Formulating the Support Vector Machine Optimization Problem

The hypothesis and the setup

This blog post has an interactive demo (mostly used toward the end of the post). The source for this demo is available in a Github repository.

Last time we saw how the inner product of two vectors gives rise to a decision rule: if w is the normal to a line (or hyperplane) L, the sign of the inner product \langle x, w \rangle tells you whether x is on the same side of L as w.

Let’s translate this to the parlance of machine-learning. Let x \in \mathbb{R}^n be a training data point, and y \in \{ 1, -1 \} is its label (green and red, in the images in this post). Suppose you want to find a hyperplane which separates all the points with -1 labels from those with +1 labels (assume for the moment that this is possible). For this and all examples in this post, we’ll use data in two dimensions, but the math will apply to any dimension.

problem_setup

Some data labeled red and green, which is separable by a hyperplane (line).

The hypothesis we’re proposing to separate these points is a hyperplane, i.e. a linear subspace that splits all of \mathbb{R}^n into two halves. The data that represents this hyperplane is a single vector w, the normal to the hyperplane, so that the hyperplane is defined by the solutions to the equation \langle x, w \rangle = 0.

As we saw last time, w encodes the following rule for deciding if a new point z has a positive or negative label.

\displaystyle h_w(z) = \textup{sign}(\langle w, x \rangle)

You’ll notice that this formula only works for the normals w of hyperplanes that pass through the origin, and generally we want to work with data that can be shifted elsewhere. We can resolve this by either adding a fixed term b \in \mathbb{R}—often called a bias because statisticians came up with it—so that the shifted hyperplane is the set of solutions to \langle x, w \rangle + b = 0. The shifted decision rule is:

\displaystyle h_w(z) = \textup{sign}(\langle w, x \rangle + b)

Now the hypothesis is the pair of vector-and-scalar w, b. A trickier but perhaps cleaner way to do the same thing is to make the normal w \in \mathbb{R}^{n+1}, i.e., add one dimension, and then augment all your data with a new dimension whose coordinate is always 1. This makes the normal inner product output the same formula above. We’ll do this, and call the new coordinate x_0, so that the original data is x = (x_1, \dots, x_n).

Now, the key intuitive idea behind the formulation of the SVM problem is that there are many possible separating hyperplanes for a given set of labeled training data. For example, here is a gif showing infinitely many choices.

svm_lots_of_choices.gif

The question is: how can we find the separating hyperplane that not only separates the training data, but generalizes as well as possible to new data? The assumption of the SVM is that a hyperplane which separates the points, but is also as far away from any training point as possible, will generalize best.

optimal_example.png

While contrived, it’s easy to see that the separating hyperplane is as far as possible from any training point.

More specifically, fix a labeled dataset of points (x_i, y_i), or more precisely:

\displaystyle D = \{ (x_i, y_i) \mid i = 1, \dots, m, x_i \in \mathbb{R}^{n}, y_i \in \{1, -1\}  \}

And a hypothesis defined by the normal w \in \mathbb{R}^{n}. Let’s also suppose that w defines a hyperplane that correctly separates all the training data into the two labeled classes, and we just want to measure its quality. That measure of quality is the length of its margin.

Definition: The geometric margin of a hyperplane w with respect to a dataset D is the shortest distance from a training point x_i to the hyperplane defined by w.

The best hyperplane has the largest possible margin.

This margin can even be computed quite easily using our work from last post. The distance from x to the hyperplane defined by w is the same as the length of the projection of x onto w. And this is just computed by an inner product.

decision-rule-3

If the tip of the x arrow is the point in question, then a is the dot product, and the distance from x to the hyperplane L defined by w.

A naive optimization objective

If we wanted to, we could stop now and define an optimization problem that would be very hard to solve. It would look like this:

\displaystyle \begin{aligned} & \max_{w} \min_{x_i} \left | \left \langle x_i, \frac{w}{\|w\|} \right \rangle \right | & \\ \textup{subject to \ \ } & \textup{sign}(\langle x_i, w \rangle) = \textup{sign}(y_i) & \textup{ for every } i = 1, \dots, m \end{aligned}

The reasons this is hard boil down to this formulation being horrifyingly nonlinear. In more detail:

  1. The constraints are nonlinear due to the sign comparisons.
  2. There’s a min and a max! A priori, we have to do this because we don’t know which point is going to be the closest to the hyperplane.
  3. The objective is nonlinear in two ways: the absolute value and the projection requires you to take a norm and divide.

The rest of this post (and indeed, a lot of the work in grokking SVMs) is dedicated to converting this optimization problem to one in which the constraints are all linear inequalities and the objective is a single, quadratic polynomial we want to minimize or maximize.

Along the way, we’ll notice some neat features of the SVM.

Trick 1: linearizing the constraints

To solve the first problem, we can use a trick. We want to know whether \textup{sign}(\langle x_i, w \rangle) = \textup{sign}(y_i) for a labeled training point (x_i, y_i). The trick is to multiply them together. If their signs agree, then their product will be positive, otherwise it will be negative.

So each constraint becomes:

\displaystyle \langle x_i, w \rangle \cdot y_i \geq 0

This is still linear because y_i is a constant (input) to the optimization problem. The variables are the coefficients of w.

The left hand side of this inequality is often called the functional margin of a training point, since, as we will see, it still works to classify x_i, even if w is scaled so that it is no longer a unit vector. Indeed, the sign of the inner product is independent of how w is scaled.

Trick 1.5: the optimal solution is midway between classes

This small trick is to notice that if w is the supposed optimal separating hyperplane, i.e. its margin is maximized, then it must necessarily be exactly halfway in between the closest points in the positive and negative classes.

In other words, if x_+ and x_- are the closest points in the positive and negative classes, respectively, then \langle x_{+}, w \rangle = -\langle x_{-}, w \rangle . If this were not the case, then you could adjust the bias entry of w until it they are exactly equal, and you will have increased the margin. The closest point, say x_+ will have gotten farther away, and the closest point in the opposite class, x_- will have gotten closer, but will not be closer than x_+.

Trick 2: getting rid of the max + min

Resolving this problem essentially uses the fact that the hypothesis, which comes in the form of the normal vector w, has a degree of freedom in its length.

Indeed, in the animation below, I can increase or decrease the length of w without changing the decision boundary.

svm_w_length.gif

I have to keep my hand very steady (because I was too lazy to program it so that it only increases/decreases in length), but you can see the point. The line is perpendicular to the normal vector, and it doesn’t depend on the length.

Let’s combine this with tricks 1 and 1.5. If we increase the length of w, that means the absolute values of the dot products \langle x_i, w \rangle used in the constraints will all increase by the same amount (without changing their sign). Indeed, for any vector a we have \langle a, w \rangle = \|w \| \cdot \langle a, w / \| w \| \rangle.

In this world, the inner product measurement of distance from a point to the hyperplane is no longer faithful. The true distance is \langle a, w / \| w \| \rangle, but the distance measured by \langle a, w \rangle is measured in units of 1 / \| w \|.

units.png

In this example, the two numbers next to the green dot represent the true distance of the point from the hyperplane, and the dot product of the point with the normal (respectively). The dashed lines are the solutions to <x, w> = 1. The magnitude of w is 2.2, the inverse of that is 0.46, and indeed 2.2 = 4.8 * 0.46 (we’ve rounded the numbers).

Now suppose we had the optimal hyperplane and its normal w. No matter how near (or far) the nearest positively labeled training point x is, we could scale the length of w to force \langle x, w \rangle = 1. This is the core of the trick. One consequence is that the actual distance from x to the hyperplane is \frac{1}{\| w \|} = \langle x, w / \| w \| \rangle.

units2.png

The same as above, but with the roles reversed. We’re forcing the inner product of the point with w to be 1. The true distance is unchanged.

In particular, if we force the closest point to have inner product 1, then all other points will have inner product at least 1. This has two consequences. First, our constraints change to \langle x_i, w \rangle \cdot y_i \geq 1 instead of \geq 0. Second, we no longer need to ask which point is closest to the candidate hyperplane! Because after all, we never cared which point it was, just how far away that closest point was. And now we know that it’s exactly 1 / \| w \| away. Indeed, if the optimal points weren’t at that distance, then that means the closest point doesn’t exactly meet the constraint, i.e. that \langle x, w \rangle > 1 for every training point x. We could then scale w shorter until \langle x, w \rangle = 1, hence increasing the margin 1 / \| w \|.

In other words, the coup de grâce, provided all the constraints are satisfied, the optimization objective is just to maximize 1 / \| w \|, a.k.a. to minimize \| w \|.

This intuition is clear from the following demonstration, which you can try for yourself. In it I have a bunch of positively and negatively labeled points, and the line in the center is the candidate hyperplane with normal w that you can drag around. Each training point has two numbers next to it. The first is the true distance from that point to the candidate hyperplane; the second is the inner product with w. The two blue dashed lines are the solutions to \langle x, w \rangle = \pm 1. To solve the SVM by hand, you have to ensure the second number is at least 1 for all green points, at most -1 for all red points, and then you have to make w as short as possible. As we’ve discussed, shrinking w moves the blue lines farther away from the separator, but in order to satisfy the constraints the blue lines can’t go further than any training point. Indeed, the optimum will have those blue lines touching a training point on each side.

svm_solve_by_hand

 

I bet you enjoyed watching me struggle to solve it. And while it’s probably not the optimal solution, the idea should be clear.

The final note is that, since we are now minimizing \| w \|, a formula which includes a square root, we may as well minimize its square \| w \|^2 = \sum_j w_j^2. We will also multiply the objective by 1/2, because when we eventually analyze this problem we will take a derivative, and the square in the exponent and the 1/2 will cancel.

The final form of the problem

Our optimization problem is now the following:

\displaystyle \begin{aligned} & \min_{w}  \frac{1}{2} \| w \|^2 & \\ \textup{subject to \ \ } & \langle x_i, w \rangle \cdot y_i \geq 1 & \textup{ for every } i = 1, \dots, m \end{aligned}

This is much simpler to analyze. The constraints are all linear equalities (which, because of linear programming, we know are tractable to optimize). The objective to minimize, however, is a convex quadratic function of the input variables—a sum of squares of the inputs.

Such problems are generally called quadratic programming problems (or QPs, for short). There are general methods to find solutions! However, they often suffer from numerical stability issues and have less-than-satisfactory runtime. Luckily, the form in which we’ve expressed the support vector machine problem is specific enough that we can analyze it directly, and find a way to solve it without appealing to general-purpose numerical solvers.

We will tackle this problem in a future post (planned for two posts sequel to this one). Before we close, let’s just make a few more observations about the solution to the optimization problem.

Support Vectors

In Trick 1.5 we saw that the optimal separating hyperplane has to be exactly halfway between the two closest points of opposite classes. Moreover, we noticed that, provided we’ve scaled \| w \| properly, these closest points (there may be multiple for positive and negative labels) have to be exactly “distance” 1 away from the separating hyperplane.

Another way to phrase this without putting “distance” in scare quotes is to say that, if w is the normal vector of the optimal separating hyperplane, the closest points lie on the two lines \langle x_i, w \rangle = \pm 1.

Now that we have some intuition for the formulation of this problem, it isn’t a stretch to realize the following. While a dataset may include many points from either class on these two lines \langle x_i, w \rangle = \pm 1, the optimal hyperplane itself does not depend on any of the other points except these closest points.

This fact is enough to give these closest points a special name: the support vectors.

We’ll actually prove that support vectors “are all you need” with full rigor and detail next time, when we cast the optimization problem in this post into the “dual” setting. To avoid vague names, the formulation described in this post called the “primal” problem. The dual problem is derived from the primal problem, with special variables and constraints chosen based on the primal variables and constraints. Next time we’ll describe in brief detail what the dual does and why it’s important, but we won’t have nearly enough time to give a full understanding of duality in optimization (such a treatment would fill a book).

When we compute the dual of the SVM problem, we will see explicitly that the hyperplane can be written as a linear combination of the support vectors. As such, once you’ve found the optimal hyperplane, you can compress the training set into just the support vectors, and reproducing the same optimal solution becomes much, much faster. You can also use the support vectors to augment the SVM to incorporate streaming data (throw out all non-support vectors after every retraining).

Eventually, when we get to implementing the SVM from scratch, we’ll see all this in action.

Until then!

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The Reasonable Effectiveness of the Multiplicative Weights Update Algorithm

papad

Christos Papadimitriou, who studies multiplicative weights in the context of biology.

