On the Tree Conjecture for the Network Creation Game
Abstract
Selfish Network Creation focuses on modeling real world networks from a gametheoretic point of view. One of the classic models by Fabrikant et al. [PODC’03] is the network creation game, where agents correspond to nodes in a network which buy incident edges for the price of per edge to minimize their total distance to all other nodes. The model is wellstudied but still has intriguing open problems. The most famous conjectures state that the price of anarchy is constant for all and that for all equilibrium networks are trees.
We introduce a novel technique for analyzing stable networks for high edgeprice and employ it to improve on the best known bounds for both conjectures. In particular we show that for all equilibrium networks must be trees, which implies a constant price of anarchy for this range of . Moreover, we also improve the constant upper bound on the price of anarchy for equilibrium trees.
1 Introduction
Many important networks, e.g. the Internet or social networks, have been created in a decentralized way by selfishly acting agents. Modeling and understanding such networks is an important challenge for researchers in the fields of Computer Science, Network Science, Economics and Social Sciences. A significant part of this research focuses on assessing the impact of the agents’ selfish behavior on the overall network quality measured by the price of anarchy [31]. Clearly, if there is no or little coordination among the egoistic agents, then it cannot be expected that the obtained networks minimize the social cost. The reason for this is that each agent aims to improve the network quality for herself while minimizing the spent cost. However, empirical observations, e.g. the famous smallworld phenomenon [39, 29, 9], suggest that selfishly built networks are indeed very efficient in terms of the overall cost and of the individually perceived service quality. Thus, it is a main challenge to justify these observations analytically.
A very promising approach towards this justification is to model the creation of a network as a strategic game which yields networks as equilibrium outcomes and then to investigate the quality of these networks. For this, a thorough understanding of the structural properties of such equilibrium networks is the key.
We contribute to this endeavor by providing new insights into the structure of equilibrium networks for one of the classical models of selfish network creation [23]. In this model, agents correspond to nodes in a network and can buy costly links to other nodes to minimize their total distance in the created network. Our insights yield improved bounds on the price of anarchy and significant progress towards settling the socalled tree conjecture [23, 35].
1.1 Model and Definitions
We consider the classical network creation game as introduced by Fabrikant et al. [23]. There are agents , which correspond to nodes in a network, who want to create a connected network among themselves. Each agent selfishly strives for minimizing her cost for creating network links while maximizing her own connection quality. All edges in the network are undirected and unweighted and agents can create any incident edge for the price of , where is a fixed parameter of the game. The strategy of an agent denotes which edges are bought by this agent, that is, agent is willing to create (and pay for) all the edges , for all . Let be the dimensional vector of the strategies of all agents. The strategyvector induces an undirected network , where for each edge we have or . If , then we say that agent is the owner of edge or that agent buys the edge , otherwise, if , then agent owns or buys the edge .^{1}^{1}1No edge can have two owners in any equilibrium network. Hence, we will assume throughout the paper that each edge in has a unique owner. Since the created networks will heavily depend on we emphasize this by writing instead of . The cost of an agent in the network is the sum of her cost for buying edges, called the creation cost, and her cost for using the network, called the distance cost, which depends on agent ’s distances to all other nodes within the network. The cost of is defined as
where the distance cost is defined as
Here denotes the length of a shortest path between and in the network . We will mostly omit the reference to the strategy vector, since it is clear that a strategy vector directly induces a network and vice versa. Moreover, if the network is clear from the context, then we will also omit the reference to the network, e.g. writing instead of .
A network is in pure Nash equilibrium (NE), if no agent can unilaterally change her strategy to strictly decrease her cost. That is, in a NE network no agent can profit by a strategy change if all other agents stick to their strategies. Since in a NE network no agent wants to change the network, we call them stable.
The social cost, denoted , of a network is the sum of the cost of all agents, that is, . Let OPT be the minimum social cost of a agent network and let maxNE be the maximum social cost of any NE network on agents. The price of anarchy (PoA) [31] is the maximum over all of the ratio .
