The Price of Anarchy in Network Creation Games

14 Pages·2009·0.25 MB·English

The Price of Anarchy in Network Creation Games
ERIK D. DEMAINE
MIT, Cambridge, Massachusetts
MOHAMMADTAGHI HAJIAGHAYI1
AT&T Labs — Research, Florham Park, New Jersey
HAMID MAHINI
Sharif University of Technology, Tehran, Iran
MORTEZA ZADIMOGHADDAM2
MIT, Cambridge, Massachusetts
1 Introduction
Network design is a fundamental family of problems at the intersection between
computerscienceandoperationsresearch,amplifiedinimportancebythesustained
growthofcomputernetworkssuchastheInternet. Traditionally,thegoalistofind
a minimum-cost (sub) network that satisfies some specified property such as k-
connectivity or connectivity on terminals (as in the classic Steiner tree problem).
Suchaformulationcapturesthe(possiblyincremental)creationcostofthenetwork,
but does not incorporate the cost of actually using the network. By contrast,
network routing has the goal of optimizing the usage cost of the network, but
assumes that the network has already been created.
The network creation game attempts to unify the network design and network
routing problems by modeling both creation and usage costs. Specifically, each
node in the system can create a link (edge) to any other node, at a cost of α. In
addition to these creation costs, each node incurs a usage cost related to the dis-
tances to the other nodes. In the model introduced by Fabrikant, Luthra, Maneva,
Papadimitriou, and Shenker [2003] at PODC 2003, the usage cost incurred by a
node is the sum of distances to all other nodes. Equivalently, if we divide the cost
ApreliminaryversionofthispaperappearedintheProceedingsofthe26thAnnualACMSIGACT-
SIGOPS Symposium on Principles of Distributed Computing (PODC2007),August2007,pages
292–298. This work was done while M. Hajiaghayi was at MIT and CMU and M. Zadimoghad-
dam was at Sharif University of Technology. H. Mahini was supported in part by Institute for
TheoreticalPhysicsandMathematics(IPM)undergrantnumberCS1384-2-01.
Author’s addresses: E. Demaine, M. Zadimoghaddam, Computer Science and Artifi-
cial Intelligence Laboratory, MIT, 32 Vassar St., Cambridge, MA 02139, USA, e-mail:
{edemaine,morteza}@mit.edu; M. Hajiaghayi, AT&T Labs — Research, 180 Park Ave.,
Florham Park, NJ 07932, USA, e-mail: [email protected]; H. Mahini, Department
of Computer Engineering, Sharif University of Technology, Azadi St., Tehran, Iran, e-mail:
[email protected].
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2 DEMAINE, HAJIAGHAYI, MAHINI, ZADIMOGHADDAM
(and thus α) by the number n of nodes, the usage cost is the average distance to
other nodes. In another natural model, the usage cost incurred by a node is the
maximum distance to all other nodes: this model captures the worst-case instead
of average-case behavior of routing.
To model the dominant behavior of large-scale networking scenarios such as the
Internet,weconsidereachnodetobeanagent(player)[Jackson2003]thatselfishly
triestominimizeitsowncreationandusagecost[Fabrikantetal.2003;Albersetal.
2006; Corbo and Parkes 2005]. In this context, the price of anarchy [Koutsoupias
and Papadimitriou 1999; Papadimitriou 2001; Roughgarden 2002b] is the worst
possible ratio of the total cost found by some independent selfish behavior and
theoptimaltotalcostpossiblebyacentralized, socialwelfaremaximizingsolution.
Thepriceofanarchyisawell-studiedconceptinalgorithmicgametheoryforprob-
lems such as load balancing, routing, and network design; see, e.g., Papadimitriou
[2001; Czumaj and V¨ocking [2002; Roughgarden [2002a; Fabrikant et al. [2003;
Anshelevich et al. [2003; Anshelevich et al. [2004; Chun et al. [2004; Corbo and
Parkes [2005; Albers et al. [2006].
Previous work. Three papers consider the price of anarchy in the sum network
√
creation game. Fabrikant et al. [2003] prove an upper bound of O( α) on the
price of anarchy for all α. Lin [2003] prove that the price of anarchy is constant
√
for two ranges of α: α = O( n) and α ≥ cn3/2 for some c > 0. Independently,
√
Albers et al. [2006] prove that the price of anarchy is constant for α = O( n),
as well as for the larger range α ≥ 12n(cid:100)lgn(cid:101).3 In addition, Albers et al. prove
(cid:16) (cid:17)
a general upper bound of 15 1+(min{α2,n2})1/3 . The latter bound shows the
n α
first sublinear worst-case bound, O(n1/3), for all α. But no better bound is known
√
for α between ω( n) and o(nlgn). Yet α ≈ n is perhaps the most interesting
range, for it corresponds to considering the average distance (instead of the sum
of distances) to other nodes to be roughly on par with link creation (effectively
dividing α by n).