Hard to believe

Sanjeev Arora and his coauthors consider it “a basic tool [that should be] taught to all algorithms students together with divide-and-conquer, dynamic programming, and random sampling.” Christos Papadimitriou calls it “so hard to believe that it has been discovered five times and forgotten.” It has formed the basis of algorithms in machine learning, optimization, game theory, economics, biology, and more.

What mystical algorithm has such broad applications? Now that computer scientists have studied it in generality, it’s known as the Multiplicative Weights Update Algorithm (MWUA). Procedurally, the algorithm is simple. I can even describe the core idea in six lines of pseudocode. You start with a collection of n objects, and each object has a weight.

Set all the object weights to be 1.
For some large number of rounds:
   Pick an object at random proportionally to the weights
   Some event happens
   Increase the weight of the chosen object if it does well in the event
   Otherwise decrease the weight

The name “multiplicative weights” comes from how we implement the last step: if the weight of the chosen object at step t is w_t before the event, and G represents how well the object did in the event, then we’ll update the weight according to the rule:

\displaystyle w_{t+1} = w_t (1 + G)

Think of this as increasing the weight by a small multiple of the object’s performance on a given round.

Here is a simple example of how it might be used. You have some money you want to invest, and you have a bunch of financial experts who are telling you what to invest in every day. So each day you pick an expert, and you follow their advice, and you either make a thousand dollars, or you lose a thousand dollars, or something in between. Then you repeat, and your goal is to figure out which expert is the most reliable.

This is how we use multiplicative weights: if we number the experts 1, \dots, N, we give each expert a weight w_i which starts at 1. Then, each day we pick an expert at random (where experts with larger weights are more likely to be picked) and at the end of the day we have some gain or loss G. Then we update the weight of the chosen expert by multiplying it by (1 + G / 1000). Sometimes you have enough information to update the weights of experts you didn’t choose, too. The theoretical guarantees of the algorithm say we’ll find the best expert quickly (“quickly” will be concrete later).

In fact, let’s play a game where you, dear reader, get to decide the rewards for each expert and each day. I programmed the multiplicative weights algorithm to react according to your choices. Click the image below to go to the demo.

mwua

This core mechanism of updating weights can be interpreted in many ways, and that’s part of the reason it has sprouted up all over mathematics and computer science. Just a few examples of where this has led:

  1. In game theory, weights are the “belief” of a player about the strategy of an opponent. The most famous algorithm to use this is called Fictitious Play, and others include EXP3 for minimizing regret in the so-called “adversarial bandit learning” problem.
  2. In machine learning, weights are the difficulty of a specific training example, so that higher weights mean the learning algorithm has to “try harder” to accommodate that example. The first result I’m aware of for this is the Perceptron (and similar Winnow) algorithm for learning hyperplane separators. The most famous is the AdaBoost algorithm.
  3. Analogously, in optimization, the weights are the difficulty of a specific constraint, and this technique can be used to approximately solve linear and semidefinite programs. The approximation is because MWUA only provides a solution with some error.
  4. In mathematical biology, the weights represent the fitness of individual alleles, and filtering reproductive success based on this and updating weights for successful organisms produces a mechanism very much like evolution. With modifications, it also provides a mechanism through which to understand sex in the context of evolutionary biology.
  5. The TCP protocol, which basically defined the internet, uses additive and multiplicative weight updates (which are very similar in the analysis) to manage congestion.
  6. You can get easy \log(n)-approximation algorithms for many NP-hard problems, such as set cover.

Additional, more technical examples can be found in this survey of Arora et al.

In the rest of this post, we’ll implement a generic Multiplicative Weights Update Algorithm, we’ll prove it’s main theoretical guarantees, and we’ll implement a linear program solver as an example of its applicability. As usual, all of the code used in the making of this post is available in a Github repository.

The generic MWUA algorithm

Let’s start by writing down pseudocode and an implementation for the MWUA algorithm in full generality.

In general we have some set X of objects and some set Y of “event outcomes” which can be completely independent. If these sets are finite, we can write down a table M whose rows are objects, whose columns are outcomes, and whose i,j entry M(i,j) is the reward produced by object x_i when the outcome is y_j. We will also write this as M(x, y) for object x and outcome y. The only assumption we’ll make on the rewards is that the values M(x, y) are bounded by some small constant B (by small I mean B should not require exponentially many bits to write down as compared to the size of X). In symbols, M(x,y) \in [0,B]. There are minor modifications you can make to the algorithm if you want negative rewards, but for simplicity we will leave that out. Note the table M just exists for analysis, and the algorithm does not know its values. Moreover, while the values in M are static, the choice of outcome y for a given round may be nondeterministic.

The MWUA algorithm randomly chooses an object x \in X in every round, observing the outcome y \in Y, and collecting the reward M(x,y) (or losing it as a penalty). The guarantee of the MWUA theorem is that the expected sum of rewards/penalties of MWUA is not much worse than if one had picked the best object (in hindsight) every single round.

Let’s describe the algorithm in notation first and build up pseudocode as we go. The input to the algorithm is the set of objects, a subroutine that observes an outcome, a black-box reward function, a learning rate parameter, and a number of rounds.

def MWUA(objects, observeOutcome, reward, learningRate, numRounds):
   ...

We define for object x a nonnegative number w_x we call a “weight.” The weights will change over time so we’ll also sub-script a weight with a round number t, i.e. w_{x,t} is the weight of object x in round t. Initially, all the weights are 1. Then MWUA continues in rounds. We start each round by drawing an example randomly with probability proportional to the weights. Then we observe the outcome for that round and the reward for that round.

# draw: [float] -> int
# pick an index from the given list of floats proportionally
# to the size of the entry (i.e. normalize to a probability
# distribution and draw according to the probabilities).
def draw(weights):
    choice = random.uniform(0, sum(weights))
    choiceIndex = 0

    for weight in weights:
        choice -= weight
        if choice <= 0:
            return choiceIndex

        choiceIndex += 1

# MWUA: the multiplicative weights update algorithm
def MWUA(objects, observeOutcome, reward, learningRate numRounds):
   weights = [1] * len(objects)
   for t in numRounds:
      chosenObjectIndex = draw(weights)
      chosenObject = objects[chosenObjectIndex]

      outcome = observeOutcome(t, weights, chosenObject)
      thisRoundReward = reward(chosenObject, outcome)

      ...

Sampling objects in this way is the same as associating a distribution D_t to each round, where if S_t = \sum_{x \in X} w_{x,t} then the probability of drawing x, which we denote D_t(x), is w_{x,t} / S_t. We don’t need to keep track of this distribution in the actual run of the algorithm, but it will help us with the mathematical analysis.

Next comes the weight update step. Let’s call our learning rate variable parameter \varepsilon. In round t say we have object x_t and outcome y_t, then the reward is M(x_t, y_t). We update the weight of the chosen object x_t according to the formula:

\displaystyle w_{x_t, t} = w_{x_t} (1 + \varepsilon M(x_t, y_t) / B)

In the more general event that you have rewards for all objects (if not, the reward-producing function can output zero), you would perform this weight update on all objects x \in X. This turns into the following Python snippet, where we hide the division by B into the choice of learning rate:

# MWUA: the multiplicative weights update algorithm
def MWUA(objects, observeOutcome, reward, learningRate, numRounds):
   weights = [1] * len(objects)
   for t in numRounds:
      chosenObjectIndex = draw(weights)
      chosenObject = objects[chosenObjectIndex]

      outcome = observeOutcome(t, weights, chosenObject)
      thisRoundReward = reward(chosenObject, outcome)

      for i in range(len(weights)):
         weights[i] *= (1 + learningRate * reward(objects[i], outcome))

One of the amazing things about this algorithm is that the outcomes and rewards could be chosen adaptively by an adversary who knows everything about the MWUA algorithm (except which random numbers the algorithm generates to make its choices). This means that the rewards in round t can depend on the weights in that same round! We will exploit this when we solve linear programs later in this post.

But even in such an oppressive, exploitative environment, MWUA persists and achieves its guarantee. And now we can state that guarantee.

Theorem (from Arora et al): The cumulative reward of the MWUA algorithm is, up to constant multiplicative factors, at least the cumulative reward of the best object minus \log(n), where n is the number of objects. (Exact formula at the end of the proof)

The core of the proof, which we’ll state as a lemma, uses one of the most elegant proof techniques in all of mathematics. It’s the idea of constructing a potential function, and tracking the change in that potential function over time. Such a proof usually has the mysterious script:

  1. Define potential function, in our case S_t.
  2. State what seems like trivial facts about the potential function to write S_{t+1} in terms of S_t, and hence get general information about S_T for some large T.
  3. Theorem is proved.
  4. Wait, what?

Clearly, coming up with a useful potential function is a difficult and prized skill.

In this proof our potential function is the sum of the weights of the objects in a given round, S_t = \sum_{x \in X} w_{x, t}. Now the lemma.

Lemma: Let B be the bound on the size of the rewards, and 0 < \varepsilon < 1/2 a learning parameter. Recall that D_t(x) is the probability that MWUA draws object x in round t. Write the expected reward for MWUA for round t as the following (using only the definition of expected value):

\displaystyle R_t = \sum_{x \in X} D_t(x) M(x, y_t)

 Then the claim of the lemma is:

\displaystyle S_{t+1} \leq S_t e^{\varepsilon R_t / B}

Proof. Expand S_{t+1} = \sum_{x \in X} w_{x, t+1} using the definition of the MWUA update:

\displaystyle \sum_{x \in X} w_{x, t+1} = \sum_{x \in X} w_{x, t}(1 + \varepsilon M(x, y_t) / B)

Now distribute w_{x, t} and split into two sums:

\displaystyle \dots = \sum_{x \in X} w_{x, t} + \frac{\varepsilon}{B} \sum_{x \in X} w_{x,t} M(x, y_t)

Using the fact that D_t(x) = \frac{w_{x,t}}{S_t}, we can replace w_{x,t} with D_t(x) S_t, which allows us to get R_t

\displaystyle \begin{aligned} \dots &= S_t + \frac{\varepsilon S_t}{B} \sum_{x \in X} D_t(x) M(x, y_t) \\ &= S_t \left ( 1 + \frac{\varepsilon R_t}{B} \right ) \end{aligned}

And then using the fact that (1 + x) \leq e^x (Taylor series), we can bound the last expression by S_te^{\varepsilon R_t / B}, as desired.

\square

Now using the lemma, we can get a hold on S_T for a large T, namely that

\displaystyle S_T \leq S_1 e^{\varepsilon \sum_{t=1}^T R_t / B}

If |X| = n then S_1=n, simplifying the above. Moreover, the sum of the weights in round T is certainly greater than any single weight, so that for every fixed object x \in X,

\displaystyle S_T \geq w_{x,T} \leq  (1 + \varepsilon)^{\sum_t M(x, y_t) / B}

Squeezing S_t between these two inequalities and taking logarithms (to simplify the exponents) gives

\displaystyle \left ( \sum_t M(x, y_t) / B \right ) \log(1+\varepsilon) \leq \log n + \frac{\varepsilon}{B} \sum_t R_t

Multiply through by B, divide by \varepsilon, rearrange, and use the fact that when 0 < \varepsilon < 1/2 we have \log(1 + \varepsilon) \geq \varepsilon - \varepsilon^2 (Taylor series) to get

\displaystyle \sum_t R_t \geq \left [ \sum_t M(x, y_t) \right ] (1-\varepsilon) - \frac{B \log n}{\varepsilon}

The bracketed term is the payoff of object x, and MWUA’s payoff is at least a fraction of that minus the logarithmic term. The bound applies to any object x \in X, and hence to the best one. This proves the theorem.

\square

Briefly discussing the bound itself, we see that the smaller the learning rate is, the closer you eventually get to the best object, but by contrast the more the subtracted quantity B \log(n) / \varepsilon hurts you. If your target is an absolute error bound against the best performing object on average, you can do more algebra to determine how many rounds you need in terms of a fixed \delta. The answer is roughly: let \varepsilon = O(\delta / B) and pick T = O(B^2 \log(n) / \delta^2). See this survey for more.

