Int J Game Theory (1998) 27:501±509

Crowding games are sequentially solvable*

Igal Milchtaich

Department of Mathematics and Center for Rationality and Interactive Decision Theory,Hebrew University of Jerusalem, Israel (e-mail: igal@math.huji.ac.il)

Received July 1997/Final version May 1998

Abstract. A sequential-move version of a given normal-form game G is anextensive-form game of perfect information in which each player chooses hisaction after observing the actions of all players who precede him and thepayo¨s are determined according to the payo¨ functions in G. A normal-formgame G is sequentially solvable if each of its sequential-move versions has asubgame-perfect equilibrium in pure strategies such that the players' actionson the equilibrium path constitute an equilibrium of G .

A crowding game is a normal-form game in which the players share acommon set of actions and the payo¨ a particular player receives for choosinga particular action is a nonincreasing function of the total number of playerschoosing that action. It is shown that every crowding game is sequentiallysolvable. However, not every pure-strategy equilibrium of a crowding gamecan be obtained in the manner described above. A su½cient, but not neces-sary, condition for the existence of a sequential-move version of the game thatyields a given equilibrium is that there is no other equilibrium that Paretodominates it.

Key words: Crowding games, congestion games, sequential solvability, pure-strategy equilibria

A crowding game (or a congestion game with player-speci®c payo¨ functions;Milchtaich, 1996) is an n-person normal-form game G where each player i

* This work is based on Chapter One of the author's Ph.D. dissertation, written at the Center forRationality and Interactive Decision Theory of the Hebrew University of Jerusalem under thesupervision of Prof. B. Peleg, Prof. U. Motro, and Prof. S. Hart. A previous version of this paperwas circulated under the title ``On backward induction paths and pure strategy Nash equilibria ofcongestion games.''

chooses one action j from a common ®nite set A of actions, and receives apayo¨ which is a nonincreasing function Sij of the total number nj of playerschoosing j. �nj�j AA is called the congestion vector. The actions chosen by theplayers constitute a (pure-strategy Nash) equilibrium of G if the action si

chosen by each player i is a best response for that player against the actionschosen by the other players. Formally, an action pro®le s� �s1; s2; . . . ; sn� AAn

is an equilibrium if and only if, for every player i and action j, Sisi�nsi�V

Sij�nj � 1�. Every symmetric crowding game is also a congestion game, asde®ned in Rosenthal (1973; see below). The class of all congestion games co-incides with the class of potential games (Monderer and Shapley, 1996). Thisimplies, in particular, that every such game has at least one Nash equilibriumin pure strategies. Nonsymmetric crowding games, however, generally do notadmit a potential function.

Given any n-person normal-form game G , a sequential-move version of Gis an extensive-form game of perfect information in which the n players areput in some ®xed order, each player chooses his action after observing thechoices of all players who precede him, and the payo¨s are determined ac-cording to the payo¨ functions in G . There are n! ways of ordering the players,and each order de®nes a di¨erent sequential-move version of G . As is wellknown, an extensive-form game of perfect information always has at least onesubgame-perfect equilibrium in pure strategies. We will say that the (simulta-neous-move) game G is sequentially solvable if each of its sequential-moveversions has at least one subgame-perfect equilibrium in pure strategies suchthat the player's actions on the equilibrium path constitute a (Nash) equilib-rium of G .1 The main result of this paper is the following:

Theorem 1. Every crowding game is sequentially solvable.

It follows, in particular, from Theorem 1 that every crowding game has aNash equilibrium in pure strategies. For a computationally e½cient algorithmfor ®nding such an equilibrium, and for a discussion of convergence to anequilibrium, see Milchtaich (1996). At least one of the (pure-strategy) equi-libria of every crowding game is a strong equilibrium (Konishi et al., 1997a).This is not generally true for congestion games. See, however, Holzman andLaw-Yone (1997), were su½cient conditions for the existence of a strongequilibrium in such games are obtained.