Let be any undirected connected graph with vertices. A cutvertex of is a vertex with the property that with vertex removed contains at least two connected components. We say that is biconnected if and contains no cutvertex. A biconnected component of is a maximal induced subgraph of which is also biconnected. Note that we rule out trivial biconnected components which contain exactly one edge. Thus, there exist at least two vertexdisjoint paths between any pair of vertices in a biconnected component , which implies that there exists a simple cycle containing and .
1.2 Related Work
Network creation games, as defined above, were introduced by Fabrikant et al. [23]. They gave the first general bound of on the PoA and they conjectured that above some constant edgeprice all NE networks are trees. This conjecture, called the tree conjecture, is especially interesting since they also showed that tree networks in NE have constant PoA. In particular, they proved that the PoA of stable tree networks is at most . Interestingly, the tree conjecture in its general version was later disproved by Albers et al. [1]. However, nontree NE networks are known only when , in particular, for every , there exist nontree NE networks with [35]. It is believed that for the tree conjecture may be true. Settling this claim is currently a major open problem and there has been a series of papers which improved bounds concerning the tree conjecture.
First, Albers et al. [1] proved that for every NE network is a tree. Then, using a technique based on the average degree of the biconnected component, this was significantly improved to by Mihalák & Schlegel [38] and even further to by Mamageishvili et al. [35]. The main idea of this average degree technique is to prove a lower and an upper bound on the average degree of the unique biconnected component in any equilibrium network. The lower bound has the form "for the average degree is at least " and the upper bound has the form "for the average degree is at most ", where are constants and is a function which monotonically decreases in . For large enough both bounds contradict each other, which proves that equilibrium networks for this cannot have a biconnected component and thus must be trees. Very recently a preprint by Àlvarez & Messegué [5] was announced which invokes the average degree technique with a stronger lower bound. This then yields a contradiction already for . For their stronger lower bound the authors use that in every minimal cycle (we call them "min cycles") of an equilibrium network all agents in the cycle buy exactly one edge of the cycle. This fact has been independently established by us [34] and we also use it.
The currently best general upper bound of on the PoA is due to Demaine et al. [21] and it is known that the PoA is constant if for any fixed [21]. Thus, the PoA was shown to be constant for almost all , except for the range between , for any fixed , and (or which is claimed in [5]). It is widely conjectured that the PoA is constant for all and settling this open question is a long standing problem in the field. A constant PoA proves that agents create socially closetooptimal networks even without central coordination. Quite recently, a constant PoA was proven by Chauhan et al. [15] for a version with nonuniform edge prices. In contrast, nonconstant lower bounds on the PoA have been proven for local versions of the network creation game by Bilò et al. [10, 12] and CordLandwehr & Lenzner [18].
1.3 Our Contribution
In this paper we introduce a new technique for analyzing stable nontree networks for high edgeprice and use it to improve on the current best lower bound for for which all stable networks must be trees. In particular, we prove that for any stable network must be a tree (see Section 2). This is a significant improvement over the known bound of by Mamageishvili et al. [35] and the recently claimed bound of by Àlvarez & Messegué [5]. Since the price of anarchy for stable tree networks is constant [23], our bound directly implies a constant price of anarchy for . Moreover, in Section 3, we also give a refined analysis of the price of anarchy of stable tree networks and thereby improve the best known constant upper bound for stable trees.
Thus, we make significant progress towards settling the tree conjecture in network creation games and we enlarge the range of for which the price of anarchy is provably constant.
Our new technique exploits properties of cycles in stable networks by focusing on critical pairs, strong critical pairs and min cycles. The latter have been introduced in our earlier work [34] and are also used in the preprint by Àlvarez & Messegué [5]. However, in contrast to the last attempts for settling the tree conjecture [38, 35, 5], we do not rely on the average degree technique. Instead we propose a more direct and entirely structural approach using (strong) critical pairs in combination with min cycles. Besides giving better bounds with a simpler technique, we believe that this approach is better suited for finally resolving the tree conjecture.