MWUA for linear programs

Now we’ll approximately solve a linear program using MWUA. Recall that a linear program is an optimization problem whose goal is to minimize (or maximize) a linear function of many variables. The objective to minimize is usually given as a dot product c \cdot x, where c is a fixed vector and x = (x_1, x_2, \dots, x_n) is a vector of non-negative variables the algorithm gets to choose. The choices for x are also constrained by a set of m linear inequalities, A_i \cdot x \geq b_i, where A_i is a fixed vector and b_i is a scalar for i = 1, \dots, m. This is usually summarized by putting all the A_i in a matrix, b_i in a vector, as

x_{\textup{OPT}} = \textup{argmin}_x \{ c \cdot x \mid Ax \geq b, x \geq 0 \}

We can further simplify the constraints by assuming we know the optimal value Z = c \cdot x_{\textup{OPT}} in advance, by doing a binary search (more on this later). So, if we ignore the hard constraint Ax \geq b, the “easy feasible region” of possible x‘s includes \{ x \mid x \geq 0, c \cdot x = Z \}.

In order to fit linear programming into the MWUA framework we have to define two things.

  1. The objects: the set of linear inequalities A_i \cdot x \geq b_i.
  2. The rewards: the error of a constraint for a special input vector x_t.

Number 2 is curious (why would we give a reward for error?) but it’s crucial and we’ll discuss it momentarily.

The special input x_t depends on the weights in round t (which is allowed, recall). Specifically, if the weights are w = (w_1, \dots, w_m), we ask for a vector x_t in our “easy feasible region” which satisfies

\displaystyle (A^T w) \cdot x_t \geq w \cdot b

For this post we call the implementation of procuring such a vector the “oracle,” since it can be seen as the black-box problem of, given a vector \alpha and a scalar \beta and a convex region R, finding a vector x \in R satisfying \alpha \cdot x \geq \beta. This allows one to solve more complex optimization problems with the same technique, swapping in a new oracle as needed. Our choice of inputs, \alpha = A^T w, \beta = w \cdot b, are particular to the linear programming formulation.

Two remarks on this choice of inputs. First, the vector A^T w is a weighted average of the constraints in A, and w \cdot b is a weighted average of the thresholds. So this this inequality is a “weighted average” inequality (specifically, a convex combination, since the weights are nonnegative). In particular, if no such x exists, then the original linear program has no solution. Indeed, given a solution x^* to the original linear program, each constraint, say A_1 x^*_1 \geq b_1, is unaffected by left-multiplication by w_1.

Second, and more important to the conceptual understanding of this algorithm, the choice of rewards and the multiplicative updates ensure that easier constraints show up less prominently in the inequality by having smaller weights. That is, if we end up overly satisfying a constraint, we penalize that object for future rounds so we don’t waste our effort on it. The byproduct of MWUA—the weights—identify the hardest constraints to satisfy, and so in each round we can put a proportionate amount of effort into solving (one of) the hard constraints. This is why it makes sense to reward error; the error is a signal for where to improve, and by over-representing the hard constraints, we force MWUA’s attention on them.

At the end, our final output is an average of the x_t produced in each round, i.e. x^* = \frac{1}{T}\sum_t x_t. This vector satisfies all the constraints to a roughly equal degree. We will skip the proof that this vector does what we want, but see these notes for a simple proof. We’ll spend the rest of this post implementing the scheme outlined above.

Implementing the oracle

Fix the convex region R = \{ c \cdot x = Z, x \geq 0 \} for a known optimal value Z. Define \textup{oracle}(\alpha, \beta) as the problem of finding an x \in R such that \alpha \cdot x \geq \beta.

For the case of this linear region R, we can simply find the index i which maximizes \alpha_i Z / c_i. If this value exceeds \beta, we can return the vector with that value in the i-th position and zeros elsewhere. Otherwise, the problem has no solution.

To prove the “no solution” part, say n=2 and you have x = (x_1, x_2) a solution to \alpha \cdot x \geq \beta. Then for whichever index makes \alpha_i Z / c_i bigger, say i=1, you can increase \alpha \cdot x without changing c \cdot x = Z by replacing x_1 with x_1 + (c_2/c_1)x_2 and x_2 with zero. I.e., we’re moving the solution x along the line c \cdot x = Z until it reaches a vertex of the region bounded by c \cdot x = Z and x \geq 0. This must happen when all entries but one are zero. This is the same reason why optimal solutions of (generic) linear programs occur at vertices of their feasible regions.

The code for this becomes quite simple. Note we use the numpy library in the entire codebase to make linear algebra operations fast and simple to read.

def makeOracle(c, optimalValue):
    n = len(c)

    def oracle(weightedVector, weightedThreshold):
        def quantity(i):
            return weightedVector[i] * optimalValue / c[i] if c[i] > 0 else -1

        biggest = max(range(n), key=quantity)
        if quantity(biggest) < weightedThreshold:
            raise InfeasibleException

        return numpy.array([optimalValue / c[i] if i == biggest else 0 for i in range(n)])

    return oracle

Implementing the core solver

The core solver implements the discussion from previously, given the optimal value of the linear program as input. To avoid too many single-letter variable names, we use linearObjective instead of c.

def solveGivenOptimalValue(A, b, linearObjective, optimalValue, learningRate=0.1):
    m, n = A.shape  # m equations, n variables
    oracle = makeOracle(linearObjective, optimalValue)

    def reward(i, specialVector):
        ...

    def observeOutcome(_, weights, __):
        ...

    numRounds = 1000
    weights, cumulativeReward, outcomes = MWUA(
        range(m), observeOutcome, reward, learningRate, numRounds
    )
    averageVector = sum(outcomes) / numRounds

    return averageVector

First we make the oracle, then the reward and outcome-producing functions, then we invoke the MWUA subroutine. Here are those two functions; they are closures because they need access to A and b. Note that neither c nor the optimal value show up here.

    def reward(i, specialVector):
        constraint = A[i]
        threshold = b[i]
        return threshold - numpy.dot(constraint, specialVector)

    def observeOutcome(_, weights, __):
        weights = numpy.array(weights)
        weightedVector = A.transpose().dot(weights)
        weightedThreshold = weights.dot(b)
        return oracle(weightedVector, weightedThreshold)

Implementing the binary search, and an example

Finally, the top-level routine. Note that the binary search for the optimal value is sophisticated (though it could be more sophisticated). It takes a max range for the search, and invokes the optimization subroutine, moving the upper bound down if the linear program is feasible and moving the lower bound up otherwise.

def solve(A, b, linearObjective, maxRange=1000):
    optRange = [0, maxRange]

    while optRange[1] - optRange[0] > 1e-8:
        proposedOpt = sum(optRange) / 2
        print("Attempting to solve with proposedOpt=%G" % proposedOpt)

        # Because the binary search starts so high, it results in extreme
        # reward values that must be tempered by a slow learning rate. Exercise
        # to the reader: determine absolute bounds for the rewards, and set
        # this learning rate in a more principled fashion.
        learningRate = 1 / max(2 * proposedOpt * c for c in linearObjective)
        learningRate = min(learningRate, 0.1)

        try:
            result = solveGivenOptimalValue(A, b, linearObjective, proposedOpt, learningRate)
            optRange[1] = proposedOpt
        except InfeasibleException:
            optRange[0] = proposedOpt

    return result

Finally, a simple example:

A = numpy.array([[1, 2, 3], [0, 4, 2]])
b = numpy.array([5, 6])
c = numpy.array([1, 2, 1])

x = solve(A, b, c)
print(x)
print(c.dot(x))
print(A.dot(x) - b)

The output:

Attempting to solve with proposedOpt=500
Attempting to solve with proposedOpt=250
Attempting to solve with proposedOpt=125
Attempting to solve with proposedOpt=62.5
Attempting to solve with proposedOpt=31.25
Attempting to solve with proposedOpt=15.625
Attempting to solve with proposedOpt=7.8125
Attempting to solve with proposedOpt=3.90625
Attempting to solve with proposedOpt=1.95312
Attempting to solve with proposedOpt=2.92969
Attempting to solve with proposedOpt=3.41797
Attempting to solve with proposedOpt=3.17383
Attempting to solve with proposedOpt=3.05176
Attempting to solve with proposedOpt=2.99072
Attempting to solve with proposedOpt=3.02124
Attempting to solve with proposedOpt=3.00598
Attempting to solve with proposedOpt=2.99835
Attempting to solve with proposedOpt=3.00217
Attempting to solve with proposedOpt=3.00026
Attempting to solve with proposedOpt=2.99931
Attempting to solve with proposedOpt=2.99978
Attempting to solve with proposedOpt=3.00002
Attempting to solve with proposedOpt=2.9999
Attempting to solve with proposedOpt=2.99996
Attempting to solve with proposedOpt=2.99999
Attempting to solve with proposedOpt=3.00001
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3  # note %G rounds the printed values
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
Attempting to solve with proposedOpt=3
[ 0.     0.987  1.026]
3.00000000425
[  5.20000072e-02   8.49831849e-09]

So there we have it. A fiendishly clever use of multiplicative weights for solving linear programs.

Discussion

One of the nice aspects of MWUA is it’s completely transparent. If you want to know why a decision was made, you can simply look at the weights and look at the history of rewards of the objects. There’s also a clear interpretation of what is being optimized, as the potential function used in the proof is a measure of both quality and adaptability to change. The latter is why MWUA succeeds even in adversarial settings, and why it makes sense to think about MWUA in the context of evolutionary biology.

This even makes one imagine new problems that traditional algorithms cannot solve, but which MWUA handles with grace. For example, imagine trying to solve an “online” linear program in which over time a constraint can change. MWUA can adapt to maintain its approximate solution.

The linear programming technique is known in the literature as the Plotkin-Shmoys-Tardos framework for covering and packing problems. The same ideas extend to other convex optimization problems, including semidefinite programming.

If you’ve been reading this entire post screaming “This is just gradient descent!” Then you’re right and wrong. It bears a striking resemblance to gradient descent (see this document for details about how special cases of MWUA are gradient descent by another name), but the adaptivity for the rewards makes MWUA different.

Even though so many people have been advocating for MWUA over the past decade, it’s surprising that it doesn’t show up in the general math/CS discourse on the internet or even in many algorithms courses. The Arora survey I referenced is from 2005 and the linear programming technique I demoed is originally from 1991! I took algorithms classes wherever I could, starting undergraduate in 2007, and I didn’t even hear a whisper of this technique until midway through my PhD in theoretical CS (I did, however, study fictitious play in a game theory class). I don’t have an explanation for why this is the case, except maybe that it takes more than 20 years for techniques to make it to the classroom. At the very least, this is one good reason to go to graduate school. You learn the things (and where to look for the things) which haven’t made it to classrooms yet.

Until next time!

Singular Value Decomposition Part 2: Theorem, Proof, Algorithm

I’m just going to jump right into the definitions and rigor, so if you haven’t read the previous post motivating the singular value decomposition, go back and do that first. This post will be theorem, proof, algorithm, data. The data set we test on is a thousand-story CNN news data set. All of the data, code, and examples used in this post is in a github repository, as usual.

We start with the best-approximating k-dimensional linear subspace.

Definition: Let X = \{ x_1, \dots, x_m \} be a set of m points in \mathbb{R}^n. The best approximating k-dimensional linear subspace of X is the k-dimensional linear subspace V \subset \mathbb{R}^n which minimizes the sum of the squared distances from the points in X to V.

Let me clarify what I mean by minimizing the sum of squared distances. First we’ll start with the simple case: we have a vector x \in X, and a candidate line L (a 1-dimensional subspace) that is the span of a unit vector v. The squared distance from x to the line spanned by v is the squared length of x minus the squared length of the projection of x onto v. Here’s a picture.

vectormax

I’m saying that the pink vector z in the picture is the difference of the black and green vectors x-y, and that the “distance” from x to v is the length of the pink vector. The reason is just the Pythagorean theorem: the vector x is the hypotenuse of a right triangle whose other two sides are the projected vector y and the difference vector z.

Let’s throw down some notation. I’ll call \textup{proj}_v: \mathbb{R}^n \to \mathbb{R}^n the linear map that takes as input a vector x and produces as output the projection of x onto v. In fact we have a brief formula for this when v is a unit vector. If we call x \cdot v the usual dot product, then \textup{proj}_v(x) = (x \cdot v)v. That’s v scaled by the inner product of x and v. In the picture above, since the line L is the span of the vector v, that means that y = \textup{proj}_v(x) and z = x -\textup{proj}_v(x) = x-y.

