On Computing Nash Equilibria of Borel's Colonel Blotto Game For

On Computing Nash Equilibria of Borel’s Colonel Blotto Game for Multiple Players including in Arbitrary Measure Spaces by Siddhartha Visveswara Jayanti ర శర జయం साथ वेर जयित B.S.E, Princeton University (2017) Submitted to the Department of Electrical Engineering and Computer Science in partial fulfillment of the requirements for the degree of Master of Science in Computer Science and Engineering at the MASSACHUSETTS INSTITUTE OF TECHNOLOGY May 2020 © Massachusetts Institute of Technology 2020. All rights reserved.

Author...... Department of Electrical Engineering and Computer Science May 15, 2020 Certified by ...... Constantinos Daskalakis Professor of Electrical Engineering and Computer Science Thesis Supervisor Accepted by...... Leslie A. Kolodziejski Professor of Electrical Engineering and Computer Science Chair, Department Committee on Graduate Students 2 On Computing Nash Equilibria of Borel’s Colonel Blotto Game for Multiple Players including in Arbitrary Measure Spaces by Siddhartha Visveswara Jayanti

ర శర జయం साथ वेर जयित

Submitted to the Department of Electrical Engineering and Computer Science on May 15, 2020, in partial fulfillment of the requirements for the degree of Master of Science in Computer Science and Engineering

Abstract The Colonel Blotto Problem proposed by Borel in 1921 has served as a widely applicable model of budget-constrained simultaneous winner-take-all competitions in the social sciences. Applications include elections, advertising, research and development, ecology and more. However, the classic Blotto problem and variants that have been studied hereto are only about two-player games on a finite set of discrete battlefields. This thesis extends the classic theory to multiplayer games over arbitrary measureable battlegrounds. Furthermore, it characterizes the symmetric mixed Nash equilibria of a wide array of these generalized Blotto games and provides efficient algorithms to sample from these equilibria; thereby unleashing the potential for quantitative analyses of a multitude of new and significant applications, such as multiparty elections and ideological battles across the surface of the earth.

Thesis Supervisor: Constantinos Daskalakis Title: Professor of Electrical Engineering and Computer Science

3 4 Acknowledgments

A significant portion of this thesis, namely Chapters 1, 2 and 4, are based on a collaboration with Enric Boix and Ben Edelman [10]. I am proud to have had such nice friends through our undergraduate years of study at Princeton and graduate years of study here in Cambridge, MA. None of the results in this thesis would have been possible if it were not for the long sleepless nights of contemplation that the three of us spent together over the course of the last year. I would like to thank my advisor, Costis Daskalakis, for his guidance over the past few years. Costis is an insightful thinker and an amazingly clear teacher. His engaging lectures in the course on Economics and Computation were principally responsible for luring my interest towards game theory. In fact, what started as the final project in his course lasted much longer, and eventually turned into this thesis. I am of course, ever thankful to my undergraduate advisor, Robert Tarjan, who introduced me to the joy and fulfillment of research. I would also like to recognize the contributions of the Hindu Students Council (HSC) to keeping me grounded in the greater meaning of life while I tread my academic path. I would also like to recall the efforts of my family in getting me to where I amtoday. If it were not for the endless hours that my mother and father spent teaching me since my childhood, I would not have gotten to MIT and this place in my education and research. But most importantly, I am supported by the unyielding love of my family—from my mother and father, to my sister and brother-in-law, to my grandparents, uncles, aunts, cousins and beyond. The sacrifices and encouragement of my family, including those that are nolonger with us, were pivotal to my happiness and successes, and to the overall well-being of the family; this, I will always remember in my heart. Finally, I would like to thank the United States Department of Defense that has financially supported my graduate studies through the NDSEG Fellowship program.

सयमेव जयते नानृतं सयेन पथा वततो देवयानः । येनामयृषयो ातकामा य तत् सयय परमं नधानम् ॥ ∼ मुडकोपनषत् ३.१.६

5 6 Chapter 1

Introduction

The Colonel Blotto game, proposed by Borel in 1921 [12], has been a staple of the game theory literature for the past century. In this game, two warring commanders, Captain Alice and Colonel Blotto, fight on n simultaneous and independent battlefields of different values v1, . . . , vn. Each commander has a finite budget of power—BAlice, BBlotto—to split over the battlefields, and for each battlefield i, the value vi, will be awarded to the commander that allocates more power there. Each commander aims to maximize the total expected value of his winnings in this game. The game is called symmetric if all players have the same budget, homogenous if all battlefields have the same value, continuous if allocations can be arbitrary positive real numbers, and discrete if allocations are restricted to be whole numbers. In this thesis, we are primarily interested in computing Nash Equilibria1 of the Colonel Blotto game. The Blotto game has proved to be a treasure trove for mathematicians, computer scientists, and social scientists. While the Blotto game is easy to describe, its equilibrium structure is surprisingly complex even in the mathematically pleasing continuous setting, due to the hard budget constraint.2 Thus, an analysis of the equilibria of the game has only

1A collection of prescribed strategies, one for each player, is in Nash equilibrium if no player can increase his expected winnings by unilaterally deviating from the prescription. If the prescribed strategies are deterministic, the equilibrium is pure; if they are stochasic, the equilibrium is mixed. 2It is well known that even the simplest Blotto games do not admit pure Nash equilibria. Consider the two-player continuous symmetric homogeneous Blotto game with n > 2 battlefields. If Alice fixes any bid vector ⃗a = (a1, . . . , an) (where a1 =6 0 without loss of generality), then Bob can maximize his winnings by ⃗ a1 picking the action b = (0, a2 + ϵ, a3 + ϵ, . . . , an + ϵ), where ϵ = n−1 , to win all but the first battlefield. This pair of actions is not in equilibrium because Alice can switch her action to a⃗′ = ⃗b in order to win half of the total value rather than 1/nth of it. Therefore, in general we are looking for mixed Nash equilibria.

7 emerged in the past few decades due to a long line of research [30, 43, 36, 27, 38, 41] which has culminated in the work of Kovenock and Roberson [28]. When the allocations must be discrete, closed form analyses have given way to fast algorithms that can output equilibria given a problem description; thus, computer scientists have taken the fore, producing several pseudo-polynomial time algorithms for various settings of the discrete Blotto problem [1, 7, 5, 6]. The Blotto game has also seen innumerable applications in the social sciences, ranging from the modeling of political elections [35, 31, 30, 33] to modeling competitions between species in ecological niches [23].

Equilibria for the (hard budget constraint) Blotto game have been developed based on solutions for the simpler soft budget constraint version of the game, called General Lotto. In the Lotto game, each player plays a distribution over allocations instead of a single allocation, and the budget constraint is relaxed so that the allocations need to sum to the budget only in expectation, rather than with probability one. Strategies in the Lotto game are a superset of mixed strategies in the Blotto game, since Lotto players can play strategies that go over- budget sometimes as long as they compensate by playing strategies that go under-budget some other times.3 It turns out that this freedom usually makes Lotto easier to solve in practice than Blotto. In particular, the distribution over strategies for each battlefield can be computed independently for Lotto games, as long as the expected values of the resultant distributions played on each battlefield add up to the budget constraint. In contrast, a mixed-strategy for Blotto with the same marginal allocation distributions per battlefield would additionally need to couple these marginals into a joint distribution whose support only consists of allocation vectors summing to the budget constraint. Demonstration of such a coupling is often the hardest part of an equilibrium computation for a Blotto game.

Modeling two-party elections is a famous application of the Blotto game [30, 36, 33]. Hoping to understand multiparty electoral systems, Myerson alluded to a Blotto game with more than two players in [35], which compares different types of multiparty election systems by studying the equilibrium strategies those systems induce. In this context, the classic plurality vote elections conducted in many parliamentary democracies such as India and the

3 1 ′ For example, if Alice has a budget of BAlice = 1 she can play ⃗a = (1, 1) with probability /2 and a⃗ = (0, 0) 1 with probability /2 in the Lotto game; but cannot in the Blotto game because, ||⃗a||1 = 1+1 = 2 > 1 = BAlice.

8 United Kingdom are naturally modeled by a multiplayer generalization of Colonel Blotto with, e.g., the multiple parties corresponding to players, voting districts corresponding to battlefields, and district advertising expenditures corresponding to the resource allocations. However, stating that “the hardest part of [the Blotto] problem was to construct joint distributions for allocations that always sum to the given total,” Myerson weakened the true budget constraint to the soft one and further assumed that all battlefields must be treated symmetrically by all players, and thus only analyzed what would nowadays be called multiplayer homogeneous symmetric General Lotto. Note that Myerson dealt with General Lotto rather than Colonel Blotto precisely because it is easier to deal with: “The advantage of my simplified formulation is that it will enable us to go beyond this ‘Colonel Blotto’ literature and get results about more complicated situations in which more than two candidates are competing.” While Lotto estimates Blotto in the regime of large n (by law of large numbers), it is a poor approximation in the regime of small n. In this thesis, we obtain results in the rich regime of Blotto that has remained open for nearly 30 years. Furthermore, the wide applicability of the Blotto game in modeling social competitions over discrete sets of n items begs the natural question: What about competitions over other sets, such as ecological competitions over the area of a forest or ideological battles over the surface of the earth? This thesis provides the mathematical underpinnings for such problems by extending the definition of the Blotto problem to arbitrary measure spaces and solving for equilibria in these settings.

1.1 Contributions

In this thesis we define significant generalizations of the Blotto game to multiple players and arbitrary battlegrounds and analyze the equilibria of the games when all players are symmetric.

1. Prior to this work, Colonel Blotto problems have only been studied for two players. In Chapter 2, we define the natural multiplayer generalization of Colonel Blotto in

which there are k ≥ 2 players with budgets B1,..., Bk. Once again, each battlefield is won by the player that makes the highest bid on it. The game serves as a natural

9 model for several of the famous applications studied in the two-player case, including the electoral competition application suggested by Myerson.

We focus on the symmetric case of multiplayer Blotto, where all players have the same budget, and construct efficiently-sampleable symmetric Nash equilibria for various settings of number of battlefields n and number of players k:

(a) We give equilibria for any number of players k whenever the battlefields can be partitioned into k sets of equal value (Theorem 2.2). Furthermore, we provide an O(n) time algorithm for sampling the mixed strategy (Algorithm 1).

(b) We give equilibria for symmetric three-player Blotto whenever no battlefield ac- counts for more than one third of the value of all battlefields (Theorem 2.4). Importantly, we again provide an O(n) time algorithm for sampling the mixed strategy (Algorithm 2).

2. In Chapter 3, we generalize the problem further so that the battleground can be any measurable set, Ω, rather than just a finite set of n battlefields. For instance, we allow the competition to be over the interval [0, 1] or over the land area of the earth. Our model equips Ω with two measures: one measures budget and the other measures the value of the battleground. The rules are modified so that each player i ∈ [k] plays a

function over Ω that integrates to at most Bi in the budget measure, while winning the value measure of the set in which his function is the largest of all those played. We show the following important results for the symmetric version of the game:

(a) We derive equilibria of the game for any number of symmetric-players k whenever the battleground can be partitioned into k sets of equal value (Theorem 3.5). Moreover, we extend Algorithm 1 to provide an O(k) time algorithm for sampling the mixed strategy (Algorithm 3). This result can be used for almost any continuous Blotto game, e.g. Blotto over the unit interval [0, 1], Blotto over the surface

of the earth (S2), or Blotto over a three-dimensional air space.

(b) We prove that the equilibria defined in Theorem 3.5 are the unique symmetric equilibria for all Blotto games over compact intervals of the real line (Theo-

10 rem 3.13). To our knowledge, this uniqueness result is the first of its kind for any type of Blotto game. (Previously uniqueness results have only been proved for General Lotto versions of games.)

3. Finally, in Chapter 4, we introduce and analyze a simple variant of the multiplayer Blotto game which we call the Boolean Colonel Blotto game. Boolean Blotto is the same as normal multiplayer Blotto except players have integer budgets and their bids on each battlefield are restricted to be 0 or 1. In other words, players choose which subset of battlefields to compete on (i.e., bid 1 on). The value of each battlefield is, as in Blotto, split evenly among the players who bid the most on it. We show the following important results:

(a) We give equilibria for all values of k for symmetric Boolean Blotto regardless of battlefield valuations (Theorem 4.9).

(b) The equilibria from (a) involve probabilities whose values maybe NP-hard to compute. So, we describe a polynomial-time approximation scheme for sampling the strategies (Algorithm 4).

1.2 Motivation

In the century after its introduction by Borel, the Blotto game has seen a multitude of applications. Many of these naturally generalize to the multiplayer setting. Some are even more natural to consider with many players. Here are just a few examples:

Elections: k candidates or parties compete across n winner-take-all districts [35, 31, 30, 33]. k = 2 corresponds to a two-party system, while k ≥ 3 corresponds to a multi-party system. Each candidate or party must decide how to allocate campaign funds, or candidate time, across districts. One could also consider individual voters in a single- district election to be battlefields, as Myerson did35 in[ ].

R&D: k companies have the ability to use their fixed R&D budgets to research and develop n potential drugs [23, 29]. If the first company to develop the drug will receive the

11 patent and all the profits for that drug, then this is a Blotto game.