2 Improving the Range of of the Tree Conjecture
In this section we prove our main result, that is, we show that for , every NE network with nodes must be a tree.
We proceed by first establishing properties of cycles in stable networks. Then we introduce the key concepts called critical pairs, strong critical pairs and min cycles. Finally, we provide the last ingredient, which is a critical pair with a specific additional property, and combine all ingredients to obtain the claimed result.
2.1 Properties of Cycles in Stable Networks
We begin by showing that for large values of , stable networks cannot contain cycles of length either 3 or 4.
Lemma 1.
For , no stable network contains a cycle of length 3.
Proof.
Let be a stable network for a fixed value of . For the sake of contradiction, assume that contains a cycle of length . Assume that and that contains the three edges , , and . Let . Observe that, for every we have , as . Furthermore, all the ’s are pairwise disjoint. W.l.o.g., assume that . Furthermore, w.l.o.g., assume that buys the edge . Consider the strategy change in which agent deletes the edge . The building cost of the agent decreases by while her distance cost increases by at most . Since , from we obtain . Since is stable, , i.e., , a contradiction. ∎
Lemma 2.
For , no stable network contains a cycle of length 4.
Proof.
Let be a stable network for a fixed value of . For the sake of contradiction, assume that contains a cycle of length . Assume that and that contains the four edges , , , and . For the rest of this proof, we assume that all indices are modulo 4 in order to simplify notation. Let . Observe that for every we have , as . Let . Observe that in the families of the sets and every pair of sets is pairwise disjoint.
We now rule out the case in which an agent owns two edges of . W.l.o.g., assume that agent owns the two edges and . Consider the strategy change in which agent swaps^{2}^{2}2A swap of edge to edge by agent who owns edge consists of deleting edge and buying edge . the edge with the edge and, at the same time, deletes the edge . The creation cost of agent decreases by , while her distance cost increases by . Since is stable, agent has no incentive in deviating from her current strategy. Therefore, , i.e., , where the last but one inequality follows from the pairwise disjointness of all sets, which implies . Since, , no agent can own two edges of . Therefore, to prove the claim, we need to show that no agent can own a single edge of .
W.l.o.g., assume that for every agent owns the edge . Moreover, w.l.o.g., assume that . Since , we have that .
Consider the strategy change in which agent deletes the edge . The creation cost of agent decreases by , while her distance cost increases by .
Since is stable, agent has no incentive to deviate from her current strategy. Therefore, , i.e., , when . We have obtained a contradiction. ∎
Definition 3 (Directed Cycle).
Let be a cycle of of length . We say that is directed if there is an ordering of its vertices such that, for every , is an edge of which is bought by agent .
We now show that if is large enough, then directed cycles cannot be contained in a stable network as a biconnected component.
Lemma 4.
For , no stable network with vertices contains a biconnected component which is also a directed cycle.
Proof.
Let be a stable network for a fixed value of . Let be a biconnected component of . For the sake of contradiction, assume that is a directed cycle of length . We can apply Lemma 2 to exclude the case in which . Similarly, since for every , we can use Lemma 1 to exclude the case in which .
Let be the vertices of and, w.l.o.g., assume that every agent is buying an edge towards agent . To simplify notation, in the rest of this proof we assume that all indices are modulo . Let . Observe that is a partition of . We divide the proof into two cases.
The first case occurs when is a cycle of length . W.l.o.g., assume that . In this case, consider the strategy change of agent when she swaps the edge with the edge . The distance cost of agent increases by . Thus, agent has an improving strategy, a contradiction.
