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!

Concrete Examples of Quantum Gates

So far in this series we’ve seen a lot of motivation and defined basic ideas of what a quantum circuit is. But on rereading my posts, I think we would all benefit from some concreteness.

“Local” operations

So by now we’ve understood that quantum circuits consist of a sequence of gates A_1, \dots, A_k, where each A_i is an 8-by-8 matrix that operates “locally” on some choice of three (or fewer) qubits. And in your head you imagine starting with some state vector v and applying each A_i locally to its three qubits until the end when you measure the state and get some classical output.

But the point I want to make is that A_i actually changes the whole state vector v, because the three qubits it acts “locally” on are part of the entire basis. Here’s an example. Suppose we have three qubits and they’re in the state

\displaystyle v = \frac{1}{\sqrt{14}} (e_{001} + 2e_{011} - 3e_{101})

Recall we abbreviate basis states by subscripting them by binary strings, so e_{011} = e_0 \otimes e_1 \otimes e_1, and a valid state is any unit vector over the 2^3 = 8 possible basis elements. As a vector, this state is \frac{1}{\sqrt{14}} (0,1,0,2,0,-3,0,0)

Say we apply the gate A that swaps the first and third qubits. “Locally” this gate has the following matrix:

\displaystyle \begin{pmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{pmatrix}

where we index the rows and columns by the relevant strings in lexicographic order: 00, 01, 10, 11. So this operation leaves e_{00} and e_{11} the same while swapping the other two. However, as an operation on three qubits the operation looks quite different. And it’s sort of hard to describe a general way to write it down as a matrix because of the choice of indices. There are three different perspectives.

Perspective 1: if the qubits being operated on are sequential (like, the third, fourth, and fifth qubits), then we can write the matrix as I_{2^{a}} \otimes A \otimes I_{2^{b}} where a tensor product of matrices is the Kronecker product and a + b + \log \textup{dim}(A) = n (the number of qubits adds up). Then the final operation looks like a “tiled product” of identity matrices by A, but it’s a pain to write out. Let me hurt my self for your sake, dear reader.

tensormatrix1

And each copy of A \otimes I_{2^b} looks like

tiled-piece

 

That’s a mess, but if you write it out for our example of swapping the first and third qubits of a three-qubit register you get the following:

example-3swap2

And this makes sense: the gate changes any entry of the state vector that has values for the first and third qubit that are different. This is what happens to our state:

\displaystyle v = \frac{1}{\sqrt{14}} (0,1,0,2,0,-3,0,0) \mapsto \frac{1}{\sqrt{14}} (0,0,0,0,1,-3,2,0)

Perspective 2: just assume every operation works on the first three qubits, and wrap each operation A in between an operation that swaps the first three qubits with the desired three. So like BAB for B a swap operation. Then the matrix form looks a bit simpler, and it just means we permute the columns of the matrix form we gave above so that it just has the form A \otimes I_a. This allows one to retain a shred of sanity when trying to envision the matrix for an operation that acts on three qubits that are not sequential. The downside is that to actually use this perspective in an analysis you have to carry around the extra baggage of these permutation matrices. So one might use this as a simplifying assumption (a “without loss of generality” statement).

Perspective 3: ignore matrices and write things down in a summation form. So if \sigma is the permutation that swaps 1 and 3 and leaves the other indices unchanged, we can write the general operation on a state v = \sum_{x \in \{ 0,1 \}^n } a_x e_{x} as Av = \sum_{x \in \{ 0,1 \}^n} a_x e_{\sigma(x)}.

The third option is probably the nicest way to do things, but it’s important to keep the matrix view in mind for many reasons. Just one quick reason: “errors” in quantum gates (that are meant to approximately compute something) compound linearly in the number of gates because the operations are linear. This is a key reason that allows one to design quantum analogues of error correcting codes.

So we’ve established that the basic (atomic) quantum gates are “local” in the sense that they operate on a fixed number of qubits, but they are not local in the sense that they can screw up the entire state vector.

