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.

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:

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!

Linear Programming and the Simplex Algorithm

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

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

Slack variables and equality constraints

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

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

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

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

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

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

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

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

$\displaystyle x_7 + x_3 \leq 10,$

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

$\displaystyle x_7 + x_3 + \xi = 10$

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

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

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

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

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

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

What happened to the slackness conditions?

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

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

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

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

Extreme points and basic solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The simplex tableau

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

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

and solving the equation for $x_B$ gives

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

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

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

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

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

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

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

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

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

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

The simplex algorithm

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

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

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

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

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

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

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

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

return primalSolution(tableau), objectiveValue(tableau)


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

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


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


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

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

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


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

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


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

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

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


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

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


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

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

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

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


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

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

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

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


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

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


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

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

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


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

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

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

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

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

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

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

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

When Greedy Algorithms are Good Enough: Submodularity and the (1 – 1/e)-Approximation

Greedy algorithms are among the simplest and most intuitive algorithms known to humans. Their name essentially gives their description: do the thing that looks best right now, and repeat until nothing looks good anymore or you’re forced to stop. Some of the best situations in computer science are also when greedy algorithms are optimal or near-optimal. There is a beautiful theory of this situation, known as the theory of matroids. We haven’t covered matroids on this blog (edit: we did), but in this post we will focus on the next best thing: when the greedy algorithm guarantees a reasonably good approximation to the optimal solution.

This situation isn’t hard to formalize, and we’ll make it as abstract as possible. Say you have a set of objects $X$, and you’re looking to find the “best” subset $S \subset X$. Here “best” is just measured by a fixed (known, efficiently computable) objective function $f : 2^X \to \mathbb{R}$. That is, $f$ accepts as input subsets of $X$ and outputs numbers so that better subsets have larger numbers. Then the goal is to find a subset maximizing $f$.

In this generality the problem is clearly impossible. You’d have to check all subsets to be sure you didn’t miss the best one. So what conditions do we need on either $X$ or $f$ or both that makes this problem tractable? There are plenty you could try, but one very rich property is submodularity.

The Submodularity Condition

I think the simplest way to explain submodularity is in terms of coverage. Say you’re starting a new radio show and you have to choose which radio stations to broadcast from to reach the largest number of listeners. For simplicity say each radio station has one tower it broadcasts from, and you have a good estimate of the number of listeners you would reach if you broadcast from a given tower. For more simplicity, say it costs the same to broadcast from each tower, and your budget restricts you to a maximum of ten stations to broadcast from. So the question is: how do you pick towers to maximize your overall reach?

The hidden condition here is that some towers overlap in which listeners they reach. So if you broadcast from two towers in the same city, a listener who has access to both will just pick one or the other. In other words, there’s a diminished benefit to picking two overlapping towers if you already have chosen one.

In our version of the problem, picking both of these towers has some small amount of “overkill.”

This “diminishing returns” condition is a general idea you can impose on any function that takes in subsets of a given set and produces numbers. If $X$ is a set, then for a strange reason we denote by $2^X$ the set of all subsets of $X$. So we can state this condition more formally,

Definition: Let $X$ be a finite set. A function $f: 2^X \to \mathbb{R}$ is called submodular if for all subsets $S \subset T \subset X$ and all $x \in X \setminus T$,

$\displaystyle f(S \cup \{ x \}) - f(S) \geq f(T \cup \{ x \}) - f(T)$

In other words, if $f$ measures “benefit,” then the marginal benefit of adding $x$ to $S$ is at least as high as the marginal benefit of adding it to $T$. Since $S \subset T$ and $x$ are all arbitrary, this is as general as one could possibly make it: adding $x$ to a bigger set can’t be better than adding it to a smaller set.

Before we start doing things with submodular functions, let’s explore some basic properties. The first is an equivalent definition of submodularity

Proposition: $f$ is submodular if and only if for all $A, B \subset X$, it holds that

$\displaystyle f(A \cap B) + f(A \cup B) \leq f(A) + f(B)$.

