Convergence Dynamics of Resource-Homogeneous Congestion Games: Technical Report

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1 1 Convegence Dynamics of Resouce-Homogeneous Congestion Games: Technical Repot Richad Southwell and Jianwei Huang Abstact Many esouce shaing scenaios can be modeled using congestion games A nice popety of congestion games is that simple dynamics ae guaanteed to convege to Nash equilibia Loose bounds on the convegence time ae well known, but exact esults ae difficult to obtain in geneal We exploe a class of congestion games whee the esouces ae homogeneous but can be playe specific These games ae easy to study because individuals always pefe less used esouces We deive exact conditions fo the longest and shotest convegence times Late we extend ou studies to games on gaphs, whee individuals only cause congestion to thei neighbos and can access diffeent sets of esouces We apply ou esults to study cognitive adio netwoks, whee selfish uses shae wieless spectum oppotunities that ae constantly changing In paticula, we demonstate how fast the uses need to be able to switch channels in ode to tack the time-vaiant channel availabilities I INTRODUCTION Congestion games can be used to model a myiad of systems in biology, engineeing, and the social sciences In congestion games, playes select esouces to maximize thei payoffs, while consideing the congestion due to esouce shaing A key featue of congestion games is the finite impovement popety (FIP): asynchonous playe updates (ie, switching the choices of esouces to inceases payoffs) always conveges to a Nash equilibium [1] The FIP is vey attactive due to the wide applicability of congestion games [], [3] In many situations, such as dives choosing outes [], [5], [6] o wieless uses picking channels [7], it is useful to know that distibuted and selfish behavios always lead to a stable system state As convegence is guaanteed by the FIP, we shall focus on the following question: how long does convegence take? Undestanding this issue is citical fo eal-time esouce allocation Unfotunately, detemining the convegence time of congestion games can be a computationally had poblem [8], [9] The wost case convegence time can be exponential in the numbe of playes n [10], although a polynomial convegence in n is possible fo cetain games [11], [1] In this pape, we will analyze the convegence speed of a kind of simple congestion game, whee evey esouce has the same intinsic value The playes may have complicated and diffeent payoff functions, but this does not effect the convegence dynamics All that mattes is that playes pefe less congested esouces Despite the simplicity, this type of game can model a wide ange of scenaios whee selfish individuals shae homogenous esouces In Section V, we will illustate why ou games ae paticulaly in useful analyzing cognitive adio netwoks To study the convegence of congestion games, the usual appoach is to gain loose bounds by constucting a potential function The special stuctue of ou games allows us to take a moe geometic appoach, and deive many new (and sometimes exact) esults about convegence (see Section III) In Section IV, we will analyze moe geneal congestion games, whee the playes ae spatially distibuted and can access diffeent set of esouces Ou main esults include the following: 1) An exact fomula fo the fastest convegence time, fom any given initial state (Section III-B); ) A simple updating mechanism, which guaantees the fastest convegence time (Section III-B); 3) A linea bound on the slowest possible convegence time (Section III-D); ) Deep esults about the Nash equilibia and convegence ate of ou spatially extended models (Section IV) As a concete application, in Section V we discuss in details how the analytical famewok can be applied to spectum shaing in cognitive adio netwoks As the available esouces (channels) come and go apidly in cognitive adio netwoks (eg, [13] [15]), it is citical to undestand how selfish uses behave and whethe they can adapt fast enough compaed with the envionment dynamics Infomation Engineeing Depatment, the Chinese Univesity of Hong Kong; ichadsouthwell5@gmailcom, jwhuang@iecuhkeduhk

2 A About the contents of this document One of the main puposes of this document is to act as a moe igoous and compehensive vesion of the confeence pape Convegence Dynamics of Resouce-Homogeneous Congestion Games which is published via the GameNets 011 confeence The cental theoems in this document ae Theoems 1, 3, 7, 9, 10, and 5 We pove these esults in the main text II THE MODEL In an n-playe congestion game, each playe s stategy involves a set of esouces The payoff a playe p gets fom using a esouce i is descibed by a stictly deceasing function f p i (x i), whee x i is the numbe of playes using i In this pape, we ae only concened with singleton congestion games, whee each playe uses exactly one esouce at any given time [11] We futhe assume that esouces ae symmetic, ie, f p i = f p j = f p fo any esouces i, j and any playe p, so all esouces ae equivalent fom a playe s pespective Note, howeve, that the payoffs ae playe-specific This means playes could have diffeent tastes of the same esouce, due to technological, psychological, o economic easons A cental idea in the congestion game dynamics is the bette esponse switch, ie, a playe inceases its payoff by switching to a bette (less congested) esouce A common way to study congestion games is to imagine that playes stategies evolve though time via asynchonous bette esponse switches (ie, one playe switches to a bette decision at each discete time step) 1 When no playe can incease thei payoff by switching, the system eaches a Nash equilibium To summaize, ou systems ae defined by a set of n playes, a set of esouces, and a stictly deceasing payoff function f p fo each playe p Evey playe uses one esouce fom the set {1,,, } at any time step Often we denote the set of esouces {1,,, } as R That is R = {1,,, } The system state vecto is x = (x 1, x,, x ), whee x i is the numbe of playes using esouce i We let denote the set of all state vectos In othe wods, { } = x N + : x i = n, whee N + denotes the set of non-negative integes When a playe using esouce i switches to esouce j, then x i deceases by 1 and x j inceases by 1 We efe to this action as an i j switch Let δ v = (δ 1,v, δ,v,, δ,v ) be the dimensional unit vecto Hee δ u,v denotes the Konecke delta In this technical epot we often efe to an i j switch as a δ i + δ j switch, because the esult of pefoming an i j switch upon x is x δ i + δ j We suppose that one and only one playe switches to a bette esponse in evey time step A simple and cental fact about ou games is Theoem 1 i=1 Theoem 1 (Bette Response Switch): A switch i j is a bette esponse switch if and only if x j + 1 < x i Poof: Conside a playe p using esouce i with a payoff f p (x i ) Switching to j is a bette esponse if and only if f p (x j + 1) > f p (x i ) Now, since f p is stictly deceasing, we have f p (x j + 1) > f p (x i ) if and only if x j + 1 < x i Theoem 1 has poweful implications: the bette esponse switches (though which ou system evolves) ae independent of the payoff functions and the identity of the playes All that mattes (with espect to the dynamics) is that one playe deceases its congestion level by switching A state x is a Nash equilibium if and only if no bette esponse switches can be pefomed Theoem 1 implies that x is a Nash equilibium if and only if x i x j 1, i, j R Recall that R = {1,,, } denotes the set of esouces n Theoem (Nash Equilibium): A state vecto x is a Nash equilibium if and only if (n mod ) of x s enties ae equal to, whilst the emaining (n mod ) enties of x ae equal to n 1 We ae subscibing to the elementay step hypothesis [10], that one and only one playe impoves thei stategy in each time step This is commonly used to model situations whee simultaneous updating is unlikely The analysis needs to be significantly modified fo othe types of updates

3 3 Fig 1 The left-hand side figue shows the state space of a game with n = 5 playes and = 3 esouces The ight-hand side figue shows the state space of a game with n = 7 playes and = esouces The points epesent state vectos and the aows epesent the state tansitions which can be achieved though bette esponse switches Fo example, in the left figue, thee is an aow fom (5, 0, 0) to (, 1, 0) because the bette esponse switch 1 convets (5, 0, 0) into (, 1, 0) In this case the Nash equilibia ae (1,, ), (, 1, ), and (,, 1) Poof: Suppose (n mod ) of x s enties ae equal to n, whilst the emaining (n mod ) enties of x ae equal to n In this case i=1 x i = (n mod ) n + ( (n mod )) n = n so x Also n n 1, and so we have, xi x j 1 i, j R,which means x is a Nash equilibium To see the convese suppose x is a Nash equilibium In this case x i x j 1, i, j R so thee must exist some subset K of R such that i R we have that i K implies x i = k and i / K implies x i = k 1 Also note that n = i=1 x i = k K + (k 1)( K ) = k + K must be satisfied Now suppose that k < n In this case k n (with equality if and only if k = n ) It follows that k + K = n is only satisfied when K = R and k = n, which means that x has evey enty equal to n Now suppose that k > n In this case k 1 n and so k = (k 1) n (with equality if and only if k 1 = n ) It follows that k + K = n is only satisfied when K is empty and k 1 = n, which means that x has evey enty equal to n Now suppose k = n In this case k + K = n implies K = n + n = (n mod ) We have shown that condition, k + K = n, implies that x has (n mod ) enties equal to n and the othe (n mod ) enties equal to n III CONVERGENCE TIME TO A NASH EQUILIBRIUM Stating fom an abitay state x, the playes can each a Nash equilibium though seveal outes (see Figue 1) This means the convegence time depends upon the ways playes choose to switch The convegence ate is an impotant measue of how quickly the playes oganize themselves Since playes identities and payoff functions ae ielevant to the convegence dynamics, we will only wok with the numbe of playes n, the numbe of esouces, and system state x fom now on A The geomety of switching We want to detemine the best way a goup of playes can switch thei choices of esouces to each a Nash equilibium To help undestand this, it is woth studying how many switches (of any kind) ae equied to convet one state into anothe Recall fom Theoem 1 that an i j switch is a bette esponse switch if and only if x i > x j + 1