The dot-product formula is useful for us because it allows us to compute the squared length of the projection by taking a dot product |x \cdot v|^2. So then a formula for the distance of x from the line spanned by the unit vector v is

\displaystyle (\textup{dist}_v(x))^2 = \left ( \sum_{i=1}^n x_i^2 \right ) - |x \cdot v|^2

This formula is just a restatement of the Pythagorean theorem for perpendicular vectors.

\displaystyle \sum_{i} x_i^2 = (\textup{proj}_v(x))^2 + (\textup{dist}_v(x))^2

In particular, the difference vector we originally called z has squared length \textup{dist}_v(x)^2. The vector y, which is perpendicular to z and is also the projection of x onto L, it’s squared length is (\textup{proj}_v(x))^2. And the Pythagorean theorem tells us that summing those two squared lengths gives you the squared length of the hypotenuse x.

If we were trying to find the best approximating 1-dimensional subspace for a set of data points X, then we’d want to minimize the sum of the squared distances for every point x \in X. Namely, we want the v that solves \min_{|v|=1} \sum_{x \in X} (\textup{dist}_v(x))^2.

With some slight algebra we can make our life easier. The short version: minimizing the sum of squared distances is the same thing as maximizing the sum of squared lengths of the projections. The longer version: let’s go back to a single point x and the line spanned by v. The Pythagorean theorem told us that

\displaystyle \sum_{i} x_i^2 = (\textup{proj}_v(x))^2 + (\textup{dist}_v(x))^2

The squared length of x is constant. It’s an input to the algorithm and it doesn’t change through a run of the algorithm. So we get the squared distance by subtracting (\textup{proj}_v(x))^2 from a constant number,

\displaystyle \sum_{i} x_i^2 - (\textup{proj}_v(x))^2 = (\textup{dist}_v(x))^2

which means if we want to minimize the squared distance, we can instead maximize the squared projection. Maximizing the subtracted thing minimizes the whole expression.

It works the same way if you’re summing over all the data points in X. In fact, we can say it much more compactly this way. If the rows of A are your data points, then Av contains as each entry the (signed) dot products x_i \cdot v. And the squared norm of this vector, |Av|^2, is exactly the sum of the squared lengths of the projections of the data onto the line spanned by v. The last thing is that maximizing a square is the same as maximizing its square root, so we can switch freely between saying our objective is to find the unit vector v that maximizes |Av| and that which maximizes |Av|^2.

At this point you should be thinking,

Great, we have written down an optimization problem: \max_{v : |v|=1} |Av|. If we could solve this, we’d have the best 1-dimensional linear approximation to the data contained in the rows of A. But (1) how do we solve that problem? And (2) you promised a k-dimensional approximating subspace. I feel betrayed! Swindled! Bamboozled!

Here’s the fantastic thing. We can solve the 1-dimensional optimization problem efficiently (we’ll do it later in this post), and (2) is answered by the following theorem.

The SVD Theorem: Computing the best k-dimensional subspace reduces to k applications of the one-dimensional problem.

We will prove this after we introduce the terms “singular value” and “singular vector.”

Singular values and vectors

As I just said, we can get the best k-dimensional approximating linear subspace by solving the one-dimensional maximization problem k times. The singular vectors of A are defined recursively as the solutions to these sub-problems. That is, I’ll call v_1 the first singular vector of A, and it is:

\displaystyle v_1 = \arg \max_{v, |v|=1} |Av|

And the corresponding first singular value, denoted \sigma_1(A), is the maximal value of the optimization objective, i.e. |Av_1|. (I will use this term frequently, that |Av| is the “objective” of the optimization problem.) Informally speaking, (\sigma_1(A))^2 represents how much of the data was captured by the first singular vector. Meaning, how close the vectors are to lying on the line spanned by v_1. Larger values imply the approximation is better. In fact, if all the data points lie on a line, then (\sigma_1(A))^2 is the sum of the squared norms of the rows of A.

Now here is where we see the reduction from the k-dimensional case to the 1-dimensional case. To find the best 2-dimensional subspace, you first find the best one-dimensional subspace (spanned by v_1), and then find the best 1-dimensional subspace, but only considering those subspaces that are the spans of unit vectors perpendicular to v_1. The notation for “vectors v perpendicular to v_1” is v \perp v_1. Restating, the second singular vector v _2 is defined as

\displaystyle v_2 = \arg \max_{v \perp v_1, |v| = 1} |Av|

And the SVD theorem implies the subspace spanned by \{ v_1, v_2 \} is the best 2-dimensional linear approximation to the data. Likewise \sigma_2(A) = |Av_2| is the second singular value. Its squared magnitude tells us how much of the data that was not “captured” by v_1 is captured by v_2. Again, if the data lies in a 2-dimensional subspace, then the span of \{ v_1, v_2 \} will be that subspace.

We can continue this process. Recursively define v_k, the k-th singular vector, to be the vector which maximizes |Av|, when v is considered only among the unit vectors which are perpendicular to \textup{span} \{ v_1, \dots, v_{k-1} \}. The corresponding singular value \sigma_k(A) is the value of the optimization problem.

As a side note, because of the way we defined the singular values as the objective values of “nested” optimization problems, the singular values are decreasing, \sigma_1(A) \geq \sigma_2(A) \geq \dots \geq \sigma_n(A) \geq 0. This is obvious: you only pick v_2 in the second optimization problem because you already picked v_1 which gave a bigger singular value, so v_2‘s objective can’t be bigger.

If you keep doing this, one of two things happen. Either you reach v_n and since the domain is n-dimensional there are no remaining vectors to choose from, the v_i are an orthonormal basis of \mathbb{R}^n. This means that the data in A contains a full-rank submatrix. The data does not lie in any smaller-dimensional subspace. This is what you’d expect from real data.

Alternatively, you could get to a stage v_k with k < n and when you try to solve the optimization problem you find that every perpendicular v has Av = 0. In this case, the data actually does lie in a k-dimensional subspace, and the first-through-k-th singular vectors you computed span this subspace.

Let’s do a quick sanity check: how do we know that the singular vectors v_i form a basis? Well formally they only span a basis of the row space of A, i.e. a basis of the subspace spanned by the data contained in the rows of A. But either way the point is that each v_{i+1} spans a new dimension from the previous v_1, \dots, v_i because we’re choosing v_{i+1} to be orthogonal to all the previous v_i. So the answer to our sanity check is “by construction.”

Back to the singular vectors, the discussion from the last post tells us intuitively that the data is probably never in a small subspace.  You never expect the process of finding singular vectors to stop before step n, and if it does you take a step back and ask if something deeper is going on. Instead, in real life you specify how much of the data you want to capture, and you keep computing singular vectors until you’ve passed the threshold. Alternatively, you specify the amount of computing resources you’d like to spend by fixing the number of singular vectors you’ll compute ahead of time, and settle for however good the k-dimensional approximation is.

Before we get into any code or solve the 1-dimensional optimization problem, let’s prove the SVD theorem.

Proof of SVD theorem.

Recall we’re trying to prove that the first k singular vectors provide a linear subspace W which maximizes the squared-sum of the projections of the data onto W. For k=1 this is trivial, because we defined v_1 to be the solution to that optimization problem. The case of k=2 contains all the important features of the general inductive step. Let W be any best-approximating 2-dimensional linear subspace for the rows of A. We’ll show that the subspace spanned by the two singular vectors v_1, v_2 is at least as good (and hence equally good).

Let w_1, w_2 be any orthonormal basis for W and let |Aw_1|^2 + |Aw_2|^2 be the quantity that we’re trying to maximize (and which W maximizes by assumption). Moreover, we can pick the basis vector w_2 to be perpendicular to v_1. To prove this we consider two cases: either v_1 is already perpendicular to W in which case it’s trivial, or else v_1 isn’t perpendicular to W and you can choose w_1 to be \textup{proj}_W(v_1) and choose w_2 to be any unit vector perpendicular to w_1.

Now since v_1 maximizes |Av|, we have |Av_1|^2 \geq |Aw_1|^2. Moreover, since w_2 is perpendicular to v_1, the way we chose v_2 also makes |Av_2|^2 \geq |Aw_2|^2. Hence the objective |Av_1|^2 + |Av_2|^2 \geq |Aw_1|^2 + |Aw_2|^2, as desired.

For the general case of k, the inductive hypothesis tells us that the first k terms of the objective for k+1 singular vectors is maximized, and we just have to pick any vector w_{k+1} that is perpendicular to all v_1, v_2, \dots, v_k, and the rest of the proof is just like the 2-dimensional case.

\square

Now remember that in the last post we started with the definition of the SVD as a decomposition of a matrix A = U\Sigma V^T? And then we said that this is a certain kind of change of basis? Well the singular vectors v_i together form the columns of the matrix V (the rows of V^T), and the corresponding singular values \sigma_i(A) are the diagonal entries of \Sigma. When A is understood we’ll abbreviate the singular value as \sigma_i.

To reiterate with the thoughts from last post, the process of applying A is exactly recovered by the process of first projecting onto the (full-rank space of) singular vectors v_1, \dots, v_k, scaling each coordinate of that projection according to the corresponding singular values, and then applying this U thing we haven’t talked about yet.

So let’s determine what U has to be. The way we picked v_i to make A diagonal gives us an immediate suggestion: use the Av_i as the columns of U. Indeed, define u_i = Av_i, the images of the singular vectors under A. We can swiftly show the u_i form a basis of the image of A. The reason is because if v = \sum_i c_i v_i (using all n of the singular vectors v_i), then by linearity Av = \sum_{i} c_i Av_i = \sum_i c_i u_i. It is also easy to see why the u_i are orthogonal (prove it as an exercise). Let’s further make sure the u_i are unit vectors and redefine them as u_i = \frac{1}{\sigma_i}Av_i

If you put these thoughts together, you can say exactly what A does to any given vector x. Since the v_i form an orthonormal basis, x = \sum_i (x \cdot v_i) v_i, and then applying A gives

\displaystyle \begin{aligned}Ax &= A \left ( \sum_i (x \cdot v_i) v_i \right ) \\  &= \sum_i (x \cdot v_i) A_i v_i \\ &= \sum_i (x \cdot v_i) \sigma_i u_i \end{aligned}

If you’ve been closely reading this blog in the last few months, you’ll recognize a very nice way to write the last line of the above equation. It’s an outer product. So depending on your favorite symbols, you’d write this as either A = \sum_{i} \sigma_i u_i \otimes v_i or A = \sum_i \sigma_i u_i v_i^T. Or, if you like expressing things as matrix factorizations, as A = U\Sigma V^T. All three are describing the same object.

Let’s move on to some code.

A black box example

Before we implement SVD from scratch (an urge that commands me from the depths of my soul!), let’s see a black-box example that uses existing tools. For this we’ll use the numpy library.

Recall our movie-rating matrix from the last post:

movieratings

The code to compute the svd of this matrix is as simple as it gets:

from numpy.linalg import svd

movieRatings = [
    [2, 5, 3],
    [1, 2, 1],
    [4, 1, 1],
    [3, 5, 2],
    [5, 3, 1],
    [4, 5, 5],
    [2, 4, 2],
    [2, 2, 5],
]

U, singularValues, V = svd(movieRatings)

Printing these values out gives

[[-0.39458526  0.23923575 -0.35445911 -0.38062172 -0.29836818 -0.49464816 -0.30703202 -0.29763321]
 [-0.15830232  0.03054913 -0.15299759 -0.45334816  0.31122898  0.23892035 -0.37313346  0.67223457]
 [-0.22155201 -0.52086121  0.39334917 -0.14974792 -0.65963979  0.00488292 -0.00783684  0.25934607]
 [-0.39692635 -0.08649009 -0.41052882  0.74387448 -0.10629499  0.01372565 -0.17959298  0.26333462]
 [-0.34630257 -0.64128825  0.07382859 -0.04494155  0.58000668 -0.25806239  0.00211823 -0.24154726]
 [-0.53347449  0.19168874  0.19949342 -0.03942604  0.00424495  0.68715732 -0.06957561 -0.40033035]
 [-0.31660464  0.06109826 -0.30599517 -0.19611823 -0.01334272  0.01446975  0.85185852  0.19463493]
 [-0.32840223  0.45970413  0.62354764  0.1783041   0.17631186 -0.39879476  0.06065902  0.25771578]]
[ 15.09626916   4.30056855   3.40701739]
[[-0.54184808 -0.67070995 -0.50650649]
 [-0.75152295  0.11680911  0.64928336]
 [ 0.37631623 -0.73246419  0.56734672]]

Now this is a bit weird, because the matrices U, V are the wrong shape! Remember, there are only supposed to be three vectors since the input matrix has rank three. So what gives? This is a distinction that goes by the name “full” versus “reduced” SVD. The idea goes back to our original statement that U \Sigma V^T is a decomposition with U, V^T both orthogonal and square matrices. But in the derivation we did in the last section, the U and V were not square. The singular vectors v_i could potentially stop before even becoming full rank.