Sequential-move equilibria

By Theorem 1, every sequential-move version of a given crowding game G hasa subgame-perfect equilibrium in pure strategies such that the action pro®leplayed on the equilibrium path is an equilibrium of G . Such an action pro®lewill be said to be a sequential-move equilibrium of G . The question naturallyarises, whether all equilibria of a given crowding game are sequential-moveequilibria. The following example shows that the answer, in general, is ``no.''

1 An alternative term that comes to mind is ``Stackelberg solvable''. However, this term was usedin another meaning by d'Aspremont and GeÂrard-Varet (1980). Stackelberg solvability, as de®nedby these authors, neither implies nor is implied by sequential solvability.

502 I. Milchtaich

Consider a two-person, two-action crowding game where S11�1� >S12�1� > 0, S22�1� > S21�1� > 0, and 0 > Sij�2� for every i and j. This gamehas two equilibria: (1, 2) and (2, 1). However, only the ®rst equilibrium is asequential-move equilibrium; there is no subgame-perfect equilibrium of asequential-move version of this game in which the second action pro®le isplayed.

The second equilibrium of the above game is Pareto dominated by the®rst one.2 As we will presently see, every equilibrium of a crowding gamewhich is not Pareto dominated by another equilibrium is a sequential-moveequilibrium.

In every crowding game, an equilibrium which is not Pareto dominated byanother equilibrium is strong in the sense of Aumann (1959): there exists nogroup of players who can increase their payo¨s by simultaneously changingtheir actions, if the rest of the players do not change their actions. This followsfrom the fact that if an equilibrium s is not strong, so that there is an actionpro®le s00 s such that, for every player i for which s0i0si, player i 's payo¨when s0 is played is strictly higher than his payo¨ when s is played, then s0 isan equilibrium and it Pareto dominates s. To prove this fact, note ®rst thatthe congestion vector �n0j�j AA corresponding to s0 is equal to the congestionvector �nj�j AA corresponding to s. Indeed, n0j > nj for some j would imply thatthere is a player, i, for which s0i � j and si 0 j. But it follows from our as-sumption concerning s0 and from the monotonicity of Sij that in such a caseSisi�nsi� < Sij�n0j�USij�nj � 1� ± a contradiction to the assumption that s is

an equilibrium. Since the two congestion vectors are equal, for every player ifor which s0i � si, player i 's payo¨ when s0 is played is equal to his payo¨when s is played. It follows that s0 Pareto dominates s. Also, since s is anequilibrium, Sis 0

i�n0s 0

i�VSisi

�nsi�VSij�nj � 1� � Sij�n0j � 1� for every player i

and action j. Hence, s0, too, is an equilibrium.For generic3 crowding games the converse is also true: an equilibrium s

that is Pareto dominated by another equilibrium s0 is not strong. The reason isthat, in such a case, for every player i for which s0i 0 si, player i 's payo¨ whens 0 is played cannot be equal to, and is therefore strictly higher than, his payo¨when s is played.

Theorem 2. Every strong equilibrium of a crowding game G is a sequential-moveequilibrium of G .

The converse of Theorem 2 is false; not every sequential-move equilibriumof a crowding game is strong. Consider, for example, the sequential-moveversion of a three-person, three-action generic crowding game where S11�1� >S12�1� > S12�2� > 0, S22�1� > S23�1� > 0, S32�1� > S31�1� > S31�2� > 0, and0 > Sij�m� otherwise and the players move in the order 1, 2, 3. If player 1plays 1, then player 2 will play 2 and player 3 will play 1. If player 1 plays 2,then player 2 will play 3 and player 3 will play 1. Player 1's payo¨ is negative

2 An action pro®le is Pareto dominated by another action pro®le if there is no player whose payo¨is strictly higher when the ®rst action pro®le is played and there is at least one player whose payo¨is strictly higher when the second action pro®le is played.3 A crowding game will be said to be generic if, for every, i, j, j 0, and 1Um, m0Un, j 0 j 0 impliesSij�m�0Sij 0 �m0�.

Sequentially solvable games 503

in the former case and positive in the latter case. The unique sequential-moveequilibrium corresponding to this sequential-move version is therefore �2; 3; 1�.However, this equilibrium is not strong; and it is Pareto dominated by theother equilibrium of the simultaneous-move game, �1; 3; 2�.