Proof. If we assume $f$ has the condition from this proposition, then we can set $A=T, B=S \cup \{ x \}$, and the formula just works out. Conversely, if we have the condition from the definition, then using the fact that $A \cap B \subset B$ we can inductively apply the inequality to each element of $A \setminus B$ to get

$\displaystyle f(A \cup B) - f(B) \leq f(A) - f(A \cap B)$

$\square$

Next, we can tweak and combine submodular functions to get more submodular functions. In particular, non-negative linear combinations of sub-modular functions are submodular. In other words, if $f_1, \dots, f_k$ are submodular on the same set $X$, and $\alpha_1, \dots, \alpha_k$ are all non-negative reals, then $\alpha_1 f_1 + \dots + \alpha_k f_k$ is also a submodular function on $X$. It’s an easy exercise in applying the definition to see why this is true. This is important because when we’re designing objectives to maximize, we can design them by making some simple submodular pieces, and then picking an appropriate combination of those pieces.

The second property we need to impose on a submodular function is monotonicity. That is, as your sets get more elements added to them, their value under $f$ only goes up. In other words, $f$ is monotone when $S \subset T$ then $f(S) \leq f(T)$. An interesting property of functions that are both submodular and monotone is that the truncation of such a function is also submodular and monotone. In other words, $\textup{min}(f(S), c)$ is still submodular when $f$ is monotone submodular and $c$ is a constant.

Submodularity and Monotonicity Give 1 – 1/e

The wonderful thing about submodular functions is that we have a lot of great algorithmic guarantees for working with them. We’ll prove right now that the coverage problem (while it might be hard to solve in general) can be approximated pretty well by the greedy algorithm.

Here’s the algorithmic setup. I give you a finite set $X$ and an efficient black-box to evaluate $f(S)$ for any subset $S \subset X$ you want. I promise you that $f$ is monotone and submodular. Now I give you an integer $k$ between 1 and the size of $X$, and your task is to quickly find a set $S$ of size $k$ for which $f(S)$ is maximal among all subsets of size $k$. That is, you design an algorithm that will work for any $k, X, f$ and runs in polynomial time in the sizes of $X, k$.

In general this problem is NP-hard, meaning you’re not going to find a solution that works in the worst case (if you do, don’t call me; just claim your million dollar prize). So how well can we approximate the optimal value for $f(S)$ by a different set of size $k$? The beauty is that, if your function is monotone and submodular, you can guarantee to get within 63% of the optimum. The hope (and reality) is that in practice it will often perform much better, but still this is pretty good! More formally,

Theorem: Let $f$ be a monotone, submodular, non-negative function on $X$. The greedy algorithm, which starts with $S$ as the empty set and at every step picks an element $x$ which maximizes the marginal benefit $f(S \cup \{ x \}) - f(S)$, provides a set $S$ that achieves a $(1- 1/e)$-approximation of the optimum.

We’ll prove this in just a little bit more generality, and the generality is quite useful. If we call $S_1, S_2, \dots, S_l$ the sets chosen by the greedy algorithm (where now we might run the greedy algorithm for $l > k$ steps), then for all $l, k$, we have

$\displaystyle f(S_l) \geq \left ( 1 - e^{-l/k} \right ) \max_{T: |T| \leq k} f(T)$

This allows us to run the algorithm for more than $k$ steps to get a better approximation by sets of larger size, and quantify how much better the guarantee on that approximation would be. It’s like an algorithmic way of hedging your risk. So let’s prove it.