In order to get to square matrices, what people sometimes do is take the two bases v_1, \dots, v_k and u_1, \dots, u_k and arbitrarily choose ways to complete them to a full orthonormal basis of their respective vector spaces. In other words, they just make the matrix square by filling it with data for no reason other than that it’s sometimes nice to have a complete basis. We don’t care about this. To be honest, I think the only place this comes in useful is in the desire to be particularly tidy in a mathematical formulation of something.

We can still work with it programmatically. By fudging around a bit with numpy’s shapes to get a diagonal matrix, we can reconstruct the input rating matrix from the factors.

Sigma = np.vstack([
    np.diag(singularValues),
    np.zeros((5, 3)),
])

print(np.round(movieRatings - np.dot(U, np.dot(Sigma, V)), decimals=10))

And the output is, as one expects, a matrix of all zeros. Meaning that we decomposed the movie rating matrix, and built it back up from the factors.

We can actually get the SVD as we defined it (with rectangular matrices) by passing a special flag to numpy’s svd.

U, singularValues, V = svd(movieRatings, full_matrices=False)
print(U)
print(singularValues)
print(V)

Sigma = np.diag(singularValues)
print(np.round(movieRatings - np.dot(U, np.dot(Sigma, V)), decimals=10))

And the result

[[-0.39458526  0.23923575 -0.35445911]
 [-0.15830232  0.03054913 -0.15299759]
 [-0.22155201 -0.52086121  0.39334917]
 [-0.39692635 -0.08649009 -0.41052882]
 [-0.34630257 -0.64128825  0.07382859]
 [-0.53347449  0.19168874  0.19949342]
 [-0.31660464  0.06109826 -0.30599517]
 [-0.32840223  0.45970413  0.62354764]]
[ 15.09626916   4.30056855   3.40701739]
[[-0.54184808 -0.67070995 -0.50650649]
 [-0.75152295  0.11680911  0.64928336]
 [ 0.37631623 -0.73246419  0.56734672]]
[[-0. -0. -0.]
 [-0. -0.  0.]
 [ 0. -0.  0.]
 [-0. -0. -0.]
 [-0. -0. -0.]
 [-0. -0. -0.]
 [-0. -0. -0.]
 [ 0. -0. -0.]]

This makes the reconstruction less messy, since we can just multiply everything without having to add extra rows of zeros to \Sigma.

What do the singular vectors and values tell us about the movie rating matrix? (Besides nothing, since it’s a contrived example) You’ll notice that the first singular vector \sigma_1 > 15 while the other two singular values are around 4. This tells us that the first singular vector covers a large part of the structure of the matrix. I.e., a rank-1 matrix would be a pretty good approximation to the whole thing. As an exercise to the reader, write a program that evaluates this claim (how good is “good”?).

The greedy optimization routine

Now we’re going to write SVD from scratch. We’ll first implement the greedy algorithm for the 1-d optimization problem, and then we’ll perform the inductive step to get a full algorithm. Then we’ll run it on the CNN data set.

The method we’ll use to solve the 1-dimensional problem isn’t necessarily industry strength (see this document for a hint of what industry strength looks like), but it is simple conceptually. It’s called the power method. Now that we have our decomposition of theorem, understanding how the power method works is quite easy.

Let’s work in the language of a matrix decomposition A = U \Sigma V^T, more for practice with that language than anything else (using outer products would give us the same result with slightly different computations). Then let’s observe A^T A, wherein we’ll use the fact that U is orthonormal and so U^TU is the identity matrix:

\displaystyle A^TA = (U \Sigma V^T)^T(U \Sigma V^T) = V \Sigma U^TU \Sigma V^T = V \Sigma^2 V^T

So we can completely eliminate U from the discussion, and look at just V \Sigma^2 V^T. And what’s nice about this matrix is that we can compute its eigenvectors, and eigenvectors turn out to be exactly the singular vectors. The corresponding eigenvalues are the squared singular values. This should be clear from the above derivation. If you apply (V \Sigma^2 V^T) to any v_i, the only parts of the product that aren’t zero are the ones involving v_i with itself, and the scalar \sigma_i^2 factors in smoothly. It’s dead simple to check.

Theorem: Let x be a random unit vector and let B = A^TA = V \Sigma^2 V^T. Then with high probability, \lim_{s \to \infty} B^s x is in the span of the first singular vector v_1. If we normalize B^s x to a unit vector at each s, then furthermore the limit is v_1.

Proof. Start with a random unit vector x, and write it in terms of the singular vectors x = \sum_i c_i v_i. That means Bx = \sum_i c_i \sigma_i^2 v_i. If you recursively apply this logic, you get B^s x = \sum_i c_i \sigma_i^{2s} v_i. In particular, the dot product of (B^s x) with any v_j is c_i \sigma_j^{2s}.

What this means is that so long as the first singular value \sigma_1 is sufficiently larger than the second one \sigma_2, and in turn all the other singular values, the part of B^s x  corresponding to v_1 will be much larger than the rest. Recall that if you expand a vector in terms of an orthonormal basis, in this case B^s x expanded in the v_i, the coefficient of B^s x on v_j is exactly the dot product. So to say that B^sx converges to being in the span of v_1 is the same as saying that the ratio of these coefficients, |(B^s x \cdot v_1)| / |(B^s x \cdot v_j)| \to \infty for any j. In other words, the coefficient corresponding to the first singular vector dominates all of the others. And so if we normalize, the coefficient of B^s x corresponding to v_1 tends to 1, while the rest tend to zero.

Indeed, this ratio is just (\sigma_1 / \sigma_j)^{2s} and the base of this exponential is bigger than 1.

\square

If you want to be a little more precise and find bounds on the number of iterations required to converge, you can. The worry is that your random starting vector is “too close” to one of the smaller singular vectors v_j, so that if the ratio of \sigma_1 / \sigma_j is small, then the “pull” of v_1 won’t outweigh the pull of v_j fast enough. Choosing a random unit vector allows you to ensure with high probability that this doesn’t happen. And conditioned on it not happening (or measuring “how far the event is from happening” precisely), you can compute a precise number of iterations required to converge. The last two pages of these lecture notes have all the details.

We won’t compute a precise number of iterations. Instead we’ll just compute until the angle between B^{s+1}x and B^s x is very small. Here’s the algorithm

import numpy as np
from numpy.linalg import norm

from random import normalvariate
from math import sqrt

def randomUnitVector(n):
    unnormalized = [normalvariate(0, 1) for _ in range(n)]
    theNorm = sqrt(sum(x * x for x in unnormalized))
    return [x / theNorm for x in unnormalized]

def svd_1d(A, epsilon=1e-10):
    ''' The one-dimensional SVD '''

    n, m = A.shape
    x = randomUnitVector(m)
    lastV = None
    currentV = x
    B = np.dot(A.T, A)

    iterations = 0
    while True:
        iterations += 1
        lastV = currentV
        currentV = np.dot(B, lastV)
        currentV = currentV / norm(currentV)

        if abs(np.dot(currentV, lastV)) > 1 - epsilon:
            print("converged in {} iterations!".format(iterations))
            return currentV

We start with a random unit vector x, and then loop computing x_{t+1} = Bx_t, renormalizing at each step. The condition for stopping is that the magnitude of the dot product between x_t and x_{t+1} (since they’re unit vectors, this is the cosine of the angle between them) is very close to 1.

And using it on our movie ratings example:

if __name__ == "__main__":
    movieRatings = np.array([
        [2, 5, 3],
        [1, 2, 1],
        [4, 1, 1],
        [3, 5, 2],
        [5, 3, 1],
        [4, 5, 5],
        [2, 4, 2],
        [2, 2, 5],
    ], dtype='float64')

    print(svd_1d(movieRatings))

With the result

converged in 6 iterations!
[-0.54184805 -0.67070993 -0.50650655]

Note that the sign of the vector may be different from numpy’s output because we start with a random vector to begin with.

The recursive step, getting from v_1 to the entire SVD, is equally straightforward. Say you start with the matrix A and you compute v_1. You can use v_1 to compute u_1 and \sigma_1(A). Then you want to ensure you’re ignoring all vectors in the span of v_1 for your next greedy optimization, and to do this you can simply subtract the rank 1 component of A corresponding to v_1. I.e., set A' = A - \sigma_1(A) u_1 v_1^T. Then it’s easy to see that \sigma_1(A') = \sigma_2(A) and basically all the singular vectors shift indices by 1 when going from A to A'. Then you repeat.

If that’s not clear enough, here’s the code.

def svd(A, epsilon=1e-10):
    n, m = A.shape
    svdSoFar = []

    for i in range(m):
        matrixFor1D = A.copy()

        for singularValue, u, v in svdSoFar[:i]:
            matrixFor1D -= singularValue * np.outer(u, v)

        v = svd_1d(matrixFor1D, epsilon=epsilon)  # next singular vector
        u_unnormalized = np.dot(A, v)
        sigma = norm(u_unnormalized)  # next singular value
        u = u_unnormalized / sigma

        svdSoFar.append((sigma, u, v))

    # transform it into matrices of the right shape
    singularValues, us, vs = [np.array(x) for x in zip(*svdSoFar)]

    return singularValues, us.T, vs

And we can run this on our movie rating matrix to get the following

>>> theSVD = svd(movieRatings)
>>> theSVD[0]
array([ 15.09626916,   4.30056855,   3.40701739])
>>> theSVD[1]
array([[ 0.39458528, -0.23923093,  0.35446407],
       [ 0.15830233, -0.03054705,  0.15299815],
       [ 0.221552  ,  0.52085578, -0.39336072],
       [ 0.39692636,  0.08649568,  0.41052666],
       [ 0.34630257,  0.64128719, -0.07384286],
       [ 0.53347448, -0.19169154, -0.19948959],
       [ 0.31660465, -0.0610941 ,  0.30599629],
       [ 0.32840221, -0.45971273, -0.62353781]])
>>> theSVD[2]
array([[ 0.54184805,  0.67071006,  0.50650638],
       [ 0.75151641, -0.11679644, -0.64929321],
       [-0.37632934,  0.73246611, -0.56733554]])

Checking this against our numpy output shows it’s within a reasonable level of precision (considering the power method took on the order of ten iterations!)

>>> np.round(np.abs(npSVD[0]) - np.abs(theSVD[1]), decimals=5)
array([[ -0.00000000e+00,  -0.00000000e+00,   0.00000000e+00],
       [  0.00000000e+00,  -0.00000000e+00,   0.00000000e+00],
       [  0.00000000e+00,  -1.00000000e-05,   1.00000000e-05],
       [  0.00000000e+00,   0.00000000e+00,  -0.00000000e+00],
       [  0.00000000e+00,  -0.00000000e+00,   1.00000000e-05],
       [ -0.00000000e+00,   0.00000000e+00,  -0.00000000e+00],
       [  0.00000000e+00,  -0.00000000e+00,   0.00000000e+00],
       [ -0.00000000e+00,   1.00000000e-05,  -1.00000000e-05]])
>>> np.round(np.abs(npSVD[2]) - np.abs(theSVD[2]), decimals=5)
array([[  0.00000000e+00,   0.00000000e+00,  -0.00000000e+00],
       [ -1.00000000e-05,  -1.00000000e-05,   1.00000000e-05],
       [  1.00000000e-05,   0.00000000e+00,  -1.00000000e-05]])
>>> np.round(np.abs(npSVD[1]) - np.abs(theSVD[0]), decimals=5)
array([ 0.,  0., -0.])

So there we have it. We added an extra little bit to the svd function, an argument k which stops computing the svd after it reaches rank k.

CNN stories

One interesting use of the SVD is in topic modeling. Topic modeling is the process of taking a bunch of documents (news stories, or emails, or movie scripts, whatever) and grouping them by topic, where the algorithm gets to choose what counts as a “topic.” Topic modeling is just the name that natural language processing folks use instead of clustering.