The second and last case occurs when is a cycle of length . If for some , then there exists an such that . W.l.o.g., let . The distance cost of agent when she swaps the edge with the edge increases by . Thus agent has an improving strategy, a contradiction. If , then the increase in the overall cost incurred by agent when she deletes the edge would be equal to . Since is stable and is a multiple of , , i.e., , for every , a contradiction. ∎
2.2 Critical Pairs
The next definition introduces the concept of a (strong) critical pair. As we will see, (strong) critical pairs are the first key ingredient for our analysis. Essentially, we will show that stable networks cannot have critical pairs, if is large enough.
Definition 5 (Critical Pair).
Let be a nontree network and let be a biconnected component of . We say that is a critical pair if all of the following five properties hold:

Agent buys two distinct nonbridge edges, say and , with ;

Agent , with buys at least one edge with and ;

;

there is a shortest path between and in which uses the edge ;

there is a shortest path between and in which does not use the edge .
The critical pair is strong if there is a shortest path between and which does not use the edge . See Fig. 1 for an illustration.
In the rest of this section, when we say that two vertices and of form a critical pair, we will denote by , and the vertices corresponding to the critical pair that satisfy all the conditions given in Definition 5. We can observe the following.
Observation 6.
If is a critical pair of a network , then there exists a shortest path tree of rooted at , where either the edge is not an edge of or is the parent of in .
Observation 7.
If is a critical pair, then for every shortest path tree of rooted at , either the edge is not an edge of or is the parent of in . Furthermore, if is a strong critical pair, then there is a shortest path tree of rooted at such that the edge is not contained in the shortest path tree.
The next technical lemma provides useful bounds on the distance cost of the nodes involved in a critical pair.
Lemma 8.
Let be a stable network and let be two distinct vertices of such that buys an edge , with . If and there exists a shortest path tree of rooted at such that either is not an edge of or is the parent of in , then . Furthermore, if is buying also the edge , with , , and is not an edge of , then .
Proof.
Consider the strategy change in which agent swaps the edge with the edge and deletes any other edge she owns and which is not contained in , if any. Let be a shortest path tree rooted at of the graph obtained after the swap. Observe that , for every . Furthermore, as , while , we have . Therefore, . Moreover, the distance from to every is at most . Finally, the distance from to herself, which is clearly 0, is exactly 1 less than the distance from to in . Therefore the distance cost of in is less than or equal to .
If besides performing the mentioned swap agent additionally saves at least in cost by deleting at least one additional edge which is not in , then . This is true since is stable, which implies that the overall cost of in cannot be larger than the overall cost of after the strategy change. ∎
Now we employ Lemma 8 to prove the structural property that stable networks cannot contain strong critical pairs if is large enough.
Lemma 9.
For , no stable network contains a strong critical pair.
Proof.
Let be a nontree stable network for a fixed value of and, for the sake of contradiction, let be a strong critical pair. Using Observation 6 together with Lemma 8 (where , and ), we have that
Furthermore, using Observation 7 together with Lemma 8 (where , and ), we have that
By summing up both the lefthand and the righthand side of the two inequalities we obtain , i.e., , a contradiction. ∎
2.3 Min Cycles
We now introduce the second key ingredient for our analysis: min cycles.
Definition 10 (Min Cycle).
Let be a nontree network and let be a cycle in . We say that is a min cycle if, for every two vertices , .
First, we show that every edge of every biconnected graph is contained in some min cycle. This was also proven in [34] and [5].
Lemma 11.
Let be a biconnected graph. Then, for every edge of , there is a min cycle that contains the edge .
Proof.
Since is biconnected, there exists at least a cycle containing the edge . Among all the cycles in that contain the edge , let be a cycle of minimum length. We claim that is a min cycle. For the sake of contradiction, assume that is not a min cycle. This implies that there are two vertices such that . Among all pairs of vertices such that , let be the one that minimizes the value (ties are broken arbitrarily). Let be a shortest path between and in . By the choice of and , is edge disjoint from .
