# Earthmover Distance

Problem: Compute distance between points with uncertain locations (given by samples, or differing observations, or clusters).

For example, if I have the following three “points” in the plane, as indicated by their colors, which is closer, blue to green, or blue to red?

It’s not obvious, and there are multiple factors at work: the red points have fewer samples, but we can be more certain about the position; the blue points are less certain, but the closest non-blue point to a blue point is green; and the green points are equally plausibly “close to red” and “close to blue.” The centers of masses of the three sample sets are close to an equilateral triangle. In our example the “points” don’t overlap, but of course they could. And in particular, there should probably be a nonzero distance between two points whose sample sets have the same center of mass, as below. The distance quantifies the uncertainty.

All this is to say that it’s not obvious how to define a distance measure that is consistent with perceptual ideas of what geometry and distance should be.

Solution (Earthmover distance): Treat each sample set $A$ corresponding to a “point” as a discrete probability distribution, so that each sample $x \ in A$ has probability mass $p_x = 1 / |A|$. The distance between $A$ and $B$ is the optional solution to the following linear program.

Each $x \in A$ corresponds to a pile of dirt of height $p_x$, and each $y \in B$ corresponds to a hole of depth $p_y$. The cost of moving a unit of dirt from $x$ to $y$ is the Euclidean distance $d(x, y)$ between the points (or whatever hipster metric you want to use).

Let $z_{x, y}$ be a real variable corresponding to an amount of dirt to move from $x \in A$ to $y \in B$, with cost $d(x, y)$. Then the constraints are:

• Each $z_{x, y} \geq 0$, so dirt only moves from $x$ to $y$.
• Every pile $x \in A$ must vanish, i.e. for each fixed $x \in A$, $\sum_{y \in B} z_{x,y} = p_x$.
• Likewise, every hole $y \in B$ must be completely filled, i.e. $\sum_{y \in B} z_{x,y} = p_y$.

The objective is to minimize the cost of doing this: $\sum_{x, y \in A \times B} d(x, y) z_{x, y}$.

In python, using the ortools library (and leaving out a few docstrings and standard import statements, full code on Github):

from ortools.linear_solver import pywraplp

def earthmover_distance(p1, p2):
dist1 = {x: count / len(p1) for (x, count) in Counter(p1).items()}
dist2 = {x: count / len(p2) for (x, count) in Counter(p2).items()}
solver = pywraplp.Solver('earthmover_distance', pywraplp.Solver.GLOP_LINEAR_PROGRAMMING)

variables = dict()

# for each pile in dist1, the constraint that says all the dirt must leave this pile
dirt_leaving_constraints = defaultdict(lambda: 0)

# for each hole in dist2, the constraint that says this hole must be filled
dirt_filling_constraints = defaultdict(lambda: 0)

# the objective
objective = solver.Objective()
objective.SetMinimization()

for (x, dirt_at_x) in dist1.items():
for (y, capacity_of_y) in dist2.items():
amount_to_move_x_y = solver.NumVar(0, solver.infinity(), 'z_{%s, %s}' % (x, y))
variables[(x, y)] = amount_to_move_x_y
dirt_leaving_constraints[x] += amount_to_move_x_y
dirt_filling_constraints[y] += amount_to_move_x_y
objective.SetCoefficient(amount_to_move_x_y, euclidean_distance(x, y))

for x, linear_combination in dirt_leaving_constraints.items():
solver.Add(linear_combination == dist1[x])

for y, linear_combination in dirt_filling_constraints.items():
solver.Add(linear_combination == dist2[y])

status = solver.Solve()
if status not in [solver.OPTIMAL, solver.FEASIBLE]:
raise Exception('Unable to find feasible solution')

return objective.Value()

Discussion: I’ve heard about this metric many times as a way to compare probability distributions. For example, it shows up in an influential paper about fairness in machine learning, and a few other CS theory papers related to distribution testing.