A side note on the meaning of “local”

When I was chugging through learning this stuff (and I still have far to go), I wanted to come up with an alternate characterization of the word “local” so that I would feel better about using the word “local.” Mathematicians are as passionate about word choice as programmers are about text editors. In particular, for a long time I was ignorantly convinced that quantum gates that act on a small number of qubits don’t affect the marginal distribution of measurement outcomes for other qubits. That is, I thought that if A acts on qubits 1,2,3, then Av and v have the same probability of a measurement producing a 1 in index 4, 5, etc, conditioned on fixing a measurement outcome for qubits 1,2,3. In notation, if x is a random variable whose values are binary strings and v is a state vector, I’ll call x \sim v the random process of measuring a state vector v and getting a string x, then my claim was that the following was true for every b_1, b_2, b_3 \in \{0,1\} and every 1 \leq i \leq n:

\displaystyle \begin{aligned}\Pr_{x \sim v}&[x_i = 1 \mid x_1 = b_1, x_2 = b_2, x_3 = b_3] = \\ \Pr_{x \sim Av}&[x_i = 1 \mid x_1 = b_1, x_2 = b_2, x_3 = b_3] \end{aligned}

You could try to prove this, and you would fail because it’s false. In fact, it’s even false if A acts on only a single qubit! Because it’s so tedious to write out all of the notation, I decided to write a program to illustrate the counterexample. (The most brazenly dedicated readers will try to prove this false fact and identify where the proof fails.)

import numpy

H = (1/(2**0.5)) * numpy.array([[1,1], [1,-1]])
I = numpy.identity(4)
A = numpy.kron(H,I)

Here H is the 2 by 2 Hadamard matrix, which operates on a single qubit and maps e_0 \mapsto \frac{e_0 + e_1}{\sqrt{2}}, and e_1 \mapsto \frac{e_0 - e_1}{\sqrt{2}}. This matrix is famous for many reasons, but one simple use as a quantum gate is to generate uniform random coin flips. In particular, measuring He_0 outputs 1 and 0 with equal probability.

So in the code sample above, A is the mapping which applies the Hadamard operation to the first qubit and leaves the other qubits alone.

Then we compute some arbitrary input state vector w

def normalize(z):
   return (1.0 / (sum(abs(z)**2) ** 0.5)) * z

v = numpy.arange(1,9)
w = normalize(v)

And now we write a function to compute the probability of some query conditioned on some fixed bits. We simply sum up the square norms of all of the relevant indices in the state vector.

def condProb(state, query={}, fixed={}):
   num = 0
   denom = 0
   dim = int(math.log2(len(state)))

   for x in itertools.product([0,1], repeat=dim):
      if any(x[index] != b for (index,b) in fixed.items()):
         continue

      i = sum(d << i for (i,d) in enumerate(reversed(x)))
      denom += abs(state[i])**2
      if all(x[index] == b for (index, b) in query.items()):
         num += abs(state[i]) ** 2

   if num == 0:
      return 0 

   return num / denom

So if the query is query = {1:0} and the fixed thing is fixed = {0:0}, then this will compute the probability that the measurement results in the second qubit being zero conditioned on the first qubit also being zero.

And the result:

Aw = A.dot(w)
query = {1:0}
fixed = {0:0}
print((condProb(w, query, fixed), condProb(Aw, query, fixed)))
# (0.16666666666666666, 0.29069767441860467)

So they are not equal in general.

Also, in general we won’t work explicitly with full quantum gate matrices, since for n qubits the have size 2^{2n} which is big. But for finding counterexamples to guesses and false intuition, it’s a great tool.

Some important gates on 1-3 qubits

Let’s close this post with concrete examples of quantum gates. Based on the above discussion, we can write out the 2 x 2 or 4 x 4 matrix form of the operation and understand that it can apply to any two qubits in the state of a quantum program. Gates are most interesting when they’re operating on entangled qubits, and that will come out when we visit our first quantum algorithm next time, but for now we will just discuss at a naive level how they operate on the basis vectors.

Hadamard gate:

We introduced the Hadamard gate already, but I’ll reiterate it here.

Let H be the following 2 by 2 matrix, which operates on a single qubit and maps e_0 \mapsto \frac{e_0 + e_1}{\sqrt{2}}, and e_1 \mapsto \frac{e_0 - e_1}{\sqrt{2}}.

\displaystyle H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

One can use H to generate uniform random coin flips. In particular, measuring He_0 outputs 1 and 0 with equal probability.

Quantum NOT gate:

Let X be the 2 x 2 matrix formed by swapping the columns of the identity matrix.

\displaystyle X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

This gate is often called the “Pauli-X” gate by physicists. This matrix is far too simple to be named after a person, and I can only imagine it is still named after a person for the layer of obfuscation that so often makes people feel smarter (same goes for the Pauli-Y and Pauli-Z gates, but we’ll get to those when we need them).

