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The Blum-Blum-Shub Pseudorandom Generator

Problem: Design a random number generator that is computationally indistinguishable from a truly random number generator.

Solution (in Python): note this solution uses the Miller-Rabin primality tester, though any primality test will do. See the github repository for the referenced implementation.

from randomized.primality import probablyPrime
import random

def goodPrime(p):
    return p % 4 == 3 and probablyPrime(p, accuracy=100)

def findGoodPrime(numBits=512):
    candidate = 1
    while not goodPrime(candidate):
        candidate = random.getrandbits(numBits)
    return candidate

def makeModulus():
    return findGoodPrime() * findGoodPrime()

def parity(n):
    return sum(int(x) for x in bin(n)[2:]) % 2

class BlumBlumShub(object):
    def __init__(self, seed=None):
        self.modulus = makeModulus()
        self.state = seed if seed is not None else random.randint(2, self.modulus - 1)
        self.state = self.state % self.modulus

    def seed(self, seed):
        self.state = seed

    def bitstream(self):
        while True:
            yield parity(self.state)
            self.state = pow(self.state, 2, self.modulus)

    def bits(self, n=20):
        outputBits = ''
        for bit in self.bitstream():
            outputBits += str(bit)
            if len(outputBits) == n:
                break

        return outputBits

Discussion:

An integer x is called a quadratic residue of another integer N if it can be written as x = a^2 \mod N for some a. That is, if it’s the remainder when dividing a perfect square by N. Some numbers, like N=8, have very special patterns in their quadratic residues, only 0, 1, and 4 can occur as quadratic residues.

The core idea behind this random number generator is that, for a specially chosen modulus N, telling whether a number x is a quadratic residue mod N is hard. In fact, one can directly convert an algorithm that can predict the next bit of this random number generator (by even a slight edge) into an arbitrarily accurate quadratic-residue-decider. So if computing quadratic residues is even mildly hard, then predicting the next bit in this random number generator is very hard.

More specifically, the conjectured guarantee about this random number generator is the following: if you present a polynomial time adversary with two sequences:

  1. A truly random sequence of bits of length k,
  2. k bits from the output of the pseudorandom generator when seeded with a starting state shorter than k bits.

Then the adversary can’t distinguish between the two sequences with probability “significantly” more than 1/2, where by “significantly” I mean 1/k^c for any c>0 (i.e., the edge over randomness vanishes faster than any inverse polynomial). It turns out, due to a theorem of Yao, that this is equivalent to not being able to guess the next bit in a pseudorandom sequence with a significant edge over a random guess, even when given the previous \log(N)^{10} bits in the sequence (or any \textup{poly}(\log N) bits in the sequence).

This emphasizes a deep philosophical viewpoint in theoretical computer science, that whether some object has a property (randomness) really only depends on the power of a computationally limited observer to identify that property. If nobody can tell the difference between fake randomness and real randomness, then the fake randomness is random. Offhand I wonder whether you can meaningfully apply this view to less mathematical concepts like happiness and status.

Anyway, the modulus N is chosen in such a way that every quadratic residue of N has a unique square root which is also a quadratic residue. This makes the squaring function a bijection on quadratic residues. In other words, with a suitably chosen N, there’s no chance that we’ll end up with N=8 where there are very few quadratic residues and the numbers output by the Blum-Blum-Shub generator have a short cycle. Moreover, the assumption that detecting quadratic residues mod N is hard makes the squaring function a one-way permutation.

Here’s an example of how this generator might be used:

generator = BlumBlumShub()

hist = [0] * 2**6
for i in range(10000):
    value = int(generator.bits(6), 2)
    hist[value] += 1

print(hist)

This produces random integers between 0 and 64, with the following histogram:

bbs-hist

See these notes of Junod for a detailed exposition of the number theory behind this random number generator, with full definitions and proofs.

Zero Knowledge Proofs — A Primer

In this post we’ll get a strong taste for zero knowledge proofs by exploring the graph isomorphism problem in detail. In the next post, we’ll see how this relates to cryptography and the bigger picture. The goal of this post is to get a strong understanding of the terms “prover,” “verifier,” and “simulator,” and “zero knowledge” in the context of a specific zero-knowledge proof. Then next time we’ll see how the same concepts (though not the same proof) generalizes to a cryptographically interesting setting.

