1 Network Creation Game* Presented by Miriam Allalouf On a Network Creation Game by A.Fabrikant, A. Luthra, E. Maneva, C. H. Papadimitriou, and S. Shenker, [FLMPS], PODC 2003 *Part of the Slides are taken from Alex Fabrikant PPT presentation On Nash Equilibria for a Network Creation Game by Albers, S. Eilts, E. Even-Dar, Y. Mansour and L. Roditty. [AEEMR] to appear in SODA 2006.

U C B E R K E L E Y C O M P U T E R S C I E N C E 2 Context The internet has over 20,000 Autonomous Systems (AS) Everyone picks their own upstream and/or peers MACHBA wants to be close to everyone else on the network, but doesn ’ t care about the network at large

Question: What is the “ penalty ” in terms of poor network structure incurred by having the “ users ” create the network, without centralized control?

U C B E R K E L E Y C O M P U T E R S C I E N C E 4 In this talk we … Introduce a simple model of network creation by self-interested agents Briefly review game-theoretic concepts Talk about related work Show bounds on the “ price of anarchy ” in the model – using both papers results Disprove the tree conjecture A weighted network creation game Cost sharing Discuss extensions and open relevant problems.

U C B E R K E L E Y C O M P U T E R S C I E N C E 5 A Simple Model N agents, each represented by a vertex and can buy (undirected) links to a set of others (s i ) One agent buys a link, but anyone can use it Undirected graph G is built Cost to agent: Pay $  for each link you buy Pay $1 for every hop to every node (  may depend on n)

U C B E R K E L E Y C O M P U T E R S C I E N C E 6 Example  (Convention: arrow from the node buying the link) ++ c(i)=  +13 c(i)=2  +9

U C B E R K E L E Y C O M P U T E R S C I E N C E 8 Definitions V={1..n} set of players A strategy for v is a set of vertices S v  V\{v}, such that v creates an edge to every w  S v. G(S)=(V,E) is the resulted graph given a combination of strategies S=(S 1,..,Sn), V set of plyaers / nodes and E the laid edges. Social optimum: combination of strategies that minimizes the social cost “ What a dictator would do ” Not necessarily palatable to any given agent Social cost: The simplest notion of “ global benefit ”

U C B E R K E L E Y C O M P U T E R S C I E N C E 9 Definitions: Nash Equilibria Nash equilibrium: a situation such that no single player can change what he is doing and benefit Presumes complete rationality and knowledge on behalf of each agent Not guaranteed to exist, but they do for our model The cost of player i under s:

U C B E R K E L E Y C O M P U T E R S C I E N C E 10 Definitions: Nash Equilibria A combination of strategies S forms Nash equilibrium, if for any player i and any other strategy U, such that U differs from S only in i ’ s component > G(S) is the equilibrium graph Strong Nash equilibrium is when for any i Otherwise, it is a weak Nash equilibrium, where at least one player can change its strategy without affecting its cost. Transient Nash equilibria is a weak equilibrium from which there exists a sequence of single-player strategy changes, which do not change the deviator ’ s cost, leading to a non-equilibrium position.

U C B E R K E L E Y C O M P U T E R S C I E N C E Example Set  =5, and consider: ?!

U C B E R K E L E Y C O M P U T E R S C I E N C E 12 Definitions: Price of Anarchy Price of Anarchy (Koutsoupias & Papadimitriou, 1999): the ratio between the worst-case social cost of a Nash equilibrium network and the optimum network over all Nash equilibria S We bound the worst-case price of anarchy to evaluate “ the price we pay ” for operating without centralized control

U C B E R K E L E Y C O M P U T E R S C I E N C E 13 The presented papers On a Network Creation Game by A.Fabrikant, A. Luthra, E. Maneva, C. H. Papadimitriou, and S. Shenker, [FLMPS], PODC 2003 On Nash Equilibria for a Network Creation Game by Albers, S. Eilts, E. Even-Dar, Y. Mansour and L. Roditty. [AEEMR] to appear in SODA 2006.