The SVD can help one model topics as follows. First you construct a matrix A called a document-term matrix whose rows correspond to words in some fixed dictionary and whose columns correspond to documents. The (i,j) entry of A contains the number of times word i shows up in document j. Or, more precisely, some quantity derived from that count, like a normalized count. See this table on wikipedia for a list of options related to that. We’ll just pick one arbitrarily for use in this post.

The point isn’t how we normalize the data, but what the SVD of A = U \Sigma V^T means in this context. Recall that the domain of A, as a linear map, is a vector space whose dimension is the number of stories. We think of the vectors in this space as documents, or rather as an “embedding” of the abstract concept of a document using the counts of how often each word shows up in a document as a proxy for the semantic meaning of the document. Likewise, the codomain is the space of all words, and each word is embedded by which documents it occurs in. If we compare this to the movie rating example, it’s the same thing: a movie is the vector of ratings it receives from people, and a person is the vector of ratings of various movies.

Say you take a rank 3 approximation to A. Then you get three singular vectors v_1, v_2, v_3 which form a basis for a subspace of words, i.e., the “idealized” words. These idealized words are your topics, and you can compute where a “new word” falls by looking at which documents it appears in (writing it as a vector in the domain) and saying its “topic” is the closest of the v_1, v_2, v_3. The same process applies to new documents. You can use this to cluster existing documents as well.

The dataset we’ll use for this post is a relatively small corpus of a thousand CNN stories picked from 2012. Here’s an excerpt from one of them

$ cat data/cnn-stories/story479.txt
3 things to watch on Super Tuesday
Here are three things to watch for: Romney's big day. He's been the off-and-on frontrunner throughout the race, but a big Super Tuesday could begin an end game toward a sometimes hesitant base coalescing behind former Massachusetts Gov. Mitt Romney. Romney should win his home state of Massachusetts, neighboring Vermont and Virginia, ...

So let’s first build this document-term matrix, with the normalized values, and then we’ll compute it’s SVD and see what the topics look like.

Step 1 is cleaning the data. We used a bunch of routines from the nltk library that boils down to this loop:

    for filename, documentText in documentDict.items():
        tokens = tokenize(documentText)
        tagged_tokens = pos_tag(tokens)
        wnl = WordNetLemmatizer()
        stemmedTokens = [wnl.lemmatize(word, wordnetPos(tag)).lower()
                         for word, tag in tagged_tokens]

This turns the Super Tuesday story into a list of words (with repetition):

["thing", "watch", "three", "thing", "watch", "big", ... ]

If you’ll notice the name Romney doesn’t show up in the list of words. I’m only keeping the words that show up in the top 100,000 most common English words, and then lemmatizing all of the words to their roots. It’s not a perfect data cleaning job, but it’s simple and good enough for our purposes.

Now we can create the document term matrix.

def makeDocumentTermMatrix(data):
    words = allWords(data)  # get the set of all unique words

    wordToIndex = dict((word, i) for i, word in enumerate(words))
    indexToWord = dict(enumerate(words))
    indexToDocument = dict(enumerate(data))

    matrix = np.zeros((len(words), len(data)))
    for docID, document in enumerate(data):
        docWords = Counter(document['words'])
        for word, count in docWords.items():
            matrix[wordToIndex[word], docID] = count

    return matrix, (indexToWord, indexToDocument)

This creates a matrix with the raw integer counts. But what we need is a normalized count. The idea is that a common word like “thing” shows up disproportionately more often than “election,” and we don’t want raw magnitude of a word count to outweigh its semantic contribution to the classification. This is the applied math part of the algorithm design. So what we’ll do (and this technique together with SVD is called latent semantic indexing) is normalize each entry so that it measures both the frequency of a term in a document and the relative frequency of a term compared to the global frequency of that term. There are many ways to do this, and we’ll just pick one. See the github repository if you’re interested.

So now lets compute a rank 10 decomposition and see how to cluster the results.

    data = load()
    matrix, (indexToWord, indexToDocument) = makeDocumentTermMatrix(data)
    matrix = normalize(matrix)
    sigma, U, V = svd(matrix, k=10)

This uses our svd, not numpy’s. Though numpy’s routine is much faster, it’s fun to see things work with code written from scratch. The result is too large to display here, but I can report the singular values.

>>> sigma
array([ 42.85249098,  21.85641975,  19.15989197,  16.2403354 ,
        15.40456779,  14.3172779 ,  13.47860033,  13.23795002,
        12.98866537,  12.51307445])

Now we take our original inputs and project them onto the subspace spanned by the singular vectors. This is the part that represents each word (resp., document) in terms of the idealized words (resp., documents), the singular vectors. Then we can apply a simple k-means clustering algorithm to the result, and observe the resulting clusters as documents.

    projectedDocuments = np.dot(matrix.T, U)
    projectedWords = np.dot(matrix, V.T)

    documentCenters, documentClustering = cluster(projectedDocuments)
    wordCenters, wordClustering = cluster(projectedWords)

    wordClusters = [
        [indexToWord[i] for (i, x) in enumerate(wordClustering) if x == j]
        for j in range(len(set(wordClustering)))
    ]

    documentClusters = [
        [indexToDocument[i]['text']
         for (i, x) in enumerate(documentClustering) if x == j]
        for j in range(len(set(documentClustering)))
    ]

And now we can inspect individual clusters. Right off the bat we can tell the clusters aren’t quite right simply by looking at the sizes of each cluster.

>>> Counter(wordClustering)
Counter({1: 9689, 2: 1051, 8: 680, 5: 557, 3: 321, 7: 225, 4: 174, 6: 124, 9: 123})
>>> Counter(documentClustering)
Counter({7: 407, 6: 109, 0: 102, 5: 87, 9: 85, 2: 65, 8: 55, 4: 47, 3: 23, 1: 15})

What looks wrong to me is the size of the largest word cluster. If we could group words by topic, then this is saying there’s a topic with over nine thousand words associated with it! Inspecting it even closer, it includes words like “vegan,” “skunk,” and “pope.” On the other hand, some word clusters are spot on. Examine, for example, the fifth cluster which includes words very clearly associated with crime stories.

>>> wordClusters[4]
['account', 'accuse', 'act', 'affiliate', 'allegation', 'allege', 'altercation', 'anything', 'apartment', 'arrest', 'arrive', 'assault', 'attorney', 'authority', 'bag', 'black', 'blood', 'boy', 'brother', 'bullet', 'candy', 'car', 'carry', 'case', 'charge', 'chief', 'child', 'claim', 'client', 'commit', 'community', 'contact', 'convenience', 'court', 'crime', 'criminal', 'cry', 'dead', 'deadly', 'death', 'defense', 'department', 'describe', 'detail', 'determine', 'dispatcher', 'district', 'document', 'enforcement', 'evidence', 'extremely', 'family', 'father', 'fear', 'fiancee', 'file', 'five', 'foot', 'friend', 'front', 'gate', 'girl', 'girlfriend', 'grand', 'ground', 'guilty', 'gun', 'gunman', 'gunshot', 'hand', 'happen', 'harm', 'head', 'hear', 'heard', 'hoodie', 'hour', 'house', 'identify', 'immediately', 'incident', 'information', 'injury', 'investigate', 'investigation', 'investigator', 'involve', 'judge', 'jury', 'justice', 'kid', 'killing', 'lawyer', 'legal', 'letter', 'life', 'local', 'man', 'men', 'mile', 'morning', 'mother', 'murder', 'near', 'nearby', 'neighbor', 'newspaper', 'night', 'nothing', 'office', 'officer', 'online', 'outside', 'parent', 'person', 'phone', 'police', 'post', 'prison', 'profile', 'prosecute', 'prosecution', 'prosecutor', 'pull', 'racial', 'racist', 'release', 'responsible', 'return', 'review', 'role', 'saw', 'scene', 'school', 'scream', 'search', 'sentence', 'serve', 'several', 'shoot', 'shooter', 'shooting', 'shot', 'slur', 'someone', 'son', 'sound', 'spark', 'speak', 'staff', 'stand', 'store', 'story', 'student', 'surveillance', 'suspect', 'suspicious', 'tape', 'teacher', 'teen', 'teenager', 'told', 'tragedy', 'trial', 'vehicle', 'victim', 'video', 'walk', 'watch', 'wear', 'whether', 'white', 'witness', 'young']

As sad as it makes me to see that ‘black’ and ‘slur’ and ‘racial’ appear in this category, it’s a reminder that naively using the output of a machine learning algorithm can perpetuate racism.

Here’s another interesting cluster corresponding to economic words:

>>> wordClusters[6]
['agreement', 'aide', 'analyst', 'approval', 'approve', 'austerity', 'average', 'bailout', 'beneficiary', 'benefit', 'bill', 'billion', 'break', 'broadband', 'budget', 'class', 'combine', 'committee', 'compromise', 'conference', 'congressional', 'contribution', 'core', 'cost', 'currently', 'cut', 'deal', 'debt', 'defender', 'deficit', 'doc', 'drop', 'economic', 'economy', 'employee', 'employer', 'erode', 'eurozone', 'expire', 'extend', 'extension', 'fee', 'finance', 'fiscal', 'fix', 'fully', 'fund', 'funding', 'game', 'generally', 'gleefully', 'growth', 'hamper', 'highlight', 'hike', 'hire', 'holiday', 'increase', 'indifferent', 'insistence', 'insurance', 'job', 'juncture', 'latter', 'legislation', 'loser', 'low', 'lower', 'majority', 'maximum', 'measure', 'middle', 'negotiation', 'offset', 'oppose', 'package', 'pass', 'patient', 'pay', 'payment', 'payroll', 'pension', 'plight', 'portray', 'priority', 'proposal', 'provision', 'rate', 'recession', 'recovery', 'reduce', 'reduction', 'reluctance', 'repercussion', 'rest', 'revenue', 'rich', 'roughly', 'sale', 'saving', 'scientist', 'separate', 'sharp', 'showdown', 'sign', 'specialist', 'spectrum', 'spending', 'strength', 'tax', 'tea', 'tentative', 'term', 'test', 'top', 'trillion', 'turnaround', 'unemployed', 'unemployment', 'union', 'wage', 'welfare', 'worker', 'worth']

One can also inspect the stories, though the clusters are harder to print out here. Interestingly the first cluster of documents are stories exclusively about Trayvon Martin. The second cluster is mostly international military conflicts. The third cluster also appears to be about international conflict, but what distinguishes it from the first cluster is that every story in the second cluster discusses Syria.

>>> len([x for x in documentClusters[1] if 'Syria' in x]) / len(documentClusters[1])
0.05555555555555555
>>> len([x for x in documentClusters[2] if 'Syria' in x]) / len(documentClusters[2])
1.0

Anyway, you can explore the data more at your leisure (and tinker with the parameters to improve it!).

Issues with the power method

Though I mentioned that the power method isn’t an industry strength algorithm I didn’t say why. Let’s revisit that before we finish. The problem is that the convergence rate of even the 1-dimensional problem depends on the ratio of the first and second singular values, \sigma_1 / \sigma_2. If that ratio is very close to 1, then the convergence will take a long time and need many many matrix-vector multiplications.

One way to alleviate that is to do the trick where, to compute a large power of a matrix, you iteratively square B. But that requires computing a matrix square (instead of a bunch of matrix-vector products), and that requires a lot of time and memory if the matrix isn’t sparse. When the matrix is sparse, you can actually do the power method quite quickly, from what I’ve heard and read.

But nevertheless, the industry standard methods involve computing a particular matrix decomposition that is not only faster than the power method, but also numerically stable. That means that the algorithm’s runtime and accuracy doesn’t depend on slight changes in the entries of the input matrix. Indeed, you can have two matrices where \sigma_1 / \sigma_2 is very close to 1, but changing a single entry will make that ratio much larger. The power method depends on this, so it’s not numerically stable. But the industry standard technique is not. This technique involves something called Householder reflections. So while the power method was great for a proof of concept, there’s much more work to do if you want true SVD power.

Until next time!

Big Dimensions, and What You Can Do About It

Data is abundant, data is big, and big is a problem. Let me start with an example. Let’s say you have a list of movie titles and you want to learn their genre: romance, action, drama, etc. And maybe in this scenario IMDB doesn’t exist so you can’t scrape the answer. Well, the title alone is almost never enough information. One nice way to get more data is to do the following:

  1. Pick a large dictionary of words, say the most common 100,000 non stop-words in the English language.
  2. Crawl the web looking for documents that include the title of a film.
  3. For each film, record the counts of all other words appearing in those documents.
  4. Maybe remove instances of “movie” or “film,” etc.

