Algorithmic Issues in Coalitional and Dynamic

Algorithmic Issues in Coalitional and Dynamic Network Games ? Dimitris Fotakis3 , Spyros Kontogiannis1,2 , Panagiota Panagopoulou2 , Christoforos Raptopoulos2 , and Paul Spirakis2 1

Computer Science Department, University of Ioannina, 45110 Ioannina, Greece. [email protected]

2

R.A. Computer Technology Institute, N. Kazantzaki Str., Patras University Campus, P.O. Box 1382, 26500 Patras, Greece. {kontog,panagopp,spirakis}@cti.gr,[email protected] 3

Dept. of Information and Communication Systems Engineering, University of the Aegean, 83200 Samos, Greece. [email protected]

November 4, 2005

Abstract. We discuss some new algorithmic and complexity issues in coalitional and dynamic/evolutionary games, related to the understanding of modern selfish and Complex networks. In particular: (a) We examine the achievement of equilibria via natural distributed and greedy approaches in networks. (b) We present a model of a coalitional game in order to capture the anarchy cost and complexity of constructing equilibria in such situations. (c) We propose a stochastic approach to some kinds of local interactions in networks, that can be viewed also as extensions of the classical evolutionary game theoretic setting.

1

Introduction

The new research field of Algorithmic Game Theory, till now, has its focus mostly on non–cooperative games in strategic form. Indeed, the cases where dynamic game elements are studied are rather rare. In addition, the main concept of Algorithmic Game Theory, namely, the price of anarchy (PoA), has been applied mostly to non–cooperative selfish players with a variety of pure strategy sets and payoff functions. Most of the examples are motivated from network traffic and congestion problems. On the other hand, the new generation of Complex Systems depicts not well understood characteristics arising from dynamic and evolving behavior, coali?

Invited talk at HNI 2005, presented by Paul Spirakis. This work was partially supported by the EU within 6th Framework Programme under contract 001907 (DELIS).

tional antagonism and also time-changing component micro–interactions of a large number. This contrast of paradigms poses some challenges to Algorithmic Game Theory. In this work, we explore and present models to meet such challenges, in three directions: (a) dynamic, greedy game evolution, (b) coalitional games, and (c) local and dynamic interactions in a changing population mixture. We are aware of the vast and deep research conducted in the so–called Continuous Dynamic Game Theory (see eg [2] for a nice overview of that field) and its relation to Nonlinear Dynamics and Control Theory. We are also aware of the recent and emerging field of Evolutionary Game Theory (see e.g. [25, 14]). In contrast, our proposed approaches are based on a discrete–time, discrete– space setting, and the interactions involve a finite set of players or coalitions. All our models take an underlying network structure into account. Thus, for example, our approach in the direction of antagonistic particles’ interactions assumes the existence of a certain network where particles “move” (thus, it departs from the amorphus population mixture of types in classical Evolutionary Game Theory). In addition, both the discrete time element and the description of local interactions as games, differentiate our approach from the Physics–oriented research in continuous–time, non–antagonistic systems of interacting particles (as e.g. in [17]). The purpose of the present work is to motivate new research in these directions. We do this via concrete examples (or case studies) that are designed to be simple, in order to capture only the essential parts of the added complexity. Our work is a modest and new attempt to examine dynamic changes and evolution from an algorithmic point of view.

2

The GBR Model

Network congestion games provide a sound model for selfish routing of unsplittable traffic and have recently been the subject of intensive research. The prevailing questions in recent work have to do with the performance degradation due to lack of users’ coordination (e.g., [23, 12, 10, 1, 4]) and the efficient computation of pure Nash equilibria (e.g., [7, 11, 10]). A natural greedy approach for computing a pure Nash equilibrium (PNE) is Greedy Best Response (GBR). Let us consider a dynamic setting with new users arriving in the network. The users play only once and irrevocably choose their strategy upon arrival. Each new user routes her traffic on the minimum delay path given the paths of the users currently in the network. Hopefully the existing users are not affected by the new one and the network configuration remains at a PNE without any defections taking place. This approach is not only intuitive and computationally efficient, but also resembles how things work in practice. 2

A natural question is whether there are any interesting classes of networks for which Greedy Best Response maintains a PNE. GBR can be regarded as a generalization of Graham’s LPT algorithm [13]. The restriction of GBR to parallel-link networks is known to maintain a PNE for arbitrary non-decreasing latency functions and weighted users arriving in nonincreasing order of weights [18, 8]. In this work, we prove that GBR maintains a PNE for symmetric congestion games in series-parallel networks. This result is extended to weighted congestion games with a certain notion of symmetry, namely that the users have the same best response strategies for any initial network traffic. Contribution. If the users are identical, GBR behaves as an online algorithm. For weighted users, GBR is the most natural greedy algorithm since it determines a fixed order in which the users are considered and each user makes an irrevocable greedy choice given the choices of the previous users. In [9] we essentially characterize the class of network congestion games for which GBR maintains a PNE. More specifically, we prove that GBR maintains a PNE for symmetric congestion games in series-parallel networks. This is extended to weighted congestion games with the common best response property. This property requires that the users have the same best response strategies for any initial network traffic. In addition to symmetric network congestion games, this class includes weighted congestion games in layered networks with identical edges (i.e., edge delays are given by a common linear latency function). We also prove that the restriction to series-parallel networks and games with the common best response property is essentially necessary for GBR to maintain a PNE. 2.1

Definitions and Preliminaries

The Model. A network congestion game is a tuple (N, G, (de )e∈E ), where N = {1, . . . , n} is the set of users controlling a unit of traffic demand each, G = (V, E) is a directed graph representing the communication network, and de is the latency function associated with edge e ∈ E. We assume that de ’s are non-negative and non-decreasing functions of the edge loads. If the edge delays are given by a common linear latency function, we say that the edges are identical. For identical edges, we assume w.l.o.g. that the edge delays are given by the identity function, i.e. ∀e ∈ E, de (x) = x. We restrict our attention to single-commodity network congestion games, where the network G has a single source s and destination t and the set of users’ strategies is the set of s − t paths, denoted P. W.l.o.g. we assume that G is connected and every vertex of G lies on a directed s − t path. We also consider weighted single-commodity network congestion games, where user i controls wi units of traffic demand4 . The users are indexed in nonincreasing order of weights, i.e., w1 > w2 > . . . > wn > 0. Single-commodity network congestion games are symmetric 5 . However, weighted games are non4 5

In (unweighted) congestion games, w1 = w2 = . . . = wn = 1. A game is symmetric if all users have the same strategy set and the users’ costs are given by identical symmetric functions of other users’ strategies. In congestion games, the users are identical and a common strategy set implies symmetry.