Local Monopolies: k competing companies in the same industry want to become the dom- inant player in each of n new local markets. If each market will tend to be dominated by the company that allocates the most resources to the market (due to network effects, for example) then this is a Blotto game.

Advertising: k companies compete to advertise a substitute good to n consumers [22]. Each consumer will probably only purchase one of the substitutes, so each battlefield (consumer) is indeed winner-take-all.

Ecology: k species in a habitat compete to fill n distinct ecological niches [23]. In this setting, if each niche can only be filled by one species, we can potentially think ofthe species as evolving Blotto strategies through natural selection.

There are also substantial mathematical connections between Blotto and simultaneous all-pay auctions [36, 37]. It is natural to consider these in the multiplayer setting. The generalized multiplayer Blotto game is newly introduced in this thesis, but has applications to new and exciting areas of study in the social sciences including:

Data Providers: k mobile data plans compete over the surface of the earth. Consumers in a given place will opt for the data plan that has better signal and service in that place. The cost of providing a given signal strength varies with the terrain and accessibility across the world, and the value of being the best provider increases with population density.

War of Ideologies: k different ideologies compete over the surface of the earth. Thebudget is determined by how difficult it is to compete in a given region of the earth, while value is assigned by how important a given piece of land is (by population density, for instance).

Finally, the Boolean Blotto game is a good model for any Blotto-type situation where whether to compete in a battlefield is a binary decision. For example:

12 Advertising Substitutes: consider an election—perhaps a local election, or party primary— in which there are n issues and the k candidates distinguish themselves by choosing some subset of issues to focus on. Or consider k companies each marketing substitute products (e.g. medications) by highlighting certain features.

Beyond its immediate applications, we introduce Boolean Blotto because it is a simple variation of the standard Blotto game that requires completely different mathematical tech- niques to analyze. We also hope that our solution to this game will help solve discrete valued multiplayer Blotto.

1.3 Prior work

All of the prior work on Blotto deals with only two players and a finite set of n battlefields. The Colonel Blotto game has been the subject of a considerable body of work over the course of a century. The game (both the discrete budget and continuous budget variations) was first introduced, without a general solution, by Émile Borel in1921[12]. This paper was, notably, the first ever in the nascent game theory literature to describe the concepts ofpure and mixed strategies. Borel referred to Blotto as among the simplest games “for which the manners of playing form a doubly infinite continuum.” In 1938, Borel and Ville [13] found equilibria for symmetric homogeneous three-battlefield Blotto. In a pair of papers in 1950, Gross and Wagner [25, 24] found equilibria for all n, including for the heterogeneous setting. During the postwar period, there was a sizable classified military literature on Blotto9 [ ]. In 1981, with applications to financial investment in mind, Bell and Cover introduced what, in modern terminology, would be called the one-battlefield General Lotto game [8]. Myerson, apparently independently, described one-battlefield General Lotto in 1993, inthe context of political economy [35]. Myerson’s paper is very relevant to our work because it appears to be the only prior work that considers generalizing Blotto (or rather, the easier-to- analyze Lotto) to a multiplayer setting. Myerson considers an infinite family of multiplayer generalizations corresponding to different voting systems; the natural multiplayer game we consider corresponds in this taxonomy to the plurality voting system. Myerson derived the unique symmetric Nash equilibria for these multiplayer General Lotto games; these

13 correspond to the marginals of equilibrium strategies in our setting, as in Lemma 2.8. Note that Myerson dealt with General Lotto rather than Colonel Blotto precisely because it is easier to deal with: “The advantage of my simplified formulation is that it will enable usto go beyond this ‘Colonel Blotto’ literature and get results about more complicated situations in which more than two candidates are competing.” In this thesis, we obtain results in these complicated situations in the rich regime of Blotto.

The current century has seen a resurgence of interest in the Blotto game [31, 30]. In a landmark 2006 paper, Roberson found equilibria for all n for the homogeneous non- symmetric setting [36]. A string of recent works has worked towards the still incomplete goal of characterizing solutions to Blotto in the heterogeneous non-symmetric setting [38, 28, 41]. Kovenock and Roberson’s paper [28] includes a survey of the progress on this question.

A simultaneous recent line of work has dealt with the discrete version of Colonel Blotto. In 2008, Hart solved homogeneous symmetric discrete Blotto, and gave the General Lotto game its name [26]. In 2012, [21] solved non-symmetric discrete General Lotto. Also in the discrete setting, Hortala-Vallve and Llorente-Saguer introduced a variant of Blotto in which the two players can value battlefields differently, and identified some pure strategy equilibria for this case [27]. Many other variations of Blotto have been introduced over the decades in both the continuous and discrete settings [42, 39, 29, 23].

Several recent papers have given algorithms for variations of the discrete Blotto game, typically in time polynomial in n and the size of players’ budgets [1, 7, 5, 6]. Our algorithms, on the other hand, only depend polylogarithmically on the budget size.

Another side of the Blotto literature applies the Blotto model to various social science settings. In addition to the early military applications and later political economy and finance applications, Blotto has also been used to study topics such as U.S. presidential elections [33], terrorism [3], phishing [18], and advertising [22]. It is closely related to the study of all-pay auctions [4]. Still another line of work, experimental in nature, tries to determine what strategies people will actually use in real-life Blotto games—see [20] for a survey.

14 1.4 Organization of this Thesis

The remainder of this thesis contains four chapters. In Chapter 2 we formally define and solve cases of the multiplayer continuous Blotto game for a finite discrete set of battlefields. In Chapter 3 we extend the definition of multiplayer Blotto to measure spaces, extend our Nash equilibrium solutions to these spaces, and show some natural spaces in which our Nash equilibrium solutions are unique. In Chapter 4 we formally define the multiplayer Boolean Blotto game and solve the symmetric version of the game. We end with some remarks and open problems in Chapter 5.

15 16 Chapter 2

Colonel Blotto equilibria

In this chapter, we formally define the General Lotto and Colonel Blotto games for multiple players, solve for their equilibria and construct efficient sampling methods for the equilibrium strategies. The Colonel Blotto equilibria are presented in Theorems 2.2 and Theorem 2.4. We begin by formally defining the Blotto game.

Definition 2.1. The multiplayer Colonel Blotto game is specified by a tuple

∈ N ∈ N B⃗ ∈ Rn ∈ Rn k , n , ≥0,⃗v ≥0 ,

where k is the number of players, n is the number of battlefields, Bi is the budget of player i ∈ [k], and vj is the value of the battlefield j ∈ [n]. We denote the sum total of the battlefield P k k n values by V = ⃗v 1 = j=1 vj. ∈ ∈ Rn Each player i [k] plays a bid vector Ai,∗ = (Ai,1,...,Ai,n) ≥0 satisfying the budget constraint X k k ≤ B Ai,∗ 1 = Ai,j i. j∈[n]

Let the bid matrix A = (Ai,j)(i,j)∈[k]×[n] be the matrix whose ith row is Ai,∗. For each i ∈ [k], the payoff function for player i given the bid matrix A of all players is

X X 1(i ∈ arg maxi′∈[k] Ai′,j) Ui(A) := Ui,j(A) := vj · . | arg maxi′ Ai′,j| j∈[n] j∈[n]

17 In words: each battlefield’s value is split evenly among the players who tied for the highest bid on that battlefield. The game is called symmetric if all the player budgets are equal, and homogeneous if all battlefield values are equal.

A result of Dasgupta and Maskin establishes the existence of Nash equilibria for all values of k and n, and guarantees the existence of symmetric equilibria in the symmetric-budget setting [19]. In this thesis we give explicit symmetric equilibria for the symmetric setting. Our first theorem holds for any number of players, but restricted battlefield values:

Theorem 2.2. Suppose that in the Colonel Blotto game with equal budgets Bi = 1, we are given a k-partition π :[n] → [k] of the battlefields such that there is equal value on each set of the partition:

X 1 Xn V v = v = ∀m ∈ [k]. l k l k l∈π−1(m) l=1 Then if each of the players independently runs Algorithm 1, the players will be in Nash equilibrium. Moreover, Algorithm 1 runs in O(n) time.

The following important special case of this theorem immediately follows by defining π(j) := (j mod k) + 1.

Corollary 2.3. Suppose that in the Colonel Blotto game with equal budgets Bi = 1 there are

n = mk battlefields of equal value vi = V /n, for some m ∈ N. Then Algorithm 1 gives a Nash equilibrium in O(n) time.

Our second main theorem holds only for three player games (k = 3 case), but allows us to handle a much wider range of battlefield valuations:

Theorem 2.4. Suppose that in the 3-player Colonel Blotto game with equal budgets Bi = 1, the valuations satisfy V v ≤ , ∀j ∈ [n]. j 3 Then if each of the players independently runs Algorithm 2, the players will be in Nash equilibrium. Moreover, Algorithm 2 runs in O(n) time.

18 The main difficulty in proving Theorems 2.2 and 2.4 is that the strict budget constraint of the Blotto game generally means that a given player’s bid distributions on the various battlefields have to be correlated, so that any sample bid-vector from this joint distribution sums to one. That is, each player i’s bids must be coupled in some potentially complicated P n ≤ B way so that the budget constraint j=1 Ai,j i holds with probability 1 over player i’s mixed strategy. In order to overcome this difficulty, we follow the meta-approach of[26] and prove both theorems by first analyzing the simpler General Lotto game. This is a variant of the Colonel Blotto game in which the budget constraints are relaxed to hold only in expectation over each player’s bids, instead of almost surely:

Definition 2.5. An instance of the General Lotto game is specified by a tuple (k, n, B⃗,⃗v), as in the Colonel Blotto game. However, instead of playing a real-valued bid for each battlefield, each player plays a distribution of bids. For each i ∈ [k] and j ∈ [n], player i plays a

distribution Di,j over R≥0 such that the budget constraint is met in expectation:

Xn E ≤ B Ai,j ∼Di,j [Ai,j] i. j=1

The payoff function for player i ∈ [k] given the bids of all the players is EAUi(A), where for ′ ′ each i ∈ [k], j ∈ [k] the bids Ai′,j′ ∼ Di′,j′ are drawn independently.

Given a Nash equilibrium (Di,j)i∈[k],j∈[n] of the General Lotto problem, our approach

will be to try to convert (Di,j)i,j into a Nash equilibrium of the Colonel Blotto problem. ∈ Rk×n Our objective will be to construct a random variable A ≥0 such that the rows Ai,∗ are

independent of each other, such that Ai,j ∼ Di,j for each i ∈ [k], j ∈ [n], and such that the k k ≤ B ∈ budget constraint Ai,∗ 1 i holds for each i [k] almost surely. These conditions will ensure that A is a mixed Nash equilibrium for the Colonel Blotto problem. We realize this program as follows: in Section 2.1, we characterize symmetric General Lotto equilibria, in Section 2.2 we derive a sufficient condition for symmetric Colonel Blotto equilibria, andin Sections 2.3 and 2.4 we use this sufficient condition to prove Theorems 2.2 and 2.4.

19 2.1 General Lotto equilibria

We now construct symmetric multiplayer General Lotto equilibria. Our construction is similar to Myerson [35], who constructed equilibria for the homogeneous case and proved that they were unique. Similar arguments to [35] would prove uniqueness of our construction in the heterogeneous case, but for the sake of brevity we omit these arguments since they are not necessary in order to obtain sufficient conditions for Colonel Blotto equilibria. First, recall the definition of the Kumaraswamy distribution:

Definition 2.6. For any a, b > 0, the Kumaraswamy(a, b) distribution is the distribution supported on the interval (0, 1) with CDF P(X ≤ x) = 1 − (1 − xa)b. In particular, when b = 1, the CDF reduces of Kumaraswamy(a, 1) is P(X ≤ x) = 1 − xa. For those familiar with the Beta-distribution, it is noteworthy, that Kumaraswamy(a, 1) ≡ Beta(a, 1).

We now describe, informally, how we arrive at the General Lotto equilibria in Lemma 2.7 below, assuming for simplicity that we are in the homogeneous setting considered by Myerson [35], so the battlefields have value 1. We are looking for an equilibrium that exploits the symmetry of the game across players and across battlefields. It is natural to guess that this can be achieved by all k players playing the same single-variable distribution of bids on each of the n battlefields. Denote the cumulative distribution function (CDF) of this distribution by F . In order to derive F , we guess that F has no atoms and is supported in a finite interval [0, θ]. Then we consider what happens once players 1, . . . , k − 1 have fixed their General Lotto strategies to playing F on all n battlefields. Suppose that player k deviates and plays distributions G1,...,Gn on the n battlefields. Since F has no atoms, a tie between the players is a measure-zero event, so the utility derived by player k on battlefield j is

P[Ak,j > maxi∈[k−1] Ai,j], where A1,j,...,Ak−1,j ∼ F and Ak,j ∼ Gj are independent. Hence player k’s payoff on battlefield j depends only on their bid relative to the maximum bid value Mj = maxi∈[k−1] Ai,j of all the other players.

Now, if Mj is not uniform over [0, θ] for some θ, then player k can strictly gain over the other players by playing a slight perturbation F˜ of the distribution F , where ϵ probability mass is moved from values of x where P[M < x]/x is lower to values of x where P[M < x]/x

20 P k−1 x is higher. Therefore, if the players are in equilibrium, [Mj < x] = (F (x)) = min(1, θ ), which implies that for all i ∈ [k − 1] we have

1 F (x) = min 1, (x/θ) k−1 .

One can solve for the scaling parameter θ by requiring that the budget constraint be tightly P n E ∈ − enforced: j=1 [Ai,j] = 1 for any i [k 1]. We note that F is a scaling of the Kumaraswamy(1/(k − 1), 1) distribution.