The bilateral variation on the network creation game, considered by Corbo and
Parkes[2005]inPODC2005, requiresbothnodestoagreebeforetheycancreatea
link between them. In the sum bilateral network creation game, Corbo and Parkes
√
provethatthepriceofanarchyisbetweenΩ(lgα)andO( α)forα≤n. Although
notclaimedintheirpaper,theproofoftheirupperboundalsoestablishesanupper
√
bound of O(n/ α) for α≥n.
Ourresults. Inthesum(unilateral)networkcreationgameofFabrikantetal.[2003],
√
weprovethefirsto(nε)upperboundforgeneralα,namely2O( lgn). Wealsoprove
aconstantupperboundforα=O(n1−ε)foranyfixedε>0,substantiallyreducing
the range of α for which constant bounds have not been obtained. Along the way,
we also improve the constant upper bound by Albers et al. [2006] (with the lead
constant of 15) to 6 for α<(n/2)1/2 and to 4 for α<(n/2)1/3.
Inthemax(unilateral)version,weprovethatthepriceofanarchyisatmost2for
√ √ √
α≥n, O(min{4 lgn,(n/α)1/3}) for 2 lgn≤α≤n, and O(n2/α) for α<2 lgn.
Our primary proof technique can be most closely viewed as a kind of “region
3Asusual,lgndenoteslog n.
2
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The Price of Anarchy in Network Creation Games 3
α= 0 1 2 2√lgnp3 n/2 pn/2 O(n1−ε) n 12n(cid:100)lgn(cid:101) n2
√
Sumunilateral 1 4 ∗ ≤4(§4) ≤6(§4) Θ(1)(§5) 2O( lgn) (§6) Θ(1)†
3 √
Maxunilateral O(n2/α)(§7) O(min{4 lgn,(n/α)1/3})(§7) ≤2(§7)
√
Sumbilateral 1 Θ( α)(O in‡,Ωin§8) Θ(√n )(O in¶,Ωin§8)
α
Maxbilateral Θ( n )(§9) ≤2(§9)
α+1
∗[Fabrikantetal.2003]
†[Albersetal.2006]
‡[CorboandParkes2005]
√
¶[Corbo and Parkes 2005, Proposition 4] claims O( α) for α < n2 but in fact proves
√ √
O(min{ α,n/ α})
Table I. Summary of best known results for all three models of network creation
games and for all ranges of α.
growing” from approximation algorithms; see, e.g., Leighton and Rao [1999].
Inthe(sum)bilateralnetworkcreationgameofCorboandParkes[2005],weshow
√ √
that their O(min{ α,n/ α}) upper bound is tight by proving a matching lower
√ √
bound of Ω(min{ α,n/ α}). In the max (bilateral) version, we prove matching
upper and lower bounds of Θ( n ) for α≤n, and an upper bound of 2 for α>n.
α+1
Table I summarizes the best bounds known for each of the three models and for
all ranges of α. Interestingly, these results reveal that the price of anarchy can
depend significantly on α.
2 Models
Formally, we define four games depending on the objective (sum or max) and the
consent (unilateral or bilateral). In all versions, we have n players; call them
1,2,...,n. The strategy of player i is specified by a subset s of {1,2,...,n}−{i},
i
which corresponds to the set of neighbors to which player i forms a link. Together,
let s=(cid:104)s ,s ,...,s (cid:105) denote the strategies of all players.
1 2 n
Todefinethecostofastrategy,weintroduceanundirectedgraphG withvertex
s
set {1,2,...,n}. In the unilateral game, G has an edge {i,j} if either i ∈ s or
s j
j ∈s . Inthebilateralgame,G hasanedge{i,j}ifbothi∈s andj ∈s . Define
i s j i
d (i,j)tobethedistance(thenumberofedgesinashortestpath)betweenvertices
s
i and j in graph G . In the sum game, the cost incurred by player i is
s
n
(cid:88)
c (s)=α|s |+ d (i,j),
i i s
j=1
and in the max game, the cost incurred by player i is
n
c (s)=α|s |+maxd (i,j).
i i s
j=1
In both cases, the total cost incurred by strategy s is c(s)=(cid:80)n c (s).
i=1 i
Intheunilateralgame,a(pure)Nashequilibrium isastrategyssuchthatc (s)≤
i
c (s(cid:48))forallstrategiess(cid:48)thatdifferfromsinonlyoneplayeri. Thepriceofanarchy
i
is then the maximum cost of a Nash equilibrium divided by the minimum cost of
any strategy (called the social optimum).