4 Fig The state space when we have n = 5 playes and = 5 esouces The points epesent state vectos and the aows epesent the state tansitions which can be achieved though bette esponse switches Fomally we efe to the numbe of switches (of any kind) equied to convet x into y as the switching distance, d(x, y) Theoem 3 (The Switching Distance): Fo any two states x and y, the minimal numbe of switches (of any kind), d(x, y), equied to convet x into y is i=1 d(x, y) = max {0, y i x i } = x i y i i=1 Poof: Recall that R = {1,,, } denotes the set of esouces Let ψ(x, y) = max{0, x i y i } = x i y i i=1 i R:x i>y i Now ( ) ( ) 0 = x i y i = x i y i = x i y i + x i y i + x i y i i R i R i R i R:x i>y i i R:y i>x i i R:y i=x i

5 5 Now x i y i = 0 i R:y i=x i, x i y i = ψ(x, y) i R:x i>y i, x i y i = ψ(y, x) i R:x i>y i gives 0 = ψ(x, y) ψ(y, x) so ψ(x, y) = ψ(y, x) Also i=1 x i y i = ( ) ( ) x i R:xi>yi i y i + y i R:yi>xi i x i = ψ(x, y) + ψ(y, x) = ψ(x, y) A switch coesponds to the addition of a vecto δ a + δ b fo some a, b R : a b If x = y then d(x, y) = 0 switches ae equied to convet x into y (note how ψ(x, y) = 0 agees with this) Now suppose that x y Let A denote the multi-set consisting of x i y i copies of each i R such that x i > y i Let B denote the multi-set consisting of y i x i copies of each i R such that y i > x i Clealy A = ψ(x, y) = B so we can constuct a pai of bijections l : {1,,, ψ(x, y)} A and : {1,,, ψ(x, y)} B Now we shall constuct the sequence of switches δ l (1) + δ (1), δ l () + δ (),, δ l (ψ(x,y)) + δ (ψ(x,y)) Stating with x and pefoming this sequence of switches yields y The eason is that the vecto which esults fom pefoming these switches upon x satisfies fo each i R x ψ(x,y) δ l(t) t=1 + ψ(x,y) δ (t) t=1 = x i {i A} + {i B} = y i, We have hence constucted a sequence of ψ(x, y) switches that convets x into y and shown d(x, y) ψ(x, y) To see that d(x, y) ψ(x, y) suppose that i δ a1 + δ b1, δ a + δ b,, δ aq + δ bq is a sequence of Q switches that convets x to y It follows that i R we must have ( ( Q ) ( Q )) y i x i = δ at + δ bt = {t {1,,, Q} : a t = i} + {t {1,,, Q} : b t = i} t=1 t=1 i Fom this it follows that {t {1,,, Q} : a t = i} x i y i, i R Fom this we get ψ(x, y) = x i y i {t {1,,, Q} : a t = i} Q, i R:x i>y i i R:x i>y i so evey Q length sequence of switches that convets x into y must have Q ψ(x, y) so d(x, y) ψ(x, y) The Manhattan distance is a well known metic, defined as i=1 x i y i Theoem 3 states that d(x, y) is half of the Manhattan distance fom x to y This means d is a metic on

6 6 Lemma (When switches move close): x, y a, b {1,,, }, suppose we pefom the switch δ a + δ b upon x to obtain the vecto x δ a + δ b In this case the following statements ae equivalent 1) d(x δ a + δ b, y) = d(x, y) 1 ) d(x δ a + δ b, y) < d(x, y) 3) x a > y a and x b < y b Poof: Let ψ(x, y) = i R:x i>y i x i y i Accoding to theoem 3, d(u, v) = ψ(u, v), u, v Suppose x = x δ a + δ b is the esult of pefoming the switch δ a + δ b on x Now ψ(x, y) = x i y i = x i δ a,i + δ b,i y i i R:x i >yi i R:x i δ a,i+δ b,i >y i Fo an inequality I let us define α(i) to be equal to 1 if I is tue and equal to 0 othewise Now we get ψ(x, y) = x i y i α(x a > y a ) + α(x b y b ) = ψ(x, y) α(x a > y a ) + α(x b y b ) i R:x i>y i Fom this we get that the condition; ψ(x, y) < ψ(x, y), is equivalent to the condition α(x a > y a ) = 1 and α(x b y b ) = 0, which is equivalent to the condition x a > y a and x b < y b, which is equivalent to the condition ψ(x, y) = ψ(x, y) 1 Now ψ(x, y) = d(x, y) and ψ(x, y) = d(x, y) implies the lemma We define the Nash distance, T (x), of a vecto x to be the minimal numbe of switches equied to convet x into a Nash equilibium In othe wods T (x) is the minimal value of d(x, y) such that y is a Nash equilibium We see late how undestanding T (x) is impotant fo undestanding the fastest convegence time Lemma 5 (Nash Distance): x the Nash distance of x will be Poof: T (x) = ( i=1 { n } ) { { n } } max 0, x i min i {1,,, } : x i, n mod Recall R = {1,,, } is the set of esouces When n mod = 0 the only Nash equilibium (accoding to Theoem ) will be the vecto y with evey enty equal to n It follows fom Theoem 3 that the Nash distance will be T (x) = d(x, y ) = i R:x i> x n i n which is equal to the stated expession in this case Now suppose that n mod 0 and let y be any Nash equilibium Also u let H(u) = {i R : u i n }

7 7 Accoding to Theoem we have, i R, that i H(y) implies y i = n and i / H(y) implies y i = n Now d(x, y) = i R:x i>y i x i y i by Theoem 3 It follows that d(x, y) = i H(x) H(y) n x i + i H(x) H(y) n x i Also i H(x) H(y) we have x i n 0 so x i > n and x i n = x i n 1 It follows that So i H(x) H(y) n x i = d(x, y) = i H(x) i H(x) H(y) Also i R we have x i > n if and only if i H(x) and so d(x, y) = n x i i R:x i> n n x i H(x) H(y) n x i H(x) H(y) H(x) H(y) Accoding to Theoem we have that H(y) = n mod and the set of pemutations of y is the set of Nash equilibia Let ζ denote the set of all size n mod subsets X of R In othe wods ζ = {X R : X = n mod } Now X ζ let y X be the vecto such that i X implies (y X ) i = n and i / X implies (yx ) i = n Fo example when n = 8, = 3 and X = {1, 3} we have y X = (3,, 3) Now Theoem implies that {y X : X ζ} is pecisely the set of all Nash equilibia It follows that fo any X ζ, the Nash equilibium y X distance fom x to the Nash equilibium y X will be will be such that H(y X ) = X It follows that the switching d(x, y X ) = n x i H(x) X i R:x i> n This distance will be minimized (and hence equal to the Nash distance T (x)) when we select a Nash equilibium y X such that H(x) X is maximized If (n mod ) H(x) then we can maximize H(x) X by selecting ou X ζ such that X H(x) This gives H(x) X = X = (n mod ) and so the Nash distance will be T (x) = n x i (n mod ) i R:x i> n If (n mod ) > H(x) then we can maximize H(x) X by selecting ou X ζ such that H(x) X This gives H(x) X = H(x) and so the Nash distance will be T (x) = n x i H(x) i R:x i> n B The fastest convegence Next we look at the least numbe of bette esponse switches that ae equied to convet the state vecto x into a Nash equilibium We efe to this as the best convegence time of x, b(x) To explain how to compute b(x), we will define a new concept, the stong switch Suppose x is not a Nash equilibium We say an i j switch is a stong switch of x if x i is one of x s maximal enties and x j is one of x s minimal enties Fo example, the switch is a stong switch