Let and be the two edgedisjoint paths between and in and, w.l.o.g., assume that is contained in . Let and be the length of and , respectively. See Fig. 2. Clearly, the length of is equal to . Since , we obtain . Therefore, the cycle obtained by concatenating and has a length equal to , and therefore, it is strictly shorter than , a contradiction. ∎
Now we proceed with showing that stable networks contain only min cycles which are directed and not too short. For this, we employ our knowledge about strong critical pairs.
Lemma 12.
For , every min cycle of a nontree stable network with vertices is directed and has a length of at least 5.
Proof.
Let be a nontree stable network for a fixed and let be a min cycle of . Since for every , using Lemma 1, we have that cannot be a cycle of length equal to 3. Furthermore, Since for every , using Lemma 2, we have that cannot be either a cycle of length equal to 4. Therefore, is a cycle of length greater than or equal to 5.
For the sake of contradiction, assume that is not directed. This means that contains a agent, say , that is buying both her incident edges in . We prove the contradiction thanks to Lemma 9, by showing that contains a strong critical pair.
If is an oddlength cycle, then has two distinct antipodal vertices which are also adjacent in .^{3}^{3}3In a cycle of length , two vertices of the cycles are antipodal if their distance is equal to . W.l.o.g., assume that is buying the edge towards . Clearly, . Furthermore, since is a min cycle, it is easy to check that is a strong critical pair.
If is an evenlength cycle, then let be the (unique) antipodal vertex of and let be a vertex that is adjacent to in . Observe that . Again using the fact that is a min cycle, we have the following:

If is buying the edge towards , then is a strong critical pair.

If is buying the edge towards , then is a strong critical pair.
In both cases, we have proved that contains a strong critical pair. ∎
Let be a nontree stable network with vertices for a fixed and let be a biconnected component of . Since for every , Lemma 4 implies that cannot be a directed cycle. At the same time, if is a cycle, then it is also a min cycle and therefore, Lemma 12 implies that must be directed, which contradicts Lemma 4. Therefore, we have proved the following.
Corollary 13.
For , no nontree stable network with vertices contains a cycle as one of its biconnected components.
2.4 Combining the Ingredients
Towards our main result, we start with proving that every stable network must contain a critical pair which satisfies an interesting structural property. This lemma is the third and last ingredient that is used in our analysis.
Lemma 14.
For , every nontree stable network with vertices contains a critical pair . Furthermore, there exists a path between and in such that (a) the length of is at most and (b) uses none of the edges and .
Proof.
Let be a network of vertices which is stable for a fixed , and let be any biconnected component of . By Corollary 13, we have that cannot be a cycle. As a consequence, contains at least edges and, therefore, it has a vertex, say , that buys at least two edges of .
Let and be the two distinct vertices of such that buys the edges and . Let be the min cycle that contains the edge , whose existence is guaranteed by Lemma 11. Lemma 12 implies that is a directed cycle of length greater than or equal to 5. Therefore, since is an edge of bought by agent , cannot contain the edge , which is also bought by . Similarly, since is an edge of bought by agent , cannot contain the edge , which is also bought by .
Let be a shortest path tree rooted at which gives priority to the shortest paths using the edges or . More precisely, for every vertex , if there is a shortest path from to containing the edge , then is a descendant of in . Furthermore, if no shortest path from to contains the edge , but there is a shortest path from to containing the edge , then is a descendant of in .
Consider the directed version of in which each edge is directed from their owner agent towards the other end vertex. Let be, among the vertices of which are also descendants of in , the one which is in maximum distance from w.r.t. the directed version of . Finally, let be the edge of which is bought by agent . Clearly, is not a descendant of in . Therefore, by construction of , , otherwise would have been a descendant of in , or there would have been a min cycle containing both edges and (which are both bought by agent ), thus contradicting Lemma 12.
W.l.o.g., assume that . Let and . We show that is a critical pair. By Lemma 12, is a cycle of length . As is a min cycle, . Moreover, since , we have that . Therefore . Next, the shortest path in between and uses the edge which is owned by agent . Furthermore, the shortest path in between and does not use the edge . Therefore, is a critical pair.