U C B E R K E L E Y C O M P U T E R S C I E N C E 14 Related Work Corbo and Parkes (PODS 05) Study the P.O.A of the network creation game assuming the edges are bought by both players Anshelevich, et al. (STOC 2003) Agents are “ global ” and pick from a set of links to connect between their own terminals, observed the “ price of stability ” A body of similar work on social networks in the econometrics literature (e.g. Bala&Goyal 2000, Dutta&Jackson 2000) Earning by forming links Players heterogenity, etc

U C B E R K E L E Y C O M P U T E R S C I E N C E 15 In this talk we … Introduce a simple model of network creation by self- interested agents Briefly review game-theoretic concepts Talk about related work Show bounds on the “ price of anarchy ” in the model – using both papers results Disprove the tree conjecture A weighted network creation game Cost sharing Discuss extensions and open problems we believe to be relevant and potentially tractable.

U C B E R K E L E Y C O M P U T E R S C I E N C E 16 Social optima - clique When  <2, any missing edge can be added at cost  and subtract at least 2 from social cost

U C B E R K E L E Y C O M P U T E R S C I E N C E 17 Social optima - star When  2, consider a star. Any extra edges are too expensive.

U C B E R K E L E Y C O M P U T E R S C I E N C E 18 Equilibria: very small  (<2) For  <1, the clique is the only N.E. For 1 2 Then, the star is the worst N.E., can be seen to yield P.o.A. of at most 4/3 ++ -2

U C B E R K E L E Y C O M P U T E R S C I E N C E 19 P.O.A for very small  (<2) The star is also a Nash equilibrium, but there may be worse Nash equilibrium.

U C B E R K E L E Y C O M P U T E R S C I E N C E 20 General Upper Bound [FLMPS] Assume  >2 (the interesting case) Lemma: if G is a N.E., Generalization of the above: … ++ -(d-1)-(d-3)-(d-5)=-Θ(d 2 )

U C B E R K E L E Y C O M P U T E R S C I E N C E 21 General Upper Bound (cont.) A counting argument then shows that for every edge present in a Nash equilibrium, Ω( ) others are absent Then: C(star)= Ω(n 2 ), thus P.o.A. is O( )

U C B E R K E L E Y C O M P U T E R S C I E N C E 22 RESULTS: p.o.a Bounds for different  values Constant P.O.A Upper bounds: [AEEMR] For Not larger than 1.5 Goes to 1 as  increases For any other , Constant for Increases for Maximum at  =n : 

U C B E R K E L E Y C O M P U T E R S C I E N C E 23 Nash Equilibrium Characteristics [FLMPS] Tree Conjecture: For all  >A (A constant), all non-transient Nash equilibria are for trees [AEEMR] disproves it and show that for any positive integer n0, there exists a graph built by n ≥ n0 players that contains cycles and forms a strong Nash equilibrium, But If  every Nash equilibrium is a tree

U C B E R K E L E Y C O M P U T E R S C I E N C E 24 p.o.a Upper Bound [AEEMR] Price of Anarchy Upper Bound [AEEMR]

U C B E R K E L E Y C O M P U T E R S C I E N C E 25 Constant p.o.a Upper Bound for α ≥ 12n log n [AEEMR] (1) Theorem 2 For α ≥ 12n log n, the price of anarchy is bounded by 1 + (6n log n / α) ≤ 1.5 and any G(S) equilibrium graph is a tree. Proof Based on Proposition 1 where proved that G(S) whose girth ≥ 12log n is a tree whose maximal depth is 6log n, and on Lemma 5 that connects between the girth length and α.

U C B E R K E L E Y C O M P U T E R S C I E N C E 26 Improved Upper Bound for α ≥ 12n log n [AEEMR] (2) Proposition 1 If G(S) is an equilibrium graph whose girth ≥ 12 log n then The diameter of G(S ) ≤ 6 log n G(S) is a tree In order to prove the above, The following graph analysis Were provided

U C B E R K E L E Y C O M P U T E R S C I E N C E 27 Definition G(S) is an equilibrium graph. T(u) in V shortest path tree rooted at u. and this vertex represents layer 0 of the tree. Given vertex layers 0 to i − 1, layer i is constructed as follows. Tree edges A node w belongs to layer i if it is not yet contained in layers 0 to i − 1 and there is a vertex v in layer i−1 such that {v,w}  E (only one is added to the the shortest path tree). Non-tree edges - all remaining edges of E that are added to T(u) T(u) a layered version of G with distinguished tree edges. Improved Upper Bound for α ≥ 12n log n [AEEMR] (3)

U C B E R K E L E Y C O M P U T E R S C I E N C E 28 A vertex v in V at a depth ≤ 6 log n in T(u), is: Expanding - If v has at least two children, each with at least one descendent in the Boundary level. Neutral - If v has exactly one child with at least one descendent in the Boundary level. Degenerate - If v does not have any descendent in the Boundary level Improved Upper Bound for α ≥ 12n log n [AEEMR] (4)