3

asymmetric in general because the users’ cost functions are different and asymmetric due to different user weights. A vector P = (p1 , . . . , pn ) consisting P of an s − t path pi for each user i is a pure strategies profile. Let `e (P ) ≡ i:e∈pi wi denote the load of edge e in P . The cost λip (P ) of user i for routing her demand on path p in the profile P is λip (P ) ≡

P e∈p∩pi

de (`e (P )) +

P e∈p\pi

de (`e (P ) + wi )

The cost λi (P ) of user i in P is λipi (P ), namely the total delay along her path. A vector Q = (q1 , . . . , qn ) consisting of a probability distribution qi over P for each user i is a mixed strategies profile. For each path p, qP i (p) denotes the n probability that user i routes her demand on p. Let `p (Q) ≡ j=1 qj (p)wj be the expected load routed on path p in Q, and let `p (Q−i ) ≡ `p (Q) − qi (p)wi be the expected load on p excluding of user i. Similarly, let `e (Q) ≡ P P the contribution −i −i p:e∈p `p (Q) and `e (Q ) ≡ p:e∈p `p (Q ) be the expected load on edge e with and without user i respectively. The cost λip (Q) of user i for routing her demand on path p in the mixed strategies profile Q is the expectation according to Q−i of λip (P −i ⊕ p) over all pure strategies profiles6 P −i . The cost λi (Q) of user i in Q is the expectation according to Q of λi (P ) over all pure strategies profiles P . For a strategies profile Q, let λmax (Q) ≡ maxi∈N {λi (Q)} be the maximum user cost in Q. Here we consider mixed strategies profiles only forP identical users and linear latency functions de (x) = ae x. Then, simply λip (Q) ≡ e∈p ae (`e (Q−i ) + 1) and P λi (Q) ≡ p∈P qi (p)λip (Q) by linearity of expectation. A (mixed in general) strategies profile Q is a Nash equilibrium if for every user i and every p, p0 ∈ P with qi (p) > 0, λip (Q) 6 λip0 (Q). Therefore, if Q is a Nash equilibrium, λip (Q) = λip0 (Q) = λi (Q) for every user i and every p, p0 ∈ P with both qi (p), qi (p0 ) > 0. We evaluate strategies profiles using the objective of maximum latency. The maximum latency L(P ) of a pure strategies profile P is the maximum user cost in P , L(P ) ≡ λmax (P ). The maximum latency L(Q) of a mixed strategies profile Q is the P expectation according to Q of L(P ) over Q all pure strategies profiles P , n L(Q) ≡ P ∈P n IP(P, Q)L(P ), where IP(P, Q) = i=1 qi (pi ) is the occurrence probability of P in Q. The optimal solution, denoted P ∗ , corresponds to a pure strategies profile and the optimal cost is L(P ∗ ). The Price of Anarchy (PoA) is defined as worst-case ratio L(Q)/L(P ∗ ) over all Nash equilibria Q. Greedy Best Response Property. GBR considers the users one-by-one in non-increasing order of weight. Each user adopts her best response strategy given the strategies of previous users. The choice is irrevocable since no user can change her strategy in the future. In simple words, each user plays only once and selects its best response strategy at the moment she is considered by the algorithm. 6

For a n-dimensional vector X, X −i ≡ (x1 , . . . , xi−1 , xi+1 , . . . , xn ) and X −i ⊕ x ≡ (x1 , . . . , xi−1 , x, xi+1 , . . . , xn ).

4

Formally, let pi be the path of user i, and let P i = (p1 , . . . , pi ) be the pure strategies profile for users 1, . . . , i. Then, the path pi+1 of user i + 1 is P pi+1 = arg minp∈P { e∈p de (`e (P i ) + wi+1 )} (1) We say that GBR succeeds if every profile P i is a Nash equilibrium. Common Best Response. The single-commodity network congestion game ((wi )i∈N , G, (de )e∈E ) has the common best response property if for every initial flow f (not necessarily feasible), all users have the same set of best response strategies w.r.t. the edge loads induced by f . In other words, if a path p is a best response w.r.t. f for some user, then the following inequality holds for all users j and all paths p0 : P P e∈p0 de (fe + wj ) > e∈p de (fe + wj ) Furthermore, every segment π of a best response path p is a best response for routing the demand of any user between π’s endpoints. We should highlight that in the definition above, best responses are computed without taking into account that some users may already contribute to the initial flow f . The common best response property requires a notion of symmetry between the users, namely that all of them have the same topmost preferences for any initial traffic conditions. This notion of symmetry is weaker than that of a symmetric game but still strong enough to make GBR work in series-parallel networks. Layered and Series-Parallel Graphs. A directed (multi)graph G = (V, E) with a distinguished source s and destination t is layered if all directed s − t paths have exactly the same length and each vertex lies on a directed s − t path. A multigraph is series-parallel with terminals (s, t) if it is either a single edge (s, t) or can be obtained from two series-parallel graphs with terminals (s1 , t1 ) and (s2 , t2 ) connected either in series or in parallel. In a series connection, t1 is identified with s2 , s1 becomes s, and t2 becomes t. In a parallel connection, s1 is identified with s2 and becomes s, and t1 is identified with t2 and becomes t. A directed graph with terminals (s, t) is series-parallel if and only if it does not contain a θ-graph with degree-2 terminals as a topological minor (e.g. the graph in Fig. 1.b is not series-parallel) [6]. Proposition 1. Let G = (V, E) be a series-parallel graph with terminals (s, t), and let vertices u, v connected by two disjoint paths, denoted π and π 0 , only sharing their endpoints. Every s − t path having at least one edge in common with π 0 contains both u and v. 2.2