Proof. Let’s set up some notation first. Fix your $l$ and $k$, call $S_i$ the set chosen by the greedy algorithm at step $i$, and call $S^*$ the optimal subset of size $k$. Further call $\textup{OPT}$ the value of the best set $f(S^*)$. Call $x_1^*, \dots, x_k^*$ the elements of $S^*$ (the order is irrelevant). Now for every $i < l$ monotonicity gives us $f(S^*) \leq f(S^* \cup S_i)$. We can unravel this into a sum of marginal gains of adding single elements. The first step is

$\displaystyle f(S^* \cup S_i) = f(S^* \cup S_i) - f(\{ x_1^*, \dots, x_{k-1}^* \} \cup S_i) + f(\{ x_1^*, \dots, x_{k-1}^* \} \cup S_i)$

The second step removes $x_{k-1}^*$, from the last term, the third removes $x_{k-2}^*$, and so on until we have removed all of $S^*$ and get this sum

$\displaystyle f(S^* \cup S_i) = f(S_i) + \sum_{j=1}^k \left ( f(S_i \cup \{ x_1^*, \dots, x_j^* \}) - f(S_i \cup \{ x_1^*, \dots, x_{j-1}^* \} ) \right )$

Now, applying submodularity, we can change all of these marginal benefits of “adding one more $S^*$ element to $S_i$ already with some $S^*$ stuff” to “adding one more $S^*$ element to just $S_i$.” In symbols, the equation above is at most

$\displaystyle f(S_i) + \sum_{x \in S^*} f(S_i \cup \{ x \}) - f(S_i)$

and because $S_{i+1}$ is greedily chosen to maximize the benefit of adding a single element, so the above is at most

$\displaystyle f(S_i) + \sum_{x \in S^*} f(S_{i+1}) - f(S_i) = f(S_i) + k(f(S_{i+1}) - f(S_i))$

Chaining all of these together, we have $f(S^*) - f(S_i) \leq k(f(S_{i+1}) - f(S_i))$. If we call $a_{i} = f(S^*) - f(S_i)$, then this inequality can be rewritten as $a_{i+1} \leq (1 - 1/k) a_{i}$. Now by induction we can relate $a_l \leq (1 - 1/k)^l a_0$. Now use the fact that $a_0 \leq f(S^*)$ and the common inequality $1-x \leq e^{-x}$ to get

$\displaystyle a_l = f(S^*) - f(S_l) \leq e^{-l/k} f(S^*)$

And rearranging gives $f(S_l) \geq (1 - e^{-l/k}) f(S^*)$.

$\square$

Setting $l=k$ gives the approximation bound we promised. But note that allowing the greedy algorithm to run longer can give much stronger guarantees, though it requires you to sacrifice the cardinality constraint. $1 - 1/e$ is about 63%, but doubling the size of $S$ gives about an 86% approximation guarantee. This is great for people in the real world, because you can quantify the gains you’d get by relaxing the constraints imposed on you (which are rarely set in stone).

So this is really great! We have quantifiable guarantees on a stupidly simple algorithm, and the setting is super general. And so if you have your problem and you manage to prove your function is submodular (this is often the hardest part), then you are likely to get this nice guarantee.

Extensions and Variations

This result on monotone submodular functions is just one part of a vast literature on finding approximation algorithms for submodular functions in various settings. In closing this post we’ll survey some of the highlights and provide references.

What we did in this post was maximize a monotone submodular function subject to a cardinality constraint $|S| \leq k$. There are three basic variations we could do: we could drop constraints and see whether we can still get guarantees, we could look at minimization instead of maximization, and we could modify the kinds of constraints we impose on the solution.

There are a ton of different kinds of constraints, and we’ll discuss two. The first is where you need to get a certain value $f(S) \geq q$, and you want to find the smallest set that achieves this value. Laurence Wolsey (who proved a lot of these theorems) showed in 1982 that a slight variant of the greedy algorithm can achieve a set whose size is a multiplicative factor of $1 + \log (\max_x f(\{ x \}))$ worse than the optimum.