Lemma 2.7. Consider the continuous symmetric multiplayer General Lotto game (k, n, B⃗ =

⃗1,⃗v) with k ≥ 2 players and equal budgets Bi = 1. Suppose that for each i ∈ [k] and j ∈ [n], D kvj · 1 player i plays distribution i,j = V Kumaraswamy( k−1 , 1) on battlefield j. Then the players are in Nash equilibrium.

Proof. For this proof, let X ∼ Kumaraswamy(1/(k − 1), 1). First, the General Lotto budget constraint is satisfied for all i ∈ [k]

Xn Xn Xn kvj vj E ∼D [A ] = E [X] = = 1 = B . Ai,j i,j i,j V V i j=1 j=1 j=1

D′ D′ Now suppose that player k deviates by playing distributions k,1,..., k,n meeting the ∈ − ∈ ∼ D ∼ D′ General Lotto budget constraint. For all i [k 1], j [n] let Ai,j i,j, and Ak,j k,j be independent random variables. The expected payoff of player k from battlefield j is

E[Uk,j(A) | Ak,∗] = vj · P[∀i ∈ [k − 1],Ai,j ≤ Ak,j | Ak,j] − − kY1 kv k 1 = v · P[A ≤ A | A ] = v · P j · X ≤ A j i,j k,j k,j j V k,j i=1 ! − k−1 V 1/(k 1) = v · min 1, · A j kv k,j j V V = vj · min 1, Ak,j ≤ Ak,j, kvj k

21 where we have used that ties are measure-zero events. Therefore,

X V X V E[U (A)] = E[U (A)] ≤ · E[A ] ≤ k k,j k k,j k j∈[n] j∈[n]

The last inequality is the General Lotto budget constraint. By symmetry between the D′ D ∈ E V players, if k,j = k,j for all j [n] then this upper bound is achieved: [Uk(A)] = k . So playing according to Lemma 2.7 is indeed a Nash equilibrium.

2.2 Sufficient Conditions for Colonel Blotto equilibrium

The General Lotto equilibria of Lemma 2.7 immediately give sufficient conditions for players to be in Colonel Blotto equilibrium:

Lemma 2.8. Consider the symmetric Colonel Blotto game (k, n, B⃗ = ⃗1,⃗v). The players are in equilibrium if each player i ∈ [k] independently bids a random vector of bids Ai,∗ =

(Ai,1,...,Ai,n) such that: P n ≤ B ∼ kvj · − (a) j=1 Ai,j 1 = i. (b) Ai,j V Kumaraswamy(1/(k 1), 1).

Proof. The budget constraints are met by (a). By linearity of expectation, the utilities only depend on the marginal distributions of the players’ bids for each battlefield. So, if any player deviates from the strategy, then by Lemma 2.7 and the fact that any Colonel Blotto strategy is also a General Lotto strategy, the deviating player’s utility cannot improve.

Therefore, we have reduced the problem of computing Colonel Blotto equilibria to the problem of coupling Kumaraswamy-distributed variables so as to satisfy the budget constraint. In the following two sections, we give computationally efficient constructions of such couplings in order to prove Theorems 2.2 and 2.4.

Remark 2.9 (Blotto =6 Lotto). The conditions in Lemma 2.8 are not necessary for players to be in Blotto equilibrium. For example, in the Colonel Blotto game specified by (k = 2, n = 1, B⃗ = 1,⃗v = 1), Lemma 2.8 would require the distribution 2 · Kumaraswamy(1, 1), which is

22 equal to Unif[0, 2], to have support ≤ 1 in order to meet condition (a). Clearly this is not the case, so the conditions of Lemma 2.8 are not satisfied, and yet the Colonel Blotto game still has an equilibrium (in which both players play all of their budget on the one battlefield).

2.3 Couplings for arbitrary numbers of players (Theo- rem 2.2)

We now prove Theorem 2.2 using the sufficient condition of Lemma 2.8. We will make use of a property of the multivariate Beta distribution—also known as the Dirichlet distribution.

Definition 2.10. The Dirichlet distribution Dir(α1, . . . , αm) is the distribution on the (m − Q − ∝ m αi 1 1)-simplex ∆m−1 with density function f(⃗x; ⃗α) i=1 xi .

The following property of the Dirichlet distribution will be useful to us in our construction of couplings.

Proposition 2.11 (folklore, e.g. [32]). Let (X1,...,Xm) ∼ Dir(α1, . . . , αm). Then

P ∈ ∼ (i) For each i [m], Xi Kumaraswamy αi, j≠ i αj .

P m (ii) i=1 Xi = 1 almost surely

Proposition 2.11 implies that the Dirichlet distribution on ∆k−1 with parameters ⃗α = 1 ⃗1 − k−1 has marginals equal to Kumaraswamy(1/(k 1), 1). This leads us to the following algorithm to sample a symmetric Nash equilibrium strategy for each player in a k-player

23 Blotto game where the battlefields can be partitioned into k sets of equal value. Algorithm 1: NashEquilThm2.2: Colonel Blotto Nash Equilibrium for Theo- rem 2.2 Input: a Colonel Blotto game (k, n, B⃗ = 1,⃗v) and a partition function π :[n] → [k] P ∈ satisfying ℓ∈π−1(m) vℓ = V /k for each m [k]. n Output: a sample (A1,...,An) ∈ R from a mixed equilibrium strategy for a single player.

1 Draw (X ,...,X ) ∼ Dir(1/(k − 1),..., 1/(k − 1)). 1 k kvj 2 ← · ∈ Aj V Xπ(j) for all j [n].

3 return (A1,...,An)

Proof of Theorem 2.2. Correctness: The output of Algorithm 1 meets the conditions of Lemma 2.8 and therefore the players are in Nash equilibrium:

(a) Budget constraint:   Xn Xn kv Xk X kv Xk A = j X = X  l  = X i,j V i,π(j) i,m V i,m j=1 j=1 m=1 l∈π−1(m) m=1

which is 1 by Proposition 2.11(ii). ∼ kvj · − (b) Marginal constraint: Ai,j V Kumaraswamy(1/(k 1), 1) by Proposition 2.11(i).

Running time: we can sample the Dirichlet variable in O(n) time, using the method of [2]

∑ Yi to sample n i.i.d variables Yi ∼ Gamma(1/(k − 1), 1) and letting Xi = n for all i ∈ [n]. l=1 Yl

2.4 Couplings for 3 players (Theorem 2.4)

We prove Theorem 2.4, which vastly improves over Theorem 2.2 (from the previous section) in the k = 3 case. The proof of this theorem is much more involved, and is inspired by the

relationship between the Dirichlet distribution and the Lp-norm uniform distribution defined in [17].

24 In particular, [40] proves that given (U1,...,Um) drawn from the m-dimensional Lp-norm p p uniform distribution, then (|U1| ,..., |Um| ) is distributed as Dir(1/p, . . . , 1/p). Therefore,

for the construction of Theorem 2.2, in order to draw (X1,...,Xk) from the Dir(1/(k − k−1 k−1 1),..., 1/(k − 1)) distribution, we could have set (X1,...,Xk) = (|U1| ,..., |Uk| ) for

(U1,...,Uk) drawn from the Lk−1-norm uniform distribution. The k = 3 case is very special,

because the Lk−1 = L2-norm uniform distribution is the uniform distribution on the unit L2 sphere, and therefore it is rotationally symmetric. We will take advantage of the rotational

symmetry of the uniform distribution on the L2 sphere in order to handle a much wider range of battlefield valuations in Theorem 2.4 when k = 3. We summarize this intuition by stating the following remarkable geometric fact:

Proposition 2.12. Let U ∈ R3 be a point drawn uniformly at random from the surface of P 3 2 ∈ R3 · the unit sphere l=1 Ul = 1. Let c . Then the inner product c U is distributed as

c · U ∼ Unif[−kck, kck], and so (c · U)2 ∼ kck2 · Kumaraswamy(1/2, 1).

c · Proof. By the rotational symmetry of U, the inner product ∥c∥ U is equal in distribution

to U1. Since U1 ∼ Unif[−1, 1] (see e.g., Theorem 2.1 of [40]), c · U ∼ Unif[−kck, kck] follows. √ (c·U)2 P (c·U)2 ≤ ∈ Therefore the CDF of ∥c∥2 is [ ∥c∥2 a] = a for any a [0, 1]. This proves that (c·U)2 ∼ ∥c∥2 Kumaraswamy(1/2, 1).

The analysis of Algorithm 2, which constructs the equilibrium for Theorem 2.4, will depend on this proposition. In short, the algorithm samples a vector U ∈ R3 uniformly from

3 n the unit sphere S2 ⊂ R . It then maps U into R with a linear isometry described by a matrix M. Finally, it outputs the coordinate-wise square of this point. In order to ensure correctness, the algorithm must use an isometry M that has squared row norms proportional to the battlefield valuations. Finding such an M is the core technical challenge, and it is accomplished by the helper algorithm ConstructM, which constructs an M that has the following guarantee (proof deferred): P ≤ ≤ ∈ Z n Claim 2.13. Given values 0 s1, . . . , sn 1 and m such that j=1 sj = m, the n×m T method ConstructM returns in O(nm) time a matrix M ∈ R such that M M = Im and 2 kMj,∗k = sj, for all j ∈ [n]. (Here Mj,∗ denotes the jth row of M.)

25 Assuming Claim 2.13, we prove the correctness of NashEquilThm2.4:

Proof of Theorem 2.4. Correctness: The inputs to ConstructM in step 2 of Algorithm 2 P ≤ ≤ n satisfy the prerequisites 0 s1, . . . , sn 1 and j=1 sj = m = 3. Therefore, the matrix n×3 T M ∈ R is guaranteed to have the following properties by Claim 2.13: M M = I3 and k k2 3vj ∈ ∈ Mj,∗ = V for all j [n]. So if each player i [3] bids (Ai,1,...,Ai,n) by independently 3 running Algorithm 2 with a random sphere point Ui ∈ R , then the sufficient conditions of Lemma 2.8 are met: P P n n · 2 k k2 T T k k2 (a) Budget constraint: j=1 Ai,j = j=1(Mj,∗ Ui) = MUi = Ui M MUi = Ui = T 1, using M M = I3 and the fact that Ui is on the unit sphere.

(b) Marginal constraint: We see that

2 2 3vj A = (M ∗ · U ) ∼ kM ∗k · Kumaraswamy(1/2, 1) = · Kumaraswamy(1/2, 1) i,j j, i j, V

2 by Proposition 2.12 and kMj,∗k = 3vj/V .

Running time: The call to ConstructM with m = 3 in step 2 takes O(n) time by Claim 2.13. Sampling U ∈ R3 in step 3 takes O(1) time, for example using the algorithm of [34]. And finally step 4 takes 6n multiplications and additions. So the total running time is O(n).

2.5 Proof of Claim 2.13 (ConstructM correctness)

The algorithm ConstructM greedily updates a matrix M ∈ Rn×m using the helper algorithm RotatePair until the desired properties of M are achieved. M is initialized to the matrix

T (Im, 0) . Each application of RotatePair applies a linear rotation transformation to a pair of rows from M such that at least one of these rows becomes scaled correctly, while the column-orthogonality of M is maintained. We prove the correctness of the RotatePair subroutine, and then give an invariant argument that demonstrates that greedily applying RotatePair works.

26 Algorithm 2: NashEquilThm2.4: Nash Equilibrium for Theorem 2.4 ≥ ≥ · · · ≥ ≥ Input: The number of battlefields n and the battlefieldP valuations v1 v2 vn 0 such ≤ 1 ∈ n that vj 3 V for all j [n]. Recall that V = k=1 vk. T n Output: A bid vector A = (A1,...,An) ∈ R for the n battlefields that is sampled from a distribution satisfying Lemma 2.8 for the Blotto game G = (3, n, ⃗1,⃗v).

1 Function SampleBid(⃗v) is n×3 3 2 ∈ R ← · Construct M by running: M ConstructM V ⃗v; m = 3 3 3 Sample U ∈ R uniformly at random from the unit ℓ2-sphere S2 = {x | kxk2 = 1}. 2 4 return Aj ← (Mj,∗ · U) for all j ∈ [n]. 5 end P ≤ ≤ ∈ N n Input: Values 0 s1, s2, . . . , sn 1 and m such that j=1 sj = m. n×m T 2 Output: M ∈ R such that M M = Im and kMj,∗k = sj, for all j ∈ [n].