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4 DEMAINE, HAJIAGHAYI, MAHINI, ZADIMOGHADDAM
In the bilateral game, Nash equilibria are not so interesting because the game
requirescoalitionbetweentwoplayerstocreateanedge(ingeneral). Forexample,if
everyplayerichoosestheemptystrategys =∅,thenweobtainaNashequilibrium
i
inducing an empty graph G , which has an infinite cost c(s). To address this issue,
s
CorboandParkes[2005]usethenotionofpairwise stability [JacksonandWolinsky
1996]: astrategyispairwisestableif(1)foranyedge{i,j}ofG ,bothc (s)≤c (s(cid:48))
s i i
and c (s)≤c (s(cid:48)) where s(cid:48) differs from s only in deleting edge {i,j} from G ; and
j j s
(2) for any nonedge {i,j} of G , either c (s) < c (s(cid:48)) or c (s) < c (s(cid:48)) where s(cid:48)
s i i j j
differs from s only in adding edge {i,j} to G .4 The price of anarchy is then the
s(cid:48)
maximum cost of a pairwise-stable strategy divided by the social optimum (the
minimum cost of any strategy).
We spend the bulk of this paper (Sections 3–6) on the original version of the
network creation game, sum unilateral. Then we consider the max unilateral game
in Section 7, the sum bilateral game in Section 8, and the max bilateral game in
Section 9. Thus we default to the unilateral game, often omitting the term, and
within the unilateral game, we default to the sum version.
3 Preliminaries
In this section, we define some helpful notation and prove some basic results. Call
a graph G corresponding to a Nash equilibrium s an equilibrium graph. In such
s
a graph, let d (u,v) be the length of the shortest path from u to v. Let N (u)
s k
denote the set of vertices with distance at most k from vertex u, and let N =
k
min |N (v)|.
v∈G k
We start with a lemma proved in Albers et al. [2006] about the sum unilateral
game:
Lemma 1. [Albers et al. 2006, proof of Theorem 3.2] For any Nash equilibrium
s and any vertex v in G , the cost c(s) is at most 2α(n−1)+nDist(v )+(n−1)2
0 s 0
(cid:80)
where Dist(v )= d (v ,v).
0 v∈V(Gs) s 0
Now we use this lemma to relate the price of anarchy to depth in a breadth-
first search (BFS) tree, a connection also used in Albers et al. [2006]. We assume
henceforththatα≥2,becauseotherwisethepriceofanarchyforthesumunilateral
game is already known from Fabrikant et al. [2003]. In this case, it is known that
the social optimum is attained by a star graph [Fabrikant et al. 2003].
Lemma 2. If the BFS tree in a equilibrium graph G rooted at vertex u has
s
depth d, then price of anarchy is at most d+1.
Proof: ByLemma1andbecausethecostofthesocialoptimum(astar)isatleast
α(n−1)+n(n−1),thepriceofanarchyisatmost 2α(n−1)+nDist(u)+(n−1)2. Because
α(n−1)+n(n−1)
the height of the BFS tree rooted at u is d, we have Dist(u) ≤ (n−1)d. Hence
the price of anarchy is at most 2α(n−1)+n(n−1)d+(n−1)2 < 2α(n−1)+n(n−1)(d+1) ≤
α(n−1)+n(n−1) α(n−1)+n(n−1)
max{2α(n−1),n(n−1)(d+1)}=max{2,d+1}=d+1. (cid:50)
α(n−1) n(n−1)
4Tocorrectlyhandlethecaseofdisconnectedgraphs,wemustdefinec(s)toconsistoftwoparts,
afinitepartandaninfinitepart,andobservethataddinganedgetoadisconnectedgraphreduces
theinfinitepart.
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The Price of Anarchy in Network Creation Games 5
Lemma 3. For any equilibrium graph G , |N (u)| > n/(2α) for every vertex u
s 2
and α≥1.