After this process you have a length-100,000 vector of integers associated with each movie title. IMDB’s database has around 1.5 million listed movies, and if we have a 32-bit integer per vector entry, that’s 600 GB of data to get every movie.

One way to try to find genres is to cluster this (unlabeled) dataset of vectors, and then manually inspect the clusters and assign genres. With a really fast computer we could simply run an existing clustering algorithm on this dataset and be done. Of course, clustering 600 GB of data takes a long time, but there’s another problem. The geometric intuition that we use to design clustering algorithms degrades as the length of the vectors in the dataset grows. As a result, our algorithms perform poorly. This phenomenon is called the “curse of dimensionality” (“curse” isn’t a technical term), and we’ll return to the mathematical curiosities shortly.

A possible workaround is to try to come up with faster algorithms or be more patient. But a more interesting mathematical question is the following:

Is it possible to condense high-dimensional data into smaller dimensions and retain the important geometric properties of the data?

This goal is called dimension reduction. Indeed, all of the chatter on the internet is bound to encode redundant information, so for our movie title vectors it seems the answer should be “yes.” But the questions remain, how does one find a low-dimensional condensification? (Condensification isn’t a word, the right word is embedding, but embedding is overloaded so we’ll wait until we define it) And what mathematical guarantees can you prove about the resulting condensed data? After all, it stands to reason that different techniques preserve different aspects of the data. Only math will tell.

In this post we’ll explore this so-called “curse” of dimensionality, explain the formality of why it’s seen as a curse, and implement a wonderfully simple technique called “the random projection method” which preserves pairwise distances between points after the reduction. As usual, and all the code, data, and tests used in the making of this post are on Github.

Some curious issues, and the “curse”

We start by exploring the curse of dimensionality with experiments on synthetic data.

In two dimensions, take a circle centered at the origin with radius 1 and its bounding square.

circle.png

The circle fills up most of the area in the square, in fact it takes up exactly \pi out of 4 which is about 78%. In three dimensions we have a sphere and a cube, and the ratio of sphere volume to cube volume is a bit smaller, 4 \pi /3 out of a total of 8, which is just over 52%. What about in a thousand dimensions? Let’s try by simulation.

import random

def randUnitCube(n):
   return [(random.random() - 0.5)*2 for _ in range(n)]

def sphereCubeRatio(n, numSamples):
   randomSample = [randUnitCube(n) for _ in range(numSamples)]
   return sum(1 for x in randomSample if sum(a**2 for a in x) <= 1) / numSamples 

The result is as we computed for small dimension,

 >>> sphereCubeRatio(2,10000)
0.7857
>>> sphereCubeRatio(3,10000)
0.5196

And much smaller for larger dimension

>>> sphereCubeRatio(20,100000) # 100k samples
0.0
>>> sphereCubeRatio(20,1000000) # 1M samples
0.0
>>> sphereCubeRatio(20,2000000)
5e-07

Forget a thousand dimensions, for even twenty dimensions, a million samples wasn’t enough to register a single random point inside the unit sphere. This illustrates one concern, that when we’re sampling random points in the d-dimensional unit cube, we need at least 2^d samples to ensure we’re getting a even distribution from the whole space. In high dimensions, this face basically rules out a naive Monte Carlo approximation, where you sample random points to estimate the probability of an event too complicated to sample from directly. A machine learning viewpoint of the same problem is that in dimension d, if your machine learning algorithm requires a representative sample of the input space in order to make a useful inference, then you require 2^d samples to learn.

Luckily, we can answer our original question because there is a known formula for the volume of a sphere in any dimension. Rather than give the closed form formula, which involves the gamma function and is incredibly hard to parse, we’ll state the recursive form. Call V_i the volume of the unit sphere in dimension i. Then V_0 = 1 by convention, V_1 = 2 (it’s an interval), and V_n = \frac{2 \pi V_{n-2}}{n}. If you unpack this recursion you can see that the numerator looks like (2\pi)^{n/2} and the denominator looks like a factorial, except it skips every other number. So an even dimension would look like 2 \cdot 4 \cdot \dots \cdot n, and this grows larger than a fixed exponential. So in fact the total volume of the sphere vanishes as the dimension grows! (In addition to the ratio vanishing!)

def sphereVolume(n):
   values = [0] * (n+1)
   for i in range(n+1):
      if i == 0:
         values[i] = 1
      elif i == 1:
         values[i] = 2
      else:
         values[i] = 2*math.pi / i * values[i-2]

   return values[-1]

This should be counterintuitive. I think most people would guess, when asked about how the volume of the unit sphere changes as the dimension grows, that it stays the same or gets bigger.  But at a hundred dimensions, the volume is already getting too small to fit in a float.

>>> sphereVolume(20)
0.025806891390014047
>>> sphereVolume(100)
2.3682021018828297e-40
>>> sphereVolume(1000)
0.0

The scary thing is not just that this value drops, but that it drops exponentially quickly. A consequence is that, if you’re trying to cluster data points by looking at points within a fixed distance r of one point, you have to carefully measure how big r needs to be to cover the same proportional volume as it would in low dimension.

Here’s a related issue. Say I take a bunch of points generated uniformly at random in the unit cube.

from itertools import combinations

def distancesRandomPoints(n, numSamples):
   randomSample = [randUnitCube(n) for _ in range(numSamples)]
   pairwiseDistances = [dist(x,y) for (x,y) in combinations(randomSample, 2)]
   return pairwiseDistances

In two dimensions, the histogram of distances between points looks like this

2d-distances.png

However, as the dimension grows the distribution of distances changes. It evolves like the following animation, in which each frame is an increase in dimension from 2 to 100.

distances-animation.gif

The shape of the distribution doesn’t appear to be changing all that much after the first few frames, but the center of the distribution tends to infinity (in fact, it grows like \sqrt{n}). The variance also appears to stay constant. This chart also becomes more variable as the dimension grows, again because we should be sampling exponentially many more points as the dimension grows (but we don’t). In other words, as the dimension grows the average distance grows and the tightness of the distribution stays the same. So at a thousand dimensions the average distance is about 26, tightly concentrated between 24 and 28. When the average is a thousand, the distribution is tight between 998 and 1002. If one were to normalize this data, it would appear that random points are all becoming equidistant from each other.

So in addition to the issues of runtime and sampling, the geometry of high-dimensional space looks different from what we expect. To get a better understanding of “big data,” we have to update our intuition from low-dimensional geometry with analysis and mathematical theorems that are much harder to visualize.

The Johnson-Lindenstrauss Lemma

Now we turn to proving dimension reduction is possible. There are a few methods one might first think of, such as look for suitable subsets of coordinates, or sums of subsets, but these would all appear to take a long time or they simply don’t work.

Instead, the key technique is to take a random linear subspace of a certain dimension, and project every data point onto that subspace. No searching required. The fact that this works is called the Johnson-Lindenstrauss Lemma. To set up some notation, we’ll call d(v,w) the usual distance between two points.

Lemma [Johnson-Lindenstrauss (1984)]: Given a set X of n points in \mathbb{R}^d, project the points in X to a randomly chosen subspace of dimension c. Call the projection \rho. For any \varepsilon > 0, if c is at least \Omega(\log(n) / \varepsilon^2), then with probability at least 1/2 the distances between points in X are preserved up to a factor of (1+\varepsilon). That is, with good probability every pair v,w \in X will satisfy

\displaystyle \| v-w \|^2 (1-\varepsilon) \leq \| \rho(v) - \rho(w) \|^2 \leq \| v-w \|^2 (1+\varepsilon)

Before we do the proof, which is quite short, it’s important to point out that the target dimension c does not depend on the original dimension! It only depends on the number of points in the dataset, and logarithmically so. That makes this lemma seem like pure magic, that you can take data in an arbitrarily high dimension and put it in a much smaller dimension.

On the other hand, if you include all of the hidden constants in the bound on the dimension, it’s not that impressive. If your data have a million dimensions and you want to preserve the distances up to 1% (\varepsilon = 0.01), the bound is bigger than a million! If you decrease the preservation \varepsilon to 10% (0.1), then you get down to about 12,000 dimensions, which is more reasonable. At 45% the bound drops to around 1,000 dimensions. Here’s a plot showing the theoretical bound on c in terms of \varepsilon for n fixed to a million.

boundplot

 

But keep in mind, this is just a theoretical bound for potentially misbehaving data. Later in this post we’ll see if the practical dimension can be reduced more than the theory allows. As we’ll see, an algorithm run on the projected data is still effective even if the projection goes well beyond the theoretical bound. Because the theorem is known to be tight in the worst case (see the notes at the end) this speaks more to the robustness of the typical algorithm than to the robustness of the projection method.

A second important note is that this technique does not necessarily avoid all the problems with the curse of dimensionality. We mentioned above that one potential problem is that “random points” are roughly equidistant in high dimensions. Johnson-Lindenstrauss actually preserves this problem because it preserves distances! As a consequence, you won’t see strictly better algorithm performance if you project (which we suggested is possible in the beginning of this post). But you will alleviate slow runtimes if the runtime depends exponentially on the dimension. Indeed, if you replace the dimension d with the logarithm of the number of points \log n, then 2^d becomes linear in n, and 2^{O(d)} becomes polynomial.

Proof of the J-L lemma

Let’s prove the lemma.

Proof. To start we make note that one can sample from the uniform distribution on dimension-c linear subspaces of \mathbb{R}^d by choosing the entries of a c \times d matrix A independently from a normal distribution with mean 0 and variance 1. Then, to project a vector x by this matrix (call the projection \rho), we can compute

\displaystyle \rho(x) = \frac{1}{\sqrt{c}}A x

Now fix \varepsilon > 0 and fix two points in the dataset x,y. We want an upper bound on the probability that the following is false

\displaystyle \| x-y \|^2 (1-\varepsilon) \leq \| \rho(x) - \rho(y) \|^2 \leq \| x-y \|^2 (1+\varepsilon)

Since that expression is a pain to work with, let’s rearrange it by calling u = x-y, and rearranging (using the linearity of the projection) to get the equivalent statement.

\left | \| \rho(u) \|^2 - \|u \|^2 \right | \leq \varepsilon \| u \|^2

And so we want a bound on the probability that this event does not occur, meaning the inequality switches directions.

Once we get such a bound (it will depend on c and \varepsilon) we need to ensure that this bound is true for every pair of points. The union bound allows us to do this, but it also requires that the probability of the bad thing happening tends to zero faster than 1/\binom{n}{2}. That’s where the \log(n) will come into the bound as stated in the theorem.

Continuing with our use of u for notation, define X to be the random variable \frac{c}{\| u \|^2} \| \rho(u) \|^2. By expanding the notation and using the linearity of expectation, you can show that the expected value of X is c, meaning that in expectation, distances are preserved. We are on the right track, and just need to show that the distribution of X, and thus the possible deviations in distances, is tightly concentrated around c. In full rigor, we will show

\displaystyle \Pr [X \geq (1+\varepsilon) c] < e^{-(\varepsilon^2 - \varepsilon^3) \frac{c}{4}}

Let A_i denote the i-th column of A. Define by X_i the quantity \langle A_i, u \rangle / \| u \|. This is a weighted average of the entries of A_i by the entries of u. But since we chose the entries of A from the normal distribution, and since a weighted average of normally distributed random variables is also normally distributed (has the same distribution), X_i is a N(0,1) random variable. Moreover, each column is independent. This allows us to decompose X as

X = \frac{k}{\| u \|^2} \| \rho(u) \|^2 = \frac{\| Au \|^2}{\| u \|^2}

Expanding further,

X = \sum_{i=1}^c \frac{\| A_i u \|^2}{\|u\|^2} = \sum_{i=1}^c X_i^2

Now the event X \leq (1+\varepsilon) c can be expressed in terms of the nonegative variable e^{\lambda X}, where 0 < \lambda < 1/2 is parameter, to get

\displaystyle \Pr[X \geq (1+\varepsilon) c] = \Pr[e^{\lambda X} \geq e^{(1+\varepsilon)c \lambda}]

This will become useful because the sum X = \sum_i X_i^2 will split into a product momentarily. First we apply Markov’s inequality, which says that for any nonnegative random variable Y, \Pr[Y \geq t] \leq \mathbb{E}[Y] / t. This lets us write

\displaystyle \Pr[e^{\lambda X} \geq e^{(1+\varepsilon) c \lambda}] \leq \frac{\mathbb{E}[e^{\lambda X}]}{e^{(1+\varepsilon) c \lambda}}

Now we can split up the exponent \lambda X into \sum_{i=1}^c \lambda X_i^2, and using the i.i.d.-ness of the X_i^2 we can rewrite the RHS of the inequality as

\left ( \frac{\mathbb{E}[e^{\lambda X_1^2}]}{e^{(1+\varepsilon)\lambda}} \right )^c

A similar statement using -\lambda is true for the (1-\varepsilon) part, namely that

\displaystyle \Pr[X \leq (1-\varepsilon)c] \leq \left ( \frac{\mathbb{E}[e^{-\lambda X_1^2}]}{e^{-(1-\varepsilon)\lambda}} \right )^c

The last thing that’s needed is to bound \mathbb{E}[e^{\lambda X_i^2}], but since X_i^2 \sim N(0,1), we can use the known density function for a normal distribution, and integrate to get the exact value \mathbb{E}[e^{\lambda X_1^2}] = \frac{1}{\sqrt{1-2\lambda}}. Including this in the bound gives us a closed-form bound in terms of \lambda, c, \varepsilon. Using standard calculus the optimal \lambda \in (0,1/2) is \lambda = \varepsilon / 2(1+\varepsilon). This gives

\displaystyle \Pr[X \geq (1+\varepsilon) c] \leq ((1+\varepsilon)e^{-\varepsilon})^{c/2}

Using the Taylor series expansion for e^x, one can show the bound 1+\varepsilon < e^{\varepsilon - (\varepsilon^2 - \varepsilon^3)/2}, which simplifies the final upper bound to e^{-(\varepsilon^2 - \varepsilon^3) c/4}.

