0 \leq a, b \leq 1 or -1 < a,b < 0. Again, imagine every point is the end of a different arrow for the corresponding complex number. The animation is for a single squaring, and the points move along the arc they would travel if one rotated/stretched them smoothly.

complex-squaring

So that’s pretty, but this is by all accounts a well-behaved transformation. It’s “predictable,” because for example we can always tell which complex numbers will get bigger and bigger (in length) and which will get smaller.

What if, just for the sake of tinkering, we changed the transformation a little bit? That is, instead of sending z = a+bi to z^2 (I’ll often write this z \mapsto z^2), what if we sent

\displaystyle z \mapsto z^2 + 1

Now it’s not so obvious: which numbers will grow and which will shrink? Notice that it’s odd because adding 1 only changes the real part of the number. So a number whose length is greater than 1 can become small under this transformation. For example, i is sent to 0, so something slightly larger would also be close to zero. Indeed, 5i/4 \mapsto -9/16.

So here’s an interesting question: are there any complex numbers that will stay small even if I keep transforming like this forever? Specifically, if I call f(z) = z^2, and I call f^2(z) = f(f(z)), and likewise call f^k(z) for k repeated transformations of z, is there a number z so that for every k, the value f^k(z) < 2? “Obvious” choices like z=0 don’t work, and neither do random guesses like z=i or z=1. So should we guess the answer is no?

Before we jump to conclusions let’s write a program to see what happens for more than our random guesses. The program is simple: we’ll define the “square plus one” function, and then repeatedly apply that function to a number for some long number of times (say, 250 times). If the length of the number stays under 2 after so many tries, we’ll call it “small forever,” and otherwise we’ll call it “not small forever.”

def squarePlusOne(z):
   return z*z + 1

def isSmallForever(z, f):
   k = 0

   while abs(z) < 2: z = f(z) k += 1 if k > 250:
         return True

   return False

This isSmallForever function is generic: you can give it any function f and it will repeatedly call f on z until the result grows bigger than 2 in length. Note that the abs function is a built-in Python function for computing the length of a complex number.

Then I wrote a classify function, which you can give a window and a small increment, and it will produce a grid of zeros and ones marking the results of isSmallForever. The details of the function are not that important. I also wrote a function that turns the grid into a picture. So here’s an example of how we’d use it:

from plotcomplex.plot import gridToImage

def classifySquarePlusOne(z):
   return isSmallForever(z, squarePlusOne)

grid = classify(classifySquarePlusOne) # the other arguments are defaulted to [-2,2], [-2,2], 0.1
gridToImage(grid)

And here’s the result. Points colored black grow beyond 2, and white points stay small for the whole test.

Looks like they'll always grow big.

Looks like they’ll always grow big.

So it looks like repeated squaring plus one will always make complex numbers grow big. That’s not too exciting, but we can always make it more exciting. What happens if we replace the 1 in z^2 + 1 with a different complex number? For example, if we do z^2 - 1 then will things always grow big?

You can randomly guess and see that 0 will never grow big, because 0^2 - 1 = -1 and (-1)^2 - 1 = 0. It will just oscillate forever. So with -1 some numbers will grow and some will not! Let’s use the same routine above to see which:

def classifySquareMinusOne(z):
      return isSmallForever(z, squareMinusOne)

grid = classify(classifySquareMinusOne)
gridToImage(grid)

And the result:

second-attempt

Now that’s a more interesting picture! Let’s ramp up the resolution

grid = classify(classifySquareMinusOne, step=0.001)
gridToImage(grid)

second-attempt-zoomed

Gorgeous. If you try this at home you’ll notice, however, that this took a hell of a long time to run. Speeding up our programs is very possible, but it’s a long story for another time. For now we can just be patient.

Indeed, this image has a ton of interesting details! It looks almost circular in the middle, but if we zoom in we can see that it’s more like a rippling wave

second-attempt-zoomed2

It’s pretty incredible, and a huge question is jumping out at me: what the heck is causing this pattern to occur? What secret does -1 know that +1 doesn’t that makes the resulting pattern so intricate?

But an even bigger question is this. We just discovered that some values of c make z \mapsto z^2 + c result in interesting patterns, and some do not! So the question is which ones make interesting patterns? Even if we just, say, fix the starting point to zero: what is the pattern in the complex numbers that would tell me when this transformation makes zero blow up, and when it keeps zero small?

Sounds like a job for another program. This time we’ll use a nice little Python feature called a closure, which we define a function that saves the information that exists when it’s created for later. It will let us write a function that takes in c and produces a function that transforms according to z \mapsto z^2+c.

def squarePlusC(c):
   def f(z):
      return z*z + c

   return f

And we can use the very same classification/graphing function from before to do this.

def classifySquarePlusC(c):
   return isSmallForever(0, squarePlusC(c))

grid = classify(classifySquarePlusC, xRange=(-2, 1), yRange=(-1, 1), step=0.005)
gridToImage(grid)

And the result:

mandelbrot

Stunning. This wonderful pattern, which is still largely not understood today, is known as the Mandelbrot set. That is, the white points are the points in the Mandlebrot set, and the black points are not in it. The detail on the border of this thing is infinitely intricate. For example, we can change the window in our little program to zoom in on a particular region.

mandelbrot-zoomed

And if you keep zooming in you keep getting more and more detail. This was true of the specific case of z^2 - 1, but somehow the patterns in the Mandelbrot set are much more varied and interesting. And if you keep going down eventually you’ll see patterns that look like the original Mandelbrot set. We can already kind of see that happening above. The name for this idea is a fractal, and the z^2 - 1 image has it too. Fractals are a fascinating and mysterious subject studied in a field called discrete dynamical systems. Many people dedicate their entire lives to studying these things, and it’s for good reason. There’s a lot to learn and even more that’s unknown!

So this is the end of our journey for now. I’ve posted all of the code we used in the making of this post so you can continue to play, but here are some interesting ideas.

  • The Mandelbrot set (and most fractals) are usually colored. The way they’re colored is as follows. Rather than just say true or false when zero blows up beyond 2 in length, you return the number of iterations k that happened. Then you pick a color based on how big k is. There’s a link below that lets you play with this. In fact, adding colors shows that there is even more intricate detail happening outside the Mandelbrot set that’s too faint to see in our pictures above. Such as this.
  • Some very simple questions about fractals are very hard to answer. For example, is the Mandelbrot set connected? That is, is it possible to “walk” from every point in the Mandelbrot set to every other point without leaving the set? Despite the scattering of points in the zoomed in picture above that suggest the answer is no, the answer is actually yes! This is a really difficult thing to prove, however.
  • The patterns in many fractals are often used to generate realistic looking landscapes and generate pseudo randomness. So fractals are not just mathematical curiosities.
  • You should definitely be experimenting with this stuff! What happens if you change the length threshold from 2 to some bigger number? What about a smaller number? What if you do powers different than 2? There’s so much to explore!
  • The big picture thing to take away from this is that it’s not the numbers themselves that are particularly interesting, it’s the transformations of the numbers that generate these patterns! The interesting questions are what kinds of things are the same under these transformations, and what things are different. This is a very general idea in mathematics, and the more math you do the more you’ll find yourself wondering about useful and bizarre transformations.

For the chance to keep playing with the Mandelbrot set, check out this Mandelbrot grapher that works in your browser. It lets you drag rectangles to zoom further in on regions of interest. It’s really fun.

Until next time!