8 8 of x = (5, 5, 3, 1), while the 3 switch is not A stong switch coesponds to a playe fom one of the most congested esouces switching to one of the least congested esouces, and thus is always a bette esponse switch We will see (Coollay 8) that pefoming stong switches is the fastest way to each a Nash equilibium Ou expession fo b(x) is given in Theoem 7 Essentially it is the same as the expession fo T (x) given in Lemma 5 In othe wods, the numbe of bette esponse switches equied to each a Nash equilibium is equal to the numbe of switches of any kind equied to each a Nash equilibium To establish this we will show that, wheneve the state is not in Nash equilibium, one can always pefom a stong switch which moves the state close to the closest Nash equilibium (with espect to the switching distance d) This is the subject of the following Lemma Lemma 6 (Stong switches always decease Nash distance): Suppose x is not a Nash equilibium, and suppose that x is the esult of pefoming a stong switch x In this case T (x ) = T (x) 1 Poof: Suppose we pefom the stong switch i j upon x to yield x In this case we must have x i = max{x v : v R}, x j = min{x v : v R} and x = x δ i + δ j, accoding to the definition of a stong switch Let N denote the set of all Nash equilibia of the system (see theoem ) Let us define K(x) = {y N : d(x, y) = T (x)} In othe wods K(x) is the set of Nash equilibia which ae closest to x (the Nash equilibia which can be eached within the least possible numbe of switches, T (x)) The way we shall pove ou esult is to show that we can always select a Nash equilibium y K(x) such that x i > y i and x j < y j Once we have established this, the fact that d(x δ i + δ j, y) = d(x, y) 1 = T (x) 1 follows tivially fom Lemma The easiest case to show that y K(x) : x i > y i, x j < y j is when n mod = 0 In this case Theoem implies that K(x) = N just consists of a single point, y, which has each of it s enties equal to n This means that to show y K(x) : x i > y i, x j < y j we must simply show x i > n and x j < n We will pove x i > n by contadiction Suppose x i n Now since x i is one of x s maximal enties we must have x v n, v R Moeove, since x is not a Nash equilibium we must have that x i x j > 1 This implies x j < n So now we know that some of x s tems ae less than n and none of x s tems ae geate than n It follows that v=1 x v < n which contadicts ou assumption that x Next we will pove x j < n by contadiction Suppose x j n Now since x j is one of x s minimal enties we must have x v n, v R Moeove, since x is not a Nash equilibium we must have that x i x j > 1 This implies x i > n So now we know that some of x s tems ae geate than n and none of x s tems ae less than n It follows that v=1 x v > n which contadicts ou assumption that x So now we have poved that x i > n and x j < n, we have shown that y K(x) : x i > y i, x j < y j fo the case whee n mod = 0 Now we will deal with the emaining case whee n mod 0 In these cases (accoding to Theoem ) N is the set of all vectos with n mod enties equal to n c and (n mod ) enties equal to n c We will begin by showing that x i n and x j n We will pove x i n by contadiction Suppose x i < n Now since x i is one of the maximal enties in x, this means x v n, v R Now this means that i=1 x i < n, which gives ou contadiction We will pove x j n by contadiction Suppose x j > n Now since x i is one of the minimal enties in x, this means x v n, v R Now this means that i=1 x i > n, which gives ou contadiction So now we have poved that x i n and x j n Now we will deal with the thee diffeent possible cases The fist case we will conside is when x i > n and x j < n Hee the fact that y K(x) : x i > y i, x j < y j is obvious because fo any Nash equilibium y K(x) we must have that y v { n, n }, v R and so clealy when x i > n and x j < n we must have x i > y i and x j < y j

9 9 The two emaining cases ae moe complicated, so we ecall some notation used in the poof of Lemma 5 Recall that u, we have H(u) = {v R : u v n } Also ecall that ζ is defined to be the set of all size n mod subsets of R In othe wods ζ = {X R : X = n mod } Also ecall that X ζ we define y X to be such that v R we have v X (y X ) v = n and v / X (yx ) v = n Note that now, accoding to Theoem we have N = {y X : X ζ} is the set of all Nash equilibia With this notation in place we shall begin to discuss the next case Suppose x i = n Now since x is not a Nash equilibium (by ou assumption) we must have that x i x j > 1 In othe wods we must have x j < n Also note that, since x i is one of x s maximal enties, we must have that x v n, v R Now in this case we must have H(x) > n mod We will pove this by contadiction Suppose, to the contay, that H(x) n mod Now (since N = {y X : X ζ} and H(x) X, X ζ) we can find a Nash equilibium, y X N, such that H(x) X In this case we have v R we have that; n n x v = (y X ) v =, and, n n x v x v (y X ) v x v This means (y X ) v x v, v R Moeove (y X ) j > x j So this means v=1 (yx ) v = n > v=1 x v, which is clealy a contadiction Ou eductio absudum agument implies that it must, in fact, be tue that H(x) > n mod This means it must be the case that X < H(x), X ζ This means that thee must exist some X ζ such that X H(x) {i} Now accoding to the poof of Lemma 5, in this case (whee H(x) > n mod ), we have that K(x) = {y N : d(x, y) = T (x)} = {y X : X ζ, X H(x)} So clealy the Nash equilibium y X will be such that y X K(x) In othe wods y X will be one of the closest Nash equilibia to x Also, since i / X we have (y X ) i = n < n = x i Now since (y X ) j { n, n } and x j < n we will also have x j < (y X ) j So we have shown that y X K(x) and x i > (y X ) i and x j < (y X ) j So we have poved that y K(x) : x i > y i, x j < y j fo this (ou second case) Now we will move on to ou thid (and final) case Suppose x j = n Now since x is not a Nash equilibium (by ou assumption) we must have that x i x j > 1 In othe wods we must have x i > n Also note that, since x j is one of x s minimal enties, we must have that x v n, v R Now in this case we must have H(x) < n mod We will pove this by contadiction Suppose, to the contay, that H(x) n mod Now in this case we have H(x) (n mod ) Now, since N = {y X : X ζ}, we can find a Nash equilibium, y X N, such that { n } v R : x v = R H(x) R X = { v R : (x X ) v = n } In this case we have v R we have that; n x v = (y X ) v = n, and, n n x v x v (y X ) v x v This means (y X ) v x v, v R Moeove (y X ) i < x i So this means v=1 (yx ) v = n < v=1 x v, which is clealy a contadiction Ou eductio absudum agument implies that it must, in fact, be tue that H(x) < n mod This means it must be the case that X > H(x), X ζ This means that thee must exist some X ζ such that j X and H(x) X {j}

10 10 Now accoding to the poof of Lemma 5, in this case (whee H(x) < n mod ), we have that K(x) = {y N : d(x, y) = T (x)} = {y X : X ζ, H(x) X} So clealy the Nash equilibium y X will be such that y X K(x) Also, since j X and j / H(x) we have (y X ) j = n > n = x j Now since (y X ) i { n, n } and x i > n we will also have x i > (y X ) i So we have shown that y X K(x) and x i > (y X ) i and x j < (y X ) j So we have poved that y K(x) : x i > y i, x j < y j fo this (ou thid case) So now we have shown that, whateve the value of n mod, fo any x that is not a Nash equilibium we have y K(x) : x i > y i, x j < y j Now we can complete the poof Fo any x that is not a Nash equilibium let y be such that y K(x) : x i > y i, x j < y j (we have just poved the existence of such a y) Now by the definition of K we have d(x, y) = T (x) Also, since x i > y i, x j < y j, Lemma implies that T (x ) = d(x, y) = d(x δ i + δ j, y) = d(x, y) 1 = T (x) 1, which poves ou esult Now, with this Lemma in place we can show the cental esult of this section, which is Theoem 7 Theoem 7 (Fastest Convegence Time): Fo any state x, the best (minimum) convegence time to any Nash equilibium is ( { n } ) { { n } } b(x) = max 0, x i min i {1,,, } : x i, n mod i=1 Poof: By using ou othe esults the tuth of this theoem is tivial Recall fom lemma 5 that the minimal numbe of switches (of any kind) equied to each a Nash equilibium fom x is ( { n } ) { { n } } T (x) = max 0, x i min i {1,,, } : x i, n mod This means b(x) T (x) i=1 Now suppose T (x) = t > 0 In this case lemma 6 implies that we can pefom a stong switch (which is a type of bette esponse switch) upon x to yield x such that T (x ) = t 1 Using this fact with induction, implies that one can find a sequence of t bette esponse switches, which tansfoms x into a Nash equilibium It follows that b(x) T (x) Note that ou esult not only poves that a Nash equilibium can be eached within T (x) bette esponse switches, it also poves that a Nash equilibium can be eached within T (x) stong switches (Coollay 8) The expession in Theoem 7 implies that the maximum value of b(x) (ove all states x) occus when all the playes use the same esouce (eg, x = (n, 0, 0,, 0)) In othe wods, b((n, 0, 0,, 0)) = n n is the lagest numbe of bette esponse switches eve equied to each a Nash equilibium An impotant esult (which comes fom ou poof Theoem 7) is Coollay 8 This states that stong switching is optimal, in the sense that it leads to a Nash equilibium within the minimal numbe of switches, b(x) Coollay 8: Staing fom an abitay state x, any sequence of b(x) stong switches leads to a Nash equilibium C The aveage fastest convegence fom andom initial conditions Often we do not have a choice of the initial system state Thus it is useful to undestand how fast we can each a Nash equilibium fom andom initial conditions This gives a global aveage measue of the best case pefomance of ou systems Suppose the playes select thei initial esouces fom {1,,, } unifomly at andom In this case, we let β (n) denote the expected numbe of stong switches equied to each Nash equilibium (fo a game with esouces and n playes) In othe wods, β (n) is the expected value of b(x), when x is geneated by allocating esouces andomly and unifomly