Greedy Best Response in Series-Parallel Networks

We first show that GBR succeeds if the network is series-parallel and the game has the common best response property. Theorem 1. If G is a series-parallel graph with terminals (s, t) and the game ((wi )i∈N , G, (de )e∈E ) has the common best response property, GBR succeeds and computes a pure Nash equilibrium in time O(nm log m). 5

Proof. The proof is by induction on the number of users considered by the algorithm. The claim holds for the first user, since she adopts her best response strategy and is the only user in the network. We inductively assume that after user i has been considered, P i = (p1 , . . . , pi ) is a Nash equilibrium. Let pi+1 be the path chosen by user i + 1 according to (1). To reach a contradiction, we assume that P i+1 = (p1 , . . . , pi , pi+1 ) is not a Nash equilibrium. Consequently, there is a user j, j 6 i, preferring another path p to her path pj . Let u be a split point where p departs from pj (u may be s). Any pair of different paths has at least one split point because they have a common source. Let v be the first merge point after u where p joins pj again (v may be t). Each split point is followed by a merge point because the paths have a common destination. For simplicity of notation, let π and πj denote the segments of p and pj respectively between u and v. By the definition of v, π and πj are edge disjoint and have only their endpoints u and v in common. Since j wants to defect from πj in P i+1 but not in P i , it is pi+1 that shares some edges with πj and makes it inferior to π for user j. Since pi+1 and πj have at least one edge in common, pi+1 contains both u and v by Proposition 1. Let πi+1 be the segment of pi+1 between u and v (Fig. 1.a). The path pi+1 is a best response for user i + 1 w.r.t. the flow induced by P i . Since the game has the common best response property, pi+1 is also a best response for user j w.r.t. the flow induced by P i (ignoring that wj already contributes to the flow). Therefore, the path segment πi+1 is a best response w.r.t. P i for routing the demand of user j from u to v: X X de (`e (P i ) + wj ) > de (`e (P i ) + wj ) (2) e∈π

e∈πi+1

Since j prefers π to πj after user i + 1 routing her traffic on πi+1 , X X de (`e (P i )) + de (`e (P i ) + wi+1 ) > e∈πj ∩πi+1

e∈πj \πi+1

X

X

de (`e (P i ) + wj ) >

e∈π

de (`e (P i ) + wj ) >

e∈πi+1

X

de (`e (P i ) + wj ) +

e∈πi+1 \πj

X

de (`e (P i ) + wi+1 )

e∈πi+1 ∩πj

The second inequality follows from Ineq. (2). The last inequality holds because the latency functions are non-decreasing and wj > wi+1 . If πj = πi+1 , the contradiction is immediate. If πj 6= πi+1 , user j prefers the path segment πi+1 \ πj to the path segment πj \ πi+1 even in P i : X X λjπj \πi+1 (P i ) = de (`e (P i )) > de (`e (P i ) + wj ) = λjπi+1 \πj (P i ) e∈πj \πi+1

e∈πi+1 \πj

This contradicts to the inductive hypothesis that P i is a Nash equilibrium. Therefore, p and pj does not have any split points and p coincides with pj . Consequently, P i+1 is a Nash equilibrium. 6

GBR performs n s − t shortest path computations in a graph of m edges. This can be done in time O(nm log m) using Dijkstra’s algorithm. u t Single-commodity network congestion games with identical users have the common best response property because the users’ cost functions are identical functions of the edge loads. We are also aware of a class of weighted singlecommodity network congestion games with the common best response property. Proposition 2. A weighted single-commodity congestion game in a layered network with identical edges has the common best response property for any set of user weights. Corollary 1. GBR succeeds for single-commodity congestion games in seriesparallel networks: 1. if the users are identical (for arbitrary non-decreasing edge delays). 2. if the graph is layered and the edges are identical (for arbitrary user weights). GBR has a natural distributed implementation based on a leader election algorithm. There is a process corresponding to each player. We assume that the processes know the network and the edge latency functions. We also assume a message passing subsystem and an underlying synchronization mechanism (e.g. logical timestamps) allowing a distributed algorithm to proceed in logical rounds. Initially, all processes are active. In each round, they run a leader election algorithm and determine the active process of largest weight. This process routes its demand on its best response path, announces its strategy to the remaining active processes, and becomes passive. Notice that all processes can compute their best responses locally. In the offline setting, the algorithm terminates as soon as there are no active processes. In the online setting, new users/processes may enter the system at any point in time. We conclude the study of GBR by providing some simple examples demonstrating that GBR may not succeed in maintaining a PNE if either the network is not series-parallel or the game does not have the common best response property. Hence both conditions of Theorem 1 are necessary for GBR to succeed. If the network is not series-parallel, the simplest symmetric game for which GBR fails consists of two identical users and the 3-layered equivalent of the θgraph with identical edges (Fig. 1.b). The pure Nash equilibrium assigns one user to π1 and the other to π3 . If GBR assigns the first user to π2 , there is no strategy for the second user that yields a Nash equilibrium. We can force GBR to assign the first user to π2 by slightly decreasing the latency function of the second edge to (1 − ²)x, where ² is a small positive constant. The common best response property is also necessary for series-parallel networks other than a sequence of parallel-link graphs connected in series7 . For example, let us consider the 2-layered series-parallel graph of Fig. 1.c and three 7

If the network consists of bunches of parallel-link connected in series, a pure Nash equilibrium can be computed by independently applying GBR to each bunch of parallel links.

7

  

 

 






 




  


  



 


  
  

















Fig. 1. (a) The graph in the proof of Theorem 1. GBR may fail if (b) the network is not series-parallel (even if the game is symmetric) and (c) the game does not have the common best response property (even if the network is series-parallel).

users of weights w1 = 100, w2 = 10, and w3 = 4. The corresponding congestion game does not have the common best response property. GBR assigns the first user to the path π1 , the second user to π2 , and the third user to π3 , while in every pure Nash equilibrium the first two users are assigned to π1 .