U C B E R K E L E Y C O M P U T E R S C I E N C E 29 v  T(u) is a Neutral vertex. Du(v) is the set of its Degenerate children and their descendants at T(u). Lemma 3 G(S) an equilibrium graph whose girth ≥12 log n. Every path from x  Du(v) to y  V\Du(v) in G(S) must go through v neutral vertex. Neutral edge An edge on the shortest path from u to v that both of its endpoints are Neutral vertices. Lemma 4 G(S) an equilibrium graph whose girth ≥ 12 log n. The total number of Neutral edges is 2 log n. Proof It is more beneficial to buy a link to a neutral node than to a degenerated mode. This decision can be taken no more than log n Improved Upper Bound for α ≥ 12n log n [AEEMR] (5)

U C B E R K E L E Y C O M P U T E R S C I E N C E 30 Improved Upper Bound for α ≥ 12n log n [AEEMR] (6) Proposition 1 If G(S) is an equilibrium graph whose girth ≥ 12 log n then The diameter of G(S ) ≤ 6 log n G(S) is a tree Proof By contradiction, assume that the diameter is at least 6 log n. Let u  V on one of the endpoints of the diameter, and look on T(u). Since U is either Neutral or Expanding vertex (one of the diameter endpoints). Goal: show that the number of descendants at the Boundary level is at least n.  leads to contradiction and implies that the maximal depth is at most 6 log n and that there are no cycles. : see details …

U C B E R K E L E Y C O M P U T E R S C I E N C E 31 Improved Upper Bound for α ≥ 12n log n [AEEMR] (7) Proposition 1 Proof more details Let v  V, d the depth of v in T(u), b the number of Neutral edges on the path from u to v. (d, b) is a label per vertex [example: for u it is (0,0)] Let v be a non-Degenerate vertex whose label is (d, b) N(d, b) be a lower bound on the number of its descendants at the Boundary level. Note: two vertices might have the same label, but have different number of descendants at the boundary level. We claim that and for the root : thus proving the claim will lead to the desired contradiction.

U C B E R K E L E Y C O M P U T E R S C I E N C E 32 Improved Upper Bound for α ≥ 12n log n [AEEMR] (8) Lemma 5 If G(S) is an equilibrium graph and c be any positive constant. If α>cn log n then the length of the girth of G(S ) ≥ c log n. Proof Assume by contradiction that minimal girth is clog n. U on the cycle wants to buy an edge: Benefit by distance reduction: (clog n -1)n Loss by edge addition: α = cnlog n  It is not an equilibia, contradiction.

U C B E R K E L E Y C O M P U T E R S C I E N C E 33 Theorem 3 Let α > 0. For any Nash equilibrium N, the price of anarchy is bounded by Proof Fix an arbitrary v0  V, such that v0 built only tree edges in T(v0). For any vertex v  V, let Ev be the number of tree edges built by v in T(v0). for any v  V, v  v0, Cost(v) ≤ α(Ev + 1) + Dist(v0) + n − 1  Cost(N) ≤ 2α(n − 1) + nDist(v0) + (n − 1) 2. Need to analyze Dist(v0) …. More details … Improved Upper Bound for any α (α < 12n log n) [AEEMR] (1)

U C B E R K E L E Y C O M P U T E R S C I E N C E 34 Improved Upper Bound for any α (α < 12n log n) [AEEMR] (2) For α<1, Dist(v0)≤n-1 (complete graph) Cost(N) ≤ 2α(n−1)+2n(n−1), Cost(OPT) ≥ α(n−1)+n(n−1)  p.o.a ≤ 2 For α > n 2, Dist(v0)≤(n-1) 2 Cost(N) ≤ 2α(n−1)+2n(n−1 )2, Cost(OPT) > α(n−1)> n 2 (n−1)  p.o.a ≤ 4

U C B E R K E L E Y C O M P U T E R S C I E N C E 35 Improved Upper Bound for any α (α < 12n log n) [AEEMR] (3) For 1≤α≤ n 2 Cost(OPT)> α(n−1)+ 2(n-1) 2 > α(n−1)+ n 2 (star), for n ≥ 2 players For d ≤ 9, Dist(v0)≤9n, Cost(N) ≤ 2α(n−1)+10n 2 For d ≥ 10,Dist(v0)≤(n-1)15α/n c ≤15αn 1-c ), Cost(N) ≤ 2α(n−1)+ 15αn 2-c +n 2, For α≤ n, n c =(αn) 1/3 For α> n, n c= (αn) 1/3, [α(n-1)+n 2 >αn because α ≤ n 2 ]