11 11 Theoem 9 (Aveage Fastest Convegence Time): Suppose n is divisibly by, in this case we have, ( β (n) = 1 1 ) n n ( ) n 1 ( n n ) n! (n n )! n!, lim β (n) = n n( 1) π Poof: Suppose n is divisibly by, then the only Nash equilibium is ( n, n,, n ) Conside a geneic esouce i The numbe of playes using esouce i, in a andomly geneated x, follows a binomial distibution: We have n tials (one fo each playe) and the chance of success (the playe being andomly allocated to choice i) is 1 The mean of this distibution is n It follows that the expected value, E ( xi n ), is equal to the expected absolute deviation of the binomial distibution with n tials and success pobability 1 This is well known [16] to be ( xi E n ) ( = 1 1 ) n n ( ) n 1 +1 ( n ) + 1 n! ( n + 1)! ( n n 1)! Since β (n) is the expected value of b(x) ove the ealization of x, we can use Theoem 7 to get, ( ) β (n) = 1 E x i n = E ( x j n ), i=1 which gives the stated esult Applying Stiling s appoximation to this expession yields the asymptotic esult This is shown in moe detail as follows Since n is divisible by we shall define m = n Now accoding to ou esult we have β (n) = β (m) = ( 1 1 ) m( 1) When m is lage we may apply Stiling s appoximation of to each factoial tem in this expession to get v! (πv) 1 ( v e ( ) m 1 m( 1) ) v ( ) 1 ( 1)m β (m), π now making the substitution n = m yields that ( ) 1 ( 1)n β (n) = β (m) π (m)! m!(m( 1))! This asymptotic fom is nice because it gives us a simple view of this complex equation The question is does the appoximation also hold when n mod 0 One can see by making plots that the value of β (n) goes up and down as n changes by small amounts In fact it seems as though β (n) has a local maxima wheeve n is divisible Although we ) 1 cannot pove it yet it seems likely that n implies β (n), even when n mod 0 ( ( 1)n π Ou eason fo thinking this is that this asymptotic esult holds always tue is the following bound, which we conjectue holds n, we believe that β (n) is always bounded below by and bounded above by n l (n) = i=0 ( ) i ( ) n i ( n i ) ( n ) i

12 1 n h (n) = i=0 ( ) i ( In othe wods we conjectue l (n) β (n) h (n), n, ) n i ( n i ) ( n ) i Fig 3 A plot showing convegence time elated quantities when = The gold line shows the conjectued uppe bound h (n) The puple line shows the eal value of β (n) The blue line shows the conjectued lowe bound l (n) The conjectue is inteesting because l (n) and h (n) convege when n is lage When n mod = 0 we have l (n) = β (n) = h (n) When n mod 0 we have that n h (n) l (n) = i=0 ( ) i ( ) n i ( ) n i ( ( Now Hoeffding s inequality implies h (n) l (n) exp n ) n ) and so h (n) l (n) when n is lage Now since β (n) goes up as n 1/ the diffeence between l (n) and h (n) becomes negligible fo n n D The wost possible pefomance In eality, playes may not switch in the optimal way Often it is moe likely that andom playes pefom andom bette esponse switches In this case, thee ae many ways the system can evolve In this section we discuss the longest convegence time w (n) fo a system with esouces and n playes (fo any choice of initial vecto x) Hee w (n) is defined as the longest sequence of bette esponse switches which can be pefomed upon a vecto, in ode to each a Nash equilibium This means w (n) is equal to the length of the longest path in G 3 Now we shall state ou cental esults about w (n) We will pove these esults in Theoem 10 individually with theoems 15, 17, 18, 19, 3 and coollay 16 late on in this section Theoem 10 (Slowest Convegence Time): The wost (slowest) convegence time, w (n), has the following popeties; ( ) n( 1) 1) w (n) n( 1) ) lim n w (n) = n( 1) 3) w (n) = n 3 Actually w (n) is finite, because evey sufficiently long path in G teminates at a Nash equilibium -this is a consequence of the finite impovement popety, although ou poofs do not ely on this fact

13 13 ) w 3 (n) = n 1 5) w (n) = δ n,1 + δ n, + 3n (n mod ) 5 (n + 1 mod ) ( + ( n mod )) whee δ i,j is the Konecke delta 6) If n then w (n) = 1+8k 1 n 1 k=0 The emainde of this section is devoted to the poof of Theoem 10 To undestand w (n), we can think of the state space as a diected gaph G as in Figue 1 Paths in G coespond to sequences of states which can be obtained by pefoming bette esponse switches The longest convegence time w (n) is the length of the longest path in the state space Recall that the set of all state vectos is { } = x N + : x i = n, whee N + denotes the set of non-negative integes is essentially the set of all compositions of n into non-negative pats The diected state space gaph G = (, E ) has a vetex set The edge set E of this gaph is defined such that fo each x, y, we have (x, y) E if and only if y can be obtained by pefoming a single bette esponse switch upon x Again, see Figue 1 Moe fomally we could wite E = {(x, y) : i, j R : x i > x j + 1, y = x δ i + δ j } w (n) is equal to the length of the longest path in G Clealy G has a lot of symmety States which ae pemutations of one anothe have symmetic positions in G This means they ae equivalent with espect to the dynamics The symmeties in G allow us to conside a simple system: one which is defined upon the set, ρ = {x : x 1 x x } In othe wods ρ is the set of all patitions of n into non-negative pats (with thei enties in descending ode) Let us define the diected educed state space gaph G ρ = (ρ, E ρ ) to be the subgaph of G induced upon ρ (see Figue ) In othe wods E ρ = {(x, y) ρ : i, j R : x i > x j + 1, y = x δ i + δ j } Now, since we ae esticted to ρ we ae only concened with the bette esponse switches that yield othe vectos in ρ Fo example, thee is an edge ((3, 3, 1, 1), (3,,, 1)) of G ρ coesponding to pefoming the bette esponse switch 3, upon (3, 3, 1, 1), but thee is no edge of G ρ coesponding to the bette esponse switch 1 3, because pefoming the switch 1 3 upon (3, 3, 1, 1) yields (, 3,, 1), which is not in ρ Moe pecisely, the edges of G ρ coespond to the bette esponse switches which keep the vecto enties in descending ode So thee will be an edge (x, y) E ρ if and only if thee is an i j switch fom x to y such that x i > x j + 1, x i x i+1 and x j x j 1 (in this case the descending ode will be peseved 5 ) We can give the following (equivalent) expession of the edge set of E ρ which makes these constains moe explicit: E ρ = {(x, y) ρ : i, j R : i < j, x i x i+1, x j x j 1, x i > x j + 1, y = x δ i + δ j } To show that G ρ holds essentially the same infomation as G, we intoduce the mapping od : ρ The mapping od(x) is the pemutation of the state x with its enties in a descending ode (eg, od((3, 1, 5)) = (5, 3, 1)) Notice that od is a stuctue peseving mapping fom G to G ρ This is illustated by Figue (ight) and poved by Lemma 11 6 i=1 Lemma 11: Fo all states x, y, if (x, y) E, then (od(x), od(y)) E ρ Evey sufficiently long path in G teminates at a Nash equilibium (ie G is acyclic and w (n) is finite), this is a consequence of the finite impovement popety Ou poof does not ely on this fact 5 The eason the descending ode will be peseved is that x ρ, we have x i x i+1 x i > x i+1 x i 1 x i+1 and x j x j 1 x j < x j 1 x j + 1 x j 1 and so x δ i + δ j ρ 6 Technically speaking, Lemma 11 shows that od is a etactive homomophism [17] fom G to G ρ, such details ae not elevant to ou discussion, but we shall give the definitions nevetheless A homomophism, fom a diected gaph G = (V, E) to anothe diected gaph H = (W, F ) is a mapping f : V W such that u, v V we have (u, v) E (f(u), f(v)) F Such an f is a etactive homomophism if H is a subgaph of G