The second kind of constraint is a generalization of a cardinality constraint called a knapsack constraint. This means that each item $x \in X$ has a cost, and you have a finite budget with which to spend on elements you add to $S$. One might expect this natural extension of the greedy algorithm to work: pick the element which maximizes the ratio of increasing the value of $f$ to the cost (within your available budget). Unfortunately this algorithm can perform arbitrarily poorly, but there are two fun caveats. The first is that if you do both this augmented greedy algorithm and the greedy algorithm that ignores costs, then at least one of these can’t do too poorly. Specifically, one of them has to get at least a 30% approximation. This was shown by Leskovec et al in 2007. The second is that if you’re willing to spend more time in your greedy step by choosing the best subset of size 3, then you can get back to the $1-1/e$ approximation. This was shown by Sviridenko in 2004.

Now we could try dropping the monotonicity constraint. In this setting cardinality constraints are also superfluous, because it could be that the very large sets have low values. Now it turns out that if $f$ has no other restrictions (in particular, if it’s allowed to be negative), then even telling whether there’s a set $S$ with $f(S) > 0$ is NP-hard, but the optimum could be arbitrarily large and positive when it exists. But if you require that $f$ is non-negative, then you can get a 1/3-approximation, if you’re willing to add randomness you can get 2/5 in expectation, and with more subtle constraints you can get up to a 1/2 approximation. Anything better is NP-hard. Fiege, Mirrokni, and Vondrak have a nice FOCS paper on this.

Next, we could remove the monotonicity property and try to minimize the value of $f(S)$. It turns out that this problem always has an efficient solution, but the only algorithm I have heard of to solve it involves a very sophisticated technique called the ellipsoid algorithm. This is heavily related to linear programming and convex optimization, something which I hope to cover in more detail on this blog.

Finally, there are many interesting variations in the algorithmic procedure. For example, one could require that the elements are provided in some order (the streaming setting), and you have to pick at each step whether to put the element in your set or not. Alternatively, the objective functions might not be known ahead of time and you have to try to pick elements to jointly maximize them as they are revealed. These two settings have connections to bandit learning problems, which we’ve covered before on this blog. See this survey of Krause and Golovin for more on the connections, which also contains the main proof used in this post.

Indeed, despite the fact that many of the big results were proved in the 80’s, the analysis of submodular functions is still a big research topic. There was even a paper posted just the other day on the arXiv about its relation to ad serving! And wouldn’t you know, they proved a $(1-1/e)$-approximation for their setting. There’s just something about $1-1/e$.

Until next time!

Linear Programming and the Most Affordable Healthy Diet — Part 1

Optimization is by far one of the richest ways to apply computer science and mathematics to the real world. Everybody is looking to optimize something: companies want to maximize profits, factories want to maximize efficiency, investors want to minimize risk, the list just goes on and on. The mathematical tools for optimization are also some of the richest mathematical techniques. They form the cornerstone of an entire industry known as operations research, and advances in this field literally change the world.

The mathematical field is called combinatorial optimization, and the name comes from the goal of finding optimal solutions more efficiently than an exhaustive search through every possibility. This post will introduce the most central problem in all of combinatorial optimization, known as the linear program. Even better, we know how to efficiently solve linear programs, so in future posts we’ll write a program that computes the most affordable diet while meeting the recommended health standard.

Generalizing a Specific Linear Program

Most optimization problems have two parts: an objective function, the thing we want to maximize or minimize, and constraints, rules we must abide by to ensure we get a valid solution. As a simple example you may want to minimize the amount of time you spend doing your taxes (objective function), but you certainly can’t spend a negative amount of time on them (a constraint).

The following more complicated example is the centerpiece of this post. Most people want to minimize the amount of money spent on food. At the same time, one needs to maintain a certain level of nutrition. For males ages 19-30, the United States National Institute for Health recommends 3.7 liters of water per day, 1,000 milligrams of calcium per day, 90 milligrams of vitamin C per day, etc.

