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We study network formation with

In network formation, a multitude of individuals, called

We study the price of anarchy in an

Although it appears limiting that the adversary can only destroy one link, this model already is challenging to analyze. It is a contribution to the understanding of how networks are formed when it is important that every vertex can reach every other vertex, for example for data transmission or delivery of goods.

After preparations and discussion of related work (Sections 2 to 5) we start out with a simple

Tight bounds on the price of anarchy for other adversaries, or for a general adversary are left for future work. As one of the most intriguing open problems, we leave the case of an adversary removing more than one link. Since our proofs rely heavily on the restriction of only one link being removed, this is expected to be a new challenge.

This work's focus is on unilateral link formation and Nash equilibrium. Bilateral link formation with its appropriate equilibrium concepts is planned to be studied in a separate publication. Currently, for bilateral link formation we can show a bound of

We give a rigorous description of the model framework that will be used in the following. Let _{v}^{n}_{vw}_{v}_{1},…, _{n}^{n×n}

The graph which is actually built is called the

So the wish of one endpoint, either

For _{vw}_{wv}

We speak of

Fix parameters _{v}_{v}_{v}

The indirect cost _{v}_{v}_{w∈V} dist_{G}

We call indirect cost _{v}_{ϕ(}_{v}_{)}(_{v}

The _{v∈V} _{v}_{s}_{∈}_{s}_{(}_{n}_{)}

A strategy profile

In other words, an essential strategy profile does not contain _{vϵV}_{v}

For each strategy profile

Recall that we can specify strategy profiles as directed graphs. Furthermore, since the social cost is fully determined by the final graph (since we restrict to essential strategy profiles), it suffices to consider the final graph (which is an undirected graph) in places where only the social cost is relevant.

A strategy profile

We call a NE _{v}_{1}) + … + (_{k}_{v}_{1}},…, {_{k}

A NE _{v}

We require some basic graph-theoretic notions. Let an undirected graph _{0},…, _{l}_{i−1},_{i}_{o},…,_{l}_{i−1},_{i}_{o},…, _{ℓ−1}}| = ℓ) and the walk is closed (_{o} = _{l}_{o}, _{1}, _{1}, _{2}, …, _{l},v_{l}

One might suggest using multigraphs instead of graphs, since in our adversary model, connectivity under removal of edges is relevant. However, none of our results becomes false when we allow multigraphs. Where not obvious, a remark on this is made. So we can stick to the simpler notion of graphs.

In order to not have to introduce names for all occurring constants, we use “

The “

An _{G}(_{G}_{e}_{∈E} rel_{G}(_{G}(

Since ∞ is assigned to disconnected final graphs, optima and NE are connected.

It is clear for optima. For NE note that since a connected graph has finite indirect cost, a player would always choose to build enough links in order to make the graph connected.

The _{ν∈V}rel(

A symmetric adversary induces anonymous disconnection cost

Let _{1}, _{2} be the two components of _{i}_{i}

Let

There is a vast body of literature on game-theoretic network formation, by far not limited to studies of the price of anarchy. A good starting point is the survey by Jackson [

Bilateral link formation (BLF) follows a concept given by Myerson [

Pairwise stability (PS) is an equilibrium concept suited for BLF. Essentially, it introduces a minimum of cooperation between players: a link not being in the final graph requires the additional justification that building the link would be an impairment for at least one of its endpoints. On the other hand, PS is only concerned with single-link deviations. Jackson and Wolinsky discussed several variations of PS, including what would later be known as

Watts [

Bloch and Jackson [

Bala and Goyal [_{vw}

Anshelevich, Dasgupta, Tardos, and Wexler [

Chun, Fonseca, Stoica, and Kubiatowicz [

Johari, Mannor, and Tsitsiklis [

The work of Fabrikant, Luthra, Maneva, Papadimitriou, and Shenker [_{0}, there exists a non-transient NE on _{0} vertices containing cycles, for any

Corbo and Parkes [^{2} on the price of anarchy. As noticed later in 2007 by Demaine

Albers

An ^{1−}^{ε}

Moscibroda, Schmid, and Wattenhofer [_{w∈V}dist(

Brandes, Hoefer, and Nick [

Baumann and Stiller [

Our adversary model addresses robustness in a way that, to the best of the author's knowledge, has not been studied theoretically before. We compare our approach to previous work that also addresses robustness.

