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Second case study: Network Creation Games (a.k.a. Local Connection Games)

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)

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  1. Second case study: Network Creation Games (a.k.a. Local Connection Games)

  2. 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

  3. 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)

  4. 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 + vV distG(S)(u,v)

  5. 2 1 -1 3 -3 4 2 1 cu=2+9 Cost of a player: an example u +  cu=+13 Convention: arrow from the node buying the link Notice that if <4 this is an improving move for u

  6. 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,vV distG(S)(u,v)

  7. 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

  8. Some (bad) computational aspects of NCG • NCG are not potential games • Computing a best moves for a player is NP-hard (as in the GCG) • The complexity of establishing the existence of an improving moves for a player is open

  9. +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!

  10. How does an optimal network look like?

  11. Some notation Kn:completegraph with n nodes A star is a tree with height at most 1 (when rooted at its center)

  12. 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,vV 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(n-1)/2) and to LB(n-1), respectively,

  13. 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≥ LB(m) ≥ ≤ 2 max m LB(n(n-1)/2) = SC(Kn)

  14. Are the complete graph and stars stable?

  15. 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

  16. 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.

  17. 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

  18. 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 ?

  19. State-of-the-art

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