Now, consider the path which is obtained from by removing the edge . Recalling that does not contain the edge , it follows that is a path between and which uses none of the two edges and . Therefore, recalling that , the overall length of is less than or equal to
Finally, we prove our main result. For this and in the rest of the paper, given a vertex of a network and a subset of vertices of , we denote by .
Theorem 15.
For , every stable network with vertices is a tree.
Proof.
First of all, it is easy to check that for every stable network with vertices is a tree. Moreover, the same holds true for for .
Let be a fixed value and let be a stable network with vertices. Since , for every , we have that if is not a tree, then, by Lemma 14, it contains a critical pair satisfying the conditions stated in Lemma 14. Moreover, Lemma 9 implies that cannot be a strong critical pair. As a consequence, every shortest path from to uses the edge . Since is a critical pair, this implies that there is a shortest path from to which uses both the edges and . To finish our proof, we show that this contradicts the assumed stability of . This implies that must be a tree.
Let be a shortest path tree of rooted at having as the parent of and as the parent of . Observe that, by definition of a critical pair, there is a shortest path between and containing the edge . Therefore, is well defined. Furthermore, let be the set of vertices which are descendants of in . Note that since , we have .
Since is a critical pair, thanks to Observation 6, we can use Lemma 8 (where , and ) to obtain
(1) 
Furthermore, observe that
(2)  
Therefore, by substituting in (1) with (2) we obtain the following
(3) 
Let be the tree obtained from by the the swap of the edge with the edge . The distance cost incurred by agent if she swaps the edge with the edge is at most
Since is stable, agent cannot decrease her distance cost by swapping any of the edges she owns. Therefore, we obtain
(4) 
By summing both the lefthand and the righthand sides of the two inequalities (3) to (4) and simplifying we obtain
(5) 
Consider the network induced by the strategy vector in which agent deviates from her current strategy by swapping the edge with the edge and, at the same time, by deleting the edge. By Lemma 14, there exists a path between and in , of length at most , such that uses none of the edges and . As a consequence, using both (1) and (5) in the second to last inequality of the following chain, the distance cost of w.r.t. is upper bounded by
By her strategy change, agent will save in edge cost and her distance cost will increase by at most . Thus, if , then this yields a strict cost decrease for agent which contradicts the stability of . ∎
Corollary 16.
For the PoA is at most .
In Section 3 we improve the upper bound of on the PoA for stable tree networks from Fabrikant et al. [23]. With this, we establish the following:
Corollary 17.
For every the PoA is at most .
3 Improved Price of Anarchy for Stable Tree Networks
In this section we show a better bound on the PoA of stable tree networks. To prove the bound, we need to introduce some new notation first. Let be a tree on vertices and, for a vertex of , let be the forest obtained by removing vertex together with all its incident edges from . We say that is a centroid of if every tree in has at most vertices. It is well known that every tree has at least one centroid vertex [27].
Lemma 18.
Let be a stable tree network rooted at a centroid of , and let , with , be two vertices such that buys the edge towards in . Then , i.e. is the parent of in . Furthermore, if denotes the subtree of rooted at , then is a centroid of .
Proof.
We show that by proving that if , then is not stable. So, assume that . Consider the forest obtained from after the removal of the edge and let be the tree of the forest that contains vertex . Since is closer to than in and is not a vertex of , it follows that is a vertex of . Consider the strategy change in which player swaps the edge with the edge . Since and are both in the same tree, say , of the forest , it follows that the tree induced by all the vertices of which are not contained in , say , is entirely contained in and has vertices, as is a centroid of . Observe that after the swap of the edge with the edge , the distance from each of the vertices in decreases by , while the distance from each of all the other vertices of increases by at most . Therefore, if we denote by the number of vertices of , then the usage cost of player increases by at most
Therefore, is not stable.
We now prove that is a centroid of . Let be the set of vertices of . Observe that the claim trivially holds if . Therefore, we assume that . Notice that
Since