Proof: We can assume that the number vertices with distance more than 2 from
u is at least n/2; otherwise |N (u)| > n/2 ≥ n/(2α). Let S be the vertices whose
2
distance is exactly 2 from u. For each vertex v with d (u,v)≥2, we pick any one
s
of its shortest paths to u and assign v to the only vertex in this path that is in S.
Thenumberofverticesassignedtoavertexw ∈S canbenomorethanα, because
otherwise vertex u could buy edge {u,w} and decrease its distance to each vertex
assigned to w by at least 1. There is one assignment for each vertex with distance
at least 2 from u, so the total number of assignments to vertices in S is more than
n/2. Therefore, |S|>(n/2)/α=n/(2α). (cid:50)
√
4 Improved Constant Upper Bounds for α=O( n)
In this section, we use the basic results from the previous section to improve the
√
O(1) bounds for the sum unilateral game with α = O( n) obtained by Albers et
al. [2006]. Recall that their bound has a lead constant of (and is thus at least) 15.
We prove an upper bound of 6 for α<(n/2)1/2 and 4 for α<(n/2)1/3.
Theorem 4. For α<(n/2)1/2, the price of anarchy is at most 6.
Proof: We prove that the height of the BFS tree rooted at an arbitrary vertex
u in an equilibrium graph is at most 5. Suppose for contradiction that there is
a vertex v at distance at least 6 from u. Vertex v can buy {u,v} to decrease its
distance from all vertices in N (u) by at least 1. Because vertex v has not bought
2
the edge {u,v}, we conclude that |N (u)| < α. By Lemma 3, |N (u)| > n/(2α).
2 2
Hence α > n/(2α), which contradicts the hypothesis of the lemma. Therefore the
height of the BFS tree rooted at u is at most 5. By Lemma 2, the price of anarchy
is at most 5+1=6. (cid:50)
Theorem 5. For α<(n/2)1/3, the price of anarchy is at most 4.
Proof: We prove that there is a choice of root vertex v in an equilibrium graph
such that the height of the BFS tree is at most 3. Let ∆ be the maximum degree
of a vertex in the graph and let u be an arbitrary vertex. Certainly |N (u)| ≤
2
1+∆+∆(∆−1) = 1+∆2. On the other hand, by Lemma 3, |N (u)| > n/(2α).
2
Bythehypothesisofthelemma,1+∆2 >n/(2α)>α2. Hence∆>α−1. Letv be
a vertex with degree equal to ∆. Suppose for contradiction that there is vertex u
at distance at least 4 from v. Vertex u can buy edge {u,v} to decrease its distance
to all vertices in N (v) by at least 1 and thus decrease its total cost by at least
1
|N (v)| = ∆+1 > α, contradicting equilibrium. Therefore the height of the BFS
1
treerootedatvisatmost3. ByLemma2,thepriceofanarchyisatmost3+1=4.
(cid:50)
5 Constant Upper Bound for α=O(n1−ε)
In this section, we extend the range for which we know a constant upper bound on
the price of anarchy in the sum unilateral game. Specifically, for 1≤α<n1−ε and
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6 DEMAINE, HAJIAGHAYI, MAHINI, ZADIMOGHADDAM
Fig.1. ProofofLemma9.
ε≥1/lgn, we prove an upper bound of 2O(1/ε). We start with some lemmas that
will be useful in this section and the next.
Lemma 6. For any vertex u in an equilibrium graph G , if |N (u)|>n/2, then
s k
|N (u)|≥n.
2k+2α/n
Proof: We prove the contrapositive. Suppose |N (u)|<n. Then there is a
2k+2α/n
vertexvwithd (u,v)≥2k+1+2α/n. Foreveryvertexx∈N (u),d (u,x)≤k. By
s k s
thetriangleinequalityd (u,x)+d (x,v)≥d (u,v),wehaved (x,v)≥k+1+2α/n.
s s s s
Ifvertexvboughttheedge{v,u},thenthedistancebetweenvandxwoulddecrease
by at least 2α/n, so Dist(v) would decrease by at least N (u)·2α/n. Because v
k
has not bought the edge {v,u}, we have α≥|N (u)|·2α/n, i.e., |N (u)|≤n/2. (cid:50)
k k
By setting α = n/2 and α = 12nlgn, we conclude the following corollaries,
which we will use in Theorems 10 and 12.