If we’re thinking of e_0 as the boolean value “false” and e_1 as the boolean value “true”, then the quantum NOT gate simply swaps those two states. In particular, note that composing a Hadamard and a quantum NOT gate can have interesting effects: XH(e_0) = H(e_0), but XH(e_1) \neq H(e_1). In the second case, the minus sign is the culprit. Which brings us to…

Phase shift gate:

Given an angle \theta, we can “shift the phase” of one qubit by an angle of \theta using the 2 x 2 matrix R_{\theta}.

\displaystyle R_{\theta} = \begin{pmatrix} 1 & 0 \\ 0 & e^{i \theta} \end{pmatrix}

“Phase” is a term physicists like to use for angles. Since the coefficients of a quantum state vector are complex numbers, and since complex numbers can be thought of geometrically as vectors with direction and magnitude, it makes sense to “rotate” the coefficient of a single qubit. So R_{\theta} does nothing to e_0 and it rotates the coefficient of e_1 by an angle of \theta.

Continuing in our theme of concreteness, if I have the state vector v = \frac{1}{\sqrt{204}} (1,2,3,4,5,6,7,8) and I apply a rotation of pi to the second qubit, then my operation is the matrix I_2 \otimes R_{\pi} \otimes I_2 which maps e_{i0k} \mapsto e_{i0k} and e_{i1k} \mapsto -e_{i1k}. That would map the state v to (1,2,-3,-4,5,6,-7,-8).

If we instead used the rotation by \pi/2 we would get the output state (1,2,3i, 4i, 5, 6, 7i, 8i).

Quantum AND/OR gate:

In the last post in this series we gave the quantum AND gate and left the quantum OR gate as an exercise. Rather than write out the matrix again, let me remind you of this gate using a description of the effect on the basis e_{ijk} where i,j,k \in \{ 0,1 \}. Recall that we need three qubits in order to make the operation reversible (which is a consequence of all unitary gates being unitary matrices). Some notation: \oplus is the XOR of two bits, and \wedge is AND, \vee is OR. The quantum AND gate maps

\displaystyle e_{ijk} \mapsto e_{ij(k \oplus (i \wedge j))}

In words, the third coordinate is XORed with the AND of the first two coordinates. We think of the third coordinate as a “scratchwork” qubit which is maybe prepared ahead of time to be in state zero.

Simiarly, the quantum OR gate maps e_{ijk} \mapsto e_{ij(k \oplus (i \vee j))}. As we saw last time these combined with the quantum NOT gate (and some modest number of scratchwork qubits) allows quantum circuits to simulate any classical circuit.

Controlled-* gate:

The last example in this post is a meta-gate that represents a conditional branching. If we’re given a gate A acting on k qubits, then we define the controlled-A to be an operation which acts on k+1 qubits. Let’s call the added qubit “qubit zero.” Then controlled-A does nothing if qubit zero is in state 0, and applies A if qubit zero is in state 1. Qubit zero is generally called the “control qubit.”

The matrix representing this operation decomposes into blocks if the control qubit is actually the first qubit (or you rearrange).

\displaystyle \begin{pmatrix} I_{2^{k-1}} & 0 \\ 0 & A \end{pmatrix}

A common example of this is the controlled-NOT gate, often abbreviated CNOT, and it has the matrix

\displaystyle \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}

Looking forward

Okay let’s take a step back and evaluate our life choices. So far we’ve spent a few hours of our time motivating quantum computing, explaining the details of qubits and quantum circuits, and seeing examples of concrete quantum gates and studying measurement. I’ve hopefully hammered into your head the notion that quantum states which aren’t pure tensors (i.e. entangled) are where the “weirdness” of quantum computing comes from. But we haven’t seen any examples of quantum algorithms yet!

Next time we’ll see our first example of an algorithm that is genuinely quantum. We won’t tackle factoring yet, but we will see quantum “weirdness” in action.

Until then!

Hashing to Estimate the Size of a Stream

Problem: Estimate the number of distinct items in a data stream that is too large to fit in memory.

Solution: (in python)

import random

def randomHash(modulus):
   a, b = random.randint(0,modulus-1), random.randint(0,modulus-1)
   def f(x):
      return (a*x + b) % modulus
   return f

def average(L):
   return sum(L) / len(L)

def numDistinctElements(stream, numParallelHashes=10):
   modulus = 2**20
   hashes = [randomHash(modulus) for _ in range(numParallelHashes)]
   minima = [modulus] * numParallelHashes
   currentEstimate = 0

   for i in stream:
      hashValues = [h(i) for h in hashes]
      for i, newValue in enumerate(hashValues):
         if newValue < minima[i]:
            minima[i] = newValue

      currentEstimate = modulus / average(minima)

      yield currentEstimate

Discussion: The technique used here is to use random hash functions. The central idea is the same as the general principle presented in our recent post on hashing for load balancing. In particular, if you have an algorithm that works under the assumption that the data is uniformly random, then the same algorithm will work (up to a good approximation) if you process the data through a randomly chosen hash function.