Doing the same thing for the (1-\varepsilon) version gives an equivalent bound, and so the total bound is doubled, i.e. 2e^{-(\varepsilon^2 - \varepsilon^3) c/4}.

As we said at the beginning, applying the union bound means we need

\displaystyle 2e^{-(\varepsilon^2 - \varepsilon^3) c/4} < \frac{1}{\binom{n}{2}}

Solving this for c gives c \geq \frac{8 \log m}{\varepsilon^2 - \varepsilon^3}, as desired.

\square

Projecting in Practice

Let’s write a python program to actually perform the Johnson-Lindenstrauss dimension reduction scheme. This is sometimes called the Johnson-Lindenstrauss transform, or JLT.

First we define a random subspace by sampling an appropriately-sized matrix with normally distributed entries, and a function that performs the projection onto a given subspace (for testing).

import random
import math
import numpy

def randomSubspace(subspaceDimension, ambientDimension):
   return numpy.random.normal(0, 1, size=(subspaceDimension, ambientDimension))

def project(v, subspace):
   subspaceDimension = len(subspace)
   return (1 / math.sqrt(subspaceDimension)) * subspace.dot(v)

We have a function that computes the theoretical bound on the optimal dimension to reduce to.

def theoreticalBound(n, epsilon):
   return math.ceil(8*math.log(n) / (epsilon**2 - epsilon**3))

And then performing the JLT is simply matrix multiplication

def jlt(data, subspaceDimension):
   ambientDimension = len(data[0])
   A = randomSubspace(subspaceDimension, ambientDimension)
   return (1 / math.sqrt(subspaceDimension)) * A.dot(data.T).T

The high-dimensional dataset we’ll use comes from a data mining competition called KDD Cup 2001. The dataset we used deals with drug design, and the goal is to determine whether an organic compound binds to something called thrombin. Thrombin has something to do with blood clotting, and I won’t pretend I’m an expert. The dataset, however, has over a hundred thousand features for about 2,000 compounds. Here are a few approximate target dimensions we can hope for as epsilon varies.

>>> [((1/x),theoreticalBound(n=2000, epsilon=1/x))
       for x in [2, 3, 4, 5, 6, 7, 8, 9, 10, 15, 20]]
[('0.50', 487), ('0.33', 821), ('0.25', 1298), ('0.20', 1901),
 ('0.17', 2627), ('0.14', 3477), ('0.12', 4448), ('0.11', 5542),
 ('0.10', 6757), ('0.07', 14659), ('0.05', 25604)]

Going down from a hundred thousand dimensions to a few thousand is by any measure decreases the size of the dataset by about 95%. We can also observe how the distribution of overall distances varies as the size of the subspace we project to varies.

The animation proceeds from 5000 dimensions down to 2 (when the plot is at its bulkiest closer to zero).

The animation proceeds from 5000 dimensions down to 2 (when the plot is at its bulkiest closer to zero).

The last three frames are for 10, 5, and 2 dimensions respectively. As you can see the histogram starts to beef up around zero. To be honest I was expecting something a bit more dramatic like a uniform-ish distribution. Of course, the distribution of distances is not all that matters. Another concern is the worst case change in distances between any two points before and after the projection. We can see that indeed when we project to the dimension specified in the theorem, that the distances are within the prescribed bounds.

def checkTheorem(oldData, newData, epsilon):
   numBadPoints = 0

   for (x,y), (x2,y2) in zip(combinations(oldData, 2), combinations(newData, 2)):
      oldNorm = numpy.linalg.norm(x2-y2)**2
      newNorm = numpy.linalg.norm(x-y)**2

      if newNorm == 0 or oldNorm == 0:
         continue

      if abs(oldNorm / newNorm - 1) &gt; epsilon:
         numBadPoints += 1

   return numBadPoints

if __name__ == &quot;__main__&quot;
   from data import thrombin
   train, labels = thrombin.load() 

   numPoints = len(train)
   epsilon = 0.2
   subspaceDim = theoreticalBound(numPoints, epsilon)
   ambientDim = len(train[0])
   newData = jlt(train, subspaceDim)

   print(checkTheorem(train, newData, epsilon))

This program prints zero every time I try running it, which is the poor man’s way of saying it works “with high probability.” We can also plot statistics about the number of pairs of data points that are distorted by more than \varepsilon as the subspace dimension shrinks. We ran this on the following set of subspace dimensions with \varepsilon = 0.1 and took average/standard deviation over twenty trials:

   dims = [1000, 750, 500, 250, 100, 75, 50, 25, 10, 5, 2]

The result is the following chart, whose x-axis is the dimension projected to (so the left hand is the most extreme projection to 2, 5, 10 dimensions), the y-axis is the number of distorted pairs, and the error bars represent a single standard deviation away from the mean.

thrombin-worst-case

This chart provides good news about this dataset because the standard deviations are low. It tells us something that mathematicians often ignore: the predictability of the tradeoff that occurs once you go past the theoretically perfect bound. In this case, the standard deviations tell us that it’s highly predictable. Moreover, since this tradeoff curve measures pairs of points, we might conjecture that the distortion is localized around a single set of points that got significantly “rattled” by the projection. This would be an interesting exercise to explore.

Now all of these charts are really playing with the JLT and confirming the correctness of our code (and hopefully our intuition). The real question is: how well does a machine learning algorithm perform on the original data when compared to the projected data? If the algorithm only “depends” on the pairwise distances between the points, then we should expect nearly identical accuracy in the unprojected and projected versions of the data. To show this we’ll use an easy learning algorithm, the k-nearest-neighbors clustering method. The problem, however, is that there are very few positive examples in this particular dataset. So looking for the majority label of the nearest k neighbors for any k > 2 unilaterally results in the “all negative” classifier, which has 97% accuracy. This happens before and after projecting.

To compensate for this, we modify k-nearest-neighbors slightly by having the label of a predicted point be 1 if any label among its nearest neighbors is 1. So it’s not a majority vote, but rather a logical OR of the labels of nearby neighbors. Our point in this post is not to solve the problem well, but rather to show how an algorithm (even a not-so-good one) can degrade as one projects the data into smaller and smaller dimensions. Here is the code.

def nearestNeighborsAccuracy(data, labels, k=10):
   from sklearn.neighbors import NearestNeighbors
   trainData, trainLabels, testData, testLabels = randomSplit(data, labels) # cross validation
   model = NearestNeighbors(n_neighbors=k).fit(trainData)
   distances, indices = model.kneighbors(testData)
   predictedLabels = []

   for x in indices:
      xLabels = [trainLabels[i] for i in x[1:]]
      predictedLabel = max(xLabels)
      predictedLabels.append(predictedLabel)

   totalAccuracy = sum(x == y for (x,y) in zip(testLabels, predictedLabels)) / len(testLabels)
   falsePositive = (sum(x == 0 and y == 1 for (x,y) in zip(testLabels, predictedLabels)) /
      sum(x == 0 for x in testLabels))
   falseNegative = (sum(x == 1 and y == 0 for (x,y) in zip(testLabels, predictedLabels)) /
      sum(x == 1 for x in testLabels))

   return totalAccuracy, falsePositive, falseNegative

And here is the accuracy of this modified k-nearest-neighbors algorithm run on the thrombin dataset. The horizontal line represents the accuracy of the produced classifier on the unmodified data set. The x-axis represents the dimension projected to (left-hand side is the lowest), and the y-axis represents the accuracy. The mean accuracy over fifty trials was plotted, with error bars representing one standard deviation. The complete code to reproduce the plot is in the Github repository.

thrombin-knn-accuracy

Likewise, we plot the proportion of false positive and false negatives for the output classifier. Note that a “positive” label made up only about 2% of the total data set. First the false positives

thrombin-knn-fp

Then the false negatives

thrombin-knn-fn

As we can see from these three charts, things don’t really change that much (for this dataset) even when we project down to around 200-300 dimensions. Note that for these parameters the “correct” theoretical choice for dimension was on the order of 5,000 dimensions, so this is a 95% savings from the naive approach, and 99.75% space savings from the original data. Not too shabby.

Notes

The \Omega(\log(n)) worst-case dimension bound is asymptotically tight, though there is some small gap in the literature that depends on \varepsilon. This result is due to Noga Alon, the very last result (Section 9) of this paper.

We did dimension reduction with respect to preserving the Euclidean distance between points. One might naturally wonder if you can achieve the same dimension reduction with a different metric, say the taxicab metric or a p-norm. In fact, you cannot achieve anything close to logarithmic dimension reduction for the taxicab (l_1) metric. This result is due to Brinkman-Charikar in 2004.

The code we used to compute the JLT is not particularly efficient. There are much more efficient methods. One of them, borrowing its namesake from the Fast Fourier Transform, is called the Fast Johnson-Lindenstrauss Transform. The technique is due to Ailon-Chazelle from 2009, and it involves something called “preconditioning a sparse projection matrix with a randomized Fourier transform.” I don’t know precisely what that means, but it would be neat to dive into that in a future post.

The central focus in this post was whether the JLT preserves distances between points, but one might be curious as to whether the points themselves are well approximated. The answer is an enthusiastic no. If the data were images, the projected points would look nothing like the original images. However, it appears the degradation tradeoff is measurable (by some accounts perhaps linear), and there appears to be some work (also this by the same author) when restricting to sparse vectors (like word-association vectors).

Note that the JLT is not the only method for dimensionality reduction. We previously saw principal component analysis (applied to face recognition), and in the future we will cover a related technique called the Singular Value Decomposition. It is worth noting that another common technique specific to nearest-neighbor is called “locality-sensitive hashing.” Here the goal is to project the points in such a way that “similar” points land very close to each other. Say, if you were to discretize the plane into bins, these bins would form the hash values and you’d want to maximize the probability that two points with the same label land in the same bin. Then you can do things like nearest-neighbors by comparing bins.

Another interesting note, if your data is linearly separable (like the examples we saw in our age-old post on Perceptrons), then you can use the JLT to make finding a linear separator easier. First project the data onto the dimension given in the theorem. With high probability the points will still be linearly separable. And then you can use a perceptron-type algorithm in the smaller dimension. If you want to find out which side a new point is on, you project and compare with the separator in the smaller dimension.

Beyond its interest for practical dimensionality reduction, the JLT has had many other interesting theoretical consequences. More generally, the idea of “randomly projecting” your data onto some small dimensional space has allowed mathematicians to get some of the best-known results on many optimization and learning problems, perhaps the most famous of which is called MAX-CUT; the result is by Goemans-Williamson and it led to a mathematical constant being named after them, \alpha_{GW} =.878567 \dots. If you’re interested in more about the theory, Santosh Vempala wrote a wonderful (and short!) treatise dedicated to this topic.

randomprojectionbook