14 1 Fig The left figue illustates the gaph G ρ with n = 7 and = (note that is the subgaph of the stuctue shown in Fig 1, esticted to the states with enties in descending ode On the ight we illustate how the od opeation foms a stuctue peseving mapping fom G to G ρ when = and n = Poof: Recall R = {1,,, } Now if (x, y) E, then thee is a bette esponse switch i j fom x to y Let us define i = {v {1,,, } : x v x i } to be the numbe of enties that ae in x and ae geate than o equal to x i Also let us define j = 1 + {v {1,,, } : x v > x j } Now we popose that i j is a bette esponse switch fom od(x) to od(y) We will show this as follows Fistly note that v R we have that od(x) v is the vth lagest enty of x This means that i = max{v R : od(x) v = x i } is the index of the ight most enty of od(x) that is equal to x i This also means that j 1 = max{v R : od(x) v > x j } and so j = min{v R : od(x) v = x j } is the index of the left most enty of od(x) that is equal to x j Now since od(x) i od(x) i +1 and od(x) j od(x) j 1 we have that the esult, od(x) δ i + δ j, of pefoming the switch i j upon od(x), is such that od(x) δ i + δ j ρ Moeove, we have od(x) i = x i and od(x) j = x j This means that od(x) δ i + δ j and y = x δ i + δ j ae pemutations of one anothe (ie they have the same numbe of enties equal to any paticula value) Now since od(y) is a pemutation of y, it follows that od(x) δ i + δ j and od(y) ae pemutations of one anothe Moeove od(x) δ i + δ j ρ and od(y) both have thei enties in descending ode Thee is clealy only one way to aange a vectos enties into descending ode This means that od(x) δ i +δ j and od(y) ae equal, because they both hold the same enties, aanged in descending ode So we have od(x) δ i + δ j = od(y), and so pefoming the switch i j upon od(x) yields od(y) Moeove, since od(x) i = x i > x j + 1 = od(x) j + 1, we have that i j is a bette esponse switch that convets od(x) into od(y) This means (od(x), od(y)) E ρ A path in G is a sequence u 0 u 1 u t such that (u i, u i+1 ) E, i {1,,, t 1} Lemma 11 implies that the mapping od peseves paths, ie, if u 0 u 1 u t is a path in G, then od(u 0 ) od(u 1 ) od(u t ) is a path in G ρ Lemma 1: w (n) is equal to the length of the longest path in G ρ Poof: By definition w (n) is the length of the longest path in G The longest gaph in G ρ cannot be longe than the longest gaph in G because G ρ is a subgaph of G Suppose u 0 u 1 u t is a path of length w (n) in G ρ Now Lemma 11 implies that od(u 0 ) od(u 1 ) od(u t ) is a length w (n) path in G ρ This shows the longest path in G ρ must be of length geate than o equal to w (n) Let us define a potential function ξ : ρ N + that will be vey useful fo easoning about the length of the longest path in

15 15 G ρ We define ξ such that x ρ we have ξ(x) = k=1 (k 1)x k Edges in G ρ coespond to bette esponse switches i j : i < j The next lemma states that evey such switch inceases the potential function ξ by j i Lemma 13: Suppose x, y ρ ae such that (x, y) E ρ Now we have that i < j such that y = x δ i +δ j Moeove ξ(y) = ξ(x)+j i Poof: The fact that, i < j such that y = x δ i + δ j, follows diectly fom the definition of G ρ and E ρ Now ( ) ξ(y) = (k 1)x k + (i 1)(x i 1) + (j 1)(x j + 1) = (k 1)x k j + j = ξ(x) i + j k {1,,,} {i,j} k=1 Let us define x = ( n n n n ),,,,, ρ, to be the only Nash equilibium that has its tems in descending ode (see Theoem ) Lemma 1: Thee is a path in G ρ, fom (n, 0, 0,, 0) to x, of length w (n) Poof: Suppose P = x 0 x 1 x t is one of the longest paths in G ρ Now P has length w (n) accoding to Lemma 1 Also, Lemma 13 implies that ξ(x e ) inceases duing evey tavesal of an edge in G ρ In othe wods ξ(x 0 ) < ξ(x 1 ) < < ξ(x t ) Now max{ξ(x) : x ρ} is finite, and so the longest path P clealy cannot continue foeve This agument implies that G ρ is acyclic and that w (n) is finite Now we will show that the stating point of the path P is x 0 = (n, 0, 0,, 0) We will pove this by contadiction Suppose x 0 (n, 0, 0,, 0) and let j = max{k : x k > 0} Now (x 0 + δ 1 δ j, x 0 ) E ρ and so we could incease the length of P by appending x 0 + δ 1 δ j to the font In othe wods x 0 + δ 1 δ j x 0 x 1 x t would be a longe path than P, but this contadicts ou assumption that P is one of the longest paths This eductio ad absudum implies that in fact we must have x 0 = (n, 0, 0,, 0) Now we will show that the ending point of the path P is x t = x We will pove this by contadiction Suppose x t x Now thee must exist a pai i < j such that x t i > xt j +1 (because x is the only vecto in ρ within which each pai of enties diffe in size by at most one) Now let i = max{k : x t k = xt i } and j = min{k : x t k = xt j } Now (xt, x t δ i + δ j ) E ρ and so we could incease the length of P by appending x t δ i + δ j to the end In othe wods x 0 x 1 x t x t δ i + δ j would be a longe path than P, but this contadicts ou assumption that P is one of the longest paths This eductio ad absudum implies that in fact we must have x t = x So we have shown that any longest path in G ρ begins at (n, 0, 0,, 0) and ends at x

16 16 To gain moe accuate esults about the value of w (n) we define the set Ω = {x ρ : x i x i+1 1, i {1,,, 1}} One way to constuct long paths in G ρ is to pefom as many switches of the type i i + 1 as possible (such switches ae called nea shifts in [18]) Ω is the set of state vectos in ρ upon which no i i + 1 type switches can be pefomed Theoem( 15: n( 1) ) w (n) n( 1) Poof: Accoding to Lemma 1 thee is a w (n) length path P = x 0 x 1 x t in G ρ that stats at x 0 = (n, 0, 0,, 0) and ends at x t = x Now accoding to Lemma 13 we have ξ(x 0 ) < ξ(x 1 ) < < ξ(x t ) Now since ξ is an intege valued function, it must incease by at least 1 along each step of the path P It follows that P cannot have length geate than ξ(x ) ξ((n, 0, 0,, 0)) In othe wods w (n) ξ(x ) ξ((n, 0, 0,, 0)) Now ξ((n, 0, 0,, 0)) = 0 Let L = n mod If L = 0 then ξ(x ) = ( k=1 (k 1) n = n( 1) If L > 0 then ξ(x ) = L ) k=1 (k 1) n + ( k=l+1 (k 1) n ), whee n = n+ L and n = n L So ξ(x ) = ( k=1 (k 1) ( )) n L + ( ) ( ) ( ) L k=1 (k 1) = ( 1)n ( 1)L + (L 1)L, now since 1 > L 1 we have ξ(x ) n( 1) Hence w (n) ξ(x ) n( 1) holds( in geneal ) Let χ = n( 1) Befoe we pove w (n) χ let us conside anothe esult We popose that u Ω we have ξ(u) χ To see this note that fo any u Ω we have ξ(u) = k=1 (k 1)u k = ( k= (k 1) u 1 ) k v= (u v 1 u v ) Now since u Ω ρ we have u 1 n and u v 1 u v 1, j {, 3,, }, which implies k v= (u v 1 u v ) k v= 1 = (k 1) So we have ξ(u) k= (k 1) (( ) ) ( ) ( n (k 1) = n 1 k=1 k 1 ) k=1 k = χ Now we will pove w (n) χ by showing how to constuct a path P = x 0 x 1 x t of length geate than o equal to χ in G ρ We constuct P by stating at x 0 = (n, 0, 0,, 0) and using the following simple algoithm; e 0 we have that if i {1,,, 1} such that x e i xe i+1 > 1 then xe+1 = x e δ i + δ i+1, othewise halt (to use the teminology fom [18] we would say P is constucted by stating at (n, 0, 0,, 0) and then pefoming as many nea shifts as possible) The final vetex x t of this path must be such that x t Ω (othewise moe nea shifts, i i + 1, could be pefomed) Moeove x 0, x 1,, x t 1 ρ Ω (by the definition of Ω) Also note that, accoding to Lemma 13 we must have ξ(x e+1 ) = ξ(x e ) + 1 Also, since ξ(x 0 ) = 0, we have that ξ(x e ) = e, e {0, 1,, t} So now we have shown how to constuct a length t path P that leads to a vetex x t Ω such that ξ(x t ) = t Now ecall that we have just poved that ξ(u) χ u Ω, so this implies that ξ(x t ) = t χ In othe wods, this implies that the path P, that we have just shown how to constuct must be of length geate than o equal to χ Now this implies that the longest path must be of length w (n) χ Coollay 16: lim n w (n) = n( 1) Poof: ( n( 1) ) ( w (n) n( 1) and as n we have n( 1) ) n( 1)