6 Function ConstructM(⃗s,m) is ′ 7 Permute the indices of ⃗s so that sr ≥ sr′ for each r ∈ [m] and r ∈ [n] \ [m]. n×m 8 Initialize M ∈ R as Mi,i = 1 for all i ∈ [m], and 0 everywhere else. 9 j ← 1, l ← m + 1. 10 while j ≤ m and l ≤ n do 11 w1, w2 ← RotatePair(u1 = Mj,∗, u2 = Ml,∗, t1 = sj, t2 = sl). 12 Mj,∗ ← w1. Ml,∗ ← w2. 2 13 if kMj,∗k = sj then j ← j + 1. 2 14 if kMl,∗k = sl then l ← l + 1. 15 end 16 Undo the row permutation from step 7. 17 return M. 18 end

m 2 2 Input: Vectors u1, u2 ∈ R , and targets t1, t2 ∈ R such that ku1k ≥ t1 ≥ t2 ≥ ku2k and u1 · u2 = 0. m Output: w1, w2 ∈ R that are (i) supported on supp(u1) ∪ supp(u2) such that (ii) m×2 m×2 T T W = ( w1 w2 ) ∈ R and U = ( u1 u2 ) ∈ R satisfy WW = UU , (iii) 2 2 2 kw1k ≥ t1 ≥ t2 ≥ kw2k and (iv) there is k ∈ [2] such that kwkk = tk.

19 Function RotatePair(u1, u2, t1, t2) is 2 2 20 if ku1k = ku2k then a ← 1, b ← 0 2 2 21 if ku k − t ≥ t − ku k then 1 q 1 2 2 √ ∥ ∥2− ← u1 t2 ← − 2 22 a 2 2 , b 1 a . ∥u1∥ −∥u2∥ 23 else q √ −∥ ∥2 ← t1 u2 ← − 2 24 a 2 2 , b 1 a . ∥u1∥ −∥u2∥ 25 end 26 return w1 = au1 − bu2, w2 = bu1 + au2. 27 end

27 Claim 2.14. RotatePair is correct and runs in O(m) time.

2 2 2 2 Proof. If ku1k = ku2k , then we must also have ku1k = t1 = t2 = ku2k , so returning

(w1, w2) ← (u1, u2) is correct. Otherwise, items (i)-(iv) still hold:

(i) supp(w1) ∪ supp(w2) ⊆ supp(u1) ∪ supp(u2) since w1, w2 are a linear combination of u1, u2.   a b ∈ Rm×2 ∈ Rm×2   (ii) W = ( w1 w2 ) and U = ( u1 u2 ) are related by W = U , −b a so       a b a −b a2 + b2 0 WW T = U     U T = U   U T = UU T , −b a b a 0 a2 + b2 since a2 + b2 = 1.

(iii and iv) There are two cases to consider. Note that since u1 · u2 = 0, we have 2 2 2 2 2 2 2 2 2 2 kw1k = a ku1k + b ku2k and kw2k = b ku1k + a ku2k :

2 2 • If ku1k − t1 ≥ t2 − ku2k , then

(ku k2 − t )ku k2 (t − ku k2)ku k2 k k2 1 2 1 2 2 2 k k2 k k2 − ≥ w1 = 2 2 + 2 2 = u1 + u2 t2 t1 ku1k − ku2k ku1k − ku2k

(t − ku k2)ku k2 (ku k2 − t )ku k2 k k2 2 2 1 1 2 2 w2 = 2 2 + 2 2 = t2. ku1k − ku2k ku1k − ku2k

2 2 • If ku1k − t1 < t2 − ku2k , then

(t − ku k2)ku k2 (ku k2 − t )ku k2 k k2 1 2 1 1 1 2 w1 = 2 2 + 2 2 = t1 ku1k − ku2k ku1k − ku2k

(ku k2 − t )ku k2 (t − ku k2)ku k2 k k2 1 1 1 1 2 2 k k2 k k2 − ≥ w2 = 2 2 + 2 2 = u1 + u2 t1 t2. ku1k − ku2k ku1k − ku2k

And in both cases conditions (iii) and (iv) hold. The running time is O(m), because we just compute the norms of two vectors of size m and output a linear combination of the vectors.

Proof of Claim 2.13. Correctness: We analyze the algorithm by proving several invariants

28 on M, j, l. These hold at step 10.

Invariant 1: sj ≥ sl.

T Invariant 2: The columns of M are orthonormal. M M = Im.

′ 2 2 Invariant 3: For all r ∈ {j, . . . , m} and r ∈ {l, . . . , n}, kMr,∗k ≥ sr and kMr′,∗k ≤ sr′ . ′ Invariant 4: For all r ∈ {j, . . . , m} and r ∈ {l, . . . , n}, Mr,∗ · Mr′,∗ = 0. The invariants clearly hold when the algorithm first reaches step 10. We prove that they are maintained on each iteration. Let M, j, l be the states of the variables before running an iteration of the while loop, and M ′, j′, l′ the states of the variables after. If M, j, l respect

2 the invariants, then the preconditions of RotatePair are met, because kMj,∗k ≥ sj ≥ sl ≥ 2 kMl,∗k by Invariants 1 and 3, and Mj,∗ · Ml,∗ = 0 by Invariant 4. Invariant 1: This follows from the preprocessing in step 7, because j ∈ [m] and l ∈ [n] \ [m]. Invariant 2: Notice that for all a, b ∈ [m],

X ′ T ′ ′ ′ ′ ′ ′ ′ − − T ((M ) M )ab = McaMcb = (MjaMjb + MlaMlb MjaMjb MlaMlb) + (M M)ab c∈[n]

T ′ ′ ′ ′ − − So since M M = Im by Invariant 2, it suffices to show that MjaMjb + MlaMlb MjaMjb

MlaMlb = 0 for all a, b. This is precisely the condition that T T ′ ′ ′ ′ Mj,∗ Ml,∗ Mj,∗ Ml,∗ = Mj,∗ Ml,∗ Mj,∗ Ml,∗ , which is guaranteed by item (ii) of RotatePair. Invariant 3: Since j and l are the only rows modified from the previous step, and j′ ≥ j, l′ ≥ l, it suffices to consider rows j and l. For these, item (iii) of RotatePair guarantees k ′ k2 ≥ ≥ k ′ k2 that Mj,∗ sj and sl Ml,∗ . Invariant 4: Item (iv) of RotatePair guarantees that at least one of j and l is incremented ′ ′ ≥ ′ on each step. If j > j, then the invariant holds, because the vectors Mr,∗ for r j are { } ′ supported on coordinates r, . . . , m , while by item (i) of RotatePair the vectors Mr′,∗ for r′ ≥ l′ ≥ l are supported on coordinates {1, . . . , k − 1}. Otherwise, if j′ = j then l′ > l, and ′ ′ ≥ ′ ′ · ′ ∈ { ′ } the vectors Mr′,∗ for r j are all 0. So in both cases Mr,∗ Mr′,∗ = 0 for all r i , . . . , m

29 and r′ ∈ {j′, . . . , n}. Therefore Invariants 1 through 4 are maintained by the algorithm. Notice that the row

2 2 index j (respectively, l) is only incremented if kMj,∗k = sj (respectively, kMl,∗k = sl), and 2 after that the row is no longer modified. So if the algorithm ever exits, then kMr,∗k = sr for all r ∈ {1, . . . , j − 1} ∪ {m + 1, . . . , l − 1}. Now, if the algorithm exits, then j = m + 1 and/or l = n + 1. If j = m + 1, we have by Invariant 2

X X X T 2 2 m = trace(M M) = kMr,∗k = sr + kMr,∗k , since j = m + 1 r∈[n] r∈[l−1] r∈[n]\[l−1] X ≤ sr = m, by Invariant 3. r∈[n]

2 If there were k ∈ [n]/[j − 1] such that kMk,∗k < sk then the inequality in the last line 2 would be strict. So we may conclude that kMk,∗k = sk for all k ∈ [n]. Combining this with T the knowledge that M M = Im by Invariant 2, we have shown that if i ever reaches n + 1, then the output is correct. Similarly, if j ever reaches m + 1, we may also argue that the output is correct. So it suffices to prove that the program terminates. This is true because item (iv) of RotatePair guarantees that either i or j is incremented on each step, and so the loop terminates after at most n iterations. Running time: The initialization steps (including the permutation of the rows in steps 7 and 16) take O(mn) time, and each of the ≤ n iterations of the loop takes O(m) time (because RotatePair takes O(m) time). So the algorithm runs in O(mn) total time.

30 Chapter 3

Generalized Blotto Game

In the standard multiplayer Colonel Blotto game, players compete over a finite discrete set of n battlefields. In this chapter, we define the generalized version of the game, whereplayers compete over arbitrary measurable sets Ω. In the Generalized Multiplayer Blotto Game, there are k-players playing over a measure- space Ω equipped with two measures β and υ that are absolutely continuous with respect to each other. For reasons that will become clear shortly hereafter, we call β the budget measure and υ the value measure. (For a first time reader, it may be most illustrative to imagine Ω to be the unit-interval [0, 1] ⊆ R with β = υ being the standard Lebesgue measure

λ on the interval.) Each player i, plays a measureable function ϕi :Ω → R+, and wins a utility Ui that is the value measure of the subset of Ω on which it’s function is the maximum. If multiple players tie for the maximum, they split the utility evenly amongst themselves.

Of course, if no constraint is placed on the function ϕi, players will choose a function that tends to infinity everywhere on the domain. So, we define a budget constraint vector B⃗, R ≤ B B and restrict each ϕi to satisfy Ω ϕidβ i, i.e., it integrates to at most i by the budget measure. We formalize this explanation in Definition 3.1.

Definition 3.1. An instance of the Generalized Blotto Game is defined by the tuple:

∈ N B⃗ ∈ Rn k , Ω, +, β, υ

Here, k is the number of players, Ω is a measure-space known as the battleground, Bi is the

31 budget of player i for each i ∈ [k], β is the finite budget measure on Ω, and υ is the finite value measure on Ω; we impose that β and υ must be absolutely continuous with respect to each other. We call the game symmetric if B⃗ = ⃗1, and homogeneous if both β and υ are the uniform measure on the space Ω. We denote the total value of the battleground to be Υ ≜ υ(Ω), and impose that this be a bounded quantity.

Each player i ∈ [k] plays a measureable bid function ϕi :Ω → R+ satisfying the budget constraint Z

ϕi(x)dβ(x) ≤ Bi. Ω

For a given bid profile ϕ = (ϕi)i∈[k], the utility of player i is given by the expression Z

1 i ∈ arg max{ϕj(x)} j∈[k] Ui(ϕ) = dυ(x)

arg max{ϕj(x)} Ω j∈[k]

Often, we will need to talk about mixed strategies for the generalized Blotto game. In

this case, we will consider players playing random variables Φ = (Φi)i∈[k], where each Φi is a distribution over valid functions ϕi.

We have seen in Chapter 2, that multiplayer Colonel Blotto games are intimately tied to their cousin-version of the General Lotto game. Even in the generalized context, we can define such a cousin for the Generalized Blotto game, which wecallthe Generalized Lotto game.

∈ N B⃗ ∈ Rn Definition 3.2. An instance of Lotto is also specified by the same tuple k , Ω, +, β, υ .

In a game of Lotto, each player i plays a distribution-over-functions, Φi, that satisfies the property that

E Φi(x)dβ(x) ≤ Bi. Ω

Given a strategy profile Φ = (Φi)i∈[k], fixed functions (ϕi ∼ Φi)i∈[k] are sampled independently from the distribution over functions and each player i’s pay off is calculated as

32 Z

1 i ∈ arg max{ϕj(x)} j∈[k] Ui(ϕ) = dυ(x).

arg max{ϕj(x)} Ω j∈[k]

And thus the expected utility for player i is Ui(Φ) = Eϕ∼Φ(Ui(ϕ)).

As a sanity check and warm-up, we first verify that Generalized Blotto is indeed a constant sum multiplayer game, and thereby that it indeed partitions the total value of the battle ground amongst the k competing players.

Lemma 3.3. For any bid profile ϕ = (ϕ , . . . , ϕ ) played in a k-player Generalized Blotto 1 k ∈ N B⃗ ∈ Rn game k , Ω, +, β, υ , the total sum of the utilities is the total value of the battleground, i.e., Xk Ui(ϕ) = Υ = υ(Ω) i=1

Proof. We prove the lemma via the linearity of expectation. Z 1 ∈ { } Xk Xk i arg max ϕj(x) j∈[k] Ui(ϕ) = dυ(x)

i=1 i=1 arg max{ϕj(x)} Ω j∈[k] Z Xk 1 i ∈ arg max{ϕj(x)} j∈[k] i=1 = dυ(x)

arg max{ϕj(x)} Ω j∈[k] Z = 1 dυ(x) Ω

= υ(Ω) = Υ

Notice that the proof of Lemma 3.3 does not depend on the functions ϕ satisfying any budget constraints, and therefore applies to General Lotto also.

33 We now turn to identifying equilibria of the generalized Blotto. Our first observation is that Generalized Blotto captures standard two-player and multiplayer Blotto games. For instance, if we let Ω = [n], let β be the counting measure on n discrete points, and let υ({j}) 7→ v for each j ∈ [b], we turn the Generalized Blotto game k, Ω, B⃗, β, υ into the j standard multiplayer Colonel Blotto game k, n, B⃗,⃗v . This observation also implies that even the simplest generalized blotto games do not admit pure Nash equilibria. So, we are interested in solving, and if possible characterizing mixed Nash equilibria for generalized blotto games.

Given a random variable over functions, Φi :Ω → R+, we can take its restriction to a given point x ∈ Ω to obtain a random variable Φi(x). We call this random variable the marginal of Φi at the point x. In the next lemma, we formulate a sufficient condition for a mixed strategy profile Φ to be an equilibrium of a symmetric Blotto or Lotto game. Quite nicely, this condition only depends upon the point-wise marginals Φi(x), but not on the full distribution of the random functions Φi themselves. We will later use the freedom inherent in this sufficient condition to find, and in some cases characterize, mixed equilibria ofBlotto games.

