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Second case study: Network Creation Games (a.k.a. Local Connection Games) Introduction Introduced in [FLMPS,PODC’03] A LCG is a game that models the exnovo creation of a network Players are nodes that: pay for the links they personally activate benefit from short paths on the created network [FLMPS,PODC’03]: A. Fabrikant, A. Luthra, E. Maneva, C.H. Papadimitriou, S. Shenker, On a network creation game, PODC’03 The model n players: nodes in a graph to be built Strategy for player u: a set of incident edges (intuitively, a player buys these edges, that will be then used bidirectionally by everybody; however, only the owner of an edge can remove it, in case he decides to change its strategy) Given a strategy vector S, the constructed network will be G(S) player u’s goal: to spend as little as possible for buying edges (building cost) to make the distance to other nodes as small as possible (usage cost) The model Each edge costs ≥0 distG(S)(u,v): length of a shortest path (in terms of number of edges) between u and v nu: number of edges bought by node u Player u aims to minimize its cost: costu(S) = nu + v distG(S)(u,v) Cost of a player: an example 2 -1 3 -3 4 1 + 2 1 u cu=+13 cu=2+9 (notice that if <4 this is an improving move for u) Convention: arrow from the node buying the link The social-choice function To evaluate the overall quality of a network, we consider the utilitarian social cost, i.e., the sum of all players’ costs. Observe that: 1. 2. In G(S) each term distG(S)(u,v) contributes to the overall quality twice Each edge (u,v) is bough at most by one player Social cost of a network G(S)=(V,E): SC(S)=|E| + u,vdistG(S)(u,v) Our goal We use Nash equilibrium (NE) as the solution concept A network is optimal or socially efficient if it minimizes the social cost A graph G=(V,E) is stable (for a value ) if there exists a strategy vector S such that: S is a NE S forms G Observe that any stable network must be connected, since the distance between to nodes is infinite whenever they are not connected We aim to bound the efficiency loss resulting from stability, by using the Price of Stability (PoS) and the Price of Anarchy (PoA) Some (bad) computational aspects of NCG NCG are not potential games Computing a best response for a player is NPhard The complexity of establishing the existence of an improving strategy for a player is open Stable networks: an example Set =5, and consider: +2 -2 -1 -5 -1 +2 -1 +5 +5 +5 -5 +4 +1 -1 -5 -5 +1 That’s a stable network! How does an optimal network look like? Some notation Kn: complete graph with n nodes A star is a tree with height at most 1 (when rooted at its center) Lemma Il ≤2 then the complete graph is an optimal solution, while if ≥2 then any star is an optimal solution. proof Let G=(V,E) be an optimal solution; |E|=m and SC(G)=OPT OPT≥ m + 2m + 2(n(n-1) -2m) =(-2)m + 2n(n-1) LB(m) Notice: LB(m) is equal to SC(Kn) when m=n(n-1)/2, and to SC of any star when m=n-1; indeed: SC(Kn) = |E| + u,vdistG(S)(u,v) = n(n-1)/2 + n(n-1) SC(star)= (n-1) + 2(n-1) + 2(n-1)(n-2) = (n-1) + 2(n-1)2 and it is easy to see that they correspond to LB(n(n1)/2) and to LB(n-1), respectively, Proof (continued) G=(V,E): optimal solution; |E|=m and SC(G)=OPT LB(m)=(-2)m + 2n(n-1) ≥ 2 min m LB(n-1) = SC of any star OPT≥ min LB(m) ≥ m ≤ 2 max m LB(n(n-1)/2) = SC(Kn) Are the complete graph and stars stable? Lemma Il ≤1 the complete graph is stable, while if ≥1 then any star is stable. proof ≤1 If a node removes any k owned edges, it saves k in the building cost, but it pays k≥k more in the usage cost Proof (continued) ≥1 c cannot change its strategy v c u If v buys any k edges it pays k more in the building cost, but it saves only k≤k in the usage cost u’s cost is +1+2(n-2); if it removes (u,c) and buys any k edges, it pays k in the building cost, and its usage cost becomes k+2+3(n-k-2), and so its total cost increases to: + (k-1)+k +2+3n-3k-6 = +1+2(n-2)+[(k-1)-2k+n] ≥ +1+2(n-2)+[k-1-2k+n] = +1+2(n-2)+[n-k-1] since the quantity in square brackets is not negative, being 1≤k≤n-1. Theorem For ≤1 and ≥2 the PoS is 1. For 1<<2 the PoS is at most 4/3 proof ≤1 and ≥2 …trivial! 1<<2 …Kn is an optimal solution, any star T is stable… maximized when 1 PoS ≤ SC(T) SC(Kn) -1(n-1) + 2n(n-1) (-2)(n-1) + 2n(n-1) = n(n-1)/2 + n(n-1) ≤ n(n-1)/2 + n(n-1) 2n - 1 4n -2 = 3/2n = 3n < 4/3 What about the Price of Anarchy? …for <1 the complete graph is the only stable network, (try to prove that formally) hence PoA=1… …for larger value of ? Some more notation The diameter of a graph G is the maximum distance between any two nodes diam=2 diam=1 diam=4 Some more notation An edge e is a cut edge of a graph G=(V,E) if G-e is disconnected G-e=(V,E\{e}) A simple property: Any graph has at most n-1 cut edges (otherwise they would induce a cycle, which is impossible…think about it) Theorem The PoA is at most O(). proof It follows from the following lemmas: Lemma 1 The diameter of any stable network is at most 2 +1 . Lemma 2 The SC of any stable network with diameter d is at most 3d times the optimum SC. proof of Lemma 1 G: stable network Consider a shortest path in G between two nodes u and v u v 2k≤ distG(u,v) ≤ 2k+1 for some k …since G is stable: ≥k2 k ≤ distG(u,v) ≤ 2 + 1 k vertices reduce their distance from u from ≥2k to 1 ≥ 2k-1 from ≥2k-1 to 2 ≥ 2k-3 ● ● ● from ≥ k+1 to k ≥1 k-1 (2i+1)=k2 i=0 Proposition 1 Let G be a network with diameter d, and let e=(u,v) be a non-cut edge. Then in G-e, every node w increases its distance from u by at most 2d Proposition 2 Let G be a stable network, and let F be the set of non-cut edges bought by a node u. Then |F|≤(n-1)2d/ Lemma 2 The SC of any stable network G=(V,E) with diameter d is at most O(d) times the optimum SC. proof OPT ≥ (n-1) + n(n-1) SC(G)= u,vdG(u,v) + |E| ≤ d OPT+2d OPT = 3d OPT ≤dn(n-1) ≤d OPT |E|=|Ecut| + |Enon-cut| ≤(n-1)+n(n-1)2d ≤ 2d OPT ≤(n-1) ≤n(n-1)2d/ Prop. 2