17 17 Theoem 17: w (n) = n Poof: When n = evey bette esponse switch is a stong switch (in othe wods thee is only eve one outcome to updating a paticula state vecto) It follows that w (n) = max{b (x) : x } = b((n, 0)) = n n = n Theoem 18: w 3 (n) = n 1 Poof: We can compute that w 3 (1) = 0, w 3 () = 1 and w 3 (3) = diectly, by constucting G ρ and using a longest path finding algoithm (actually this can easily be done by hand) This shows w 3 (n) = n 1 fo n 3 Now suppose n > 3 Recall that Ω ρ is the set of vectos with subsequent tems vaying by at most one In this case, whee = 3, evey enty in Ω must be of the fom (m, m, m), (m, m, m 1), (m, m 1, m 1) o (m, m 1, m ), fo some intege m such that the enties in the esulting vecto sum to n Now note that u {(m, m, m), (m, m 1, m )} 3 k=1 u k mod 3 = 0 and u = (m, m 1, m 1) 3 k=1 u k mod 3 = 1 and u = (m, m, m 1) 3 k=1 u k mod 3 = Now suppose n mod 3 = 0 In this case, if u Ω eithe: (1) u = (m, m, m), in which case m = n 3 and u = x, o () u = (m, m 1, m ), in which case m = n+3 3 and u = ( n+3 3, n 3, n 3 3 ) This implies Ω = {x, ( n+3 3, n 3, n 3 3 )} Now accoding to Lemma 1 thee is a path P, of length w 3 (n), fom (n, 0, 0) to x Now clealy, y ρ, we have that if (y, x ) E ρ then y = ( n+3 3, n 3, n 3 n+3 3 ) This implies that P ends with the step ( 3, n 3, n 3 3 ) x This implies that the length of P (which is w 3 (n)) is one plus the length of the path P, whee P = x 0 x 1 x t is the longest path fom x 0 = (n, 0, 0) to x t = ( n+3 3, n 3, n 3 3 ) Now Lemma 13 implies that P cannot have length less than ξ(( n+3 3, n 3, n 3 3 )) ξ((n, 0, 0)) = n Also, since x0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0) to x t = ( n+3 3, n 3, n 3 3 ) of length n 1 by letting xe+1 = x e δ i +δ i+1, fo some i such that x i > x i+1 +1, e {0, 1,, t 1} (this will yield a path of length ξ(( n+3 3, n 3, n 3 3 )) ξ((n, 0, 0)) = n ) So the longest path, P, fom x 0 = (n, 0, 0) to x t = ( n+3 3, n 3, n 3 3 ) has length n, so w 3(n) = 1 + n = n 1 in this case Next suppose n mod 3 = 1 In this case, if u Ω then u = (m, m 1, m 1) = x, whee m = n+ 3 So in this case Ω = {x } Now accoding to Lemma 1 thee is a path P = x 0 x 1 x t, of length w (n), fom (n, 0, 0) to x = ( n+ 3, n 1 3, n 1 3 ) Now Lemma 13 implies that P cannot have length less than ξ(( n+ 3, n 1 3, n 1 3 )) ξ((n, 0, 0)) = n 1 Also, since x 0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0) to x t = ( n+ 3, n 1 3, n 1 3 ) of length n 1 by letting x e+1 = x e δ i + δ i+1, fo some i such that x i > x i+1 + 1, e {0, 1,, t 1} (this will yield a path of length ξ(( n+ 3, n 1 3, n 1 3 )) ξ((n, 0, 0)) = n 1) Finally, suppose n mod 3 = In this case, if u Ω then u = (m, m, m 1) = x, whee m = n+1 3 So in this case Ω = {x } Now accoding to Lemma 1 thee is a path P = x 0 x 1 x t, of length w (n), fom (n, 0, 0) to x = ( n+1 3, n+1 3, n 3 ) Now Lemma 13 implies that P cannot have length less than ξ(( n+1 3, n+1 3, n 3 )) ξ((n, 0, 0)) = n 1 Also, since x 0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0) to x t = ( n+1 3, n+1 3, n 3 ) of length n 1 by letting

18 18 x e+1 = x e δ i + δ i+1, fo some i such that x i > x i+1 + 1, e {0, 1,, t 1} (this will yield a path of length ξ(( n+1 3, n+1 3, n 3 )) ξ((n, 0, 0)) = n 1) Theoem 19: w (n) = δ n,1 + δ n, + 3n (n mod ) 5 (n + 1 mod ) ( + ( n mod )) whee δ i,j is the Konecke delta Poof: We can easily compute that w (1) = 0, w () = 1, w (3) =, w () =, w (5) = 5 and w (6) = 6 We can compute that w (1) = 0, w () = 1, w (3) =, w () =, w (5) = 5 and w (6) = 6 diectly, by constucting G ρ and using a longest path finding algoithm (actually this can easily be done by hand) It can, hence, easily be checked that ou fomula holds tue when n 6 Now suppose n > 6 Recall that Ω ρ is the set of vectos with subsequent tems vaying by at most one In this case, whee =, evey enty in Ω must be of the fom (m, m, m, m), (m + 1, m, m, m 1), (m, m, m 1, m 1), (m, m, m 1, m ), (m, m 1, m 1, m 1), (m, m 1, m 1, m ), (m, m 1, m, m ) o (m, m 1, m, m 3), fo some intege m such that the enties in the esulting vecto sum to n Now note that; (1) 1) u {(m, m, m, m), (m, m 1, m 1, m )} k=1 u k mod = 0 ) u {(m, m 1, m 1, m 1), (m, m, m 1, m )} k=1 u k mod = 1 3) u {(m, m, m 1, m 1), (m, m 1, m, m 3)} k=1 u k mod = ) u {(m, m, m, m 1), (m, m 1, m, m )} k=1 u k mod = 3 Now suppose n mod = 0 In this case, if u Ω eithe: u = (m, m, m, m), in which case m = n and u = x, o () u = (m, m 1, m 1, m ), in which case m = n+ and u = ( n+, n, n, n ) This implies that Ω = {( n, n, n, n n+ ), (, n, n, n )} in this case Now accoding to Lemma 1 thee is a path P, of length w (n), fom (n, 0, 0, 0) to x Now clealy, y ρ, we have that if (y, x ) E ρ then x = ( n+ ) This, n, n, n implies that P ends with the step ( n+, n, n, n ) x This implies that the length of P (which is w (n)) is one plus the length of the path P, whee P = x 0 x 1 x t is the longest path fom x 0 = (n, 0, 0, 0) to x t = ( n+, n, n, n ) Now Lemma 13 implies that P cannot have length less than ξ(( n+, n, n, n )) ξ((n, 0, 0, 0)) = 3n 3 Also, since x 0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0, 0) to x t = ( n+, n, n, n ) of length 3n 3 by letting x e+1 = x e δ i + δ i+1, fo some i such that x i > x i+1 + 1, e {0, 1,, t 1} (this will yield a path of length ξ(( n+, n, n, n )) ξ((n, 0, 0, 0))) So the longest path, P, fom x 0 = (n, 0, 0, 0) to x t = ( n+, n, n, n ) has length 3n 3, so w (n) = 1 + 3n 3 = 3n in this case Now suppose n mod = 1 In this case, if u Ω eithe: (1) u = (m, m 1, m 1, m 1), in which case m = n+3 and u = x, o () u = (m, m, m 1, m ), in which case m = n+3 and u = ( n+3, n+3, n 1, n 5 ) This implies that Ω = {( n+3, n+3, n 1, n 5 ), x } in this case Now accoding to Lemma 1 thee is a path P, of length w (n), fom (n, 0, 0, 0) to x Now clealy, y ρ, we have that if (y, x ) E ρ then y = ( n+3, n+3, n 1, n 5 y = ( n+7, n 1, n 1, n 5 Now note that ( ( n+7, n 1, n 1 ( n+7, n 1, n 1, n 5 ) Now this means P must pass though eithe ( n+3, n+3, n 1, n 5 ), ( n+3, n+3, n 1, n 5 ), we will still have that P passes though ( n+3, n+3, n 1, n 5 ), n 5 ) o n+7 ) o (, n 1, n 1, n 5 ) )) E ρ Now since P is a longest path, even if P passes though