Corollary 7. For any vertex u in an equilibrium graph G with α < n/2, if
s
|N (u)|>n/2, then |N (u)|=n.
k 2k+1
Corollary 8. For any vertex u in an equilibrium graph G with α<12nlgn,
s
if |N (u)|>n/2, then |N (u)|=n.
k 2k+24lgn
Lemma 9. If |N (u)| > Y for every vertex u in an equilibrium graph G , then
k s
either|N (u)|>n/2forsomevertexuor|N (u)|>Yn/αforeveryvertexu.
2k+3 3k+3
Proof: If there is a vertex u with |N (u)| > n/2, then the claim is obvious.
2k+3
Otherwise,foreveryvertexu,|N (u)|≤n/2. Letubeanarbitraryvertex;refer
2k+3
to Figure 1. Let S be the set of vertices whose distance from u is exactly 2k+3.
We select a subset of S, called center points, by the following greedy algorithm.
First we unmark all vertices in S. Then we repeatedly select an unmarked vertex
x∈S as a center point, mark all unmarked vertices in S whose distance from x in
G is at most 2k, and assign these vertices to x.
s
Suppose that we select l vertices x ,x ,...,x as center points. We prove that
1 2 l
l≥n/α. Let C be the vertices in S assigned to x . By construction, S =(cid:83)l C .
i i i=1 i
We also assign each vertex v at distance at least 2k +3 from u to one of these
center points, as follows. Pick any one shortest path from v to u, which contains
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The Price of Anarchy in Network Creation Games 7
exactly one vertex w ∈ S, and assign v to the same center point as w. Let T be
i
the set of vertices assigned to x and whose distance from u is more than 2k+3.
i
By construction, (cid:83)l T is the set of vertices at distance more than 2k+3 from u.
i=1 i
A shortest path from v ∈ T to u uses some vertex w ∈ C . If u bought the edge
i i
{u,x }, then the distance between u and w would become at most 2k+1. Because
i
d (u,w) = 2k +3 in the current graph, buying edge {u,x } would decrease u’s
s i
distance to v by at least 2k +3−(1+2k) = 2. Because u has not bought the
edge {u,x }, we conclude that α≥2|T |. On the other hand, |N (u)|≤n/2, so
i i 2k+3
(cid:80)l |T |≥n/2. Therefore, lα≥2(cid:80)l |T |≥n and hence l≥n/α.
i=1 i i=1 i
Accordingtothegreedyalgorithm,thedistancebetweenanypairofcenterpoints
is more than 2k; hence, N (x )∩N (x ) = ∅ for i (cid:54)= j. By the hypothesis of the
k i k j
lemma, |N (x )| > Y for every vertex x ; hence |(cid:83)l N (x )| = (cid:80)l |N (x )| >
k i i i=1 k i i=1 k i
lY. For every 1 ≤ i ≤ l, we have d (u,x ) = 2k +3, so vertex u has a path of
s i
length at most 3k+3 to every vertex whose distance to x is at most k. Therefore,
i
|N (u)|≥|(cid:83)l N (x )|>lY ≥Yn/α. (cid:50)
3k+3 i=1 k i
Theorem 10. Forε≥1/lgnand1≤α<n1−ε, thepriceofanarchyisatmost
4.667·3(cid:100)1/ε(cid:101)+8.
Proof: Consider an equilibrium graph G . Let X = n/α > nε. Define a = 2
s 1
and a = 3a + 3 for all i > 1. By Lemma 3, N (u) > n/(2α) = X/2 for
i i−1 2
all u. By Lemma 9, for each i ≥ 1, either N (v) > n/2 for some vertex v or
2ai+3
N ≥(n/α)N =XN . Letj betheleastnumberforwhich|N (v)|>n/2
ai+1 ai ai 2aj+3
for some vertex v. By this definition, for each i < j, N > (n/α)N = XN .
ai+1 ai ai
Because N = N > X/2, we obtain that N > Xi/2 for every i ≤ j. On the
a1 2 ai
other hand, Xj/2 < N ≤ n, so Xj < 2n. Therefore j < 1(1+1/lgn). Because
aj ε
ε ≤ 1/lgn, we must have j < 1 +1. Because j is an integer, j ≤ (cid:100)1/ε(cid:101). So
ε
|N (v)| ≥ |N (v)| > n/2. Because α < n1−ε and ε ≥ 1/lgn, we have
2a(cid:100)1/ε(cid:101)+3 2aj+3
α < n/2, and thus we can use Corollary 7. By Corollary 7, |N (v)| = n.