U C B E R K E L E Y C O M P U T E R S C I E N C E 36 Theorem 4 In any Nash equilibrium N, the total cost incurred by the players in building edges is bounded by twice the cost of the social optimum. There exists a shortest path tree such that, for any player v, the number of non-tree edges built by v is bounded by 1 + ⌊ (n − 1)/α ⌋.   The only critical part in bounding the P.O.A is the sum of the shortest path distances between players. Improved Upper Bound any α (α < 12n log n) [AEEMR] (4)

U C B E R K E L E Y C O M P U T E R S C I E N C E 37 A Lower Bound for  =(d-1)n [FLMPS] An outward-directed complete k-ary tree of depth d, at  =(d-1)n: For large d, k, the price of anarchy approaches 3 asymptotically, so 3-  is a lower bound for any  >0

U C B E R K E L E Y C O M P U T E R S C I E N C E 38 Trees [FLMPS] Conjecture: for  >  0, some constant, all Nash equilibria are trees Benefit: a tree has a center (a node that, when removed, yields no components with more than n/2 nodes) Given a tree N.E., can use the fact that no additional nodes want to link to center to bound the depth and show that the price of anarchy is at most 5

U C B E R K E L E Y C O M P U T E R S C I E N C E 39 Disproving the Tree Conjecture [AEEMR] (1) Family of graphs construction that form strong Nash equilibria and have induced cycles of length three and five. To construct these graphs, we have to define affine planes

U C B E R K E L E Y C O M P U T E R S C I E N C E 40 Disproving the Tree Conjecture [AEEMR] (2) Definition An affine plane is a pair (A,L) A is a set (of points) L is a family of subsets of A (of lines) satisfying the following four conditions. For any two points, there is a unique line containing these points. Each line contains at least two points. Given a point x and a line L that does not contain x, there is a unique line L′ that contains x and is disjoint (parallel) from L (x  L). There exists a triangle, i.e. there are three distinct points which do not lie on a line. If A is finite, then the affine plane is called finite. Equivalence relation on the lines by parallelism L ’ s equivalence class [L].

U C B E R K E L E Y C O M P U T E R S C I E N C E 41 Disproving the Tree Conjecture [AEEMR] (3) Affine Plane definition Set Where field F=GF(q) (q prime) Set AG(2, q): affine plane of order q. The plane contains: q 2 points q*(q+1) lines q+1 equivalence classes Each has q lines Each such line has q points

U C B E R K E L E Y C O M P U T E R S C I E N C E 42 Disproving the Tree Conjecture [AEEMR] (4) Graphs represnet Strong Nash Equilibria G=(V,E) construction Set of vertices V=A  L : 2q 2 +q+1 players The edge set E : A point and a line are connected by an edge  the line contains the point. Two lines are connected by an edge  they are parallel : complete subgraph K q No two points are connected by an edge.

U C B E R K E L E Y C O M P U T E R S C I E N C E 43 Disproving the Tree Conjecture [AEEMR] (5) Line L (representative from eq. class i  q) builds edges to points x 1,x 2 such that x 1  L  L q i and x 2  L  L q i+1 To L 1 -L r (parallel to L) : same equivalence class Points x 3 -x q  L builds edges to L to other lines containing x (from different equivalence class) r(L) indegree,s(L) outdegree of L in K q Line L q (from eq. class q) does not build edges If L  [L q ] then the cost of the player representing L is (2+s)α+(2q−1)+2(2q−1)q = (s+2)α+4q2−1, s=s(L)=q−1−r. If L  [L q ], then the cost is sα + 4q 2 − 1.

U C B E R K E L E Y C O M P U T E R S C I E N C E 44 Disproving the Tree Conjecture [AEEMR] (6) Point X builds edges to all lines containing it (different equivalence class) The cost of the player representing x is (q−1)α +(q+1)+2(q + 1)(2(q − 1)) = (q − 1)α + 4q 2 +q−3.

