Boolean Logic in Polynomials

Problem: Express a boolean logic formula using polynomials. I.e., if an input variable x is set to 0, that is interpreted as false, while x=1 is interpreted as true. The output of the polynomial should be 0 or 1 according to whether the formula is true or false as a whole.

Solution: You can do this using a single polynomial.

Illustrating with an example: the formula is \neg[(a \vee b) \wedge (\neg c \vee d)] also known as

not((a or b) and (not c or d))

The trick is to use multiplication for “and” and 1-x for “not.” So a \wedge b would be x_1 x_2, and \neg z would be 1-z. Indeed, if you have two binary variables x and y then xy is 1 precisely when both are 1, and zero when either variable is zero. Likewise, 1-x = 1 if x is zero and zero if x is one.

Combine this with deMorgan’s rule to get any formula. a \vee b = \neg(\neg a \wedge \neg b) translates to 1 - (1-a)(1-b). For our example above,

\displaystyle f(x_1, x_2, x_3, x_4) = 1 - (1 - (1-a)(1-b))(1 - c(1-d))

Which expands to

\displaystyle 1 - a - b + ab + (1-d)(ac + bc - abc)

If you plug in a = 1, b = 0, c = 1, d = 0 you get True in the original formula (because “not c or d” is False), and likewise the polynomial is

\displaystyle 1 - 1 - 0 + 0 + (1-0)(1 + 0 - 0) = 1

You can verify the rest work yourself, using the following table as a guide:

0, 0, 0, 0 -> 1
0, 0, 0, 1 -> 1
0, 0, 1, 0 -> 1
0, 0, 1, 1 -> 1
0, 1, 0, 0 -> 0
0, 1, 0, 1 -> 0
0, 1, 1, 0 -> 1
0, 1, 1, 1 -> 0
1, 0, 0, 0 -> 0
1, 0, 0, 1 -> 0
1, 0, 1, 0 -> 1
1, 0, 1, 1 -> 0
1, 1, 0, 0 -> 0
1, 1, 0, 1 -> 0
1, 1, 1, 0 -> 1
1, 1, 1, 1 -> 0

Discussion: This trick is used all over CS theory to embed boolean logic within polynomials, and it makes the name “boolean algebra” obvious, because it’s just a subset of normal algebra.

Moreover, since boolean satisfiability—the problem of algorithmically determining if a boolean formula has a satisfying assignment (a choice of variables evaluating to true)—is NP-hard, this can be used to show certain problems relating to multivariable polynomials is also hard. For example, finding roots of multivariable polynomials (even if you knew nothing about algebraic geometry) is hard because you’d run into NP-hardness by simply considering the subset of polynomials coming from boolean formulas.

Here’s a more interesting example, related to the kinds of optimization problems that show up in modern machine learning. Say you want to optimize a polynomial f(x) subject to a set of quadratic equality constraints. This is NP-hard. Here’s why.

Let \varphi be a boolean formula, and f_\varphi its corresponding polynomial. First, each variable x_i used in the polynomial can be restricted to binary values via the constraint x_i(x_i - 1) = 0.

You can even show NP-hardness if the target function to optimize is only quadratic. As an exercise, one can express the subset sum problem as a quadratic programming problem using similar choices for the constraints. According to this writeup you even express subset sum as a quadratic program with linear constraints.

The moral of the story is simply that multivariable polynomials can encode arbitrary boolean logic.

The Reasonable Effectiveness of the Multiplicative Weights Update Algorithm

papad

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

Hard to believe

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

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

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

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

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

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

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

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

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

mwua

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

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

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

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

The generic MWUA algorithm

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

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

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

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

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

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

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

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

        choiceIndex += 1

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

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

      ...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 Then the claim of the lemma is:

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

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

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

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

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

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

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

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

\square

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

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

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

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

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

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

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

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

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

\square

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

MWUA for linear programs

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

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

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

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

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

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

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

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

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

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

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

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

Implementing the oracle

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

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

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

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

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

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

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

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

    return oracle

Implementing the core solver

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

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

    def reward(i, specialVector):
        ...

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

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

    return averageVector

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

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

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

Implementing the binary search, and an example

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

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

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

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

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

    return result

Finally, a simple example:

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

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

The output:

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

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

Discussion

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

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

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

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

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

Until next time!

Singular Value Decomposition Part 2: Theorem, Proof, Algorithm

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

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

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

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

vectormax

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

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

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

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

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

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

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

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

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

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

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

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

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

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

At this point you should be thinking,

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

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

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

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

Singular values and vectors

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

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

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

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

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

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

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

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

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

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

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

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

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

Proof of SVD theorem.

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

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

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

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

\square

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

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

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

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

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

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

Let’s move on to some code.

A black box example

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

Recall our movie-rating matrix from the last post:

movieratings

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

from numpy.linalg import svd

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

U, singularValues, V = svd(movieRatings)

Printing these values out gives

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

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

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

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

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

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

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

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

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

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

And the result

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

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

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

The greedy optimization routine

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

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

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

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

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

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

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

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

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

\square

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

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

import numpy as np
from numpy.linalg import norm

from random import normalvariate
from math import sqrt

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

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

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

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

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

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

And using it on our movie ratings example:

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

    print(svd_1d(movieRatings))

With the result

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

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

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

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

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

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

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

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

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

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

    return singularValues, us.T, vs

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

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

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

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

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

CNN stories

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

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

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

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

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

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

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

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

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

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

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

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

Now we can create the document term matrix.

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

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

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

    return matrix, (indexToWord, indexToDocument)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Here’s another interesting cluster corresponding to economic words:

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

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

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

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

Issues with the power method

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

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

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

Until next time!

Linear Programming and the Simplex Algorithm

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

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

Slack variables and equality constraints

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

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

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

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

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

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

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

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

\displaystyle x_7 + x_3 \leq 10,

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

\displaystyle x_7 + x_3 + \xi = 10

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

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

Screen Shot 2014-10-05 at 12.06.19 AM

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

Screen Shot 2014-10-05 at 12.06.41 AM

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

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

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

What happened to the slackness conditions?

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

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

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

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

Extreme points and basic solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The simplex tableau

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

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

and solving the equation for x_B gives

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

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

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

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

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

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

tableau

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

tableau-symbols

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

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

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

The simplex algorithm

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

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

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

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

example

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

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

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

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

   return primalSolution(tableau), objectiveValue(tableau)

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

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

And after adding slack:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Obvious questions and sad answers

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

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

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

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

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

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