3

Coalitional Network Congestion Games

In this section we depart from the classical non-cooperative approach of congestion games, and consider the case where the users are allowed to form coalitions that behave as single players of the game. Each coalitional player attempts to maximize his own private objective given the strategies of the other coalitional players, but he does not actually care about the private costs of the individual users whose coalition he represents. We initially consider the case of parallel links (the classical KP model) and consequently we deal with networks. 3.1

Coalitional Players on Parallel Links

We consider a collection M = {1, 2, . . . , m} of identical machines and a collection N = {1, . . . , n} of jobs. Associated with each job i ∈ N is a load wi ∈ ZZ + , f= which may represent e.g. the processing time required by job i. Denote by W {w1 , . . . , wn } the set of loads. For any integer k > 2, denote [k] = {1, . . . , k}. Definition 1 (Coalitions). A set of κ > 1 coalitions C1 , . . . , Cκ is a partition f , Cj 6= ∅ for all j ∈ [κ] and f into κ nonempty subsets. Hence Sκ Cj = W of W j=1 Ci ∩ Cj = ∅ for all i 6= j, i, j ∈ [κ]. Pκ n For j ∈ [κ] let Cj = {wj1 , . . . , wj j }, so that j=1 nj = n. Denote Wj = Pnj i nj 1 w . Without loss of generality, assume that w j > · · · > wj for all j ∈ [κ]. i=1 j Strategies and Assignments. A pure strategy σj = (σji )i∈[nj ] for coalition Cj defines a machine σji ∈ M for each wji ∈ Cj . Denote by Σj the set of all pure strategies available to coalition Cj . Clearly, Σj = Mnj . A mixed strategy pj for coalition Cj is a probability distribution on the set Σj of its pure strategies. 8

P Clearly, for each coalition Cj , σj ∈Σj pj (σj ) = 1 and pj (σj ) ∈ [0, 1] for all σj ∈ Σj . A pure strategy profile or pure assignment σ = (σj )j∈[κ] is a collection of pure strategies, one per coalition. Denote Σ = ×κj=1 Σj . Let (σ−j , αj ) be the pure assignment resulting from σ when coalition Cj unilaterally changes its pure strategy to αj . Similarly, a mixed strategy profile or mixed assignment p = (pj )j∈[κ] is a collection of mixed strategies, one per coalition. The support of coalition Cj in the mixed assignment p is the set Sj (p) = {σj ∈ Σj : pj (σj ) > 0}; thus Sj is the set of pure strategies that Cj chooses with non-zero probability. If Sj (p) = Σj for all j ∈ [κ] then p is a fully mixed assignment. Selfish Costs. Fix a pure assignment σ = (σj )j∈[κ] . Define the load on machine ` ∈ M, denoted θ` (σ), as the total demand on machine ` with respect to σ, that is κ X X θ` (σ) = wji . j=1 i∈[nj ]:σji =`

Then, the Selfish Cost λj (σ) of coalition Cj is the maximum load over the set of machines it uses; that is λj (σ) = max θσji (σ) . i∈[nj ]

For a mixed assignment p = (pj )j∈[κ] the load on each machine ` ∈ M becomes a random variable induced by the probability distributions pj for all j ∈ [κ]. We define the expected load on machine ` ∈ M as the expectation of the load on machine ` according to p. The Conditional Expected Selfish Cost λj (p, αj ) of coalition Cj is the conditional expectation of its Selfish Cost had Cj adopted the pure strategy αj ∈ Σj . The Expected Selfish Cost of coalition Cj is P defined as λj (p) = αj ∈Σj λj (p, αj ). Nash Equilibria. The definition of Expected Selfish Costs completes the definition of a finite non-cooperative game Γ = hP, (Σj )j∈[κ] , (λj )j∈[κ] i where the set of players P is the set of coalitions {C1 , . . . , Cκ }. We are interested in the induced Nash equilibria [21] of Γ . A Nash equilibrium is a (pure or mixed) assignment such that no coalition can reduce its Selfish Cost by unilaterally changing its strategy. Formally: Definition 2 (Pure Nash equilibrium). A pure assignment σ = (σj )j∈[κ] is a pure Nash equilibrium for Γ if, for all coalitions Cj , λj (σ) 6 λj (σ−j , αj )

∀αj ∈ Σj .

Definition 3 (Mixed Nash equilibrium). A mixed assignment p = (pj )j∈[κ] is a (mixed) Nash equilibrium if, for all coalitions Cj , for all αj ∈ Σj , pj (αj ) > 0

=⇒

αj ∈ arg min λj (p, σj ) . σj ∈Σj

9

Social Cost, Optimum and the Price of Anarchy. For any pure assignment σ = (σj )j∈[κ] we define the Social Cost, denoted SC(σ), to be the maximum load over the set of machines M with respect to σ, ie, SC(σ) = max θ` (σ) = max λj (σ) . `∈M

j∈[κ]

For any mixed assignment p = (pj )j∈[κ] the Social Cost is defined as the expectation, over all random choices of the coalitions, of the maximum load over the set of machines, i.e. SC(p) =

X σ1 ∈Σ1

···

κ X Y σκ ∈Σκ j=1

pj (σj ) max θ` (σ) . `∈M

Now let σ ∗ be a pure assignment that minimizes the Social Cost, i.e. σ ∗ ∈ f to the arg minσ SC(σ). Thus σ ∗ is an optimal assignment of the set of loads W set of machines M. Denote OPT = SC(σ ∗ ). The Price of Anarchy (also referred to as Coordination Ratio [16]), denoted R, is the maximum value, over all (pure or mixed) Nash equilibria α, of the ratio SC(α) OPT . 3.2