U C B E R K E L E Y C O M P U T E R S C I E N C E 45 Disproving the Tree Conjecture [AEEMR] (7) Lemma 1 Let q>10. For α in the range 1<α<q+1, no player associated with a line L has a different strategy that achieves a cost ≤ L ’ s original one. For α in the range 1≤ α≤q+1, L has no strategy with a smaller cost. Proof Fix a line L  [L q ]. Consider all possible strategy changes. L builds l>s+2 edges, at best there are l−s−2+2q−1 vertices at distance 1 while the other vertices are at distance 2 from L. In L ’ s original strategy there are 2q−1 vertices at distance 1 while all other vertices are at distance 2.  L ’ s original strategy has a cost at least α(l−s−2)−(l−s−2) 1.  Thus buying more than s + 2 edges does not pay off. L builds at most s + 2 edges and show it does not pay off. There is an edge building cost of α while the shortest path distance costdecreases by at least q + 1. If α<q+1, there is a net cost saving and S is worse than L ’ s original strategy given by G. If α = q + 1, then L ’ s original strategy is at least as good

U C B E R K E L E Y C O M P U T E R S C I E N C E 46 Disproving the Tree Conjecture [AEEMR] (7) Lemma 2 For α in the range 1 < α ≤ q + 1, no player associated with a point x has a different strategy that achieves a cost equal to or smaller than that of x ’ s original strategy. For α = 1, no player associated with a point has a strategy that achieves a smaller cost.  Theorem 1 Let q > 10. The graph G is a strong Nash equilibrium, for 1 < α < q + 1, and a Nash equilibrium, for 1 ≤ α ≤ q + 1.

U C B E R K E L E Y C O M P U T E R S C I E N C E 47 A Weighted Network Creation Game [AEEMR] (1) Most agents don ’ t care to connect closely to everyone else What if we know the amount of traffic between each pair of nodes and weight the distance terms accordingly? If n 2 parameters is too much, what about restricted traffic matrices?

U C B E R K E L E Y C O M P U T E R S C I E N C E 48 A Weighted Network Creation Game [AEEMR] (2) Assume at least n ≥ 2 players. Player u sends a traffic amount of w uv >0 to player v, with u  v. W = (w uv ) u,v is nxn traffic matrix. Cost of player u: w min =min u  v w uv smallest traffic entry w max =max u  v w uv largest traffic entry. The sum of the traffic values

U C B E R K E L E Y C O M P U T E R S C I E N C E 49 A Weighted Network Creation Game [AEEMR] (3) Theorem 8 (generalization of Theorem 3, up to constant factors, where w min = 1 a) For 0<α≤ w min n 2 and any Nash equilibrium N, the price of anarchy is bounded by b) For w min n 2 <α<w max n 2, the price of anarchy is bounded by c) For w max n 2 ≤ α. Then the price of anarchy is bounded by 4.

U C B E R K E L E Y C O M P U T E R S C I E N C E 50 Cost Sharing [AEEMR] (1) What if agents collaborate to create a link? Each node can pay for a fraction of a link; Link exists only if total “ investment ” is  α May yield a wider variety of equilibria

U C B E R K E L E Y C O M P U T E R S C I E N C E 51 Cost Sharing [AEEMR] (2) Theorem 9 n the unweighted scenario the bounds of Theorem 3 hold. In the weighted scenario the bound of Theorem 8 hold. Theorem 10 For n > 6 and α in the range 16n 2 +n<α<12n 2 −n, there exist strong Nash equilibria with n players that contain cycle an in which the cost is split evenly among players.

U C B E R K E L E Y C O M P U T E R S C I E N C E 52 Discussion The price tag of decentralization in network design appears modest not directly dependent on the size of the network being built The Internet is not strictly a clique, or a star, or a tree, but often resembles one of these at any given scale Many possible extensions remain to be explored

U C B E R K E L E Y C O M P U T E R S C I E N C E 53 Directions for Future Work The network is dynamic : Introduce time? Network develops in stages as new nodes arrive Assume equilibrium state is reached at every stage Other points on the spectrum between dictatorship and anarchy? Agreements,sync Measurements to assess applicability to existing real systems

U C B E R K E L E Y C O M P U T E R S C I E N C E 54 Directions for Future Work Passenger Fee: Edge users will pay for the transport, (SLA … ) Stars are efficient for hop distances, but problematic for congestion What happens when agent costs are penalized for easily-congestible networks? QoS support

U C B E R K E L E Y C O M P U T E R S C I E N C E 55 AS Graph by the DIMES project A node is linked with higher probability to a node that already has a large number of links.

Q u e s t i o n s ?