One might ask: why not use other measures of dissimilarity for probability distributions (Chi-squared statistic, Kullback-Leibler divergence, etc.)? One answer is that these other measures only give useful information for pairs of distributions with the same support. An example from a talk of Justin Solomon succinctly clarifies what Earthmover distance achieves

Also, why not just model the samples using, say, a normal distribution, and then compute the distance based on the parameters of the distributions? That is possible, and in fact makes for a potentially more efficient technique, but you lose some information by doing this. Ignoring that your data might not be approximately normal (it might have some curvature), with Earthmover distance, you get point-by-point details about how each data point affects the outcome.

This kind of attention to detail can be very important in certain situations. One that I’ve been paying close attention to recently is the problem of studying gerrymandering from a mathematical perspective. Justin Solomon of MIT is a champion of the Earthmover distance (see his fascinating talk here for more, with slides) which is just one topic in a field called “optimal transport.”

This has the potential to be useful in redistricting because of the nature of the redistricting problem. As I wrote previously, discussions of redistricting are chock-full of geometry—or at least geometric-sounding language—and people are very concerned with the apparent “compactness” of a districting plan. But the underlying data used to perform redistricting isn’t very accurate. The people who build the maps don’t have precise data on voting habits, or even locations where people live. Census tracts might not be perfectly aligned, and data can just plain have errors and uncertainty in other respects. So the data that district-map-drawers care about is uncertain much like our point clouds. With a theory of geometry that accounts for uncertainty (and the Earthmover distance is the “distance” part of that), one can come up with more robust, better tools for redistricting.

Solomon’s website has a ton of resources about this, under the names of “optimal transport” and “Wasserstein metric,” and his work extends from computing distances to computing important geometric values like the barycenter, computational advantages like parallelism.

Others in the field have come up with transparency techniques to make it clearer how the Earthmover distance relates to the geometry of the underlying space. This one is particularly fun because the explanations result in a path traveled from the start to the finish, and by setting up the underlying metric in just such a way, you can watch the distribution navigate a maze to get to its target. I like to imagine tiny ants carrying all that dirt.

Finally, work of Shirdhonkar and Jacobs provide approximation algorithms that allow linear-time computation, instead of the worst-case-cubic runtime of a linear solver.

# Binary Search on Graphs

Binary search is one of the most basic algorithms I know. Given a sorted list of comparable items and a target item being sought, binary search looks at the middle of the list, and compares it to the target. If the target is larger, we repeat on the smaller half of the list, and vice versa.

With each comparison the binary search algorithm cuts the search space in half. The result is a guarantee of no more than $\log(n)$ comparisons, for a total runtime of $O(\log n)$. Neat, efficient, useful.

There’s always another angle.

What if we tried to do binary search on a graph? Most graph search algorithms, like breadth- or depth-first search, take linear time, and they were invented by some pretty smart cookies. So if binary search on a graph is going to make any sense, it’ll have to use more information beyond what a normal search algorithm has access to.

For binary search on a list, it’s the fact that the list is sorted, and we can compare against the sought item to guide our search. But really, the key piece of information isn’t related to the comparability of the items. It’s that we can eliminate half of the search space at every step. The “compare against the target” step can be thought of a black box that replies to queries of the form, “Is this the thing I’m looking for?” with responses of the form, “Yes,” or, “No, but look over here instead.”

As long as the answers to your queries are sufficiently helpful, meaning they allow you to cut out large portions of your search space at each step, then you probably have a good algorithm on your hands. Indeed, there’s a natural model for graphs, defined in a 2015 paper of Emamjomeh-Zadeh, Kempe, and Singhal that goes as follows.

You’re given as input an undirected, weighted graph $G = (V,E)$, with weights $w_e$ for $e \in E$. You can see the entire graph, and you may ask questions of the form, “Is vertex $v$ the target?” Responses will be one of two things:

• Yes (you win!)
• No, but $e = (v, w)$ is an edge out of $v$ on a shortest path from $v$ to the true target.

Your goal is to find the target vertex with the minimum number of queries.