Graph isomorphism

Let’s start with an extended example. We are given two graphs G_1, G_2, and we’d like to know whether they’re isomorphic, meaning they’re the same graph, but “drawn” different ways.

The problem of telling if two graphs are isomorphic seems hard. The pictures above, which are all different drawings of the same graph (or are they?), should give you pause if you thought it was easy.

To add a tiny bit of formalism, a graph G is a list of edges, and each edge (u,v) is a pair of integers between 1 and the total number of vertices of the graph, say n. Using this representation, an isomorphism between G_1 and G_2 is a permutation \pi of the numbers \{1, 2, \dots, n \} with the property that (i,j) is an edge in G_1 if and only if (\pi(i), \pi(j)) is an edge of G_2. You swap around the labels on the vertices, and that’s how you get from one graph to another isomorphic one.

Given two arbitrary graphs as input on a large number of vertices n, nobody knows of an efficient—i.e., polynomial time in n—algorithm that can always decide whether the input graphs are isomorphic. Even if you promise me that the inputs are isomorphic, nobody knows of an algorithm that could construct an isomorphism. (If you think about it, such an algorithm could be used to solve the decision problem!)

A game

Now let’s play a game. In this game, we’re given two enormous graphs on a billion nodes. I claim they’re isomorphic, and I want to prove it to you. However, my life’s fortune is locked behind these particular graphs (somehow), and if you actually had an isomorphism between these two graphs you could use it to steal all my money. But I still want to convince you that I do, in fact, own all of this money, because we’re about to start a business and you need to know I’m not broke.

Is there a way for me to convince you beyond a reasonable doubt that these two graphs are indeed isomorphic? And moreover, could I do so without you gaining access to my secret isomorphism? It would be even better if I could guarantee you learn nothing about my isomorphism or any isomorphism, because even the slightest chance that you can steal my money is out of the question.

Zero knowledge proofs have exactly those properties, and here’s a zero knowledge proof for graph isomorphism. For the record, G_1 and G_2 are public knowledge, (common inputs to our protocol for the sake of tracking runtime), and the protocol itself is common knowledge. However, I have an isomorphism f: G_1 \to G_2 that you don’t know.

Step 1: I will start by picking one of my two graphs, say G_1, mixing up the vertices, and sending you the resulting graph. In other words, I send you a graph H which is chosen uniformly at random from all isomorphic copies of G_1. I will save the permutation \pi that I used to generate H for later use.

Step 2: You receive a graph H which you save for later, and then you randomly pick an integer t which is either 1 or 2, with equal probability on each. The number t corresponds to your challenge for me to prove H is isomorphic to G_1 or G_2. You send me back t, with the expectation that I will provide you with an isomorphism between H and G_t.

Step 3: Indeed, I faithfully provide you such an isomorphism. If I you send me t=1, I’ll give you back \pi^{-1} : H \to G_1, and otherwise I’ll give you back f \circ \pi^{-1}: H \to G_2. Because composing a fixed permutation with a uniformly random permutation is again a uniformly random permutation, in either case I’m sending you a uniformly random permutation.

Step 4: You receive a permutation g, and you can use it to verify that H is isomorphic to G_t. If the permutation I sent you doesn’t work, you’ll reject my claim, and if it does, you’ll accept my claim.

Before we analyze, here’s some Python code that implements the above scheme. You can find the full, working example in a repository on this blog’s Github page.

First, a few helper functions for generating random permutations (and turning their list-of-zero-based-indices form into a function-of-positive-integers form)

import random

def randomPermutation(n):
    L = list(range(n))
    random.shuffle(L)
    return L

def makePermutationFunction(L):
    return lambda i: L[i - 1] + 1

def makeInversePermutationFunction(L):
    return lambda i: 1 + L.index(i - 1)

def applyIsomorphism(G, f):
    return [(f(i), f(j)) for (i, j) in G]

Here’s a class for the Prover, the one who knows the isomorphism and wants to prove it while keeping the isomorphism secret:

class Prover(object):
    def __init__(self, G1, G2, isomorphism):
        '''
            isomomorphism is a list of integers representing
            an isomoprhism from G1 to G2.
        '''
        self.G1 = G1
        self.G2 = G2
        self.n = numVertices(G1)
        assert self.n == numVertices(G2)

        self.isomorphism = isomorphism
        self.state = None

    def sendIsomorphicCopy(self):
        isomorphism = randomPermutation(self.n)
        pi = makePermutationFunction(isomorphism)

        H = applyIsomorphism(self.G1, pi)

        self.state = isomorphism
        return H

    def proveIsomorphicTo(self, graphChoice):
        randomIsomorphism = self.state
        piInverse = makeInversePermutationFunction(randomIsomorphism)

        if graphChoice == 1:
            return piInverse
        else:
            f = makePermutationFunction(self.isomorphism)
            return lambda i: f(piInverse(i))

The prover has two methods, one for each round of the protocol. The first creates an isomorphic copy of G_1, and the second receives the challenge and produces the requested isomorphism.