Chun, Fonseca, Stoica, and Kubiatowicz [

The symmetric connections model of Jackson and Wolinsky [_{v}_{G(s)} (^{distG(S) (v,w)} if we assume stochastic independence of failures. For BLF, Baumann and Stiller [^{2}, ^{3}), which implies an ^{2}^{3}

The symmetric connections model is different from ours in many respects:

All links have the same probability of failure. In our model, links can have different probabilities, and these may even depend on the final graph. [

The failure of a link

Alternative paths are not considered; it is assumed that routing happens along a specific shortest path that is fixed before the random experiment that models the link failures is conducted. In our model,

Bala and Goyal [

Haller and Sarangi [_{vw}

The failure probability of each link {_{vw}

Failures of two different links are stochastically independent. In our model, they are mutually exclusive events. (This difference is exactly as between the symmetric connections model and ours.)

Generally, independent link failures model the unavailability of links due to, e.g., deterioration, maintenance times, or influences affecting the whole infrastructure or large parts of it (e.g., natural disasters). Our adversary model, on the other hand, models the situation when faced with an entity that is malicious but only has limited means so that it can only destroy a limited number of links (we limit this number to 1 in this work).

We conduct some preparations for the analysis of equilibria in our adversary model, which will be useful regardless of the link formation rule and the equilibrium concept. In the end, in Lemma 5.2, we will have developed a simple method to bound the sum of relevances

A set of vertices

Let

Now let

What we call “

Every vertex is contained in exactly one BCC. If

Now we introduce the bridge tree. It is the graph

Then

Whenever we speak of the number of vertices in a subgraph

In other words, we count the vertices that would be there if we expanded

On several occasions, when considering the effect of building additional edges, we treat vertices of the bridge tree as players. This is justified since edges inside BCCs have relevance 0. Hence for a strategy profile

For a path

For each

Fix

We give an upper bound on the price of anarchy for a general adversary. It holds independently of the link formation rule and the equilibrium concept, provided that equilibria have few edges.

Let

If

If

Since sep(^{2}) for all

The following corollary is obvious.

Fix any link formation rule and equilibrium concept.

If the number of edges in each equilibrium is

If the number of edges in each equilibrium is

A remark on the meaning of

The main goal of Sections 8 and 9 is to show a bound of

The

We stick to ULF and NE for the rest of this work. The aim of this section is to construct optima and NE, and finally to show how a bound on the price of stability follows easily. The adversaries considered are a general one,

An optimum has social cost Θ(

If

If

An optimum can only be the cycle or a tree, because any graph containing a cycle has already the building cost

Hence the social cost of a tree is at least

The social cost of the cycle is

The following simple remark later will help establishing concrete bounds on the price of anarchy.

Assume there are constants _{0}, _{1} > 0 such that the social cost of all NE is bounded by (_{1}_{0})

If the optimum is

The following two propositions can be proved by appropriate cost-benefit analysis.

Let

If

If

In both cases, strict inequality implies a MaxNE.

Since all edges point outwards, the center player is the only one that could sell edges, but this would make the graph disconnected. Exchanges of edges by the center cannot lead to a different strategy profile. The maximum disconnection cost is experienced by a leaf vertex when the probability measure is concentrated on the one edge that connects it to the rest. The disconnection cost is then

Disconnection cost of the center is 1. By anonymity, all leafs experience the same disconnection cost. It follows easily from this that all edges have the same probability, namely

Now we apply the same arguments as for part (i). Maximality is clear in both cases by Remark 2.1.

Let

If α ≤ 1, then S is a MaxNE.

If

Maximality in both cases is due to the cycle having minimum disconnection cost, namely 0. Buying or exchanging edges is also not beneficial since the cycle already has minimum disconnection cost. We only have to check whether it is beneficial for a player

To prove (ii), we have to establish a better lower bound on the new disconnection cost for _{1},_{1},_{2},…,_{n−1},_{n}_{1}. We claim that

We are left with proving _{i}_{i}_{n−i+1} = (_{n − i}}) + _{n} − _{i}_{n-i}_{+1} and _{i}_{+1} = _{n-i}_{i}_{n−i}}), and since
_{i}_{n-i}

For anonymous disconnection cost, this proves existence of NE for all ranges of

In the following, we assume

The following is a consequence of Propositions 7.1, 7.3, and 7.4.

For anonymous disconnection cost the price of stability is 1 +

For

For

The simple-minded adversary picks an edge uniformly at random, that is,

Social cost of a path is

We have the social cost of a path:

We estimate the benefit for a player of building or selling a particular edge. This will become useful in several places. It moreover immediately leads to a structural result on the length of cycles. The following remark is purely graph-theoretic.