Lemma 3.4. Let Φ = (Φ ) ∈ be a valid mixed strategy profile for the Generalized k-player i i [k] Blotto or Lotto game k, Ω, ⃗1, β, υ . If

k dυ 1 Φ (x) ∼ · (x) · Kumaraswamy , 1 i Υ dβ k − 1

for every i ∈ [k] and almost every x ∈ Ω, and Φ is indeed a valid strategy profile for the given Blotto or Lotto game, then Φ is a mixed Nash equilibrium of the game.

Proof. Let Φ be a mixed strategy profile satisfying the conditions of the theorem. Let X ⊆ Ω be the set of points in the battleground for which all the players satisfy the marginal constraint in the theorem. We know from the hypothesis that υ(X ) = υ(Ω) = Υ and β(X ) = β(Ω) = 1. We will show that if players in the set [k − 1] fix their strategies, then player k gains no

expected utility by changing his strategy. To this end, let Φmax ≜ max Φi be the random i∈[k−1]

34 function that player k is effectively competing against, and M be the CDF of Φmax(x) for almost every x. In the remainder of the discussion, for a given random variable R, we will use ∼ k dυ FR(x) to denote its CDF. Notice that by hypothesis, M(x) Unif[0, Υ dβ (x)], and therefore Υ dβ k dυ FM (x) = x k dυ in the domain [0, Υ dβ (x)]. We now bound the expected utility of player k if he switches to the strategy ψ. Z

1 ψ ∈ arg maxf∈{ϕ } ∪{ψ}{f(x)} E E i i∈[k−1] [Uk] = dυ(x) { } arg maxf∈{ϕi}i∈[k−1]∪{ψ} f(x) Z Ω ≤ E 1 ψ ∈ argmax {f(x)} dυ(x) f∈{ϕi}i∈[k−1]∪{ψ} Ω Z = E 1 (M(x) ≤ ψ(x)) dυ(x) Ω Z = E 1 (M(x) ≤ ψ(x)) dυ(x) X Z = E[1 (M(x) ≤ ψ(x))] dυ(x) X Z

= FM (ψ(x)) dυ(x) X Z Υ dβ ≤ ψ(x) dυ X k dυ Z Υ = ψ(x)dβ(x) k X Υ = E [ψ] k β

So, even if player k switches to a randomized strategy Ψ satisfying the weaker condition Υ B Υ of Lotto, his expected utility will be at most k k = k . However, by symmetry, we know Υ that his expected utility at the mixed strategy profile Φ is already k . So, he has no incentive to switch his strategy.

Since all player’s are symmetric, the above argument concludes the proof that Φ must be a mixed Nash Equilibrium.

35 Lemma 3.4 gives sufficient conditions for a strategy profile to be in mixed Nashequi- librium for either Generalized Lotto or Blotto. We are most interested in the solutions to the Blotto game. The lemma therefore also prompts two natural questions (1) Are there strategy profiles Φ that satisfy the marginal conditions of the lemma? (2) For which Blotto games are the conditions on the marginals of Φ also necessary conditions for an equilibrium? The remainder of this thesis is spent answering these questions in part. A comprehensive answer to the questions remains the biggest open problem about Blotto games. In the remainder of this chapter, we prove the existence of equilibrium strategy profiles Φ satisfying the sufficient conditions of Lemma 3.4 for many instantiations of the Generalized Blotto Game. The most exciting part however, is that we are able to construct algorithms that efficiently sample from the equilibrium strategies. We partially answer question (2) in Theorem 3.13, by showing that all symmetric equilibria for Blotto games on continuous battlegrounds, such as intervals of the real line, have the Kumaraswamy marginals described in Lemma 3.4. For instance, for Blotto on the unit interval [0, 1] with budget and value measures equal

to the standard Lebesgue measure, all equilibria that guarantee ϕi(x) = ϕj(x) for all i, j ∈ [k]

and for almost every x ∈ [0, 1] have the property that ϕi(x) ∼ k·k·Kumaraswamy(1/(k−1), 1) for every i ∈ [k] and almost every x ∈ [0, 1]; of course any coupling of these marginals that satisfies the Colonel Blotto budget constraint will be in equilibrium, if there aresuch couplings. We now show that such couplings indeed exist. The following theorem gives very general conditions under which we can guarantee the existence of an equilibrium strategy profile Φ that satisfies the conditions of Lemma 3.4. In fact, the proof of the theorem is by the construction of an efficient O(k) time algorithm to sample each player’s strategy in such an equilibrium profile. The algorithm requires however, that we be able to define a function that partitions the domain Ω into parts of equal value measure. To this end, we say a function π :Ω → [k] is a k-equiparition of Ω with respect to ∈ −1 1 measure µ if, for each i [k], µ(π (i)) = k µ(Ω). Theorem 3.5. Consider the Blotto game k, Ω, ⃗1, β, υ , and let π :Ω → [k] be a k- equipartition of Ω with respect to the value measure υ. Then, if each of the players independently runs Algorithm 3, the players will be in mixed Nash equilibrium. Furthermore,

36 Algorithm 3 runs in O(k) time.

Algorithm 3: Algorithm to sample from a symmetric mixed equilibrium of Gen-

eralized Blotto given a k-equiparition with respect to the value measure. Input: Generalized Blotto game k, Ω, ⃗1, β, υ , and a k-equipartition of Ω with respect to the value measure π :Ω → [k]. Output: an implicit representation of function ϕ sampled from a symmetric mixed equilibrium strategy for a single player.

1 Draw (X1,...,Xk) ∼ Dir(1/(k − 1),..., 1/(k − 1)). kdυ 2 ← · ∈ ϕ(x) Υdβ (x) Xπ(x), for all x Ω.

3 return ϕ

Proof. Correctness: Assume each player i gets his strategy ϕ from an independent run of Algorithm 3. Let Φ be the distribution of outputs that player i could have received by running the algorithm. We will show that each ϕ indeed satisfies the budget constraint and Φ indeed satisfies the conditions of Lemma 3.4 and thereby the players will be in equilibrium if they each play an output from independent runs of Algorithm 3. Now, we prove the two claims.

−1 (a) Budget constraint: For each i ∈ [k], let Ωi ≜ π (i). By the partition of π, we know

that the Ωis are disjoint and their union is the full space Ω.

Z Xk Z ϕ(x)dβ(x) = ϕ(x)dβ(x) Ω i=1 Ωi Z Xk kdυ = (x) · X dβ(x) Υdβ i i=1 Ωi Z Xk k = X · · dυ i Υ i=1 Ωi Xk = Xi i=1 = 1

Where the final equality follows from Proposition 2.11(ii).

37 ∼ k · dυ · 1 (b) Marginal constraint: Φ(x) Υ dβ (x) Kumaraswamy k−1 , 1 by Proposition 2.11(i).

Running time: we can sample the Dirichlet variable in O(k) time, using the method of ∼ 1 ∑ Yi ∈ [2] to sample k i.i.d variables Yi Gamma k−1 , 1 ) and letting Xi = k for all i [k]. ℓ=1 Yℓ

Notice that the preceding theorem and algorithm are the generalization of Theorem 2.2 and Algorithm 1. Below we demonstrate the generality and applicability of the new generalized machinery.

Example 3.6. The machinery of Algorithm 3 and Theorem 3.5 requires equipartitionable spaces. Here are several examples of such measure spaces that demonstrate the wide usability of the machinery.

Symmetric Homogeneous (non-generalized) Blotto: We consider Ω = [n], and let υ = β be the counting measure. Whenever n mod k = 0, we can define the k-equipartition function π :[n] → [k] as π(j) = (j mod k) + 1.

Blotto on the interval: We consider Ω = [0, 1] and let υ = β be the Lebesgue measure. We can define the k-equipartition function π(x) to output one plus the integer part of kx. Notice that any function ϕ produced by Algorithm 3 will be a step function with at most k steps.

Blotto on the surface of the sphere: We consider Ω = S2—the standard unit-sphere centered at the origin—let β be the Lebesgue measure, and υ be any measure for which

dυ dβ (x) is uniquely determined by the latitude of x. We can divide the sphere by k equally spaced longitudes, and define the manifolds between successive longitudes to bethe k equal partitions for function π.

We have now given a sufficient condition to calculate symmetric equilibria for the generalized Blotto problem, and given an efficient algorithm for sampling a mixed strategy that satisfies the sufficient condition for a vast array of such problems (as illustrated inExam- ple 3.6). In the remaining section of this chapter, we will show that the sufficient conditions we defined in Lemma 3.4 are indeed necessary for any equilibrium in which each player

38 treats each given point identically (on the margin) for Blotto on continuous spaces such as the interval, when the budget and value measures are also sufficiently well behaved.

3.1 Equilibrium Uniqueness

In further understanding equilibria, we will be interested in computing player k’s utility { }k−1 when he plays a particular strategy where all the other players strategies, Φi i=1 are fixed.

Thus, it is useful to define the function Φmax ≜ maxi∈[k−1] Φi. First, we consider Blotto on the unit-interval and restrict our attention to equilibria in which every player plays the same marginal distribution on every point, and we define G to be the CDF of the marginal distribution of each player’s function distribution at each point x ∈ [0, 1], i.e., for all i ∈ [k] and x ∈ [0, 1] the random variable Φi(x) is distributed k−1 identically and by the CDF G. Notice that this implies that Φmax(x) is distributed as G for all x. For convenience, we denote M = Gk−1. We will show the uniqueness property that G ≡ k · Kumaraswamy(1/(k − 1), 1) if we are in an equilibrium under this restriction. In the course of the proof, we will be interested in inverting CDFs. While a CDF has a true inverse sometimes, generally, it does not because it can be constant on some intervals thereby leading to multiple inverses. We eliminate this concern by defining the following unique inverse.

Definition 3.7. Let H(x) be a cumulative distribution function (CDF) for a random variable S −1 ≜ { ≤ } ∈ X that has the support . We define H (p) arg supθ H(θ) p for every p [0, 1]. If H is strictly monotonic and continuous, then our definition corresponds to the standard inverse, while in other cases, our definition smooths out jumps in the distribution and resolves points that have multiple inverses.

Notice that the inverse-CDF has the following nice property. R −1 E Lemma 3.8. [0,1] H (x)dx = X∼H [X]

Proof. Graphically, the lemma is proved by seeing that both expressions denote the area between the y-axis and the curve of H(x). Algebraically, we perform a change of variables

39 to see that Z Z −1 H (x)dx = xdH(x) = EX∼H [X]. [0,1] supp(X)

We now show an interesting property that arises if player k plays a (deterministic) function that is an inverse-CDF.

Lemma 3.9. Let H be a CDF and M be a distribution with no atoms. If EX∼H [X] = 1, then player k’s expected utility for defecting to the strategy H−1 when all other players i ∈ [k − 1], play Φi is PX∼H,Y ∼M (Y ≤ X).

Proof. The proof is a chain of inequalities, where the main observation is in replacing an integral over H−1 with an expectation of a variable X ∼ H. Z −1 E[Uk] = E 1(Φmax ≤ H (x))dx Z [0,1] Z −1 −1 = E[1(Φmax ≤ H (x))]dx = P(Φmax ≤ H (x))dx [0,1] [0,1]

−1 = EX∼H [P(M(x) ≤ H (x)]

= EX∼H,Y ∼M [P(Y ≤ X)]

= PX∼H,Y ∼M (Y ≤ X)

We now show that the marginals of the equilibrium from Theorem 3.5 are unique amongst all equilibria that have identical marginals for all points x ∈ [0, 1].

Theorem 3.10. Let (k, [0, 1], ⃗1, λ, λ) be the generalized Blotto game on the unit-interval with both the budget and value measures equal to the standard Lebesgue measure. Further, let Φ be a symmetric mixed strategy profile that is in equilibrium and additionally satisfies the d symmetric marginals property, i.e. Φi(x) = Φj(x) for all i, j ∈ [k] and x, y ∈ [0, 1]. Then, ∈ ∈ ∼ · 1 for all players i [k] and all points x [0, 1], Φi(x) k Kumaraswamy k−1 .

Proof. Let G be the CDF of the symmetric marginal distribution, i.e. the distribution of

Φi(x) for every i ∈ [k] and x ∈ [0, 1]. Let S be the support of G. Our proof strategy is

40 to successively restrict the possible forms of G using the fact that Φ is in equilibrium until we are left with only the Kumaraswamy distribution. In tandem with Theorem 3.5, this will show that the Kumaraswamy distribution is the unique equilibrium with the symmetry property in the hypothesis.

To that end, let players 1 through k − 1 fix their strategies Φ1,..., Φk−1, and let Φmax { }k−1 k−1 be the distribution max Φi i=1 , and let M = G be the marginal distribution of Φmax(x) (which is identical for all points x ∈ [0, 1]). Observe that by symmetry, player k wins 1/k expected utility by sticking to the prescribed startegy Φk. We will now show many properties of G, and each proof will be by constructing a (mixed) strategy Ψ that player k could deviate to if G did not satisfy the corresponding property.