Obviously this only works if $G$ is connected, but slight variations of everything in this post work for disconnected graphs. (The same is not true in general for directed graphs)

When the graph is a line, this “reduces” to binary search in the sense that the same basic idea of binary search works: start in the middle of the graph, and the edge you get in response to a query will tell you in which half of the graph to continue.

And if we make this example only slightly more complicated, the generalization should become obvious:

Here, we again start at the “center vertex,” and the response to our query will eliminate one of the two halves. But then how should we pick the next vertex, now that we no longer have a linear order to rely on? It should be clear, choose the “center vertex” of whichever half we end up in. This choice can be formalized into a rule that works even when there’s not such obvious symmetry, and it turns out to always be the right choice.

Definition: median of a weighted graph $G$ with respect to a subset of vertices $S \subset V$ is a vertex $v \in V$ (not necessarily in $S$) which minimizes the sum of distances to vertices in $S$. More formally, it minimizes

$\Phi_S(v) = \sum_{u \in S} d(v, u)$,

where $d(u,v)$ is the sum of the edge weights along a shortest path from $v$ to $u$.

And so generalizing binary search to this query-model on a graph results in the following algorithm, which whittles down the search space by querying the median at every step.

Algorithm: Binary search on graphs. Input is a graph $G = (V,E)$.

• Start with a set of candidates $S = V$.
• While we haven’t found the target and $|S| > 1$:
• Query the median $v$ of $S$, and stop if you’ve found the target.
• Otherwise, let $e = (v, w)$ be the response edge, and compute the set of all vertices $x \in V$ for which $e$ is on a shortest path from $v$ to $x$. Call this set $T$.
• Replace $S$ with $S \cap T$.
• Output the only remaining vertex in $S$

Indeed, as we’ll see momentarily, a python implementation is about as simple. The meat of the work is in computing the median and the set $T$, both of which are slight variants of Dijkstra’s algorithm for computing shortest paths.

The theorem, which is straightforward and well written by Emamjomeh-Zadeh et al. (only about a half page on page 5), is that this algorithm requires only $O(\log(n))$ queries, just like binary search.

Before we dive into an implementation, there’s a catch. Even though we are guaranteed only $\log(n)$ many queries, because of our Dijkstra’s algorithm implementation, we’re definitely not going to get a logarithmic time algorithm. So in what situation would this be useful?

Here’s where we use the “theory” trick of making up a fanciful problem and only later finding applications for it (which, honestly, has been quite successful in computer science). In this scenario we’re treating the query mechanism as a black box. It’s natural to imagine that the queries are expensive, and a resource we want to optimize for. As an example the authors bring up in a followup paper, the graph might be the set of clusterings of a dataset, and the query involves a human looking at the data and responding that a cluster should be split, or that two clusters should be joined. Of course, for clustering the underlying graph is too large to process, so the median-finding algorithm needs to be implicit. But the essential point is clear: sometimes the query is the most expensive part of the algorithm.

Alright, now let’s implement it! The complete code is on Github as always.

## Always be implementing

We start with a slight variation of Dijkstra’s algorithm. Here we’re given as input a single “starting” vertex, and we produce as output a list of all shortest paths from the start to all possible destination vertices.

We start with a bare-bones graph data structure.

from collections import defaultdict
from collections import namedtuple

Edge = namedtuple('Edge', ('source', 'target', 'weight'))

class Graph:
# A bare-bones implementation of a weighted, undirected graph
def __init__(self, vertices, edges=tuple()):
self.vertices = vertices
self.incident_edges = defaultdict(list)

for edge in edges:
self.add_edge(
edge[0],
edge[1],
1 if len(edge) == 2 else edge[2]  # optional weight
)

def add_edge(self, u, v, weight=1):
self.incident_edges[u].append(Edge(u, v, weight))
self.incident_edges[v].append(Edge(v, u, weight))

def edge(self, u, v):
return [e for e in self.incident_edges[u] if e.target == v][0]