4a(cid:100)1/ε(cid:101)+7
Hence, the height of the BFS tree rooted at vertex v is at most 4a +7. By
(cid:100)1/ε(cid:101)
Lemma2,thepriceofanarchyisatmost4a +8. Solvingtherecurrencerelation
(cid:100)1/ε(cid:101)
a = 3a +3 with a = 2, we obtain that a = 73i − 3 < 73i. Therefore the
i i−1 1 i 6 2 6
price of anarchy is at most 473(cid:100)1/ε(cid:101)+8≤4.667·3(cid:100)1/ε(cid:101)+8. (cid:50)
6
6 o(nε) Upper Bound for α<12nlgn
In this section, we prove the first o(nε) bound for the sum unilateral game with
√
α between Ω(n) and o(nlgn). Specifically, we show an upper bound of 2O( lgn).
First we need the following lemma.
Lemma 11. If |N (u)|>Y for every vertex u in an equilibrium graph G , then
k s
either |N (u)| > n/2 for some vertex u or |N (u)| > Ykn/α for every
4k+1 5k+1
vertex u.
Proof: The proof is similar to the proof of Lemma 9. If there is a vertex u
with |N (u)| > n/2, then the claim is obvious. Otherwise, for every vertex u,
4k+1
|N (u)|≤n/2. Let u be an arbitrary vertex. Let S be the set of vertices whose
4k+1
distance from u is 4k +1. We select a subset of S, called center points, by the
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8 DEMAINE, HAJIAGHAYI, MAHINI, ZADIMOGHADDAM
following greedy algorithm. First we unmark all vertices in S. Then we repeatedly
select an unmarked vertex x ∈ S as a center point, mark all unmarked vertices in
S whose distance from x is at most 2k, and assign these vertices to x.
Suppose that we select l vertices x ,x ,...,x as center points. We prove that
1 2 l
l≥kn/α. LetC betheverticesinS assignedtox . Byconstruction,S =(cid:83)l C .
i i i=1 i
We also assign each vertex v with distance at least 4k+1 from u to one of these
center points. Pick any one shortest path from v to u, which contains exactly
one vertex w ∈ S, and assign v to the same center point as w. Let T be the
i
set of vertices assigned to x and whose distance from u is more than 4k+1. By
i
construction, (cid:83)l T is the set of vertices at distance more than 4k +1 from u.
i=1 i
The shortest path from v ∈T to u uses some vertex w ∈C . If u bought the edge
i i
{u,x }, then the distance between u and w would become at most 2k+1. Because
i
d (u,w)=4k+1, buying edge {u,x } would decrease u’s distance to v by at least
s i
4k+1−(2k+1)=2k. Becauseuhasnotboughttheedge{u,x },weconcludethat
i
α≥2k|T |. On the other hand, |N (u)|≤n/2 and (cid:80)l |T |≥n/2. Therefore,
i 4k+1 i=1 i
lα≥2k(cid:80)l |T |≥kn and hence l≥kn/α.
i=1 i
Accordingtothegreedyalgorithm,thedistancebetweenanypairofcenterpoints
is more than 2k; hence, N (x )∩N (x ) = ∅ for i (cid:54)= j. By the hypothesis of the
k i k j
lemma, |N (x )| > Y for every vertex x ; hence |(cid:83)l N (x )| = (cid:80)l |N (x )| >
k i i i=1 k i i=1 k i
lY. For every i ≤ l, we have d (u,x ) = 4k+1, so vertex u has a path of length
s i
at most 5k + 1 to every vertex whose distance to x is at most k. Therefore,
i
|N (u)|≥|(cid:83)l N (x )|>lY ≥Ykn/α. (cid:50)
5k+1 i=1 k i
√
Theorem 12. For 1≤α<12nlgn, the price of anarchy is O(5 lgnlgn).
Proof: The proof is similar to the proof of Theorem 10. Let Z = 12lgn. By the
hypothesis of the theorem, α/n < 12lgn = Z. Because any equilibrium graph G
s
is connected, |N |>Z. By Lemma 11, either |N (v)|>n/2 for some vertex v
Z 4k+1
or N ≥(n/α)kN . Define a ,a ,... by the recurrence relation a =5a +1
5k+1 k 0 1 i i−1
with a = Z. By induction, a > Z5i. Suppose j is the least number for which
0 i
|N (v)| > n/2. By this definition, and because n/α > 1/Z, we obtain that
4aj+1
N ≥(n/α)a N >5iN for each i<j. From these inequalities we derive that
ai+1 i ai ai
Naj > 5Pij=−01i. But(cid:112)Naj ≤ n, so (cid:80)ij√=−11i = j(j −1)/2 ≤ log5n. This inequality
implies that j <1+ 2log n<1+ lgn. By Corollary 8, the height of the BFS
5
treerootedatvisatmost2(4a √ +1)+24lgn. Solvingtherecurrencerelation,
1+ lgn
√
we obtain that a =O(5jlgn). By Lemma 2, the price of anarchy is O(5 lgnlgn).