1. E[G] = 1: the proof simply follows from the fact that any realization of Φ1 is a function that integrates to 1. Z Z 1 1 E Φ1(x)dx = 1 =⇒ E[G]dx = 1 0 0 =⇒ E[G] = 1

2. G has no atoms: Assume to the contrary that G has an atom of weight η > 0 at

a ∈ S is an atom of largest weight. For each δ ∈ (0, 1), define the distribution Ψδ over

functions ψ : [0, 1] → R+ with integral equal to 1 as follows:

(a) Sample ψ ∼ Φk

(b) Set ψ(x) ← 0 for all x ∈ [0, δ)

(d) If ψ integrates to less than 1, increase it arbitrarily at points so it integrates to exactly 1

(e) Output ψ

We will show that player k will gain an advantage by defecting to Ψδ for some (very small) δ > 0.

41 T X k −1 Let a = i=1 Φi (a) be the set of points where all the players play exactly a. Since k G has an atom of mass η at a, the expected measure E[λ(Xa)] = η because Z

E[λ(Xa)] = E 1{x | ∀i, Φi(x) = a} [0,1] ! Z ^k = P Φi(x) = a dx Z[0,1] i=1 = ηkdx [0,1] = ηk · λ([0, 1]) = ηk

Thus, deviating from Φk to Ψδ, player k gains utility on Xa ∩ [δ, 1] and loses utility on k−1 − · k − 1 [0, δ); quantitatively the total gain is at least k (1 δ) η k δ. Since η is a fixed constant, this gain is larger than 0 whenever δ is sufficiently small. Thus, for some ∗ δ ∈ (0, 1), player k gains by deviating to Ψδ∗ ; a contradiction. Thus, G has no atoms.

3. G is strictly increasing within its support: Assume to the contrary that 0 ≤ G(a) = G(b) < 1 for b > a + η, i.e., that the CDF G is constant between a and b that are at least η far apart. Further, ensure that for any ∆ > 0,G(b + ∆) > G(b). G is monotonic and continuously non-decreasing in the interval [b, b + ∆] for any ∆ > 0 that is sufficiently small; choose any ∆ < η.

For every δ ∈ (0, ∆), define Hδ to be a distribution that modifies G as follows:

(a) Let µ = G(b + δ) − G(b)

(b) Hδ moves µ/2 mass in G from [b, b + δ) to an atom at b + ∆

ϵ > 0 is picked precisely to ensure that Hδ has expectation equal to 1 just like G. (This is indeed possible because ∆ ≤ η.)

−1 Now, define ψδ = Hδ , this a legal deterministic strategy for player k by Lemma 3.8.

The expected utility gained by player k for defecting to this strategy is P(M ≤ Hδ) − P(M ≤ G) by Lemma 3.9. Notice that whenever P([b + δ, b + ∆)) > P([b, b + δ)) > 0, this gain is positive, since the µ/2 mass that was moved up causes more gains than the

42 loss due to the µ/2 mass that was moved down. Since G is continuous in the interval from b to b + ∆, we can choose δ∗ > 0 sufficiently small such that this criterion is met, which is a contradiction. Therefore, we know that G is strictly increasing from G(0) = 0 to G(t) = 1 for some upper threshold of the support t.

4. M = Gk−1 is linear in its support [0, t]: Second, note that G−1 is a continuous function of integral 1 by Lemma 3.8, and if player k defects to ψ = G−1 he gets an expected utility of 1/k by Lemma 3.9. Now, consider any values 0 ≤ a < b ≤ t, and small ϵ > 0.

We define a deterministic strategy ψa,b,ϵ using these parameters as follows:

−1 (a) Define Xa,ϵ = G ((a − ϵ, a + ϵ)) ⊆ [0, 1].

−1 (b) Define ψa,b,ϵ to be G but modified in the interval Xa,ϵ to be a step function

taking values 0 and b such that ψa,b,ϵ integrates to 1.

Defecting to ψa,b,ϵ modifies his expected utility calculation in the interval Xa,ϵ. If he plays G−1, he gets an expected utility of "Z # Z −1 −1 E 1(Φmax(x) ≤ G (x))dx = P(M ≤ G (x))dx Xa,ϵ Xa,ϵ

′ = λ(Xa,ϵ)(M(a) + ϵ )

′ where ϵ ∈ [−ϵ, ϵ] by the mean value theorem. On the other hand, if he plays ψa,b,ϵ he gets an expected utility of "Z # a + ϵ′′ E 1(Φmax(x) ≤ ψa,b,ϵ(x))dx = λ(Xa,ϵ)P(M ≤ b) Xa,ϵ b

′′ −1 where ϵ → 0 as ϵ → 0 Notice that since ψa,b,ϵ → G as ϵ → 0, we see that M(a) = a ∀ ∈ b M(b). This implies that a, b [0, 1],M(a)/a = M(b)/b.

5. M(x) = x/k in the range [0, k]: First, observe that t > 1. Otherwise, player k could defect to the constant function ψ(x) = 1 and win the whole interval. The constant function M(x)/x must be 1/k to account for the fact that playing the function ψ(x) = 1 gets player k a utility of 1/k. This implies that M(x) = x/k in the range [0, t]. Since

43 we additionally know that E[G] = 1 and Gk−1 = M, we can specify t uniquely to be k · 1 and G to be the CDF of k Kumaraswamy k−1 , 1 . · 1 Because M(x) = x/k, we conclude that G must be distributed as k Kumaraswamy k−1 , 1 .

Now that we have seen the key features of the argument in the simpler setting where ′ ′ Φi(x) = Φi(x ) for all points x, x ∈ Ω, we show the stronger theorem of equilibrium uniqueness even when this assumption does not hold.

Theorem 3.11. Let (k, [0, 1], ⃗1, λ, λ) be the generalized Blotto game on the unit-interval with both the budget and value measures equal to the standard Lebesgue measure. Further, d let Φ be a symmetric mixed strategy profile that is in equilibrium, i.e. Φi(x) = Φj(x) for all i, j ∈ [k] and for almost every x ∈ [0, 1]. Then, for all players i ∈ [k] and for almost every ∈ ∼ · 1 point x [0, 1], Φi(x) k Kumaraswamy k−1 .

Proof. Let Gx be the CDF of Φi(x) for every i ∈ [k]. Our proof strategy is to successively restrict the possible forms of Gx using the fact that Φ is in equilibrium until we are left with only the Kumaraswamy distribution. In tandem with Theorem 3.5, this will show that the distribution with Kumaraswamy marginals is the unique equilibrium with the symmetry property in the hypothesis.

To that end, let players 1 through k − 1 fix their strategies Φ1,..., Φk−1, and let Φmax be { }k−1 k−1 the distribution max Φi i=1 , and let Mx = Gx be the marginal distribution of Φmax(x). Observe that by symmetry, player k wins 1/k expected utility by sticking to the prescribed

strategy Φk. We will now show many properties of Gx, and each proof will be by constructing

a (mixed) strategy Ψ that player k could deviate to if Gx did not satisfy the corresponding property.

1. Gx has no atoms for almost every x ∈ [0, 1]: Assume the contrary, and define

Aη = {x ∈ [0, 1] | Gx has an atom at some ax of weight at least η}

By assumption, there is some η ∈ (0, 1/2) for which λ(Aη) ≥ η. Choose a set Bδ ⊆

44 [0, 1] − Aη of measure λ(Bδ) = δ > 0 for each δ ∈ (0, 1/2). Define the distribution Ψδ over functions ψ : [0, 1] → R+ with integral equal to 1 as follows:

(a) Sample ψ ∼ Φk

(b) Set ψ(x) ← 0 for each x ∈ Bδ

(d) If ψ integrates to less than 1, increase it arbitrarily at points so it integrates to exactly 1

(e) Output ψ

We will show that player k will gain an advantage by defecting to Ψδ for some (very small) δ > 0. T X k { | } Let a = i=1 x Φi(x) = ax be the set of points where all the players play exactly the values at which we picked atoms. Since Gx has an atom of mass at least η at ax k+1 for every x ∈ Aη, the expected measure E[λ(Xa)] = η because "Z #

E[λ(Xa)] = E 1{x | ∀i, Φi(x) = ax} A η ! Z ^k = P Φi(x) = ax dx ZAη i=1 ≥ ηkdx Aη k k+1 = η · λ(Aη) = η

Thus, deviating from Φk to Ψδ, player k gains utility on Xa and loses utility on Bδ; k−1 k+1 − 1 quantitatively the total gain is at least k η k δ. Since η is a fixed constant, this gain is larger than 0 whenever δ is sufficiently small. Thus, for some δ∗ ∈ (0, 1/2), player k gains by deviating to Ψδ∗ ; a contradiction. Thus, G has no atoms.

45 −1 2. Gx (x) integrates to 1 over the interval [0, 1] We simply calculate the integral Z Z −1 Gx (x) = xdGx(x) [0,1] R+

= EX∼Unif[0,1][GX (X)]

= E ∼ [Φ (X)] XZ Unif[0,1] k

= E Φk(x)dx [0,1] = 1

The last equality follows by the fact that Φk is a valid strategy and therefore must integrate to 1 in any realization.

−1 3. If player k plays Gx (x) he gets 1/k expected utility: we simply calculate the expected utility of this strategy. Z Z −1 −1 E 1(Φmax(x) ≤ G (x))dx = P(Φmax(x) ≤ G (x))dx [0,1] [0,1]

−1 = EX∼Unif[0,1]P(MX (X) ≤ G (X))

= E ∼ P(Φ (X) ≤ Φ (X)) Z X Unif[0,1] max k

= P(Φmax(x) ≤ Φk(x))dx Z[0,1]

= 1(Φmax(x) ≤ Φk(x))dx [0,1] = 1/k

4. Gx is strictly increasing within its support for almost every x: Assume the contrary.

Define tx = arg inft{Gx(t) = 1}, and further define

Aη,∆ ≜ {x | 0 < ax < bx ≤ tx −∆∧0 ≤ G(ax) = G(bx) < 1∧∀ϵ > 0,G(bx +ϵ) > G(bx)}

By assumption, there is some parameters η > ∆ > 0 for which λ(Aη,∆) = c > 0; fix

such η and ∆. Now, fix a δ ∈ (0, ∆) such that for a set Aη,∆,δ ⊆ Aη,∆ of positive

46 ′ measure λ(Aη,∆,δ) = c > 0, we have 0 < Gx([bx, bx + δ)) < Gx([bx + δ, bx + ∆]) − δ.

We can surely find such a δ because all the CDFs, Gx, are continuous. Finally, define

Hx to be Gx modified such that half the mass between in [bx, bx + δ) is moved down ∈ −1 by ∆, and half of it is moved up by ∆ for each x Aη,∆,δ. Define ψ(x) = Hx . The −1 integral of ψ is 1 because the integral of Gx(x) is 1 and the modification made does not increase the integral. Furthermore, by defecting to ψ, there is a gain in expected utility, at every point x where we modified the measure.

k−1 ∈ 5. Mx = Gx is linear in its support [0, tx] for almost every x [0, 1]: Consider once −1 again the function ψ(p) = Gp (p), and recall that player k gets an expected utility of 1/k if he defects to this function ψ. Now, consider any point p ∈ [0, 1] that is a Lebesgue point of ψ, let a = ψ(p), and consider the ϵ-neighborhood of p. The expected utility for player k in this part of the interval is Z Z

E 1(Φmax(x) ≤ ψ(x))dx = P(Φmax(x) ≤ ψ(x))dx [p−ϵ,p+ϵ] [p−ϵ,p+ϵ]

Now note that player k could have defected by playing a step function in this part

of the domain with values 0 and b ∈ (a, tp) so the integral does not change, call this

modification ψp,b,ϵ. This modification would result in an expected utility of Z Z ′ ′ E (a/b + ϵ ) · 1(Φmax(x) ≤ b)dx = (a/b + ϵ )P(Φmax(x) ≤ b)dx [p−ϵ,p+ϵ] [p−ϵ,p+ϵ]

where ϵ′ → 0 as ϵ → 0. Note that as ϵ → 0 the two utilities computed above converge,

since ψp,b,ϵ → ψ and all relevant values are continuous. Thus, taking the limit of both a ∈ sides we get Mp(a) = b Mp(b). In other words, Mp(a)/a = Mp(b)/b for all a, b [0, tp];

or identically, Mp is linear in its support.

6. Almost all the support upper bounds of Mps, i.e. the tps, are identical: Assume −1 otherwise, and remember that playing ψ(p) = Gp (p) gets 1/k expected utility for

player k. Note that ψ(p) = tp almost nowhere and ψ(p) = 0 almost nowhere. By our initial assumption, we can construct two disjoint sets of positive measure A, B ⊂ [0, 1],

such that (x ∈ A, y ∈ B) =⇒ (tx < ty − ϵ) for some positive ϵ > 0. Then,

47 constructing a modification of ψ that reduces the function value on points of B and increasing the value on points of A will increase the expected utility of player k, which is a contradiction to the fact that Φ is in equilibrium. So, we conclude that almost all

the tps are identical.