19 19 This implies that P ends with the step ( n+3, n+3, n 1, n 5 ) x This implies that the length of P (which is w (n)) is one plus the length of the path P, whee P = x 0 x 1 x t is the longest path fom x 0 = (n, 0, 0, 0) to x t = ( n+3, n+3, n 1, n 5 ) Now Lemma 13 implies that P cannot have length less than ξ(( n+3, n+3, n 1, n 5 )) ξ((n, 0, 0, 0)) = 3n 7 Also, since x0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0, 0) to x t = ( n+3, n+3, n 1, n 5 ) of length 3n 7 by letting xe+1 = x e δ i +δ i+1, fo some i such that x i > x i+1 +1, e {0, 1,, t 1} (this will yield a path of length ξ(( n+3, n+3, n 1, n 5 )) ξ((n, 0, 0, 0))) So the longest path, P, fom x 0 = (n, 0, 0, 0) to x t = ( n+3, n+3, n 1, n 5 has length 3n 7, so w (n) = 1 + 3n 7 = 3n 5 in this case Now suppose n mod = In this case, if u Ω eithe: (1) u = (m, m, m 1, m 1), in which case m = n+ and u = x, o () u = (m, m 1, m, m 3), in which case m = n+6 and u = ( n+6, n+, n, n 6 Now let us define A = ( n+6, n+, n, n 6 n+10 ), B = (, n, n, n have that Ω = {A, x } in this case Now accoding to Lemma 1 thee is a path P, of length w (n), fom (n, 0, 0, 0) to x ) ) n+6 ) and C = (, n+, n+, n 10 ) Now we Now clealy, y ρ, we have that if (y, x ) E ρ then y {A, B, C} This means that P must pass though A, B o C Now clealy, z ρ, we have that if (A, z) E ρ then z {B, C} Now since P is a longest path, P must pass though A, then eithe B o C, then teminate at x In othe wods P must end with the steps A B x, o end with the steps A C x This implies that the length of P (which is w (n)) is two plus the length of the path P, whee P = x 0 x 1 x t is the longest path fom x 0 = (n, 0, 0, 0) to x t = A Now Lemma 13 implies that P cannot have length less than ξ(a) ξ((n, 0, 0, 0)) = 3n 5 Also, since x0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0, 0) to x t = A of length 3n 5 by letting xe+1 = x e δ i + δ i+1, fo some i such that x i > x i+1 + 1, e {0, 1,, t 1} (this will yield a path of length ξ(a) ξ((n, 0, 0, 0))) So the longest path, P, fom x 0 = (n, 0, 0, 0) to x t = A has length 3n 5, so w (n) = + 3n 5 = 3n 3 in this case Now suppose n mod = 3 In this case, if u Ω eithe: (1) u = (m, m, m, m 1), in which case m = n+1 and u = x, o () u = (m, m 1, m, m ), in which case m = n+5 and u = ( n+5, n+1, n 3, n 3 ) This implies that Ω = {( n+5, n+1, n 3, n 3 ), x } in this case Now accoding to Lemma 1 thee is a path P, of length w (n), fom (n, 0, 0, 0) to x Now clealy, y ρ, we have that if (y, x ) E ρ then y = ( n+5, n+1, n 3, n 3 y = ( n+5, n+1, n+1, n 7 Now note that ( ( n+5, n+1, n+1 ( n+5, n+1, n+1, n 7 ) Now this means P must pass though eithe ( n+5, n+1, n 3, n 7 ), ( n+5, n+1, n 3, n 3 ), we will still have that P passes though ( n+5, n+1, n 3, n 3 ), n 3 ) o n+5 ) o (, n+1, n+1, n 7 ) )) E ρ Now since P is a longest path, even if P passes though This implies that P ends with the step ( n+5, n+1, n 3, n 3 ) x This implies that the length of P (which is w (n)) is one plus the length of the path P, whee P = x 0 x 1 x t is the longest path fom x 0 = (n, 0, 0, 0) to x t = ( n+5, n+1, n 3, n 3 ) Now Lemma 13 implies that P cannot have length less than ξ(( n+5, n+1, n 3, n 3 )) ξ((n, 0, 0, 0)) = 3n 7 Also, since x0, x 1,, x t 1 ρ Ω, we can constuct a path fom x 0 = (n, 0, 0, 0) to x t = ( n+5, n+1, n 3, n 3 ) of length 3n 7 by letting xe+1 = x e δ i +δ i+1, fo some i such that x i > x i+1 +1, e {0, 1,, t 1} (this will yield a path of length ξ(( n+5, n+1, n 3, n 3 )) ξ((n, 0, 0, 0))) So the longest path, P, fom x 0 = (n, 0, 0, 0) to x t = ( n+5, n+1, n 3, n 3 ) has length 3n 7, so w (n) = 1 + 3n 7 = 3n 5 in this case So we have shown that n > 6 we have that n mod = 0 3n, n mod = 1 3n 5, n mod = 3n 3 and n mod = 3 3n 5 This infomation can be encoded moe compactly by witing w (n) = δ n,1 + δ n, + 3n (n mod ) 5 (n+1 mod ) ( + ( n mod )), and this equation also gives the coect values of w (n), whee n 6 (which we computed manually)

20 0 The -pat patition lattice [18] (ρ, ) is the set ρ, patially odeed with the dominance elation This elation is defined so that x, y ρ we have x y if and only if k v=1 x v k v=1 y v, k R (also we wite x > y if and only if x y and k R : k v=1 x v > k v=1 y v) Since (ρ, ) is a patially odeed set, we can use the teminology of patially odeed sets to descibe it An element x ρ is said to cove y ρ if and only if x > y and z ρ : x > z > y A esult fom [18] states that x, y ρ we have that x coves y if and only if i, j R such that y = x δ i + δ j and, eithe i = j + 1 o x i = x j + Lemma 0: x, y ρ, if (x, y) E ρ then x > y Poof: (x, y) E ρ implies i, j R : x i > x j + 1, y = x δ i + δ j Now since the enties of x ae in descending ode, we have i < j Now k R we have:- If k < i then k v=1 y v = k v=1 x v If i k < j then ( k v=1 y k ) v = v=1 x v 1 If k j then k v=1 y v = k v=1 x v So clealy k R we have k v=1 y v k v=1 x v, thus x > y Lemma 1: x, y ρ, if x > y then x ρ such that (x, x ) E ρ and x y Poof: Let us define v = min{v R : x v > y v } Such a v must exist since x > y (1) Now if v R is such that x v = x v we have x v = x v > y v y v (the final inequality holds because v v, and the enties of y ae in descending ode) Let i = max{v R : x v = x v } Now (1) implies that x i > y i Moeove, the definition of v implies that v R : v < v we have x v y v Also, we cannot have that x v < y v fo v < v, because that would violate x > y It follows that in fact v R : v < v we have x v = y v This, togethe with (1) implies that () k < i we have k v=1 x v k v=1 y v v R : v < i and i v=1 x v > i v=1 y v Let us define j = min{v R : x v < x i 1} We use a poof by contaction to pove such a j exists Suppose such a j does not exist Now v R : v > i we have x v = x i 1, (since the definition of i foces x i+1 < x i ), also we have x v = x i 1 y i y v Putting these conditions togethe with (1) we have x v y v, v R and x i > y i This implies v=1 x v > v=1 y v, which is a contadiction This eductio absudum agument poves that such a j must in fact exist So now we have shown that ou i and j exist, we can constuct x = x δ i + δ j Since x i x i+1 and x j x j 1 we must have x ρ Moeove, since x i > x j + 1, we have (x, x ) E ρ (in othe wods x ρ can be obtained by pefoming a bette esponse update upon x) Moeove, k R we have:- If k < i then k v=1 x v = k v=1 x v k v=1 y v If i k < j then ( k k ) v=1 x v = v=1 x v 1 k v=1 y v (fom ())

