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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 ex-novo creation of a network Players are nodes that: pay for the links they personally activate
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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 ex-novo 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)
2 1 -1 3 -3 4 2 1 cu=2+9 Cost of a player: an example u + cu=+13 (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 utilitariansocial cost, i.e., the sum of all players’ costs. Observe that: • 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 NP-hard • The complexity of establishing the existence of an improving strategy for a player is open
+2 -2 -1 -5 +2 -1 -1 +5 +5 +5 -5 -5 +4 +1 -1 -5 +1 Stable networks: an example • Set =5, and consider: That’s a stable network!
How does an optimal network look like?
Some notation Kn:completegraph 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(n-1)/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) LB(n-1) = SC of any star ≥ 2 min m 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
c Proof (continued) ≥1 c cannot change its strategy u v 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! …Kn is an optimal solution, any star T is stable… 1<<2 maximized when 1 -1(n-1) + 2n(n-1) (-2)(n-1) + 2n(n-1) SC(T) PoS ≤ = ≤ n(n-1)/2 + n(n-1) n(n-1)/2 + n(n-1) SC(Kn) 2n - 1 4n -2 = = < 4/3 3/2n 3n
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=1 diam=2 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 k vertices reduce their distance from u 2k≤ distG(u,v) ≤ 2k+1 for some k from ≥2k to 1 ≥ 2k-1 from ≥2k-1 to 2 ≥ 2k-3 …since G is stable: from ≥ k+1 to k ≥ 1 ≥k2 k ≤ k-1 (2i+1)=k2 i=0 distG(u,v) ≤ 2 + 1
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 ≤(n-1)+n(n-1)2d ≤ 2d OPT |E|=|Ecut| + |Enon-cut| ≤(n-1) ≤n(n-1)2d/ Prop. 2