Nash Equilibria

Existence and Tractability of Pure Nash Equilibria. We show that: Theorem 2. There exists at least one pure Nash equilibrium. Proof. For a given pure assignment σ, let θi (σ) be the load of machine i ∈ M and rename machines so that θ1 (σ) > θ2 (σ) > · · · > θm (σ). Now consider the universe of pure assignments. Each such assignment σ induces a sorted load vector θ(σ) = hθ1 (σ), θ2 (σ), . . . , θm (σ)i. Let σ0 be the pure assignment that corresponds to the lexicographically minimum load vector. We argue that σ0 is a pure Nash equilibrium. Assume, by way of contradiction, that σ0 is not a pure Nash equilibrium. By definition of Nash equilibrium, there exists a coalition Cj that can reallocate its loads to the machines, resulting in a new pure assignment σ1 , such that λj (σ1 ) < λj (σ0 ). For any pure assignment σ, let sj (σ) denote the set of machine indices in the vector θ(σ) where coalition Cj has loads with respect to σ0 . Let i1 = min sj (σ0 ). Then clearly θi1 (σ0 ) = λj (σ0 ) . (3) Assume that, in assignment σ1 , there is a load of coalition Cj on machine with index i < i1 in the vector θ(σ0 ). Then the Selfish Cost for coalition Cj would be λj (σ1 ) > θi (σ0 ) > θi1 (σ0 ) = λj (σ0 ) , a contradiction. So, in σ1 , all loads of coalition Cj are assigned to machines with indices i > i1 in the vector θ(σ0 ). This immediately implies that for all i < i1 , θi (σ0 ) = θi (σ1 ). We now consider two cases: 10

Case 1. Assignment σ1 results from σ0 by removing some loads of Cj from machine of index i1 in the vector θ(σ0 ) (and perhaps some other loads of Cj come to that same machine). Since λj (σ1 ) < λj (σ0 ) it can not be the case that machine i1 is again the maximum loaded machine of coalition Cj because θ(σ0 ) is the lexicographically minimum load vector. Thus, another machine i01 > i1 of sj (σ1 ) is the maximum loaded machine of Cj now. But θi01 (σ1 ) = λj (σ1 ) < λj (σ0 ) = θi1 (σ0 ) , hence θ(σ1 ) is lexicographically smaller than θ(σ0 ), a contradiction. Case 2. In assignment σ1 , all loads of Cj in machines with indices i1 , . . . , ik0 , k 0 < k, are not moved, i.e. they stay as in σ0 . Then we can repeat the argument of Case 1 above for the machine with index ik0 + 1 ∈ sj (σ0 ) to obtain again a contradiction. u t Lemma 1. It is NP-hard to find a pure Nash equilibrium. Proof. Assume that κ = 1, i.e. there is only one coalition C1 = {w1 , . . . , wn }. Then a pure Nash equilibrium is an optimal assignment of the loads w1 , . . . , wn to machines 1, . . . , m. u t Inexistence of Fully Mixed Nash Equilibria. We shall now show that: Theorem 3. There exists a fully mixed Nash equilibrium if and only if nj = 1 for all coalitions Cj , j ∈ [κ]. Proof. If nj = 1 for all j ∈ [κ] then the case reduces to the standard KP model [16], for which a fully mixed Nash equilibrium is guaranteed to exist. Now fix a coalition Cj , j ∈ [κ], with nj > 1. Suppose that there exists a fully mixed Nash equilibrium p. Let θr (p−j ) be the expected load on machine r caused by all coalitions except for Cj , with respect to p. Let ` ∈ arg maxr∈M θr (p−j ). Let σj be an optimum assignment of the loads of coalition j on the set of resources M, had j been the unique coalition. Since nj > 1, it is clear that the contribution of Cj to the load of any machine is at most OPTj < Wj . Let αj be the assignment of the demands of coalition j where each wij is assigned on resource `, i.e. αji = ` for all i ∈ [nj ]. Then clearly λj (p, σj ) 6 OPTj + θ` (p−j ) < Wj + θr (p−j ) = λj (p, αj ) , a contradiction, since p is fully mixed, hence both assignments σj and αj belong to the support of coalition Cj and should give the same Conditional Expected Selfish Cost to coalition Cj . u t On the other hand, let’s assume that each coalitional player has at most m weights at his disposal. Then we can easily determine a generalized fully mixed strategy for each coalitional player, where the links for each player’s weights are chosen uniformly at random, but with no replacement (i.e., there is no chance for 11

any two weights of the same player to be assigned to the same link). This profile is indeed a Nash Equilibrium: Observe first that for any specified player j ∈ [k], the expected load of any link ` ∈ M due to the profile p−j of all the other players W −W is θ` (p−j ) = totm j . The expected cost of player j is of course λj (p) = Wmtot due to the obvious symmetry. Finally, observe that ∀aj ∈ Sj = M nj , λj (p, aj ) = wjmax + θ` (p−j ) = wjmax +

Wtot − Wj Wtot Wj = + wjmax − m m m

Wtot = λj (p) m by a simple averaging argument (we assumed that ` is the link that hosts the max weight of player j). >

3.3

Remarks on the Price of Anarchy

f = Example 1. Assume that κ = 1, i.e. there is a single coalition C1 = W {w1 , . . . , wn }. Then, any Nash equilibrium must be an optimum assignment of the set of loads C1 to the set of machines M and vice versa, hence in any Nash equilibrium σ, SC(σ) = OPT and thus R = 1. Example 2. Assume that κ = n, i.e. Cj = {wj } for all j ∈ [n]. Then the case reduces³ to the ´ standard KP model [16], for which it is well known [15] that R = Θ logloglogmm . ³ ´ Conjecture 1. The PoA is Θ logloglogmm for any coalitional game on identical parallel links with atomic weights and k coalitions, so long as the number of coalitions is k = n² , for any constant 1 > ² > 0. For k = n1/g(n) , where g(n) increasing function the PoA is at least ³ = o(log n) is arbitrary ´ n ³ of n, ´o log m log m Ω g(m)[log log and at most min k, O . m−log g(m)] log log m 3.4

Coalitions in Single Commodity Networks

Consider a single commodity network G = (V, E), i.e. a directed acyclic graph with a distinguished source-destination pair of nodes (s, t). Denote Π the set of all s − t paths of G. We extend game Γ described in Sect. 3 to a new game Γ 0 = hP, (Σj )j∈[κ] , (λj )j∈[κ] i such that: (a) The set of players P is, as before, the set of coalitions {C1 , . . . , Cκ }. (b) The set of pure strategies Σj of coalition Cj is the set of all possible allocations of the loads of Cj on the s − t paths of G. Clearly, Σj = Π nj . Thus a pure strategy σj ∈ Σj defines an s − t path σji for each load wji ∈ Cj . (c) Let σ = (σj )j∈[κ] be a pure assignment, i.e. a combination of pure strategies, one per coalition. For each edge e ∈ E define the load of e in σ, denoted θe (σ), to be the sum of the loads that cross e given σ, i.e. θe (σ) =

k X

X

j=1 i∈[nj ]:e∈σji

12

wji .