And then, since most of the work in Dijkstra’s algorithm is tracking information that you build up as you search the graph, we define the “output” data structure, a dictionary of edge weights paired with back-pointers for the discovered shortest paths.

class DijkstraOutput:
def __init__(self, graph, start):
self.start = start
self.graph = graph

# the smallest distance from the start to the destination v
self.distance_from_start = {v: math.inf for v in graph.vertices}
self.distance_from_start[start] = 0

# a list of predecessor edges for each destination
# to track a list of possibly many shortest paths
self.predecessor_edges = {v: [] for v in graph.vertices}

def found_shorter_path(self, vertex, edge, new_distance):
# update the solution with a newly found shorter path
self.distance_from_start[vertex] = new_distance

if new_distance < self.distance_from_start[vertex]:
self.predecessor_edges[vertex] = [edge]
else:  # tie for multiple shortest paths
self.predecessor_edges[vertex].append(edge)

def path_to_destination_contains_edge(self, destination, edge):
predecessors = self.predecessor_edges[destination]
if edge in predecessors:
return True
return any(self.path_to_destination_contains_edge(e.source, edge)
for e in predecessors)

def sum_of_distances(self, subset=None):
subset = subset or self.graph.vertices
return sum(self.distance_from_start[x] for x in subset)

The actual Dijkstra algorithm then just does a “breadth-first” (priority-queue-guided) search through $G$, updating the metadata as it finds shorter paths.

def single_source_shortest_paths(graph, start):
'''
Compute the shortest paths and distances from the start vertex to all
possible destination vertices. Return an instance of DijkstraOutput.
'''
output = DijkstraOutput(graph, start)
visit_queue = [(0, start)]

while len(visit_queue) > 0:
priority, current = heapq.heappop(visit_queue)

for incident_edge in graph.incident_edges[current]:
v = incident_edge.target
weight = incident_edge.weight
distance_from_current = output.distance_from_start[current] + weight

if distance_from_current <= output.distance_from_start[v]:
output.found_shorter_path(v, incident_edge, distance_from_current)
heapq.heappush(visit_queue, (distance_from_current, v))

return output

Finally, we implement the median-finding and $T$-computing subroutines:

def possible_targets(graph, start, edge):
'''
Given an undirected graph G = (V,E), an input vertex v in V, and an edge e
incident to v, compute the set of vertices w such that e is on a shortest path from
v to w.
'''
dijkstra_output = dijkstra.single_source_shortest_paths(graph, start)
return set(v for v in graph.vertices
if dijkstra_output.path_to_destination_contains_edge(v, edge))

def find_median(graph, vertices):
'''
Compute as output a vertex in the input graph which minimizes the sum of distances
to the input set of vertices
'''
best_dijkstra_run = min(
(single_source_shortest_paths(graph, v) for v in graph.vertices),
key=lambda run: run.sum_of_distances(vertices)
)
return best_dijkstra_run.start

And then the core algorithm

QueryResult = namedtuple('QueryResult', ('found_target', 'feedback_edge'))

def binary_search(graph, query):
'''
Find a target node in a graph, with queries of the form "Is x the target?"
and responses either "You found the target!" or "Here is an edge on a shortest
path to the target."
'''
candidate_nodes = set(x for x in graph.vertices)  # copy

while len(candidate_nodes) > 1:
median = find_median(graph, candidate_nodes)
query_result = query(median)

if query_result.found_target:
return median
else:
edge = query_result.feedback_edge
legal_targets = possible_targets(graph, median, edge)
candidate_nodes = candidate_nodes.intersection(legal_targets)

return candidate_nodes.pop()

Here’s an example of running it on the example graph we used earlier in the post:

'''
Graph looks like this tree, with uniform weights

a       k
b     j
cfghi
d     l
e       m
'''
G = Graph(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
'j', 'k', 'l', 'm'],
[
('a', 'b'),
('b', 'c'),
('c', 'd'),
('d', 'e'),
('c', 'f'),
('f', 'g'),
('g', 'h'),
('h', 'i'),
('i', 'j'),
('j', 'k'),
('i', 'l'),
('l', 'm'),
])

def simple_query(v):
ans = input("is '%s' the target? [y/N] " % v)
if ans and ans.lower()[0] == 'y':
return QueryResult(True, None)
else:
print("Please input a vertex on the shortest path between"
" '%s' and the target. The graph is: " % v)
for w in G.incident_edges:
print("%s: %s" % (w, G.incident_edges[w]))

target = None
while target not in G.vertices:
target = input("Input neighboring vertex of '%s': " % v)

return QueryResult(
False,
G.edge(v, target)
)

output = binary_search(G, simple_query)
print("Found target: %s" % output)

The query function just prints out a reminder of the graph and asks the user to answer the query with a yes/no and a relevant edge if the answer is no.

An example run:

is 'g' the target? [y/N] n
Please input a vertex on the shortest path between 'g' and the target. The graph is:
e: [Edge(source='e', target='d', weight=1)]
i: [Edge(source='i', target='h', weight=1), Edge(source='i', target='j', weight=1), Edge(source='i', target='l', weight=1)]
g: [Edge(source='g', target='f', weight=1), Edge(source='g', target='h', weight=1)]
l: [Edge(source='l', target='i', weight=1), Edge(source='l', target='m', weight=1)]
k: [Edge(source='k', target='j', weight=1)]
j: [Edge(source='j', target='i', weight=1), Edge(source='j', target='k', weight=1)]
c: [Edge(source='c', target='b', weight=1), Edge(source='c', target='d', weight=1), Edge(source='c', target='f', weight=1)]
f: [Edge(source='f', target='c', weight=1), Edge(source='f', target='g', weight=1)]
m: [Edge(source='m', target='l', weight=1)]
d: [Edge(source='d', target='c', weight=1), Edge(source='d', target='e', weight=1)]
h: [Edge(source='h', target='g', weight=1), Edge(source='h', target='i', weight=1)]
b: [Edge(source='b', target='a', weight=1), Edge(source='b', target='c', weight=1)]
a: [Edge(source='a', target='b', weight=1)]
Input neighboring vertex of 'g': f
is 'c' the target? [y/N] n
Please input a vertex on the shortest path between 'c' and the target. The graph is:
[...]
Input neighboring vertex of 'c': d
is 'd' the target? [y/N] n
Please input a vertex on the shortest path between 'd' and the target. The graph is:
[...]
Input neighboring vertex of 'd': e
Found target: e

## A likely story

The binary search we implemented in this post is pretty minimal. In fact, the more interesting part of the work of Emamjomeh-Zadeh et al. is the part where the response to the query can be wrong with some unknown probability.

In this case, there can be many shortest paths that are valid responses to a query, in addition to all the invalid responses. In particular, this rules out the strategy of asking the same query multiple times and taking the majority response. If the error rate is 1/3, and there are two shortest paths to the target, you can get into a situation in which you see three responses equally often and can’t choose which one is the liar.

Instead, the technique Emamjomeh-Zadeh et al. use is based on the Multiplicative Weights Update Algorithm (it strikes again!). Each query gives a multiplicative increase (or decrease) on the set of nodes that are consistent targets under the assumption that query response is correct. There are a few extra details and some postprocessing to avoid unlikely outcomes, but that’s the basic idea. Implementing it would be an excellent exercise for readers interested in diving deeper into a recent research paper (or to flex their math muscles).

But even deeper, this model of “query and get advice on how to improve” is a classic  learning model first formally studied by Dana Angluin (my academic grand-advisor). In her model, one wants to design an algorithm to learn a classifier. The allowed queries are membership and equivalence queries. A membership is essentially, “What’s its label of this element?” and an equivalence query has the form, “Is this the right classifier?” If the answer is no, a mislabeled example is provided.