j
(cid:50)
7 Upper Bounds for Max Unilateral Game
In this section, we introduce and analyze the natural max variation on the (uni-
lateral) network creation game. This problem is motivated by players forming a
networkwithguaranteedworst-caseperformancesubjecttobudgetconstraints. We
√
prove that the price of anarchy is at most 2 for α ≥ n, O(min{4 lgn,(n/α)1/3})
√ √
for 2 lgn≤α≤n, and O(n2/α) for α<2 lgn.
Let D be the distance of the farthest vertex from v and N=(v) be the set of
v k
vertices whose distance to v is exactly k. Thus the cost incurred by vertex v is
TransactionsonAlgorithms,Vol.?,No.?,?20?.
·
The Price of Anarchy in Network Creation Games 9
αe +D , where e is the number of edges that v has bought.
v v v
Lemma 13. Any equilibrium graph G has no cycle of length less than α+2.
s
Proof: Suppose for contradiction that there is cycle C with |C| < α+2. Let
{u,v} be an edge of this cycle. Suppose by symmetry that u bought this edge. If
vertex u removed the edge, it would decrease its buying cost e by α and increase
u
its distance cost D by at most |C|−2 < α. Hence it is cost effective for u to
u
remove this edge, contradicting equilibrium. (cid:50)
Theorem 14. For α≥n, the price of anarchy is at most 2.
Proof: By Lemma 13, the equilibrium graph G has no cycle of length at most n,
s
(cid:80)
which implies that G is a tree. The cost of s is at most α(n−1)+ D ≤
v∈V(G) v
α(n−1)+n(n−1)≤2α(n−1). Ontheotherhand,thecostofthesocialoptimum
is at least α(n−1). Therefore the price of anarchy is at most 2α(n−1) =2. (cid:50)
α(n−1)
Lemma 13 gives us a lower bound on the girth (length of the shortest cycle) of
an equilibrium graph. We use the following result of Dutton and Brigham [1991]
to relate the number of edges to the girth:
Lemma 15. [Dutton and Brigham 1991] An n-vertex graph of girth at least g,
where g is odd, has O(n1+2/(g−1)) edges.
Lemma 16. The number of edges in an equilibrium graph is O(n1+2/α).
Proof: By Lemma 13, the girth of the equilibrium graph G is at least α + 2
and hence at least (cid:100)α + 2(cid:101). So by Lemma 15, the number of edges in G is
O(n1+2/((cid:100)α+2(cid:101)−1))foroddvaluesof(cid:100)α(cid:101),andthenumberofedgesisO(n1+2/((cid:100)α+1(cid:101)−1))
for even values of (cid:100)α(cid:101). (cid:50)
Lemma 17. For a vertex v in an equilibrium graph, and for k ≤D /2, we have
v
|N= (v)|≥k/α and |N (v)|>k2/(2α).
k+1 k+1
Proof: If vertex v bought the |N= (v)| edges connecting v to the vertices in
k+1
N= (v), the distance from v to all vertices outside N (v) decreases by k, while
k+1 k
possiblynotchangingthedistancefromv toverticesinN (v). Thusthenewvalue
k
of D would become at most max{k,D −k} = D −k because k ≤ D /2. Thus
v v v v
D would decrease by at least D −(D −k)=k. The cost of buying these edges
v v v
is |N= (v)|α. Because v has not bought these edges, we conclude that this cost is
k+1
at least k. Hence |N= (v)| ≥ k/α. Therefore |N (v)| ≥ (cid:80)k (i/α) > k2/(2α).
k+1 k+1 i=1
(cid:50)
Lemma 18. ThediameterofanequilibriumgraphG ,diam(G ),isO((nα2)1/3).
s s
Proof: Set k = (cid:98)diam(G )/4(cid:99)−1. Let v be some vertex with D = diam(G ).
s v s
Similar to the proof of Lemma 11, we select a subset of vertices as center points
by the following greedy algorithm. First we unmark all vertices in G . Then we
s
repeatedly select an unmarked vertex x as a center point, and mark all unmarked
vertices whose distances are at most 2k from x.