7. Mp(x) = x/k in the range [0, k] for almost every p ∈ [0, 1]: Let t be the upper bound

of the threshold tp and let M denote Mp for almost all p ∈ [0, 1]. First, observe that t > 1. Otherwise, player k could defect to the constant function ψ(x) = 1 and win the whole interval. The constant function M(x)/x must be 1/k to account for the fact that playing the function ψ(x) = 1 gets player k a utility of 1/k. This implies that M(x) = x/k in the range [0, t]. Since we additionally know that E[G] = 1 and Gk−1 = M, we · 1 can specify t uniquely to be k and G to be the CDF of k Kumaraswamy k−1 , 1 . · 1 Because M(x) = x/k, we conclude that G must be distributed as k Kumaraswamy k−1 , 1 .

Finally, in the next lemma and theorem, we generalize the uniqueness proof from the unit-interval measured in Lebesgue measure for both budget and value, to a general interval of the real line measured by more general measures. In particular, we establish uniqueness of equilibrium (amongst symmetric equilibria) for all Blotto games over intervals with arbitrary budget and value measures that are mutually absolutely integrable with the standard Lebesgue measure.

Lemma 3.12. Let G = (k, [0, 1], ⃗1, λ, λ) be the generalized Blotto game on the unit-interval with both the budget and value measures equal to the standard Lebesgue measure, and G′ = (k, [a, b], ⃗1, β, υ) be the generalized Blotto game on the interval [a, b] with some budget and value measures that are absolutely continuous with respect to each other and the standard Lebesgue measure on that interval. For any strategy Φ for G define Φ′ to be the strategy ′ ∼ ′ dυ − x−a for Φ where each player i samples ϕi Φ and plays ϕi(x) = dβ (b a)ϕi b−a , and for ′ ′ ′ ∼ ′ any strategy Φ for G define Φ for G where each player i samples ϕi Φ and plays dβ 1 ′ − ′ ϕi(x) = dυ b−a ϕi (x(b a) + a). In each case, Φ is in equilibrium if and only if Φ is in equilibrium.

48 Proof. All integrals can be converted between the two regimes by a simple change of variables.

Theorem 3.13. Let (k, [a, b], ⃗1, β, υ) where b > a be the generalized Blotto game on the interval, and assume that β, υ, and λ (the Lebesgue measure) be mutually absolutely integrable with respect to each other. If Φ be a symmetric mixed strategy profile that is in equilibrium, then for all players i ∈ [k] and for almost every point x ∈ [0, 1],

k dυ 1 Φ (x) ∼ · (x) · Kumaraswamy , 1 i Υ dβ k − 1 Proof. This theorem follows from applying Lemma 3.12 to Theorem 3.11

Remark 3.14. While the proof of Theorem 3.13 was written only for Blotto on intervals of the real-line, the analogous result also applies to many continuously measured surface—like the surface of the unit sphere. The proof is identical in its steps for the surface of the sphere for instance.

49 50 Chapter 4

Boolean-valued Colonel Blotto game

Chapter 3 was the most over-arching part of this thesis. It defined and solved broad problems in continuous Blotto games, where each player can play positive real valued bids. The Blotto literature has also focused on discrete bidding however, where bids are restricted to be non- negative integers. In this chapter, we turn back from generalized battlegrounds to a finite number of battlefields, and initiate the study of discrete Blotto with multiple players by defining and solving the simplest problem in the field—the Boolean Blotto game. Inthis game, each player i chooses whether to compete or not compete in up to Bi battlefields, and the values of battlefields are split evenly among the players who compete inthem.

Definition 4.1. The Boolean-valued Colonel Blotto game has the same payoff function as the continuous-valued Colonel Blotto game, with two additional restrictions:

(integer budget) each player i ∈ [k] has an integer-valued budget Bi ∈ {0, . . . , n}

(Boolean bids) each bid Ai,j is either 0 or 1; we say player i competes in battlefield j if

Ai,j = 1. The game is symmetric if all players have the same budget B.

Definition 4.2. In the Boolean-valued General Lotto game, each player i ∈ [k] plays a vector of probabilities (pi,1, . . . , pi,n), such that the budget constraint is met in expectation: P n ≤ B E j=1 pi,j i. The payoff function for player i given the bids of all the players is AUi(A), ′ ′ where for each i ∈ [k], j ∈ [k] the bids Ai′,j′ ∼ Ber(pi′,j′ ) are drawn independently.

51 Lemma 4.3. When there are k = 2 players, it is a maximin pure strategy for player i to

compete only in the Bi battlefields of highest value.

Proof. Regardless of the other player’s strategy, the marginal gain from competing in bat-

tlefield j is vj/2, so it is optimal to compete in the most valuable battlefields.

Boolean Blotto only becomes interesting when k > 2. We now proceed to characterize the equilibria of symmetric multiplayer Boolean Blotto.

4.1 Boolean General Lotto and sufficient conditions for Colonel Blotto

As in our analysis of continuous-valued Colonel Blotto, we first characterize the symmetric equilibria of the General Lotto analogue of the game.

For a given player, Alice, and given battlefield of value v, let u1(p, v) be the expected utility earned by Alice from competing in the battlefield if all the other k − 1 players independently compete with probability p, and let u0(p, v) be Alice’s expected utility from not competing. Let mv(p) = u1(p, v) − u0(p, v) be the marginal utility of competing. We can

write Alice’s expected utility from competing with probability q as qu1 + (1 − q)u0. If Alice doesn’t compete in the battlefield, she only gains utility when nobody competes: v − k−1 u0(p, v) = k (1 p) . The total utility earned by all players from the battlefield is v, so by · − · v symmetry, p u1(p, v) + (1 p) u0(p, v) = k . Combining these two equations yields    v − k (k 1), p = 0 mv(p) =  −  v · 1−(1−p)k 1 ≤ k p , 0 < p 1

We will show that there is a unique symmetric equilibrium. The probabilities p1, . . . , pk of the equilibrium strategy are such that the marginal utilities of competing are essentially the same for all battlefields. This means that if all players including Alice play the equilibrium strategy, then Alice will have no incentive to move ε probability mass from one battlefield

to another. To obtain these probabilities it will be useful to define an inverse of mv(p).

52 Claim 4.4. When k > 2, mv(p) is continuous and monotonically decreasing on the interval p ∈ [0, 1]. Therefore it maps [0, 1] bijectively to [v/k, (k − 1) · v/k].

For ease of presentation of the proof of this claim and some other lemmas, we define the notation:   k − 1, p = 0 µ(p) =  −  1−(1−p)k 1 ≤ p , 0 < p 1

v −1 −1 Thus, mv(p) = k µ(p). The domain of µ is extended by letting µ (x) = 1 for x < 1 and µ−1(x) = 0 for x > k − 1.

Proof of Claim 4.4. By l’Hôpital’s rule

(k − 1)(1 − p)k−2 lim µ(p) = lim = k − 1 = µ(0), p→0+ p→0+ 1

proving continuity. And for any p ∈ (0, 1),

∂µ p(k − 1)(1 − p)k−2 − 1 − (1 − p)k−1 (1 − p)k−2(1 + p(k − 2)) − 1 = = < 0, ∂p p2 p2

−t −t because for t = k − 2 > 0 we have (1 − p) ≥ (1 + pt). This holds because (1 − p) |p=0 = | ∂ − −t − −t−1 ≥ ∂ ∈ 1 = (1 + pt) p=0 and ∂t (1 p) = t(1 p) t = ∂t (1 + pt) for p [0, 1]. The bijectivity follows from continuity and monotonicity.

−1 − · By Claim 4.4, the inverse mv is uniquely defined on [v/k, (k 1) v/k], and we may extend R −1 −1 − · its domain to by letting mv (x) = 1 for x < v/k and mv (x) = 0 for x > (k 1) v/k. We are now ready to characterize the symmetric General Lotto equilibrium:

Lemma 4.5. The following is the unique symmetric Nash equilibrium of the Boolean-valued

General Lotto game with k ≥ 3 players, equal integer-valued budgets Bi = B and battlefield

valuations v1 ≥ v2 ≥ · · · ≥ vn > 0:

p = m−1(x∗) ∀j ∈ [n] (4.1) j vj

n P o where x∗ = inf x ∈ R : n m−1(x) ≤ B . j=1 vj

53 ∗ Proof. If B = n, then x = −∞ so every pj = 1; this is clearly the unique equilibrium. So, we assume that B < n henceforth. Note that we chose x∗ such that the players meet their Lotto budget constraint exactly.

Now suppose Alice deviates from the strategy by playing {qj}j∈n while all other players

play {pj}j∈n. The utility she gains by deviating is

Xn Xn (q − p ) · m (p ) = (q − p ) · m (m−1(x∗)) (4.2) j j vj j j j vj vj j=1 j=1 ! Xn Xn Xn ∗ ∗ ≤ (qj − pj)x = x qj − pj (4.3) j=1 j=1 j=1 = x∗(B − B) = 0 (4.4)

where the inequality in line 4.3 arises because x∗ may lie in the extended domain of m−1 for vj some js. Thus, Alice has no incentive to deviate, so this is an equilibrium.

Now we prove uniqueness. Let {πj}j∈n be any symmetric equilibrium. We will show that there must exist an x′ such that π = m−1(x′) for each j. Assume for contradiction that j vj There is no x′ such that π = m−1(x′) for each j. j vj We will show in cases that there are battlefields i and ℓ such that: (a) the marginal utility mvi (πi) < mvℓ (πℓ), and (b) probability πi > 0 and probability πℓ < 1. Thus, a player Alice will increase her utility by shifting ϵ probability mass from battlefield i to ℓ.

∈ ′ Case 1 Suppose πi (0, 1) for some i. Let x = mvi (πi). Assumption (*) implies that there is some ℓ ∈ [n] such that π =6 m−1(x′) > 0. Three subcases ensue: (a) if π = 0, ℓ vℓ ℓ then 0 < m−1(x′), so applying the monotonically decreasing function m to both sides of vℓ vℓ the inequality yields m (π ) = m (0) > x′ = m (π ). (b) if π = 1, then 1 > m−1(x′), vℓ ℓ vℓ vi i ℓ vℓ so applying the monotonically decreasing function mvℓ to both sides of the inequality yields ′ ∈ mvℓ (πℓ) = mvℓ (1) < x = mvi (πi). (c) if πℓ (0, 1), then either mvℓ (πℓ) > mvi (πi) or mvℓ (πℓ) < mvi (πi).

∈ { } ∈ ′ ′ Case 2 Suppose πj 0, 1 for all j [n], yet there is no x such that x = mvj (πj) for all

j ∈ [n]. Then there must be indices i, ℓ ∈ [n] such that πi = 1 and πℓ = 0 and vi < (k − 1)vℓ

54 vi k−1 so mvi (πi) = mvi (1) = k < k vℓ = mvℓ (0) = mvℓ (πℓ). In all cases, moving ϵ probability mass from the battlefield with the smaller marginal util-

ity to the larger (between i and j) strictly increases Alice’s utility and shows the (π1, . . . , πn) is not an equilibrium. Since we know there is an x′ such that π = m−1(x′) for each j, it j vj is an immediate consequence of the tightness of the budget constraint that it must be x∗, thereby completing the proof.

Remark 4.6 (Limit of large k). In order to understand the asymptotic behavior of the solution as the number of players k tends to infinity and the average utility per player stays

constant (so we increase the values vj proportionally with k). Notice that mk·vj (p) tends towards v /p for each j, so the inverse m−1 (x) tends towards min(1, v /x) for x > 0. j vj ·k j Therefore, in the limit of large k, the equilibrium strategy tends towards surely competing in some of the top-valued battlefields and competing in the rest with probabilities proportional to the values of those battlefields. Quantitatively, Lemma 4.5 prescribes this strategy: iteratively assign portions of the budget to battlefields 1, . . . , n in order of decreasing value as follows. Write B(l) to denote the budget remaining after assigning budget to battlefields 1, . . . , l, and

(0) (l−1) ∑ vl let B = B. Then battlefield l is assigned budget min(1, B n ). So roughly speaking j=l vj we assign to each battlefield a fraction of the budget equal to the fraction of the total value that the battlefield represents.

Remark 4.7 (Qualitative change in strategy as k increases). We also qualitatively observe that as the number of players increases, the players are more likely to bid on battlefields of low value.

As an example, consider two battlefields with values given by v1 = 1 and v2 > 0 and k

players with budget given by B = 1. Then (i) if k ≤ 1/v2 +1, we will have p1 = 1 and p2 = 0, meaning that if there are not enough players then no one will compete in the battlefield with

small value. On the other hand, (ii) if k > 1/v2 + 1, then we prove below that p2 > 0, meaning that if there are enough players then they will compete in the low-value battlefield with some non-zero probability. P 2 −1 ≥ −1 Proof of Remark 4.7. Part (i) follows because Bk(1) = j=1 µ (1/vj) µ (1) = 1, so ∗ −1 ∗ −1 x ≥ 1, so p2 = µ (x /v2) ≤ µ (k − 1) = 0. To prove part (ii) consider x = (k −

55 1)v2, which satisfies x > 1 by the condition on the number of players. It follows that −1 −1 −1 Bk(x) = µ (x) + µ (x/v2) = µ (x) < 1. By continuity of Bk(x), there is ε > 0 such that ∗ −1 ∗ −1 Bk(x − ε) ≤ 1, and therefore x < x. Hence p2 = µ (x /v2) > µ (k − 1) = 0.

