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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 V={1,…,n} 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=(s1,…, sn), 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) in G(S) between u and v nu: number of edges bought by node u Player u aims to minimize its cost: costu(S) = nu + vV distG(S)(u,v) Cost of a player: an example 2 -1 3 -3 4 1 + 2 1 u cu=+13 cu=2+9 Convention: arrow from the node buying the link Notice that if <4 this is an improving move for u 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,vV distG(S)(u,v) Our goal A network is optimal or socially efficient if it minimizes the social cost We aim to characterize the efficiency loss resulting from selfishness, by using the Price of Stability (PoS) and the Price of Anarchy (PoA) We use Nash equilibrium (NE) as the solution concept: A network (i.e., a formed graph) G(S)=(V,E) is stable (for the given value ) if S is a NE, while a graph G=(V,E) is stable 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 two nodes is infinite whenever they are not connected Some (bad) computational aspects of NCG NCG are not potential games Computing a best moves for a player is NPhard (as in the GCG) The complexity of establishing the existence of an improving moves 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 any 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 = |E| + u,vV distG(S)(u,v) ≥ 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 the SC of any star when m=n-1; indeed: SC(Kn) = 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≥ LB(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 a 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 ? State-of-the-art