We can set up this nutrition problem mathematically, just using a few toy variables. Say we had the option to buy some combination of oranges, milk, and broccoli. Some rough estimates [1] give the following content/costs of these foods. For 0.272 USD you can get 100 grams of orange, containing a total of 53.2mg of calcium, 40mg of vitamin C, and 87g of water. For 0.100 USD you can get 100 grams of whole milk, containing 276mg of calcium, 0mg of vitamin C, and 87g of water. Finally, for 0.381 USD you can get 100 grams of broccoli containing 47mg of calcium, 89.2mg of vitamin C, and 91g of water. Here’s a table summarizing this information:

Nutritional content and prices for 100g of three foods

Food         calcium(mg)     vitamin C(mg)      water(g)   price(USD/100g)
Broccoli     47              89.2               91         0.381
Whole milk   276             0                  87         0.100
Oranges      40              53.2               87         0.272

Some observations: broccoli is more expensive but gets the most of all three nutrients, whole milk doesn’t have any vitamin C but gets a ton of calcium for really cheap, and oranges are a somewhere in between. So you could probably tinker with the quantities and figure out what the cheapest healthy diet is. The problem is what happens when we incorporate hundreds or thousands of food items and tens of nutrient recommendations. This simple example is just to help us build up a nice formality.

So let’s continue doing that. If we denote by $b$ the number of 100g units of broccoli we decide to buy, and $m$ the amount of milk and $r$ the amount of oranges, then we can write the daily cost of food as

$\displaystyle \text{cost}(b,m,r) = 0.381 b + 0.1 m + 0.272 r$

In the interest of being compact (and again, building toward the general linear programming formulation) we can extract the price information into a single cost vector $c = (0.381, 0.1, 0.272)$, and likewise write our variables as a vector $x = (b,m,r)$. We’re implicitly fixing an ordering on the variables that is maintained throughout the problem, but the choice of ordering doesn’t matter. Now the cost function is just the inner product (dot product) of the cost vector and the variable vector $\left \langle c,x \right \rangle$. For some reason lots of people like to write this as $c^Tx$, where $c^T$ denotes the transpose of a matrix, and we imagine that $c$ and $x$ are matrices of size $3 \times 1$. I’ll stick to using the inner product bracket notation.

Now for each type of food we get a specific amount of each nutrient, and the sum of those nutrients needs to be bigger than the minimum recommendation. For example, we want at least 1,000 mg of calcium per day, so we require that $1000 \leq 47b + 276m + 40r$. Likewise, we can write out a table of the constraints by looking at the columns of our table above.

$\displaystyle \begin{matrix} 91b & + & 87m & + & 87r & \geq & 3700 & \text{(water)}\\ 47b & + & 276m & + & 40r & \geq & 1000 & \text{(calcium)} \\ 89.2b & + & 0m & + & 53.2r & \geq & 90 & \text{(vitamin C)} \end{matrix}$

In the same way that we extracted the cost data into a vector to separate it from the variables, we can extract all of the nutrient data into a matrix $A$, and the recommended minimums into a vector $v$. Traditionally the letter $b$ is used for the minimums vector, but for now we’re using $b$ for broccoli.

$A = \begin{pmatrix} 91 & 87 & 87 \\ 47 & 276 & 40 \\ 89.2 & 0 & 53.2 \end{pmatrix}$

$v = \begin{pmatrix} 3700 \\ 1000 \\ 90 \end{pmatrix}$

And now the constraint is that $Ax \geq v$, where the $\geq$ means “greater than or equal to in every coordinate.” So now we can write down the more general form of the problem for our specific matrices and vectors. That is, our problem is to minimize $\left \langle c,x \right \rangle$ subject to the constraint that $Ax \geq v$. This is often written in offset form to contrast it with variations we’ll see in a bit:

$\displaystyle \text{minimize} \left \langle c,x \right \rangle \\ \text{subject to the constraint } Ax \geq v$

In general there’s no reason you can’t have a “negative” amount of one variable. In this problem you can’t buy negative broccoli, so we’ll add the constraints to ensure the variables are nonnegative. So our final form is

$\displaystyle \text{minimize} \left \langle c,x \right \rangle \\ \text{subject to } Ax \geq v \\ \text{and } x \geq 0$

In general, if you have an $m \times n$ matrix $A$, a “minimums” vector $v \in \mathbb{R}^m$, and a cost vector $c \in \mathbb{R}^n$, the problem of finding the vector $x$ that minimizes the cost function while meeting the constraints is called a linear programming problem or simply a linear program.