Let

Let

Let

The additional edge

Let _{1} and _{2} that are connected only by _{1} with _{2}. Moreover, there are no other edges between _{1} and _{2}. It follows that any cycle that contains

For each player

We repeat the counting argument from the proof of Lemma 5.2:

Fix a player

If a player builds an additional edge and the sum of her relevances drops from

If a player sells a non-bridge and the sum of her relevances increases from

We have

If a player builds an edge creating a cycle of length £, the improvement in disconnection cost is at most

If a player sells an edge destroying a cycle of length £, the impairment in disconnection cost is at most

Let _{1}, _{1},…, _{ℓ−1}, _{ℓ}, _{ℓ} bought by _{ℓ},

all ℓ edges were bridges before and became non-bridges now, and

without the additional edge, _{1}, _{ℓ−1} is relevant for (

It follows

By Remark 8.2(ii), we may consider any cycle that is destroyed. The rest is the same calculation as for (i).

Let

If a player builds an edge creating a cycle of length ℓ, she suffers an impairment in her cost.

If a player sells an edge destroying a cycle of length ℓ, she experiences an improvement in her cost.

(i)By Proposition 8.5(i), the player suffers an impairment in her cost if

We show (ii) in almost exactly the same way, using Proposition 8.5(ii) and that

It follows the structural result:

No NE contains cycles shorter than

Bounding the Price of Anarchy

The following observation is the key to showing that a NE does not have many more edges than a tree.

A NE is chord-free.

Selling a chord

We consider this case now and show that we in fact do not need a multigraph. By selling the chord, the disconnection cost for

The next two are graph-theoretic results. The first is a straightforward adaption of a result (and its proof) on vertex-connectivity to edge-connectivity; see [

Any bridgeless connected graph can be constructed from a cycle by successively adding paths or cycles of the form (u, _{1}, _{1},…, _{k},e_{k+1},w), where u, w are vertices of the already constructed graph and _{1},…, _{k} are zero or more new vertices.

Clearly, any graph that was constructed in this manner is connected and bridgeless. Now let _{1} = _{1},…, _{k}_{i}_{i}_{i}

A chord-free graph on

Let _{0} vertices, by successively adding paths of the form (_{1}, _{1},…, _{k}_{k}_{+1}, _{1},…, _{k}, k_{0}, are zero or more new vertices. For any two vertices _{1}, _{0}≕ _{1} steps in this construction. Let _{i}_{i}_{i}_{i}_{0} + 2_{1} ≤ 2_{0} + 2_{1} = 2

If

A NE has at most 3

Follows from Prop. 8.8 and 8.10.

Now we know that the total building cost in a NE is

The disconnection cost is bounded by n diam(

We have by Lemma 5.2:

The bridge tree of a NE has its diameter bounded by

Let _{0}, _{1}, _{1},…, _{ℓ}, v_{ℓ}_{0} and _{e}

At least
_{ℓ̄}_{0}.

At least
_{ℓ̄}_{l}.

Let us assume the first; the other case can be treated alike. Let _{0} and _{l}_{1},…, _{l̄}

The disconnection cost in a NE is bounded by (12

Follows from Corollary 8.12 and Lemma 8.13.

The price of anarchy with a simple-minded adversary is bounded by

The building cost and the disconnection cost in a NE are both

A closer look at Lemma 8.13 and its proof reveals that there exists a constant

The constant in Theorem 8.15 is 15 +

Building cost of a NE is bounded by 3_{1}≔15 and c_{o}

We consider an adversary that destroys an edge which separates a maximum number of vertex pairs. If there are several such edges, one of them is chosen uniformly at random. In other words, we replace the uniform probability distribution on the edges for one that is concentrated on the edges which cause maximum overall damage. Recall that sep(_{max}≔ max_{e}_{∈}_{E}_{max} ≔ {_{max}} and _{max} ≔|_{max}|. These are the edges of which each causes a maximum number of separated vertex pairs when it is deleted. We call them the

If sep_{max} = 0, then the graph is bridgeless and all edges are critical—however, their removal does not separate any vertex pairs. If sep_{max} > 0, then there are one or more critical edges, and each of them is a bridge. Recall that if _{m}_{ax}.