This is different from the usual machine learning assumption, because the learning algorithm gets to construct an example it wants to get more information about, instead of simply relying on a randomly generated subset of data. The goal is to minimize the number of queries before the target hypothesis is learned exactly. And indeed, as we saw in this post, if you have a little extra time to analyze the problem space, you can craft queries that extract quite a lot of information.

Indeed, the model we presented here for binary search on graphs is the natural analogue of an equivalence query for a search problem: instead of a mislabeled counterexample, you get a nudge in the right direction toward the target. Pretty neat!

There are a few directions we could take from here: (1) implement the Multiplicative Weights version of the algorithm, (2) apply this technique to a problem like ranking or clustering, or (3) cover theoretical learning models like membership and equivalence queries in more detail. What interests you?

Until next time!

# Bayesian Ranking for Rated Items

Problem: You have a catalog of items with discrete ratings (thumbs up/thumbs down, or 5-star ratings, etc.), and you want to display them in the “right” order.

Solution: In Python

'''
score: [int], [int], [float] -&gt; float

Return the expected value of the rating for an item with known
ratings specified by ratings, prior belief specified by
rating_prior, and a utility function specified by rating_utility,
assuming the ratings are a multinomial distribution and the prior
belief is a Dirichlet distribution.
'''
def score(self, ratings, rating_prior, rating_utility):
ratings = [r + p for (r, p) in zip(ratings, rating_prior)]
score = sum(r * u for (r, u) in zip(ratings, rating_utility))
return score / sum(ratings)

Discussion: This deceptively short solution can lead you on a long and winding path into the depths of statistics. I will do my best to give a short, clear version of the story.

As a working example I chose merely because I recently listened to a related podcast, say you’re selling mass-market romance novels—which, by all accounts, is a predictable genre. You have a list of books, each of which has been rated on a scale of 0-5 stars by some number of users. You want to display the top books first, so that time-constrained readers can experience the most titillating novels first, and newbies to the genre can get the best first time experience and be incentivized to buy more.

The setup required to arrive at the above code is the following, which I’ll phrase as a story.

Users’ feelings about a book, and subsequent votes, are independent draws from a known distribution (with unknown parameters). I will just call these distributions “discrete” distributions. So given a book and user, there is some unknown list $(p_0, p_1, p_2, p_3, p_4, p_5)$ of probabilities ($\sum_i p_i = 1$) for each possible rating a user could give for that book.

But how do users get these probabilities? In this story, the probabilities are the output of a randomized procedure that generates distributions. That modeling assumption is called a “Dirichlet prior,” with Dirichlet meaning it generates discrete distributions, and prior meaning it encodes domain-specific information (such as the fraction of 4-star ratings for a typical romance novel).

So the story is you have a book, and that book gets a Dirichlet distribution (unknown to us), and then when a user comes along they sample from the Dirichlet distribution to get a discrete distribution, which they then draw from to choose a rating. We observe the ratings, and we need to find the book’s underlying Dirichlet. We start by assigning it some default Dirichlet (the prior) and update that Dirichlet as we observe new ratings. Some other assumptions:

1. Books are indistinguishable except in the parameters of their Dirichlet distribution.
2. The parameters of a book’s Dirichlet distribution don’t change over time, and inherently reflect the book’s value.

So a Dirichlet distribution is a process that produces discrete distributions. For simplicity, in this post we will say a Dirichlet distribution is parameterized by a list of six integers $(n_0, \dots, n_5)$, one for each possible star rating. These values represent our belief in the “typical” distribution of votes for a new book. We’ll discuss more about how to set the values later. Sampling a value (a book’s list of probabilities) from the Dirichlet distribution is not trivial, but we don’t need to do that for this program. Rather, we need to be able to interpret a fixed Dirichlet distribution, and update it given some observed votes.