And here’s the corresponding class for the verifier

class Verifier(object):
    def __init__(self, G1, G2):
        self.G1 = G1
        self.G2 = G2
        self.n = numVertices(G1)
        assert self.n == numVertices(G2)

    def chooseGraph(self, H):
        choice = random.choice([1, 2])
        self.state = H, choice
        return choice

    def accepts(self, isomorphism):
        '''
            Return True if and only if the given isomorphism
            is a valid isomorphism between the randomly
            chosen graph in the first step, and the H presented
            by the Prover.
        '''
        H, choice = self.state
        graphToCheck = [self.G1, self.G2][choice - 1]
        f = isomorphism

        isValidIsomorphism = (graphToCheck == applyIsomorphism(H, f))
        return isValidIsomorphism

Then the protocol is as follows:

def runProtocol(G1, G2, isomorphism):
    p = Prover(G1, G2, isomorphism)
    v = Verifier(G1, G2)

    H = p.sendIsomorphicCopy()
    choice = v.chooseGraph(H)
    witnessIsomorphism = p.proveIsomorphicTo(choice)

    return v.accepts(witnessIsomorphism)

Analysis: Let’s suppose for a moment that everyone is honestly following the rules, and that G_1, G_2 are truly isomorphic. Then you’ll always accept my claim, because I can always provide you with an isomorphism. Now let’s suppose that, actually I’m lying, the two graphs aren’t isomorphic, and I’m trying to fool you into thinking they are. What’s the probability that you’ll rightfully reject my claim?

Well, regardless of what I do, I’m sending you a graph H and you get to make a random choice of t = 1, 2 that I can’t control. If H is only actually isomorphic to either G_1 or G_2 but not both, then so long as you make your choice uniformly at random, half of the time I won’t be able to produce a valid isomorphism and you’ll reject. And unless you can actually tell which graph H is isomorphic to—an open problem, but let’s say you can’t—then probability 1/2 is the best you can do.

Maybe the probability 1/2 is a bit unsatisfying, but remember that we can amplify this probability by repeating the protocol over and over again. So if you want to be sure I didn’t cheat and get lucky to within a probability of one-in-one-trillion, you only need to repeat the protocol 30 times. To be surer than the chance of picking a specific atom at random from all atoms in the universe, only about 400 times.

If you want to feel small, think of the number of atoms in the universe. If you want to feel big, think of its logarithm.

Here’s the code that repeats the protocol for assurance.

def convinceBeyondDoubt(G1, G2, isomorphism, errorTolerance=1e-20):
    probabilityFooled = 1

    while probabilityFooled > errorTolerance:
        result = runProtocol(G1, G2, isomorphism)
        assert result
        probabilityFooled *= 0.5
        print(probabilityFooled)

Running it, we see it succeeds

$ python graph-isomorphism.py
0.5
0.25
0.125
0.0625
0.03125
 ...
<SNIP>
 ...
1.3552527156068805e-20
6.776263578034403e-21

So it’s clear that this protocol is convincing.

But how can we be sure that there’s no leakage of knowledge in the protocol? What does “leakage” even mean? That’s where this topic is the most difficult to nail down rigorously, in part because there are at least three a priori different definitions! The idea we want to capture is that anything that you can efficiently compute after the protocol finishes (i.e., you have the content of the messages sent to you by the prover) you could have computed efficiently given only the two graphs G_1, G_2, and the claim that they are isomorphic.

Another way to say it is that you may go through the verification process and feel happy and confident that the two graphs are isomorphic. But because it’s a zero-knowledge proof, you can’t do anything with that information more than you could have done if you just took the assertion on blind faith. I’m confident there’s a joke about religion lurking here somewhere, but I’ll just trust it’s funny and move on.