When a crowding game has only one sequential-move equilibrium, thatequilibrium may be said to be commitment robust (Rosenthal, 1991). It followsas a corollary from Theorem 2 that a necessary condition for an equilibriumof a crowding game to be commitment robust is that it Pareto dominates allother equilibria. However, as the last example shows, this condition is notsu½cient for commitment robustness.

Other games

Rosenthal (1973) introduced a class of (congestion) games that are similar insome respects to the games considered here. In this class of games, the set ofstrategies of each player is not the set A of actions but some subset of thepower set of A; each player chooses a combination of actions. A player's pay-o¨ is the sum of the payo¨s associated with each of the actions included in hischoice. The payo¨ associated with action j is an arbitrary function cj of thenumber nj of players who include j in their choice. Rosenthal (1973) provedthat each game in this class has a Nash equilibrium in pure strategies. Never-theless, the following example shows that in this class of games Theorems 1and 2 do not hold.

Consider a three-person symmetric congestion game where the set of actionsis A � f1; 2; 3; 4g and the common set of strategies is ff1g; f2g; f3g; f1; 4g;f2; 3; 4gg. Suppose that c1�1� � 10, c1�2� � 0, c2�1� � 3, c3�1� � 2, c4�1� �ÿ1, and cj�m� < ÿ10 otherwise. A sequential-move version of this game has aunique subgame-perfect equilibrium, in which the ®rst player plays f1; 4g, thesecond player plays f2g, and the third player plays f3g. This, however, is not anequilibrium of the simultaneous-move game: the ®rst player can increase hispayo¨ from 9 to 10 by unilaterally shifting to f1g. And �f1g; f2g; f3g� is astrong equilibrium of the simultaneous-move game. But if the ®rst playerplayed f1g in the sequential-move version of the game then the second playerwould play f2; 3; 4g and the third player would play {1}, and so the payo¨ ofplayer the ®rst would in this case be zero.

Another closely related class of games in which the main result of thepresent paper does not hold is games in which crowding has a positive e¨ecton players' payo¨s (Konishi et al., 1997b). Consider, for example, a three-player, three-action game where the following inequalities hold: S11�3� >S12�3� > S12�2� > S11�2� > S11�1� > S12�1� > 0, S22�3� > S23�3� > S23�2� >S22�2� > S22�1� > S23�1� > 0, S33�3� > S31�3� > S31�2� > S33�2� > S33�1� >S31�1� > 0, and 0 > Sij�m� otherwise. As can be readily veri®ed, this gamedoes not even have a pure-strategy Nash equilibrium.

Proofs

It su½ces to prove Theorems 1 and 2 for generic crowding games. The reasonis that, for every crowding game G , every sequential-move version G of G ,

504 I. Milchtaich

and every generic crowding game G 0 close enough to G ,4 the unique subgame-perfect equilibrium of the sequential-move version of G 0 in which the order ofplayers is the same as in G is also a subgame-perfect equilibrium of G , andevery equilibrium of G 0 is also an equilibrium of G . Also, for every strongequilibrium s of G , G 0 can be chosen in such a way that s is also a strongequilibrium of G 0. Notice, however, that a sequential-move version of a non-generic crowding game may possess pure-strategy subgame-perfect equilibriawhose equilibrium paths do not correspond to equilibria of the simultaneous-move game.

Every subgame of a sequential-move version G of a given genericcrowding game G is itself a sequential-move version of some generic crowdinggame. Speci®cally, suppose that the players move in the order 1; 2; . . . ; n.The subgame of G determined by a given path �s1; s2; . . . ; si� �1U iU n� isa sequential-move version of a unique �nÿ i�-person generic crowding game,which will be called the subgame of G determined by �s1; s2; . . . ; si�. Theequilibrium path �si�1; si�2; . . . ; sn� of the unique subgame-perfect equilibriumof the subgame of G determined by �s1; s2; . . . ; si� will be called the subgameequilibrium path determined by �s1; s2; . . . ; si�. The congestion vector corre-sponding to the action pro®le �s1; s2; . . . ; sn� will be referred to as the con-gestion vector determined by �s1; s2; . . . ; si�. Note that, since in a crowdinggame a player's payo¨ is a¨ected only by the number of other players playingeach action, the same subgame, subgame equilibrium path, and congestionvector are determined by �st�1�; st�2�; . . . ; st�i��, where t is an arbitrary per-mutation of f1; 2; . . . ; ig. The proof of Theorem 1 depends on the followinglemma:

Lemma. In a generic crowding game, let �nj�j AA be the congestion vector deter-mined by �s1� and let �n0j�j AA be the congestion vector determined by �s01��s1; s

01 A A�. Then, for every j, n0j U nj � 1, and equality holds for at most one j,

which is not s1.

Proof of Theorem 1. Let G be a generic crowding game. It su½ces to show thatthe action pro®le s played on the equilibrium path of the unique subgame-perfect equilibrium of the sequential-move version of G in which the playersmove in the order 1; 2; . . . ; n is an equilibrium of G . If s is not an equilibriumthen, for some i and j, Sij�nj � 1� > Sisi

�nsi�, where �nk�k AA is the congestion

vector corresponding to s. Without loss of generality, i � 1. (If not, thenconsider not G but the subgame determined by �s1; s2; . . . ; siÿ1�.) Set s01 � j,and let �n0k�k AA be the congestion vector determined by �s01�. By the lemma,and the monotonicity of S1j, S1j�n0j�VS1j�nj � 1� > S1s1

�ns1�. But this con-

tradicts the assumption that s is played on the equilibrium path: player 1could have achieved more by playing j (instead of s1). 9

Proof of the lemma. The proof proceeds by induction on the number n ofplayers. The lemma is evidently true if n � 1. Assume that it holds truefor every �nÿ 1�-person generic crowding game, and let an n-person �nV 2�

4 It su½ces to assume that strict preferences are not reversed in G 0, i.e., that Sij�m� > Sij 0 �m0�implies S0 ij�m�VS0 ij 0 �m0� for every i, j, j 0, m, and m0, where �S0 ij�j AA are the functions thatdetermine player i's payo¨ in G 0.

Sequentially solvable games 505

generic crowding game be given. It is always possible to view the game as asubgame of an �n� 1�-person generic crowding game, more speci®cally thesubgame determined by �s0�, the play of the 0-th player. Moreover, it maybe assumed that s1 and s01 are di¨erent from s0, and that s0 is such thatSis0�1� < Sij�n� 1� for every i and j 0 s0. The latter assumption implies that,

in every equilibrium of every subgame of the �n� 1�-person game, no playerplays s0.

Let �n1j �j AA be the congestion vector determined by �s0; s1�, and let

�n10j �j AA be the congestion vector determined by �s0; s

01�. For every j 0 s0,

n1j � nj and n10

j � n0j . If �s2; . . . ; sn� is the subgame equilibrium path deter-mined by �s0; s1� and �s02; . . . ; s0n� is the subgame equilibrium path determinedby �s0; s

01� then �n1

j �j AA is also the congestion vector determined by �s0; s2; s1�and �n10

j �j AA is also the congestion vector determined by �s0; s02; s

01�. If s2 � s02

then it follows from the induction hypothesis, applied to the �nÿ 1�-personsubgame determined by �s0; s2�, that

n10j U n1

j � 1 for every j; equality holds for at most one j; but n10

s1U n1

s1: �1�

It also follows from the induction hypothesis that, symmetrically,

n1j U n10

j � 1 for every j; equality holds for at most one j; but n1s 01U n10

s 01: �10�

We have to show that (1) and �10� also hold if s2 0 s02. In what follows, wewill assume that the last inequality holds. In the course of the proof we willconsider several subgames, each one determined by a particular triplet ofactions, as well as the congestion vectors determined by these triplets. Thetriplets, preceded by the index of the congestion vector they determine, areshown in Fig. 1. A straight line in this ®gure indicates that the two tripletsconnected by the line di¨er in exactly one of the three actions. The romannumber near the line refers to the equation which, in the sequel, is deduced

Fig. 1. For an explanation see text

506 I. Milchtaich

from this relation between the two triplets by means of the induction hypoth-esis, applied to the subgame determined by the two common actions.