If sep_{max} > 0 and if there are more than one critical edges, they form a subgraph that is a star in the bridge tree

Let sep_{max} > 0. For any two distinct bridges _{max}, and so also for all other bridges (since they have smaller

Let _{0}, _{1}, _{1},…, _{ℓ−1},_{l},v_{l}_{1} and _{l} being distinct critical edges. First assume that _{l}_{l}. Then _{0} is in the smaller component of _{1}. Then the smaller component of _{2} cannot contain _{0}, since otherwise _{1}) < _{2}), and _{1} would not be critical. So the component of _{2} containing _{0} is the larger one, and then the same holds for the component of _{1} containing _{0}. This contradicts that _{l} is in the larger component of _{l}_{0} and _{1}. Summarizing, now we know that the smaller component of _{l} is located “before” _{l}

If ℓ ≥ 3, then there is an edge _{1} and _{l} on _{1} or _{l}. Since _{1}) = _{2}), we have thus in particular, _{1}), a contradiction that _{1} is critical. Hence there is no such edge _{1}, _{l}) of critical edges, the set of all critical edges forms a star (in the bridge tree).

The smart adversary admits a new NE topology:

If
^{2}). If

The social cost of the path is

Now consider that a player

When two edges are exchanged,

The proof of the following is even easier than previously:

A NE is chord-free

Removing a chord does not change the relevance of any edge, nor does it change sep_{max}, hence it does not change _{max}. Selling a chord so is always beneficial.

With Proposition 8.10, it follows immediately:

A NE has at most 3

We are again left with bounding the disconnection cost of NE. This requires some effort and is accomplished in the following remark and two lemmas.

If there are _{max} = {_{1},…, _{k}_{1} is put on a cycle by an additional edge, but not _{2},…,_{k}_{2},…,_{k}_{1} and _{2} on a cycle, but not _{3},…,_{k}_{3}, …,_{k}

An additional edge _{2},…, _{k}_{3},…, _{k}

Let _{max} ≥ 3. Then we have sep_{max} ≤ 2(1 + 9

Fix two critical edges _{1} and _{2}, and set _{0} ≔ _{1}). For each _{i}_{i}_{i}_{i}_{0} and rel(_{i}_{0} for all critical edges _{i}_{1},_{2}} puts _{1} and _{2} on a cycle and leaves the other _{max} – 2 critical edges critical by Remark 9.5. For each _{i}

Since we are in a NE, this is at most _{max}_{0}, we have _{0} ≥ (_{max} − 2) _{0}, and so it follows
_{max}_{0} ≥ _{0}. With these two inequalities at hand, we can bound sep_{max}. We have

Fix a NE with _{max} ∈ {1,2}. Then

we have sep_{max} ≤ 4

or we have

First we consider the case _{max} = 2. A player can make the two critical edges part of a cycle by building an additional edge. The difficulty lies in that new critical edges, with a smaller separation value, can emerge. We will have to put some more effort into estimating the improvement in disconnection cost that a player is able to achieve by building another edge. Consider the bridge tree. There are two subtrees _{1} and _{2} that are connected to the rest by the two critical edges _{1} and _{2}, respectively. They both have _{0} ≔ _{1}) = _{2}) vertices. There may be more subtrees _{3},…, _{N}_{3},…, _{N}_{k}_{k}_{1} and _{2} on a cycle.

First assume that we can arrange _{1} ∈ _{1} and _{2} ∈ _{2} such that after building {_{1}, _{2}}, there are no critical edges in _{1} nor in _{2}. If there are no subtrees except _{1} and _{2}, i.e._{1} (and also for _{2}) of building {_{1},_{2}} is hence their original disconnection cost, _{3} + _{3},…, _{N}_{N}_{k}_{0} or
_{k}_{1} and the center vertex, and so _{k}_{0} = _{1}), in which case _{1} would not be critical. Moreover, we have |_{k}_{0} < _{k}_{1} (or _{2}), a critical edge in _{k}_{k}_{k}_{1} (and also for _{2}) gained by building {_{1}, _{2}} is hence at least the original disconnection cost minus

Now consider that for all choices of _{1} ∈ _{1} and _{2} ∈ _{2}, building { _{1}; _{2}} induces a critical edge in at least one of _{1} or _{2}. For each _{i}_{i}_{i}_{i}_{i}_{i}_{i}_{i}_{i}_{i}_{i}_{1}; _{2}}, since these paths then both are located on a cycle. However, by assumption, there is a critical edge _{1}. By construction of _{1}, we have
_{1} (and also _{2}) can reduce her disconnection cost to no more than
_{max} = 2_{0} (_{0}) ≤ 2_{0}. 2α ≤ 4n α.