To satiate the reader’s burning curiosity, the solution for our calcium/vitamin C problem is roughly $x = (1.01, 41.47, 0)$. That is, you should have about 100g of broccoli and 4.2kg of milk (like 4 liters), and skip the oranges entirely. The daily cost is about 4.53 USD. If this seems awkwardly large, it’s because there are cheaper ways to get water than milk.

100g of broccoli (image source: 100-grams.blogspot.com)

Duality

Now that we’ve seen the general form a linear program and a cute example, we can ask the real meaty question: is there an efficient algorithm that solves arbitrary linear programs? Despite how widely applicable these problems seem, the answer is yes!

But before we can describe the algorithm we need to know more about linear programs. For example, say you have some vector $x$ which satisfies your constraints. How can you tell if it’s optimal? Without such a test we’d have no way to know when to terminate our algorithm. Another problem is that we’ve phrased the problem in terms of minimization, but what about problems where we want to maximize things? Can we use the same algorithm that finds minima to find maxima as well?

Both of these problems are neatly answered by the theory of duality. In mathematics in general, the best way to understand what people mean by “duality” is that one mathematical object uniquely determines two different perspectives, each useful in its own way. And typically a duality theorem provides one with an efficient way to transform one perspective into the other, and relate the information you get from both perspectives. A theory of duality is considered beautiful because it gives you truly deep insight into the mathematical object you care about.

In linear programming duality is between maximization and minimization. In particular, every maximization problem has a unique “dual” minimization problem, and vice versa. The really interesting thing is that the variables you’re trying to optimize in one form correspond to the contraints in the other form! Here’s how one might discover such a beautiful correspondence. We’ll use a made up example with small numbers to make things easy.

So you have this optimization problem

$\displaystyle \begin{matrix} \text{minimize} & 4x_1+3x_2+9x_3 & \\ \text{subject to} & x_1+x_2+x_3 & \geq 6 \\ & 2x_1+x_3 & \geq 2 \\ & x_2+x_3 & \geq 1 & \\ & x_1,x_2,x_3 & \geq 0 \end{matrix}$

Just for giggles let’s write out what $A$ and $c$ are.

$\displaystyle A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}, c = (4,3,9), v = (6,2,1)$

Say you want to come up with a lower bound on the optimal solution to your problem. That is, you want to know that you can’t make $4x_1 + 3x_2 + 9x_3$ smaller than some number $m$. The constraints can help us derive such lower bounds. In particular, every variable has to be nonnegative, so we know that $4x_1 + 3x_2 + 9x_3 \geq x_1 + x_2 + x_3 \geq 6$, and so 6 is a lower bound on our optimum. Likewise,

\displaystyle \begin{aligned}4x_1+3x_2+9x_3 & \geq 4x_1+4x_3+3x_2+3x_3 \\ &=2(2x_1 + x_3)+3(x_2+x_3) \\ & \geq 2 \cdot 2 + 3 \cdot 1 \\ &=7\end{aligned}

and that’s an even better lower bound than 6. We could try to write this approach down in general: find some numbers $y_1, y_2, y_3$ that we’ll use for each constraint to form

$\displaystyle y_1(\text{constraint 1}) + y_2(\text{constraint 2}) + y_3(\text{constraint 3})$

To make it a valid lower bound we need to ensure that the coefficients of each of the $x_i$ are smaller than the coefficients in the objective function (i.e. that the coefficient of $x_1$ ends up less than 4). And to make it the best lower bound possible we want to maximize what the right-hand-size of the inequality would be: $y_1 6 + y_2 2 + y_3 1$. If you write out these equations and the constraints you get our “lower bound” problem written as