Let �n2j �j AA be the congestion vector determined by �s0; s

01; s2�. This is the

congestion vector that would result if player 2 did not change his action asa response to the change of player 1's action from s1 to s01. �n2

j �j AA is also thecongestion vector determined by �s0; s2; s

01�. It hence follows from the induc-

tion hypothesis that either

n2s2� n1

s2� 1 and n2

j U n1j for every j 0 s2 �ia�

or

n2s2U n1

s2: �ib�

We will consider each of these two cases separately. It also follows from theinduction hypothesis that

n1s 01U n2

s 01

�ii�

and n2s1U n1

s1. In light of the latter inequality, (ia) is possible only if s2 0 s1.

Case (ia). Let �n3j �j AA be the congestion vector determined by �s0; s

01; s0�.

Since the congestion vector determined by �s0; s01; s

02� is �n10

j �j AA, by the in-duction hypothesis

n10j U n3

j � 1 for every j; and equality holds for at most one j: �iii�

Since n3s0� n10

s0� 1, by the induction hypothesis

n3j U n10

j for every j 0 s0: �iv�

Similarly, since n3s0� n2

s0� 1,

n3j U n2

j for every j 0 s0: �v�

Let �n4j �j AA be the congestion vector determined by �s0; s

01; s1�. By the induc-

tion hypothesis,

n10s1U n4

s1: �vi�

Since n3s0� n4

s0� 1,

n3j U n4

j for every j 0 s0: �vii�

Since �n4j �j AA is also the congestion vector determined by �s0; s1; s

01�, and since

the congestion vector determined by �s0; s1; s2� is �n1j �j AA, by the induction

hypothesis

n4s2U n1

s2: �viii�

Sequentially solvable games 507

It follows from (ia), (v), (vii), and (viii) that n3j U n1

j for every j 0 s0. Hence,by (iii), n10

j U n1j � 1 for every j and equality holds for at most one j. SinceP

j n10j �

Pj n1

j � n� 1, it follows that n1j U n10

j � 1 for every j and equalityholds for at most one j. To complete the proof of (1) and �10� in the case that(ia) holds, it remains to show that n10

s1U n1

s1and n1

s 01U n10

s 01.

Let �n5j �j AA be the congestion vector determined by �s01; s1; s2�. Since n4

s0�

n5s0� 1,

n4s1U n5

s1: �ix�

Since n2s0� n5

s0� 1,

n2s2U n5

s2; �x�

and hence n5s2V n1

s2� 1 by (ia). Since, as shown above, (ia) implies s2 0 s1, by

the induction hypothesis

n5s1U n1

s1: �xi�

In conjunction with (vi) and (ix) this gives n10s1U n1

s1.

By (ia), (vii), and (viii), n2s2V n3

s2� 1. Therefore, if s01 0 s2 then by the

induction hypothesis

n2s 01U n3

s 01; �xii�

and hence n1s 01U n10

s 01by (ii) and (iv). If s01 � s2 then by the induction hypoth-

esis n2s 01� n3

s 01� 1, and by (ia) n2

s 01� n1

s 01� 1. It hence follows from (iv) that in

this case, too, n1s 01U n10

s 01.