In the next post we’ll expand on this “leakage” notion, but before we get there it should be clear that the graph isomorphism protocol will have the strongest possible “no-leakage” property we can come up with. Indeed, in the first round the prover sends a uniform random isomorphic copy of G_1 to the verifier, but the verifier can compute such an isomorphism already without the help of the prover. The verifier can’t necessarily find the isomorphism that the prover used in retrospect, because the verifier can’t solve graph isomorphism. Instead, the point is that the probability space of “G_1 paired with an H made by the prover” and the probability space of “G_1 paired with H as made by the verifier” are equal. No information was leaked by the prover.

For the second round, again the permutation \pi used by the prover to generate H is uniformly random. Since composing a fixed permutation with a uniform random permutation also results in a uniform random permutation, the second message sent by the prover is uniformly random, and so again the verifier could have constructed a similarly random permutation alone.

Let’s make this explicit with a small program. We have the honest protocol from before, but now I’m returning the set of messages sent by the prover, which the verifier can use for additional computation.

def messagesFromProtocol(G1, G2, isomorphism):
    p = Prover(G1, G2, isomorphism)
    v = Verifier(G1, G2)

    H = p.sendIsomorphicCopy()
    choice = v.chooseGraph(H)
    witnessIsomorphism = p.proveIsomorphicTo(choice)

    return [H, choice, witnessIsomorphism]

To say that the protocol is zero-knowledge (again, this is still colloquial) is to say that anything that the verifier could compute, given as input the return value of this function along with G_1, G_2 and the claim that they’re isomorphic, the verifier could also compute given only G_1, G_2 and the claim that G_1, G_2 are isomorphic.

It’s easy to prove this, and we’ll do so with a python function called simulateProtocol.

def simulateProtocol(G1, G2):
    # Construct data drawn from the same distribution as what is
    # returned by messagesFromProtocol
    choice = random.choice([1, 2])
    G = [G1, G2][choice - 1]
    n = numVertices(G)

    isomorphism = randomPermutation(n)
    pi = makePermutationFunction(isomorphism)
    H = applyIsomorphism(G, pi)

    return H, choice, pi

The claim is that the distribution of outputs to messagesFromProtocol and simulateProtocol are equal. But simulateProtocol will work regardless of whether G_1, G_2 are isomorphic. Of course, it’s not convincing to the verifier because the simulating function made the choices in the wrong order, choosing the graph index before making H. But the distribution that results is the same either way.

So if you were to use the actual Prover/Verifier protocol outputs as input to another algorithm (say, one which tries to compute an isomorphism of G_1 \to G_2), you might as well use the output of your simulator instead. You’d have no information beyond hard-coding the assumption that G_1, G_2 are isomorphic into your program. Which, as I mentioned earlier, is no help at all.

In this post we covered one detailed example of a zero-knowledge proof. Next time we’ll broaden our view and see the more general power of zero-knowledge (that it captures all of NP), and see some specific cryptographic applications. Keep in mind the preceding discussion, because we’re going to re-use the terms “prover,” “verifier,” and “simulator” to mean roughly the same things as the classes Prover, Verifier and the function simulateProtocol.

Until then!

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 column space of A, i.e. a basis of the subspace spanned by the data contained in the columns 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!

Book mailing list

I’m launching an email list for my book-in-progress, which is tentatively titled, “A Programmer’s Introduction to Mathematics.”

Sign up form

This will probably end up being a monthly or once-every-two-months newsletter with my progress updates. I’ll probably also use this email list to send out sneak peeks of certain chapters and ask for feedback. A lot about the book is still up in the air, but now that the excitement of my PhD defense has passed, I can focus deeply on the book.

A short description of the intended audience (from the comments on this post)

My target audience is a programmer who has gotten through the standard CS education, but never felt they deeply understood math. Think of it as a step up from Kalid’s Better Explained blog (I won’t explain the basics of exponents or trigonometry), but a big step below the hard posts on this blog. I’ll emphasize connections and analogies between math and programming. Think of it as a coherent, linear version of the introductory posts on this blog with cool and nontrivial applications. I will also focus on building the skills and mindset that allows one to pick up other math books and continue learning.

More than anything, I want to keep this blog free of too much non-math, so this will be my last plug for the book until the day it’s released.

book-progress-2016-04-21

A mock cover. Look closely and you’ll see it’s got 95 pages so far.