For each s − t path π ∈ Π of G define the load of π in σ as X θπ (σ) = θe (σ) . e∈π

Then, the Selfish Cost λj (σ) of coalition Cj in σ is the maximum load over all paths used by Cj , i.e. i

λj (σ) = max θσj (σ) . i∈[nj ]

Now we can easily extend the definitions of Conditional Expected Selfish Cost and Expected Selfish Cost of Sect. 3 to the case of Γ 0 . More specifically, a mixed strategy pj for coalition Cj is a probability distribution on the set Σj . A mixed assignment p = (pj )j∈[κ] is a combination of mixed strategies, one per coalition. For a mixed assignment p = (pj )j∈[κ] , the expected load on edge e ∈ E is the expectation of the load on edge e according to p. The expected load on path π ∈ Π is sum of the expected loads over all edges that belong to π. Then, the Conditional Expected Selfish Cost λj (p, αj ) for coalition Cj is the conditional expectation of its Selfish Cost had Cj adopted the P pure strategy αj ∈ Σj . The Expected Selfish Cost of coalition Cj is λj (p) = aj ∈Σj λj (p, αj ). Nash Equilibria. A pure assignment σ is a pure Nash equilibrium for Γ 0 if no coalition can reduce its Selfish Cost by unilaterally reallocating its loads, i.e. if, for all coalitions Cj , λj (σ) 6 λj (σ−j , αj ) ∀αj ∈ Σj . A mixed assignment p = (pj )j∈[κ] is a (mixed) Nash equilibrium if, for all coalitions Cj , for all αj ∈ Σj , pj (αj ) > 0

=⇒

αj ∈ arg min λj (p, σj ) . σj ∈Σj

Social Cost, Optimum and the Price of Anarchy. For any pure assignment σ = (σj )j∈[κ] we define the Social Cost, denoted SC(σ), to be the load of the maximum loaded path with respect to σ, i.e. SC(σ) = max θπ (σ) = max λj (σ) . π∈Π

j∈[κ]

For any mixed assignment p = (pj )j∈[κ] the Social Cost is defined as the expectation, over all random choices of the coalitions, of the load of the maximum loaded path, i.e. SC(p) =

X σ1 ∈Σ1

···

κ X Y σκ ∈Σκ j=1

pj (σj ) max θπ (σ) . π∈Π

Now let σ ∗ be a pure assignment that minimizes the Social Cost, i.e. σ ∗ ∈ arg minσ SC(σ). Denote OPT = SC(σ ∗ ). The Price of Anarchy, denoted R, is the maximum value, over all (pure or mixed) Nash equilibria α, of the ratio SC(α) OPT . 13

4

The Evolutionary Approach

In this section we examine the dynamics of interacting selfish particles in a network. Our approach is motivated by evolutionary game theoretic considerations. 4.1

Hawks and Doves: The General Model

Consider the following symmetric two player game Γ with set of pure strategies Σ1 = Σ2 = {H, D} and payoff matrix · ¸ (−1, −1) (0, −1) AΓ = (−1, 0) (α, α) where α > 0. The first row corresponds to strategy H for the row player and likewise the first column corresponds to strategy H for the column player. Suppose now that we have k individuals each of whom plays one of the pure strategies H or D. Individuals who play the pure strategy H are called Hawks and individuals who play the pure strategy D are called Doves. In our general Hawks-and-Doves model each of the k individuals performs a random walk on a graph G. When two of them meet on a vertex of the graph, they play the game Γ . Then, each individual that played the game gives birth to a number of individuals that play the same strategy he plays. For example, if we have a Hawk and a Dove who play the game, then the Dove will disappear (i.e. he will be removed from play) and the Hawk will remain on the vertex where the game was played. Given the initial position of the Doves and the Hawks on the graph G, we are interested in the probability that Doves survive the above process. We then say that the Doves win the game. Note 1: As we may notice, the general model described above is perhaps too general for analysis. For example, it is not specified how the individuals perform their random walk. Two scenarios may come to mind. In the first one, an individual is chosen at random and moves to a neighboring vertex with equal probability. In the second, the individuals perform their random walk simultaneously. Note 2: Even if we specify the way individuals move on the graph, we also need to specify what happens when more than two individuals meet on some vertex. For example, what happens when more than two Doves occupy the same vertex? 8 Also, what happens if we have two Hawks and a Dove occupying the same vertex at the same time? In view of the above discussion, we present the following more specific model. 8

An interesting way to deal with it is to assume that we have a function f : {2, 3, . . . , n} → Z such that when λ > 2 Doves meet on a vertex, they give birth to f (λ) Doves on the whole.

14

4.2

The Distinct Hawks-and-Doves Model

We can get the distinct Hawks-and-Doves model by the general model in the following way: 1. Each individual starts on a different vertex of G. 2. At every step a single individual is chosen uniformly at random and moves equiprobably to a neighboring vertex. 3. The game Γ has α = 0, i.e. when several Doves meet on a vertex they do not give birth to any more Doves. 4. When a Hawk meets several Doves on a vertex, he plays the game Γ with every one of them, so that at the end of the step, no Doves remain on this vertex. 9 5. The whole process stops (and we say that the game ends) when only one type of individuals remains on the graph. Notice that in this model, the problem described in Note 2 can never arise. Also, the random walk of the k individuals is uniquely determined. We will henceforth be working with the distinct Hawks-and-Doves model. Let us now discuss some interesting open problems on this model. Open Problem 1 (Absorbtion Probability) Given the initial position of every individual, what is the probability that the Hawks are eliminated before all the Doves do? It is easy to see that if the number of Hawks is odd, then the probability that the Doves win is 0 (as long as our graph is connected). The question becomes interesting only in the case where there is an even number of Hawks. As we will see, if the number of individuals k is constant, then the above probability can be computed in O(npoly(k) ) steps, i.e. pseudopolynomial time. Of course, using the same method for k = k(n) → ∞ we can get an exponential time algorithm. Is this the best we can do? This gives rise to the following obvious question: Open Problem 2 (Hardness) Given the initial position of every individual, how hard is it to compute the probability that the Hawks are eliminated before all the Doves do, in the case where k = k(n) → ∞? Even if it is not likely to compute the above probability in polynomial time, we could possibly estimate it by running the model many times and looking at the outcome after a large period of time. In order for this technique to be useful, the outcome should be reached in a polynomial (expected) number of steps. Then we would have a simple expected time polynomial algorithm that approximates the value of the probability in question. Of course, using the results of [24] we can see that an upper bound on the expected time needed so that only Hawks or only 9

An interesting variation of this is to assume that the Hawk plays the game with exactly one Dove on the vertex.