The interpretation we use for a Dirichlet distribution is its expected value, which, recall, is the parameters of a discrete distribution. In particular if $n = \sum_i n_i$, then the expected value is a discrete distribution whose probabilities are

$\displaystyle \left ( \frac{n_0}{n}, \frac{n_1}{n}, \dots, \frac{n_5}{n} \right )$

So you can think of each integer in the specification of a Dirichlet as “ghost ratings,” sometimes called pseudocounts, and we’re saying the probability is proportional to the count.

This is great, because if we knew the true Dirichlet distribution for a book, we could compute its ranking without a second thought. The ranking would simply be the expected star rating:

def simple_score(distribution):
return sum(i * p for (i, p) in enumerate(distribution))

Putting books with the highest score on top would maximize the expected happiness of a user visiting the site, provided that happiness matches the user’s voting behavior, since the simple_score is just the expected vote.

Also note that all the rating system needs to make this work is that the rating options are linearly ordered. So a thumbs up/down (heaving bosom/flaccid member?) would work, too. We don’t need to know how happy it makes them to see a 5-star vs 4-star book. However, because as we’ll see next we have to approximate the distribution, and hence have uncertainty for scores of books with only a few ratings, it helps to incorporate numerical utility values (we’ll see this at the end).

Next, to update a given Dirichlet distribution with the results of some observed ratings, we have to dig a bit deeper into Bayes rule and the formulas for sampling from a Dirichlet distribution. Rather than do that, I’ll point you to this nice writeup by Jonathan Huang, where the core of the derivation is in Section 2.3 (page 4), and remark that the rule for updating for a new observation is to just add it to the existing counts.

Theorem: Given a Dirichlet distribution with parameters $(n_1, \dots, n_k)$ and a new observation of outcome $i$, the updated Dirichlet distribution has parameters $(n_1, \dots, n_{i-1}, n_i + 1, n_{i+1}, \dots, n_k)$. That is, you just update the $i$-th entry by adding $1$ to it.

This particular arithmetic to do the update is a mathematical consequence (derived in the link above) of the philosophical assumption that Bayes rule is how you should model your beliefs about uncertainty, coupled with the assumption that the Dirichlet process is how the users actually arrive at their votes.

The initial values $(n_0, \dots, n_5)$ for star ratings should be picked so that they represent the average rating distribution among all prior books, since this is used as the default voting distribution for a new, unknown book. If you have more information about whether a book is likely to be popular, you can use a different prior. For example, if JK Rowling wrote a Harry Potter Romance novel that was part of the canon, you could pretty much guarantee it would be popular, and set $n_5$ high compared to $n_0$. Of course, if it were actually popular you could just wait for the good ratings to stream in, so tinkering with these values on a per-book basis might not help much. On the other hand, most books by unknown authors are bad, and $n_5$ should be close to zero. Selecting a prior dictates how influential ratings of new items are compared to ratings of items with many votes. The more pseudocounts you add to the prior, the less new votes count.

This gets us to the following code for star ratings.

def score(self, ratings, rating_prior):
ratings = [r + p for (r, p) in zip(ratings, rating_prior)]
score = sum(i * u for (i, u) in enumerate(ratings))
return score / sum(ratings)

The only thing missing from the solution at the beginning is the utilities. The utilities are useful for two reasons. First, because books with few ratings encode a lot of uncertainty, having an idea about how extreme a feeling is implied by a specific rating allows one to give better rankings of new books.

Second, for many services, such as taxi rides on Lyft, the default star rating tends to be a 5-star, and 4-star or lower mean something went wrong. For books, 3-4 stars is a default while 5-star means you were very happy.

The utilities parameter allows you to weight rating outcomes appropriately. So if you are in a Lyft-like scenario, you might specify utilities like [-10, -5, -3, -2, 1] to denote that a 4-star rating has the same negative impact as two 5-star ratings would positively contribute. On the other hand, for books the gap between 4-star and 5-star is much less than the gap between 3-star and 4-star. The utilities simply allow you to calibrate how the votes should be valued in comparison to each other, instead of using their literal star counts.

# The Reasonable Effectiveness of the Multiplicative Weights Update Algorithm

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.

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!