So if we assume the data in the stream consists of N uniformly random real numbers between zero and one, what we would do is the following. Maintain a single number x_{\textup{min}} representing the minimum element in the list, and update it every time we encounter a smaller number in the stream. A simple probability calculation or an argument by symmetry shows that the expected value of the minimum is 1/(N+1). So your estimate would be 1/(x_{\textup{min}}+1). (The extra +1 does not change much as we’ll see.) One can spend some time thinking about the variance of this estimate (indeed, our earlier post is great guidance for how such a calculation would work), but since the data is not random we need to do more work. If the elements are actually integers between zero and k, then this estimate can be scaled by k and everything basically works out the same.

Processing the data through a hash function h chosen randomly from a 2-universal family (and we proved in the aforementioned post that this modulus thing is 2-universal) makes the outputs “essentially random” enough to have the above technique work with some small loss in accuracy. And to reduce variance, you can process the stream in parallel with many random hash functions. This rough sketch results in the code above. Indeed, before I state a formal theorem, let’s see the above code in action. First on truly random data:

S = [random.randint(1,2**20) for _ in range(10000)]

for k in range(10,301,10):
   for est in numDistinctElements(S, k):
      pass
   print(abs(est))

# output
18299.75567190227
7940.7497160166595
12034.154552410098
12387.19432959244
15205.56844547564
8409.913113220158
8057.99978043693
9987.627098464103
10313.862295081966
9084.872639057356
10952.745228373375
10360.569781803211
11022.469475216301
9741.250165892501
11474.896038520465
10538.452261306533
10068.793492995934
10100.266495424627
9780.532155130093
8806.382800033594
10354.11482578643
10001.59202254498
10623.87031408308
9400.404915767062
10710.246772348424
10210.087633885101
9943.64709187974
10459.610972568578
10159.60175069326
9213.120899718839

As you can see the output is never off by more than a factor of 2. Now with “adversarial data.”

S = range(10000) #[random.randint(1,2**20) for _ in range(10000)]

for k in range(10,301,10):
   for est in numDistinctElements(S, k):
      pass
   print(abs(est))

# output

12192.744186046511
15935.80547112462
10167.188106011634
12977.425742574258
6454.364151175674
7405.197740112994
11247.367453263867
4261.854392115023
8453.228233608026
7706.717624577393
7582.891328643745
5152.918628936483
1996.9365093316926
8319.20208545846
3259.0787592465967
6812.252720480753
4975.796789951151
8456.258064516129
8851.10133724288
7317.348220516398
10527.871485943775
3999.76974425661
3696.2999065091117
8308.843106180666
6740.999794281012
8468.603733730935
5728.532232608959
5822.072220349402
6382.349459544548
8734.008940222673

The estimates here are off by a factor of up to 5, and this estimate seems to get better as the number of hash functions used increases. The formal theorem is this:

Theorem: If S is the set of distinct items in the stream and n = |S| and m > 100 n, then with probability at least 2/3 the estimate m / x_{\textup{min}} is between n/6 and 6n.

We omit the proof (see below for references and better methods). As a quick analysis, since we’re only storing a constant number of integers at any given step, the algorithm has space requirement O(\log m) = O(\log n), and each step takes time polynomial in \log(m) to update in each step (since we have to compute multiplication and modulus of m).

This method is just the first ripple in a lake of research on this topic. The general area is called “streaming algorithms,” or “sublinear algorithms.” This particular problem, called cardinality estimation, is related to a family of problems called estimating frequency moments. The literature gets pretty involved in the various tradeoffs between space requirements and processing time per stream element.

As far as estimating cardinality goes, the first major results were due to Flajolet and Martin in 1983, where they provided a slightly more involved version of the above algorithm, which uses logarithmic space.

Later revisions to the algorithm (2003) got the space requirement down to O(\log \log n), which is exponentially better than our solution. And further tweaks and analysis improved the variance bounds to something like a multiplicative factor of \sqrt{m}. This is called the HyperLogLog algorithm, and it has been tested in practice at Google.

Finally, a theoretically optimal algorithm (achieving an arbitrarily good estimate with logarithmic space) was presented and analyzed by Kane et al in 2010.