$\displaystyle \begin{matrix} \text{maximize} & 6y_1 + 2y_2 + y_3 & \\ \text{subject to} & y_1 + 2y_2 & \leq 4 \\ & y_1 + y_3 & \leq 3 \\ & y_1+y_2 + y_3 & \leq 9 \\ & y_1,y_2,y_3 & \geq 0 \end{matrix}$

And wouldn’t you know, the matrix providing the constraints is $A^T$, and the vectors $c$ and $v$ switched places.

$\displaystyle A^T = \begin{pmatrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix}$

This is no coincidence. All linear programs can be transformed in this way, and it would be a useful exercise for the reader to turn the above maximization problem back into a minimization problem by the same technique (computing linear combinations of the constraints to make upper bounds). You’ll be surprised to find that you get back to the original minimization problem! This is part of what makes it “duality,” because the dual of the dual is the original thing again. Often, when we fix the “original” problem, we call it the primal form to distinguish it from the dual form. Usually the primal problem is the one that is easy to interpret.

(Note: because we’re done with broccoli for now, we’re going to use $b$ to denote the constraint vector that used to be $v$.)

Now say you’re given the data of a linear program for minimization, that is the vectors $c, b$ and matrix $A$ for the problem, “minimize $\left \langle c, x \right \rangle$ subject to $Ax \geq b; x \geq 0$.” We can make a general definition: the dual linear program is the maximization problem “maximize $\left \langle b, y \right \rangle$ subject to $A^T y \leq c, y \geq 0$.” Here $y$ is the new set of variables and the superscript T denotes the transpose of the matrix. The constraint for the dual is often written $y^T A \leq c^T$, again identifying vectors with a single-column matrices, but I find the swamp of transposes pointless and annoying (why do things need to be columns?).

Now we can actually prove that the objective function for the dual provides a bound on the objective function for the original problem. It’s obvious from the work we’ve done, which is why it’s called the weak duality theorem.

Weak Duality Theorem: Let $c, A, b$ be the data of a linear program in the primal form (the minimization problem) whose objective function is $\left \langle c, x \right \rangle$. Recall that the objective function of the dual (maximization) problem is $\left \langle b, y \right \rangle$. If $x,y$ are feasible solutions (satisfy the constraints of their respective problems), then

$\left \langle b, y \right \rangle \leq \left \langle c, x \right \rangle$

In other words, the maximum of the dual is a lower bound on the minimum of the primal problem and vice versa. Moreover, any feasible solution for one provides a bound on the other.

Proof. The proof is pleasingly simple. Just inspect the quantity $\left \langle A^T y, x \right \rangle = \left \langle y, Ax \right \rangle$. The constraints from the definitions of the primal and dual give us that

$\left \langle y, b \right \rangle \leq \left \langle y, Ax \right \rangle = \left \langle A^Ty, x \right \rangle \leq \left \langle c,x \right \rangle$

The inequalities follow from the linear algebra fact that if the $u$ in $\left \langle u,v \right \rangle$ is nonnegative, then you can only increase the size of the product by increasing the components of $v$. This is why we need the nonnegativity constraints.

In fact, the world is much more pleasing. There is a theorem that says the two optimums are equal!

Strong Duality Theorem: If there are any solutions $x,y$ to the primal (minimization) problem and the dual (maximization) problem, respectively, then the two problems also have optimal solutions $x^*, y^*$, and two candidate solutions $x^*, y^*$ are optimal if and only if they produce equal objective values $\left \langle c, x^* \right \rangle = \left \langle y^*, b \right \rangle$.

The proof of this theorem is a bit more convoluted than the weak duality theorem, and the key technique is a lemma of Farkas and its variations. See the second half of these notes for a full proof. The nice thing is that this theorem gives us a way to tell if an algorithm to solve linear programs is done: maintain a pair of feasible solutions to the primal and dual problems, improve them by some rule, and stop when the two solutions give equal objective values. The hard part, then, is finding a principled and guaranteed way to improve a given pair of solutions.