As in the continuous-valued case, the General Lotto solutions in Lemma 4.5 yield sufficient conditions for the players to be in Nash equilibrium:

Lemma 4.8. Let k ≥ 3. Suppose that in the symmetric Boolean-valued k-player Colonel

Blotto game with battlefield valuations v1 ≥ v2 ≥ · · · ≥ vn > 0 and equal integer-valued

budgets Bi = B, each player i ∈ [k] independently bids a vector Ai,∗ = (Ai,1,...,Ai,n) such that for each i ∈ [k]: P n ≤ B (a) j=1 Ai,j .

(b) Ai,j ∼ Ber(pj), where pj is given in the statement of Lemma 4.5, using budget B.

Then the players are in equilibrium. Furthermore, this is the unique symmetric equilibrium.

Proof of Lemma 4.8. The proof is by linearity of expectation, as in the real-valued setting. The budget constraints are met by (a). If any player deviates from the strategy, then, by the analysis of Lemma 4.5, the player’s expected payoff cannot improve. This is because by linearity of expectation the expected payoff for Colonel Blotto only depends on the marginal distributions of the bids for the battlefields.

4.2 Colonel Blotto equilibria

We now show how to obtain an efficient Colonel Blotto strategy from the equilibrium General Lotto strategy. This consists of two tasks: (1) efficiently estimating the implicitly described

pj’s, and (2) efficiently computing a coupling of allocations that has the approximate pj’s as its marginals. The first task, estimation, can be performed with a carefully tuned binary search. The second task presents an appealing puzzle: given n Bernoulli random variables P n B ∈ Z with biases pi, . . . , pn satisfying i=1 pi = ≥0, how can they be coupled into a joint

56 P ∈ { } n B distribution such that draws x1, . . . , xn 0, 1 from the distribution satisfy i=1 xi = almost surely? Algorithm 4 is a very simple procedure for solving this puzzle. Algorithm 4: NashEquilThm4.9: Boolean-valued Blotto equilibrium for Theo- rem 4.9 Input: A symmetric Boolean Blotto game (k, n, B,⃗v) with battlefield valuations

v1 ≥ v2 ≥ · · · ≥ vn > 0.

n Output: A sample (A1,...,An) ∈ R from the equilibrium mixed strategy for a single player in the Boolean-valued Blotto game (k, n, B,⃗v).

1 For each j ∈ [n], let pj be as defined in Lemma 4.5 or Theorem 4.9. P j−1 ∈ ′ 2 For each j [n], let αj = j′=1 pj

3 Draw β ∼ Unif[0, 1]

4 For each j ∈ [n], let Aj = 1[∃m ∈ Z | β + m ∈ [αj, αj + pj)]

5 return (A1,...,An)

Theorem 4.9. Suppose that in the Boolean-valued Colonel Blotto game with equal budgets

Bi = B and k > 2 players, each player i ∈ [k] independently runs Algorithm 4. Then all of the players will be in Nash equilibrium. Moreover, given parameter ε > 0, Algorithm 4 runs in time polynomial in the problem size and log(1/ε), and produces an ε-approximate Nash equilibrium.

Proof. We verify that the sufficient conditions for a Nash equilibrium from Lemma 4.8 are met. We can assume without loss of generality that B ≤ n, because otherwise all players P n B ∈ Z compete in all battlefields, which is a Nash equilibrium. So in this case j=1 pj = ≥0. An equivalent way of applying the sampling procedure is to set

Aj = 1[{β + m}m∈Z ∩ [αj, αj + pj) =6 ∅].

Note that the intervals [αj, αj + pj) constitute a partition of the interval [0, B), and that

{β +m}m∈Z intersects this long interval B times, and finally that {β +m}m∈Z intersects each interval [αj, αj + pj) at most once. It follows that, for any β, exactly B of the Aj bids are set to 1. This proves that the budget constraint holds almost surely. And probability Aj is set to 1 is pj because the interval [αj, αi + pj) has length pj. So all the sufficient conditions

57 of Lemma 4.8 are met.

Efficient approximation of equilibrium We have constructed an exact Nash equilibrium. However, our algorithm is not yet efficient, because we have not yet described howto

compute the probabilities pj. These are defined implicitly in the statement of Lemma 4.5, but there appears to be no closed form. Nevertheless, if we could approximately compute

the pj probabilities, then we could approximate the Nash equilibrium. Indeed, the utility for a player can range from 0 to V and there are k players, so in order to compute an ε-Nash equilibrium it suffices to approximate the equilibrium strategy for each player upto (ε/kV )

error in statistical total variation. Since there are n probabilities pj, this can be achieved

by approximating each pj up to additive error (ε/kV n). We want this estimation error even

after scaling the approximate pj’s so their sum is B; for this it suffices to achieve additive error (ε/kV n2). To complete the proof, we explain how to do this with a carefully tuned binary search below.

1. First, given any x ∈ R we show how to compute an additive ε′ approximation p˜ to p = µ−1(x) in poly(log k, log(1/ε′)) operations. If x ≤ 1 or x ≥ k − 1, then p˜ = 0 or p˜ = 1 are respectively correct. Otherwise, for the case 1 < x < k − 1, recall from Claim 4.4 that µ maps [0, 1] bijectively to [1, k − 1], and is continuous and monotonically decreasing. Therefore we can binary search to find p˜ such that |p˜ − p| < ε′. This binary search requires only O(log(1/ε′)) evaluations of µ, and each evaluation of µ up to ε˜′ precision costs only poly(log k, log(1/˜ε′)) operations. We can set the precision parameter to ε˜′ = ε′/2, because ′ ′′ ∈ | ′ − ′′ | ≥ | ′ − ′′| d ′ ≤ − ′ ∈ for any p , p [0, 1], we have µ(p ) µ(p ) p p , since dp µ(p ) 1 for all p [0, 1]. Therefore the total cost of the binary search is poly(log k, log(1/ε′)).

2. Second, we show how to compute x˜ such that |x˜−x∗| < ε′′, in poly(n, log k, log(V /ε′′)) op- P ∗ { ∈ R ≤ B} n −1 erations. Recall the definition x = inf x : Bk(x) , where Bk(x) = j=1 µ (x/vj).

We will use the fact that Bk(x) is monotonically non-increasing and continuous. By the proof of Lemma 4.8, in the nontrivial case B < n it holds that x∗ ∈ [0, (k − 1)V ]. Hence we can binary search to find x˜ such that |x˜ − x∗| < ε′′. The binary search requires ′′ O(log(kV /ε )) evaluations of Bk(x). Using part (1) each evaluation of Bk up to preci-

58 sion ε˜′′ costs poly(n, log k, log(n/˜ε′′)) operations, by separately evaluating each term up to precision ε˜′′/n. We now investigate the necessary precision ε˜′′. At any point in the binary search when we query point xˆ, one of two cases arises:

′ ∗ ′ −1 ′ • Case A: For each x between xˆ and x , there is a j(x ) ∈ [n] such that µ (x /vj(x′)) ∈ (0, 1). In this case, for all x′ between xˆ and x∗,

Xn dBk(x) d −1 | ′ = µ (x/v )| ′ (4.5) dx x=x dx l x=x l=1

d −1 ≤ µ (x/v ′ )| ′ (4.6) dx j(x ) x=x ≤ − 2 2 (4.7) (k − 3k + 2)vj(x′) 1 ≤ − (4.8) V k2

∗ ∗ 2 ∗ ′′ So |Bk(x) − B| = |Bk(x) − Bk(x )| ≥ |xˆ − x |/(V k ), and so if |xˆ − x | > ε /2 it suffices ′′ ′′ 2 ∗ to compute Bk(x) up to accuracy ε˜ = ε /(2V k ) in order to determine whether xˆ ≤ x or xˆ > x∗.

′ ∗ −1 ′ • Case B: Otherwise there is x between xˆ and x such that µ (x /vj) ∈ {0, 1} for all ∗ ˜ ˜ j. In this case, if xˆ ≤ x then our approximation B(ˆx) to Bk(ˆx) satisfies B(ˆx) ≥ −1 −1 ′ ′ ∗ |{j : µ (ˆx/vj) = 1}| ≥ |{j : µ (x /vj) = 1}| = Bk(x ) ≥ Bk(x ). And by a similar ˜ ∗ ∗ ˜ ∗ ∗ argument B(ˆx) > Bk(x ) if Bk(ˆx) > Bk(x ); and B(ˆx) ≤ Bk(x ) if Bk(ˆx) ≤ Bk(x ); ˜ ∗ ∗ and B(ˆx) < Bk(x ) if Bk(ˆx) < Bk(x ).

Therefore we can set the precision parameter to ε˜′′ = ε′′/(2V k2). So the total cost of the binary search is poly(n, log k, log(V /ε′′)).

3. Third, suppose we have x˜ such that |x˜ − x∗| < ε′′. Then for each j ∈ [n] we define

−1 −1 p˜j = µ (˜x/vj). By a simple calculation, µ (x) is 1-Lipschitz over R, so we are guaranteed ′′ ′′ ′′ 2 that |p˜j − pj| ≤ ε /vj ≤ ε /vn. Letting ε = (εvn/V kn )/2 and computing x˜ with the ′ 2 procedure from (2), and approximating p˜j up to ε = (ε/V kn )/2 error with the procedure 2 from (1), we obtain an overall (ε/V kn ) approximation to pj. The total running time is poly(n, log k, log(V /ε), log(V /vn)), which is polynomial in the input size of the problem.

59 4. Finally, given approximations p˜j to the true pj probabilities, the sampling procedure takes time and space linear in n and the number of bits of precision in the p˜j probabilities. This is polynomial in the problem size and log(1/ε).

60 Chapter 5

Remarks & Open Problems

In this thesis, we significantly extended the definition of the Colonel Blotto problem from a two-player game on a finite set of battlefields to a multiplayer game over an arbitrary measurable battleground, and analyzed the equilibria of these games when all players are symmetric. The newly defined generalized Blotto games have the potential for many applications in analyzing multiparty elections, modeling competitions over continuous manifolds like the surface of the earth, understanding decisions made by businesses, and many more as described in Chapter 1. Our most general results were presented in Chapter 3, which deals with equilibria of Blotto games played on arbitrary measure spaces. We presented four results that made progress towards characterizing the equilibria of these games in the context of symmetric players (i.e., players that have the same budget): (1) Lemma 3.4, showed sufficient conditions for a mixed strategy profile of a symmetric k-player generalized Blotto game over an arbitrary measure space to be in equilibrium; (2) Theorem 3.5 showed the existence of a symmetric equilibrium satisfying those sufficient conditions if the battle ground is k-value-partitionable; (3) Algorithm 3 showed an efficient way for each player to sample from such an equilibrium; and finally (4) Theorem 3.13 completely characterized the symmetric equilibria of Blotto games played on intervals of the real-line, when the battle ground is measured in budget and value by measures that are absolutely continuous with respect to each other and the standard Lebesgue measure. To the best of our knowledge, the characterization of equilibria given in Theorem 3.13 is the first of its kind for a Blotto game, since previous uniqueness results

61 have only been for Lotto versions of games. Two immediate open directions suggested by this chapter are: (1) Extending Theorem 3.13 by characterizing the asymmetric equilibria of symmetric Blotto games, and (2) Solving for equilibria in the regime of asymmetric players (i.e., players with different budgets). We have also made substantial progress in understanding symmetric multiplayer Blotto games of the standard sort—with k players and n battlefields. These games are of particular interest in the analysis of plurality vote parliamentary elections and many other applications listed in Chapter 1. The model for these games, presented in Chapter 2, is encompassed by the generalized Blotto framework of Chapter 3. Consequently, Algorithm 3 simplifies to Algorithm 1—an efficient algorithm for generating samples from the equilibrium strategy when the values of the battlefields are k-partionable. The highlight of Chapter 2 however, is Algorithm 2, which shows how to sample from equilibrium strategies for the symmetric 3-player game on n battlefields even when the values of the battlefields cannot be partitioned into three sets of equal value (as long as no single battlefield is too valuable). In particular, this algorithm shows how to couple the marginal distributions of the cousin General Lotto equilibrium into a joint distribution supported on the surface of the ℓ1-ball using the geometric intuition of rotating a 2-sphere about the origin in hyperspace. Of course, performing such a coupling, in possibly a very different way, can lead to equilibria of a vast arrayof other Colonel Blotto games. Thus, a beautiful open question that arises in this context and promises to have an impact far beyond this problem, is the following coupling problem.

Open Problem 5.1. Characterize when marginal distributions D1,..., Dn over R can be coupled into a joint distribution D supported on some specified manifold M, and construct efficient algorithms that output such couplings given the marginal distributions and constraints as inputs.

While deciding whether a given set of marginals is coupleable in this way is weakly NP-hard even in the case of finitely-supported discrete distributions (by a simple reduction from Subset-Sum), it is an alluring problem to obtain a deeper understanding of the cases in which a budget-constrained coupling exists and can be constructed efficiently. A good starting point to understand this question may lie in earlier literature on deciding whether

62 a given set of interim allocation rules are permissible [11, 15, 14, 16]. In Chapter 4, we solved the symmetric case of the multiplayer Boolean Blotto problem, where each player can play either a 0 or a 1 on each battlefield. Of course, the generalization of this problem which allows players to make integer (not just Boolean) bids—discrete multiplayer Blotto—is a natural open problem which promises to be of particular interest to the computer science community.

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