15

Doves remain on the graph is at most O(kn3 ), which is polynomial considering the fact k 6 n. A slight improvement on this upper bound can be found in [3]. However, both these techniques do not take into consideration the strategy that each individual plays. The latter may lower considerably the upper bound on the time needed to come to a conclusion on whether the Doves or the Hawks finally win. For example, consider the extreme case where there are exactly two Hawks. Then, using the result of [24], the mean time needed is just O(n3 ). Hence, the following problem arises: Open Problem 3 (Absorbtion Time Bounds) (a) Given the initial position of every individual, give an upper (and even a lower) bound on the expected number of steps needed to specify whether the Doves win or not. (b) If possible, give a graph and a starting position of the k individuals on G that achieves this upper bound. Bounds can also be given for the probability that the Doves win. That is: Open Problem 4 (Absorbtion Probability Bounds) Given the initial position of every individual, give an upper (and even a lower) bound on the probability that the Hawks are eliminated before all the Doves do. Finally, another interesting problem would be the equivalence of the distinct Hawks-and-Doves model with other models derived from the general Hawks-andDoves model. Representation of the Evolution Dynamics It is easily seen that the evolution in the Distinct Hawks-and-Doves model can be represented by a Markov Chain. This is true, because all we need to know in order to specify the next configuration of the model at some discrete time t is the number and position of the individuals left on the graph. This can be a little tricky. The state space of the Markov Chain will need a way to describe the individuals that have not yet disappeared, as well as their exact location on G. The latter can be done perhaps with a k-tuple each element of which describes the position of a single individual. But the issue of which individuals have still remain on the graph remains. As we will see below, things are considerably simpler when we have exactly 1 Dove. Single Dove Distinct Model In this section we will derive a Markov Chain that captures the evolution of the distinct Hawks-and-Doves model in the case where we have only 1 Dove. The state space of the Markov Chain is I = {(d, h1 , h2 , . . . , hk−1 ) ∈ V (G)k } i.e. the set of all k-tuples whose first position denotes the vertex of the dove and whose i-th position denotes the vertex of the (i − 1)-th Hawk. In order to describe the transition matrix P of the Markov Chain we need some additional notation. For some s ∈ I, let v(s) be the number of hawks that 16

have v in their corresponding position in s. Then, (assuming that we have an even numberSof Hawks) the absorbing states of the chain are A = {s ∈ I : h1 = · · · = hk−1 } {s ∈ I : d(s) is even}. If the chain is absorbed in the first set, then the Dove wins, otherwise he loses. Let now, for some transient state s ∈ I k(s) = 1 + |v ∈ V : v(s) is odd| i.e. k(s) is the number of players left on the graph when the state is s. Then the representation matrix P has

Ps1 ,s2

 1 1 if s1 ∈ / A, the individual chosen is on vertex x    k(s1 ) degree(x) and ∃u ∈ V : s2 = s1 [x/u] =  1 if s1 = s2 ∈ A   0 otherwise.

The notation s1 [x/u] means that all occurrences of x in s1 have been replaced by u. Note: The state space of the Markov Chain describing the evolution of the distinct model in the case of 1 Dove seems in a way easier to work with. However, the open problems 1, 2, 3 and 4 remain interesting even in this case. 4.3

Probability of absorbtion for constant k

In this subsection, we give a general algorithm that computes the absorbtion probability in the case where we have a constant number of players initially on the graph G. As we have discussed, the evolution dynamics of the model can be described by a Markov Chain with state space I and transition matrix P . Without loss of generality, we may assume that I consists only of two absorbing states sD and sH , the first one being reached when only Doves remain (i.e. the Doves win) and the second one being reached when only Hawks remain. Also we will denote by fs the probability that the Doves win the game, given that the chain starts at state s ∈ I. Let now F = [f1 , f2 , . . . , f|I| ]T , for some ordering of the states in I. Clearly, F is the unique nonnegative solution that has fsD = 1 and fsH = 0, of the following (matrix) equation F = P · F.

(4)

Let now T be the set of all transient states of our Markov Chain, that is I = T ∪ {sD } ∪ {sH }. Let also Q be the matrix that specifies the transition probabilities from some transient state to another. Then Q is a |T | × |T | matrix with each row/column corresponding to a different transient state. Also, for i, j ∈ T we have Qi,j = Pi,j . Note that the sum of the entries of some row in Q can be less than 1. 17

Consider now the |T | × |T | matrix M , with Mi,j being the expected number of visits to (transient) state j, given that we start at (transient) state i. Conditioning on the first step of the Markov Chain we get M = I + Q · M. −1

Hence, M = (I − Q) can be computed in O(|T |3 ) time. For i, j ∈ T , let fi,j denote the probability that the chain ever reaches j given that it starts at i. Then, by the Markov property we have Mi,j = fi,j Mjj so fi,j can also be computed in O(|T |3 ) time. The probability that the Doves win given that we start at i ∈ T then is fi = Pi (Doves win) = Pi {the chain is absorbed in sD } X = Pi {the chain reaches sD via j} j∈Γ (sD )

=

X

Pj,k (fi,j + fi,j fj,j + · · · )

j∈Γ (sD )

=

X j∈Γ (sD )

Pj,k fi,j 1 − fj,j

where Γ (sD ) denotes the set of transient states that can reach sD in a single step. Since Γ (sD ) 6 |T |, the above procedure results in an algorithm that computes the probability that the Doves win given that we start at state s ∈ T in O(|T |3 ) time. Remark: By the discussion of section 4.2 we can see that |I| 6 nk when we have only one Dove. So, the above algorithm runs in O(n3k ) pseudopolynomial time. 4.4