On the other hand, you can also prove the strong duality theorem by inventing an algorithm that provably terminates. We’ll see such an algorithm, known as the simplex algorithm in the next post. Sneak peek: it’s a lot like Gaussian elimination. Then we’ll use the algorithm (or an equivalent industry-strength version) to solve a much bigger nutrition problem.

In fact, you can do a bit better than the strong duality theorem, in terms of coming up with a stopping condition for a linear programming algorithm. You can observe that an optimal solution implies further constraints on the relationship between the primal and the dual problems. In particular, this is called the complementary slackness conditions, and they essentially say that if an optimal solution to the primal has a positive variable then the corresponding constraint in the dual problem must be tight (is an equality) to get an optimal solution to the dual. The contrapositive says that if some constraint is slack, or a strict inequality, then either the corresponding variable is zero or else the solution is not optimal. More formally,

Theorem (Complementary Slackness Conditions): Let $A, c, b$ be the data of the primal form of a linear program, “minimize $\left \langle c, x \right \rangle$ subject to $Ax \geq b, x \geq 0$.” Then $x^*, y^*$ are optimal solutions to the primal and dual problems if any only if all of the following conditions hold.

• $x^*, y^*$ are both feasible for their respective problems.
• Whenever $x^*_i > 0$ the corresponding constraint $A^T_i y^* = c_i$ is an equality.
• Whenever $y^*_j > 0$ the corresponding constraint $A_j x^* = b_j$ is an equality.

Here we denote by $M_i$ the $i$-th row of the matrix $M$ and $v_i$ to denote the $i$-th entry of a vector. Another way to write the condition using vectors instead of English is

$\left \langle x^*, A^T y^* - c \right \rangle = 0$
$\left \langle y^*, Ax^* - b \right \rangle$

The proof follows from the duality theorems, and just involves pushing around some vector algebra. See section 6.2 of these notes.

One can interpret complementary slackness in linear programs in a lot of different ways. For us, it will simply be a termination condition for an algorithm: one can efficiently check all of these conditions for the nonzero variables and stop if they’re all satisfied or if we find a variable that violates a slackness condition. Indeed, in more mature optimization analyses, the slackness condition that is more egregiously violated can provide evidence for where a candidate solution can best be improved. For a more intricate and detailed story about how to interpret the complementary slackness conditions, see Section 4 of these notes by Joel Sobel.

Finally, before we close we should note there are geometric ways to think about linear programming. I have my preferred visualization in my head, but I have yet to find a suitable animation on the web that replicates it. Here’s one example in two dimensions. The set of constraints define a convex geometric region in the plane

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

Now the optimization function $f(x) = \left \langle c,x \right \rangle$ is also a linear function, and if you fix some output value $y = f(x)$ this defines a line in the plane. As $y$ changes, the line moves along its normal vector (that is, all these fixed lines are parallel). Now to geometrically optimize the target function, we can imagine starting with the line $f(x) = 0$, and sliding it along its normal vector in the direction that keeps it in the feasible region. We can keep sliding it in this direction, and the maximum of the function is just the last instant that this line intersects the feasible region. If none of the constraints are parallel to the family of lines defined by $f$, then this is guaranteed to occur at a vertex of the feasible region. Otherwise, there will be a family of optima lying anywhere on the line segment of last intersection.

In higher dimensions, the only change is that the lines become affine subspaces of dimension $n-1$. That means in three dimensions you’re sliding planes, in four dimensions you’re sliding 3-dimensional hyperplanes, etc. The facts about the last intersection being a vertex or a “line segment” still hold. So as we’ll see next time, successful algorithms for linear programming in practice take advantage of this observation by efficiently traversing the vertices of this convex region. We’ll see this in much more detail in the next post.

Until then!