Case (ib). Since the action of player 2 in the subgame equilibrium pathdetermined by �s0; s

01� is s02, it must be that S2s 0

2�n10

s 02�VS2s2

�n2s2�. Since we

have assumed that s 02 0 s2, the assumed genericity of the game implies thatthis inequality is in fact strict. Similarly, since the action of player 2 in thesubgame equilibrium path determined by �s0; s1� is s2, it must be thatS2s2�n1

s2� > S2s 0

2�n20

s 02�, where �n20

j �j AA is the congestion vector determined by

�s0; s1; s02�. Note that, formally, the last triplet is obtained from the triplet

�s0; s01; s2� that determines �n2

j �j AA by taking the prime o¨ s1 and adding aprime to s2; hence the index 20. By (ib) and the monotonicity of S2s2

,S2s 0

2�n10

s 02� > S2s2

�n2s2�VS2s2

�n1s2� > S2s 0

2�n20

s 02�. Hence, by the monotonicity of

S2s 02, n10

s 02< n20

s 02. It follows, by the induction hypothesis, that

n20s 02� n10

s 02� 1 and n20

j U n10j for every j 0 s02: �xiii�

Formally, (xiii) is a ``primed'' version of (ia). More precisely, the transfor-mation that interchanges s1 with s01 and s2 with s02 transforms (ia) into (xiii).Hence, our analysis of case (ia) implies at once that the formal transforms of(1) and �10� hold true. But these transforms are �10� and (1), respectively. Thus,(1) and �10� hold in the present case, too. 9

508 I. Milchtaich

Proof of Theorem 2. The proof proceeds by induction on the number n ofplayers. The theorem is evidently true if n � 1. Assume that it holds true forevery �nÿ 1�-person generic crowding game, and let s be a strong equilibriumof an n-person �nV 2� generic crowding game. Consider the directed graph,with n vertices corresponding one-to-one to the players in the game, where anarc is going from vertex i to vertex i0 if and only if player i envies player i0when s is played, that is, if and only if Sisi

�nsi� < Sisi0 �nsi0 �, where �nj�j AA is

the congestion vector corresponding to s. The assumption that s is a strongequilibrium implies that this graph has no cycles. Indeed, if the graph had acycle then the players involved could change their actions±each player shiftingto the action played by the player he envies±and they all would gain. There-fore, there exists a player, i, who envies no other player; Sisi

�nsi� >

Sij�maxf1; njg� holds for every j 0 si. Without loss of generality, i � 1.�s2; s3; . . . ; sn� is a strong equilibrium of the subgame determined by �s1�.Hence, by the induction hypothesis, it is also a sequential-move equilibrium ofthat subgame, say the sequential-move equilibrium corresponding to thesequential-move version in which the players move in the order 2; 3; . . . ; n. Toprove that s is a sequential-move equilibrium of the n-person game, it su½cesto show that, in the sequential-move version of that game in which the playersmove in the order 1; 2; . . . ; n, player 1 cannot gain from choosing an actionj di¨erent from s1. Let �n0k�k AA be the congestion vector determined by � j�.By the lemma, n0j V nj (and, of course, n0j V 1). It therefore follows from ourassumption concerning player 1 that S1j�n0j�US1j�maxf1; njg� < S1s1

�ns1�.

Thus, playing s1 is indeed the best option for player 1. 9

References

Aumann RJ (1959) Acceptable points in general cooperative n-person games. In: Tucker AW,Luce RD (eds) Contributions to the theory of games, IV, Princeton University Press, Princeton,pp. 287±324

D'Aspremont C, GeÂrard-Varet LA (1980) Stackelberg-solvable games and pre-play communica-tion. Journal of Economic Theory 23:201±217

Holzman R., Law-Yone N (1997) Strong equilibrium in congestion games. Games and EconomicBehavior 21:85±101

Konishi H, Le Breton M, Weber S (1997a) Equilibria in a model with partial rivalry. Journal ofEconomic Theory 72:225±237

Konishi H, Le Breton M, Weber S (1997b) Pure strategy Nash equilibrium in a group formationgame with positive externalities. Games and Economic Behavior 21:161±182

Milchtaich I (1996) Congestion games with player-speci®c payo¨ functions. Games and EconomicBehavior 13:111±124

Monderer D, Shapley LS (1996) Potential games. Games and Economic Behavior 14:124±143Rosenthal RW (1973) A class of games possessing pure-strategy Nash equilibria. International

Journal of Game Theory 2:65±67Rosenthal RW (1991) A note on robustness of equilibria with respect to commitment oppor-

tunities. Games and Economic Behavior 3:237±243

Sequentially solvable games 509