Probability of Absorbtion (some cases)

Let G denote the graph on which the individuals of the distinct hawks-and-doves model move. Let also the number of Doves be 1 and the number of Hawks be k − 1 (even). Assume for a moment that every individual chooses his starting vertex with some probability distribution on V (G). For individual i, let us call this initial distribution Fi . Then, by symmetry we can see that the following holds: Lemma 2. If Fi = Fj for all individuals i 6= j, then the probability that the Dove wins is 1 . k As we will see, when G is the complete graph, then 2 holds even in the case where every individual starts on a separate vertex. P (D wins) =

18

Complete Graph In this section we assume that the graph G on which the individuals move is the complete graph Kn . Also, assume that we work with the distinct hawks-and-doves model where the number of Hawks is k − 1 and we have only one Dove (the general case with s Doves can be easily reduced to the case with 1 Dove and is discussed in the end of the section). Finally we assume that k − 1 is even, because if it is odd, the probability that the Dove wins is 0 (G is connected and there will always remain at least one Hawk). We can prove the following Theorem 4. In the distinct hawks-and-doves model where the number of Hawks is k − 1 (even number) and we have only one Dove, the probability that the Dove wins the game is k1 . Proof. Of course, one can prove the theorem by simply solving the recurrence implied in (4)(see [3]). We will instead use a more elegant proof. Because of the symmetry of Kn and the symmetry (concerning the way different ¡ ¢individuals are handled) of the distinct model, we can see that any of the k2 pairs of the k individuals is equally likely to meet first. Hence k−1 P (D is not the first to meet) = 1 − ¡k¢ 2

Also, after the first pair of Hawks meet, then we have k − 3 Hawks and 1 Dove and we can see that the evolution of the model depends only on the current positions of the individuals left (i.e. independent of the past). So we can easily apply the above equation and Bayes formula to get k−1 2 −1

P (D wins) =

Y

Ã

k − 1 − 2i 1 − ¡k−2i¢

!

2

i=0 k−1 2 −1

= =

Y (k − 2i − 1)(k − 2i − 2) (k − 2i)(k − 2i − 1) i=0

1 k

which completes the proof. ♦ Notice that the above result is actually a special case of Lemma 2. Suppose now that we have s Doves instead of one and also, when a Hawk meets with several Doves on a vertex, it “eats” them all. Then we can see that the number of Doves that survive the process is actually a Binomial random variable X ∼ B(s, k1 ). Hence we get µ ¶s 1 P (D wins) = 1 − P (X = 0) = 1 − 1 − . k 19

Cycle In this section we assume that the graph G on which the individuals move is the cycle Cn . Also, assume that we work with the distinct hawks-and-doves model where the number of Hawks is 2 and we have only one Dove. This choice of parameters is made because of the following reasons: 1. The probability that the Doves win in the case of one Dove is a lower bound on the probability that the Doves win in the case of more than one Doves. So, a lower bound for the worst case applies also for the second (more general) case. Notice that this is true irrespectively of the graph G. 2. In the case where the graph is Cn and we have k−1 hawks and 1 Dove, a lower bound on the probability that the Dove wins is actually the probability that the Dove wins when it start from a vertex with two Hawks on its adjacent vertices. This is true because (a) only the first Hawk to the Dove’s right and the first Hawk to the Dove’s left can end the process in favor of the Hawks, i.e. “eat” the Dove and (b) all that the rest of the Hawks do is to turn these two further away from the Dove. Let us now try to construct a Markov Chain that represents the evolution dynamics in this special case. It turns out that instead of having a state space that consists of 3-tuples, we can do better. All we need is the distance of each Hawk from the Dove. So, the state space is ICn = {(i, j) : i, j > 0, i + j 6 n} − {(0, n)} − {(n, 0)}. Since i and j measure the distance of each Hawk from the Dove, it is obvious that if ij = 0, then the Dove has lost the game. Similarly, if i + j = n, then the two Hawks have met, and so the Dove wins. These two kinds of states are all absorbing. We will denote the first kind of absorbing states as AH (the Hawks win) and the second as AD (the Dove wins). For s = (i, j) ∈ ICn and s0 = (i0 , j 0 ) ∈ ICn , the transition matrix of the Markov Chain is 1 0 0  6 , if (i = i and j = j ± 1)   0   or (i = i ± 1 and j = j 0 ) or (i = i0 ± 1 and j = j 0 ∓ 1) Ps,s0 =   1, if s = s0 and s ∈ AD ∪ AH    0, otherwise. We are interested in the probability that the Dove wins, given that the initial state of the Markov Chain described is some state s0 = (i0 , j0 ), i.e. the probability of absorbtion at some state of AD given that the initial distances of the two Hawks from the Dove are i0 and j0 . Notice that this uniquely identifies the number of edges on the arc of the cycle that connects the two Hawks and does not contain the Dove. We will denote this probability by fi0 ,j0 . Solving the recurrence implied by (4) can give us the exact value of fi,j , for all (i, j) ∈ IG . But this can be very difficult at first sight. Instead we will solve an equivalent recurrence. Notice that the process ends when the first meeting 20

occurs. The Dove will have lost if either of the two Hawks does not meet anyone. Then we have P (D wins) = 1 − P (H1 does not meet) − P (H2 does not meet) and by the definition of fi,j and using the fact fi,j = fj,i , for all i, j, we get the recurrence fi,j = 1 − fi,n−i−j − fj,n−i−j

(5)

for (i, j) ∈ ICn . Note that we have fi,j = 0 if (i, j) ∈ AH and fi,j = 1 if (i, j) ∈ AD , as needed. The solution to this recurrence is µ ¶ ij 2n−i−j fi,j = 1+ (n − j)(n − i) 3 i+j which can be verified to be the absorbtion probabilities we seek. In other words we have proved the following Theorem 5. In the distinct hawks-and-doves model with 2 Hawks and one Dove, the probability that the Dove wins when the initial distances of the two Hawks from the Dove are i0 and j0 is µ ¶ i0 j0 2 n − i0 − j0 P (D wins) = 1+ . (n − j0 )(n − i0 ) 3 i0 + j0 ♦

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