The case of _{max} = 1 can be treated similarly. Let _{1} be the critical edge and _{1} the subtree with _{0} ≔ _{1}) vertices. There are zero or more additional subtrees, say _{2},…, _{N}_{2} ≔ _{1}, which consists of just one vertex in the bridge tree then (but can consist of multiple vertices in _{2}| ≥ |_{k}_{1} and _{2} in the roles of the former subtrees of the same name. Assume first that we can find _{1} ∈ _{1} and _{2} ∈ _{2} such that building {v_{1},v_{2}} does not induce any critical edges in _{1} nor _{2}. If _{1} of at least _{0}, and so sep_{max} = 2n_{0} (_{0}) ≤ 2n_{0} _{k}_{0}. Moreover, we have |_{k}_{0}) ≤ _{k}_{1}, _{2} } reduces the disconnection cost of _{1} to no more than
_{1} of at least

If each choice of _{1} and _{2} induces a critical edge in _{1} or _{2}, we can, as before, show that by a careful choice of these vertices, building {_{1}, _{2}} reduces the disconnection cost for _{1} (and _{2}) to at most
_{1} originally has disconnection cost _{0} ≥ _{0}, so she experiences an improvement of at least
_{2} originally has disconnection cost _{0}, so she as well experiences an improvement of at least
_{0} ≤ 2_{max} = 2_{0} (_{0}) ≤ 4

The price of anarchy with a smart adversary is

Let _{max} ≥ 3, then Lemma 9.6 gives sep_{max} = _{max} ∈ {1, 2}, Lemma 9.7 gives the same, since _{max} is the total disconnection cost, it so has a ratio of

If

The constant in Theorem 9.8 is 9 +

We proceed as in the proof of the theorem, but do more detailed calculations. Let
_{max} ≥ 3, then Lemma 9.6 gives sep_{max} ≤ 2 (1 + 9c) _{max} ∈ {1,2}, Lemma 9.7 gives sep_{max} ≤ 4

Next consider ^{2} on the disconnection cost yields the

In all cases disconnection cost is bounded by 6_{1} ≔9 and _{0} ≔ 0 yields a bound on the price of anarchy of

This work is based on parts of Chapter 5 of my dissertation [

The argument in the beginning of the proof of Proposition 8.8 is flawed, since

We show that the bridge tree does not change by removal of _{1} and _{2}. Since _{1} and _{2}. But then, due to the existence of _{1} and _{2}. Hence removal of the single edge _{1} from _{2}.

Let this be the final graph _{1} the subgraph to the left starting with _{2} the subgraph to the right starting with _{1} experience the same disconnection cost, and the same holds for _{2}. (Precisely, we have _{x}_{1}) and _{y}_{2}) and _{v}_{1} and _{2}, respectively.

Bridge tree construction. Vertices representing BCCs of more than 1 vertices have their number of vertices attached, here 4, 7, and 3, respectively.

NE if

Schematic view of the bridge tree with two critical edges _{1} and _{2}, drawn dashed. Subtrees are represented by triangles.

How _{1} and _{2} are put on a cycle by a new edge {_{1}; _{2}}. Paths that are part of the new cycle and located inside _{1} and _{2} are depicted as zigzag paths. New critical edges can emerge, e.g., _{3} can become critical.

Detailed view of _{i}_{i}

I thank the German Research Foundation (Deutsche Forschungsgemeinschaft (DFG)) for financial support through Priority Program 1307 “Algorithm Engineering” (Grant SR7/12-2). I thank the two anonymous referees for their very helpful and constructive comments. I thank everyone at

We use “player” and “vertex” synonymously.

We use “link” and “edge” synonymously.

Optimal networks are also called “efficient” in the literature.

The “essential” term was used in [

This means if _{vw}_{wv}

However, large parts of our analysis will be restricted to two specific cases: One in which the adversary picks a link uniformly at random (simple-minded adversary) and another in which he picks a link that causes maximum overall damage (smart adversary).

Haller and Sarangi also briefly discussed failure probabilities depending on the final graph. They considered an example where for non-increasing functions _{v}_{vw}_{vw}_{G}_{(}_{S}_{)}(_{w}_{G}_{(}_{S}_{)}(_{vw}

See [

The restriction to

It would suffice to restrict to