TransactionsonAlgorithms,Vol.?,No.?,?20?.
·
10 DEMAINE, HAJIAGHAYI, MAHINI, ZADIMOGHADDAM
Suppose that we select l vertices x ,x ,...,x as center points. By construction,
1 2 l
every vertex in the graph has distance at most 2k to some center point. If vertex
v bought the l edges {v,x },{v,x },...,{v,x }, it would decrease D by at least
1 2 l v
D −(2k+1)≥(4k+4)−(2k+1)>2k. Buying these edges costs αl. Because v
v
has not bought these edges, we have αl≥2k. On the other hand, according to the
greedy algorithm, the distance between any pair of center points is at least 2k+1.
Thus we have N (x )∩N (x ) = ∅ for 1 ≤ i < j ≤ l. Therefore, |(cid:83)l N (x )| =
k i k j i=1 k i
(cid:80)l |N (x )|,whichshouldbelessthanorequalton. ByLemma17,wehaven≥
i=1 k i
|(cid:83)l N (x )|=(cid:80)l |N (x )|≥l(k−1)2/(2α), which means that l ≤2nα/(k−
i=1 k i i=1 k i
1)2. Combiningwithαl≥2k,weobtainthat2k ≤2nα2/(k−1)2,i.e.,k(k−1)2 ≤
nα2. Therefore diam(G )=O((nα2)1/3). (cid:50)
s
Theorem 19. For 6≤α<n, the price of anarchy is O((n/α)1/3).
Proof: By Lemma 16, the number of edges in G is O(n1+2/α). By Lemma 18,
s
diam(G )=O((nα2)1/3). HencethecostofG isatmostαO(n1+2/α)+nO((nα2)1/3).
s s
On the other hand, the cost of the social optimum is Ω(nα). Therefore the price
of anarchy is O(n2/α+(nα2)1/3/α)=O(n2/α+(n/α)1/3)=O((n/α)1/3) because
α≥6. (cid:50)
Lemma 20. If |N (u)|>Y for every vertex u in an equilibrium graph G , then
k s
either D ≤5k for some vertex u or |N (u)|>Yk/α for every vertex u.
u 4k+1
Proof: The proof is similar to the proof of Lemma 11. If there is a vertex u
with D ≤ 5k, then the claim is obvious. Otherwise, for every vertex u, we have
u
D >5k. Let u be an arbitrary vertex. Let S be the set of vertices whose distance
u
from u is 3k +1. We select a subset of S, called center points, by the following
greedy algorithm. First we unmark all vertices in S. Then we select an unmarked
vertex x ∈ S as a center point, mark all unmarked vertices in S whose distance
from x is at most 2k, and assign these vertices to x.
Supposethatweselectlverticesx ,x ,...,x ascenterpoints. Weprovethatl≥
1 2 l
k/α. If vertex u bought the l edges {u,x },{u,x },...,{u,x }, it would decrease
1 2 l
D by at least D −max{D −k,3k+1}=D −(D −k)=k. Because u has not
u u u u u
bought these edges, we must have lα≥k.
Accordingtothegreedyalgorithm,thedistancebetweenanypairofcenterpoints
is more than 2k; hence N (x )∩N (x ) = ∅ for i (cid:54)= j. By the hypothesis of the
k i k j
lemma, |N (x )| > Y for every vertex x ; hence |(cid:83)l N (x )| = (cid:80)l |N (x )| >
k i i i=1 k i i=1 k i
lY. For every i ≤ l, we have d (u,x ) = 3k + 1, so vertex u has a path of
s i
length at most 4k+1 to every vertex whose distance to x is at most k. Therefore
i
|N (u)|≥|(cid:83)l N (x )|>lY ≥Yk/α. (cid:50)
4k+1 i=1 k i
√
Theorem 21. The price of anarchy is O(4 lgn+n2/α).
Proof: The proof is similar to the proof of Theorem 12. Let Z = α. Consider
an arbitrary vertex v. Because any equilibrium graph is connected, |N | > Z.
Z
By Lemma 20, either D ≤ 5k for some vertex v or N > (k/α)N for every
v 4k+1 k
vertexv. Definethenumbersa ,a ,... usingtherecurrencerelationa =4a +1
0 1 i i−1
witha =Z. Byinduction,a ≥Z4i. Supposethatj istheleastnumberforwhich
0 i
TransactionsonAlgorithms,Vol.?,No.?,?20?.