21 1 If k j then k v=1 x v = k v=1 x v k v=1 y v This implies x y Theoem : x, y ρ, thee is a path fom x to y in G ρ if and only if x y Poof: The theoem is tivially tue when x = y (because in this case x y and x y and thee is a zeo length path fom x to y in G ρ Fom now on let us suppose x y Suppose x = x 0 x 1 x x t = y is a path fom x to y in G ρ (as in (x q, x q+1 ) E ρ, q {0, 1,, t 1}) Now Lemma 0 implies that x = x 0 > x 1 > x > > x t = y Now since is a patial odeing, this clealy implies x > y and hence x y Now let us pove the convese Suppose x y Now since we ae supposing x y we have x > y Now we will use Lemma 1 to constuct a path fom x to y Let us define x 0 = x We know x 0 > y Now conside the following algoithm: Fo t fom 1 do Let x t be such that (x t 1, x t ) E ρ and x t y (a pocedue fo constucting such a x t is descibed within the poof of Lemma 1) If x t = y then halt end if end do Now we popose that this algoithm will eventually halt To see this conside the sequence x 0, x 1,, x t ρ geneated by unning this algoithm fo t time steps Clealy x 0 x 1 x t will be a path in G ρ Now Lemma 0 implies that x 0 > x 1 > > x t y This implies that the sequence (x 0, x 1,, x t ) holds no epeated elements (ie, the paths geneated by this algoithm will neve visit the same vetex twice) Now since G ρ holds only a finite numbe of vetices, and no vetex can be visited moe than once, this algoithm cannot geneate an abitaily long path, and most theefo halt at some point Suppose the algoithm halts at time t In this case we have that x = x 0 x 1 x x t = y is a path fom x to y in G ρ Theoem 3: w (n) is equal to the length of the longest chain in the -pat patition lattice (ρ, ) and, when n, this is equal to w (n) = 1+8k 1 n 1 k=0 Poof: Theoem implies the longest path in G ρ is equal to the longest chain in the -pat patition lattice (ρ, ) The length of the longest chain in (ρ, ), fo geneic n and was studied in [18] (although no closed fom esults ae given) When n the -pat patition lattice (ρ, ) is simply known as the patition lattice In this case the length of the longest chain is well known to be 1+8k 1 n 1 k=0, (see [19], [0]) E The aveage convegence time, with andom bette esponse switches In many scenaios, the playes will update in some sot of andom ode Undestanding of this andom case gives an insight into the moe ealistic systems, and tests the elevance of ou pefomance bounds In this section we conside a simple andom

22 updating model, whee a andom unsatisfied playe (ie, one that can pefom a bette switch) pefoms a andom bette switch at evey time step Let a(x) denote the expected value of the convegence time of x unde this andom bette esponse system We simulate and compae the pefomances of vaious updates ules as in Figue 5 Fig 5 Convegence times with = esouces The left figue shows the convegence times stating fom state (n, 0, 0, 0) The ight figue shows the convegence times when the playes stat with andom esouces The top puple line depicts the longest un time w (n) (calculated by exhaustive seach) The middle blue line depicts the aveage un time of andom bette esponses (aveaged ove 10 tials) The low gold line depicts the shotest un time 1) b(x) is the length of the shotest path in G going fom state x to a Nash equilibium ) β (n) is the expected value of b(x), when the playes choose thei initial esouces andomly and unifomly 3) w (n) is the length of the longest path in G ) a(x) is the expected convegence time unde the andom update s defined above 5) Also, we define α (n) as the expected value of a(x), when x is geneated by the playes choosing thei initial esouces andomly and unifomly Simulations suggest that the andom convegence time a(x) is often close to the fastest convegence time b(x), and is much smalle than the wost case convegence time w (n) (when the numbe of playes n is easonable lage) This implies that b(x) is a good estimation of the eal convegence time and w (n) is too pessimistic IV SPATIAL VARIATIONS ON OUR MODELS So fa we have assumed that each playe inteacts with all othe playes in the system In many situations, the playes ae distibuted ove space, and can only cause congestion to thei neighbos This is the case in wieless netwoks, whee only uses close-by may cause significant intefeences In this section, we look at the convegence of the spatial congestion games poposed in ou pevious wok [15], whee a gaph stuctue G is used to epesent the space elationship among playes 7 This spatial game model is a genealization of the one defined in Section II with the added dimension of space Figue 6 shows an example of gaph G Each vetex in G epesents a playe A pai of playes ae connected by an edge in G when they ae close enough to potentially cause congestion fo one anothe Each playe p has a stictly deceasing payoff function, f p (N p i + 1), whee N p i is the numbe of playe p s neighbos that use the same esouce i as playe p Just like befoe, the details of the payoff functions ae ielevant with espect to the convegence dynamics All that mattes is that playes always pefe esouces that ae used by less of thei neighbos In this geneal model, a state is an assignment of esouces on the gaph, one esouce to each vetex (playe) A conflict happens when a vetex uses the same esouce as one of its neighbos We still assume the esouces ae homogenous, so evey vetex simply wishes to minimize the numbe of conflicts they incu As befoe, the system evolves via asynchonous bette esponse switches Evey time step, one vetex switches to a esouce shaed by less of its neighbos A Nash equilibium is a system state whee no vetex can make a bette esponse switch We futhe genealize the model by assuming that diffeent playes may have diffeent sets of esouces available to them due to spatial vaiations Fomally, each playe p has a specific esouce set R(p) Despite the geneality of the model, convegence to a Nash equilibium, though bette esponse switches, 7 The gaphs, G, discussed in this section ae diffeent fom the gaphs, G and G ρ, discussed peviously

23 3 Fig 6 An example of a state (esouce allocation) of a spatial congestion game Evey vetex can access two esouces: ed and blue The only unsatisfied playe (vetex) is the cicled one in the middle This playe cuently suffes 3 conflicts Afte this playe does a bette esponse switch (changing fom blue to ed), the system will be in a Nash equilibium is guaanteed, as shown in ou pevious wok [15] Next we povide futhe chaacteization of the convegence speed and equilibium chaacteistics Theoem is an extension of a esult fom [15] Theoem : Conside a geneal spatial congestion game with heteogeneous esouce availabilities defined on a gaph G Any sequence of G asynchonous bette esponse switches leads to a Nash equilibium, whee G n(n 1) is the numbe of edges in the gaph G Poof: The wost case initial state is whee evey vetex uses the same esouce In this case thee ae G conflicts Accoding to [15], the numbe of conflicts in a given state foms a potential function, which deceases with bette esponse switches It follows that the system will convege to Nash equilibium in at most G time steps Ou next esult bounds the wost case pefomance of a playe at Nash equilibium Theoem 5: At a Nash equilibium of a geneal spatial congestion game with heteogeneous esouce availabilities, a playe p will not suffe moe than conflicts, whee d(p) is p s degee (numbe of neighbos) and R(p) is the numbe of esouces available to p d(u) R(p) Poof: Let us pove by contadiction Suppose playe p suffes K > d(p) R(p) conflicts Since the system is in a Nash equilibium, playe p cannot benefit by switching to a diffeent esouce This means fo any esouce i R(p), playe p has at least K neighbos using esouce i It follows that p has degee geate than o equal to R(p) K > d(p), which contadicts ou assumption that playe p has a degee d(p) Theoem 5 implies that when evey playe can access moe esouces than its degee, the system will always convege to a Nash equilibium which involves no conflicts whatsoeve V APPLICATION: COGNITIVE RADIO NETWORKING A good application fo the above analytical famewok is spectum shaing in cognitive adio netwoks Most of the usable wieless spectum is owned by license holdes who possess exclusive tansmission ights Howeve, extensive measuements show that many wieless channels ae heavily unde-utilized most of the time The cognitive adio technology allows unlicensed uses to oppotunistically access these channels when the license holdes ae not pesent One of the cental questions of such spectum shaing is: how should the uses access the channels in a distibuted fashion? A natual way to model the uses behavio is the congestion game, whee uses switch channels to minimize thei intefeence The models discussed in this pape ae vey useful hee due to the following easons (1) In most wieless communication standads, spectum is divided into equal bandwidth channels Inteleaving techniques can futhe homogenize the qualities of the channels This means that the channels ae homogeneous to the same use, and the channel with the least numbe of uses povides the best pefomance Diffeent uses, howeve, can achieve diffeent data ates due to diffeent channel conditions and choices of coding and modulation schemes () A wieless use only geneates significant intefeences to close-by uses, and thus the spatial infomation is impotant (3) As licensed holdes ae spatially located and may have diffeent activities, uses at diffeent locations may be able to access diffeent channels () The license holdes often have stochastic taffic, which means that the channel availability is often time vaying In the est of this section, we will conside how pevious esults can be applied in the spectum shaing setting, whee we teat uses as playes and channels as esouces A The significance of pevious esults Fast convegence is essential in cognitive adio netwoks due to the time vaiability of channels Coollay 8 shows that the fastest convegence is achieved with stong switches, in which case no use needs to switch channel moe than once This is