Let G = (V, E) be connected and ϕ : V → V a graph automorphism of G. If e = {v, w} is a non-bridge, then ϕ(e) ≔ {ϕ(v), ϕ(w)} is a non-bridge as well, and so sep(e) = 0 = sep(ϕ (e)) and rel(e, v) = 0 = rel(ϕ (e), ϕ(v)) for all v ∈ V. Let e = {v, w} be a bridge and G₁, G₂ be the two components of G – e. Then ϕ(e) is a bridge as well. Let G 1 ′, G 2 ′ be the two components of G – ϕ(e). Then ϕ(V(G_i)) = V( G i ′) for all i ∈ {1, 2}, or ϕ(V(G_i)) = V( G j ′) for all i,j ∈ {1, 2}, i ≠ j. In either case, sep (e) = sep(ϕ(e)), and also rel(e,v) = rel(ϕ(e), ϕ(v)) for all v ∈ V. In total, we have for all e ∈ E and all v ∈ V: (1) sep ( e ) = sep ( ϕ ( e ) ) (2) rel ( e , v ) = rel ( ϕ ( e ) , ϕ ( v ) )

Let v ∈ V. Then: I ϕ ( v ) ( G ) = ∑ e ∈ E rel ( e , ϕ ( v ) ) Pr ( { e } ) = ∑ e ∈ E rel ( ϕ ( e ) , ϕ ( v ) ) Pr ( { ϕ ( e ) } ) ϕ is bijective = ∑ e ∈ E rel ( e , v ) Pr ( { ϕ ( e ) } ) by ( 2 ) = ∑ e ∈ E rel ( e , v ) Pr ( { e } ) by ( 1 ) and symmetric adversary = I v ( G )The converse of Proposition 3.1 does not hold, as shown in Figure 1 on this page. Proposition 3.1 is useful since symmetry can be recognized directly from the definitions of the two special adversaries studied later, and so we know that they induce anonymous disconnection cost.

Let W be maximal under the condition of G[W] being connected and not containing any bridges of G[W], i.e., we follow the original definition given above. Clearly, G[W] does not contain any bridges of G, since if removal of some edge disconnects G, then it also disconnects G[W] if the endpoints of this edge are in W. We choose U ⊇ W maximal under the condition that G[U] is connected and G[U] does not contain any bridges of G. Suppose U ≠ W. Then G[U] contains a bridge e of G[U]. Since this is no bridge of G, it is located on a cycle C. Then V(C) ⊈ U, since e is a bridge of G[U]. But G[U ∪ V(C)] would still be connected and would contain no bridge of G. This contradicts the maximality of U.

Now let W be maximal under the condition of G[W] being connected and not containing any bridges of G. If G[W] contained a bridge e of G[W] (but not of G), we could use the cycle-argument from before to augment W and have a contradiction to its maximality. Suppose there is U ⊉W such that G[U] is connected and G[U] does not contain any bridges of G[U]. Then G[U] contains a bridge of G. As noted earlier, this is also a bridge of G[U], a contradiction.

What we call “bcc” is sometimes called “block” in the literature, and what we call “bridge tree” is then called “bridge-block tree”. We refrain from using the term “block” here, since it usually is related to vertex-connectivity; see [43] (p. 55).

Every vertex is contained in exactly one BCC. If W is a BCC, we have to remove at least 2 edges from G[W] in order to make it disconnected. A graph from which we have to remove at least 2 edges to make it disconnected is also called being “2-edge-connected” in common terminology, provided that it has more than 1 vertices; see [43] (p. 12).

Now we introduce the bridge tree. It is the graph G˜ = (V˜, E˜) defined by: V ∼ ≔ { B ⊆ V ; B is a BCC } , E ∼ ≔ { { B , B ′ } ; B , B ′ ∈ V ∼ , ∃ v ∈ B , w ∈ B ′ : { v , w } ∈ E }

Then G˜ is a tree (assuming G is connected). By Proposition 5.1 there is a 1 : 1 mapping between the edges of G and the bridges of G. We make the following special convention concerning the bridge tree:

Fix v ∈ V. We have R ( v ) = def ∑ e ∈ E rel ( e , v ) = ( 3 ) ∑ e ∈ E e is a bridge | { P ∈ P ( v ) ; e ∈ E ( P ) } | = ∑ P ∈ P ( v ) | { e ∈ E ( P ) ; e is a bridge } | = ∑ P ∈ P ( v ) | P ˜ | ≤ ( n − 1 ) diam ( G ˜ )The last estimation is true since the bridge tree is a tree and so every path is a shortest path.

Since sep(e) = O(n²) for all e, we have C ( S ) = O ( m α + n 2 ∑ e ∈ E Pr ( { e } ) ) = O ( n α + n 2 )Since an optimum is connected, the optimal social cost is Ω(nα). Dividing by this yields (i). Assertion (ii) follows from (i).

The following corollary is obvious.

An optimum can only be the cycle or a tree, because any graph containing a cycle has already the building cost nα of the cycle, and the cycle has optimal disconnection cost. So an optimum is either the cycle, or it is cycle-free. Let T be any tree. We have its indirect cost: ∑ e ∈ E ( T ) sep ( e ) Pr ( { e } ) = 2 ∑ e ∈ E ( T ) ν ( e ) ( n − ν ( e ) ) Pr ( { e } ) ≥ 2 ⋅ 1 ( n − 1 ) ∑ e ∈ E ( T ) Pr ( { e } ) = 2 ( n − 1 )

Hence the social cost of a tree is at least ( n − 1 ) α + 2 ( n − 2 ) = ( n − 1 ) ( α − 2 )

The social cost of the cycle is nα. So if α ≤ 2 (n − 1), the cycle is better or as good as any tree, hence it is an optimum. If α > 2 (n − 1), then we look for a good tree. A star has social cost (n − 1) (α + 2), which matches the lower bound given above, and is hence optimal (and better than the cycle). However, for α = 2 (n− 1), both cycle and star are optimal with social cost 2n (n − 1).

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

If the optimum is nα, we have the ratio ( c 1 n + c 0 ) α n α = c 1 + c 0 n ≤ c 1 + c 1 + c 0 n − 1. Otherwise, if the optimum is (n - 1) (α + 2), we have the ratio ( c 1 n + c 0 ) α ( n − 1 ) ( α + 2 ) < c 1 n + c 0 n − 1 = c 1 ( n − 1 ) + c 1 + c 0 n − 1 = c 1 + c 1 + c 0 n − 1.

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

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

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 v to sell her one edge. Selling the edge yields a path with v at one of its ends. This increases disconnection cost for v to at least 1, since the removal of any edge disconnects v from at least one other vertex. This proves (i).

To prove (ii), we have to establish a better lower bound on the new disconnection cost for v. Anonymity of indirect cost allows us to do so. Let the path be (v₁,e₁,v₂,…,e_n−1,v_n) with v = v₁. We claim that (4) Pr ( { e i } ) = Pr ( { e n − i } ) for all i ∈ { 1 , … , n − 1 } ,i.e., the adversary behaves like a symmetric one on this graph. From (4) it follows ∑ i = 1 ⌈ ( n − 1 ) / 2 ⌉ Pr ( { e i } ) ≥ 1 2. Since each of the edges e 1 , … , e ⌈ ( n − 1 ) / 2 ⌉ has relevance at least ⌊ ( n − 1 ) / 2 ⌋ for v, the proposition follows.

We are left with proving (4). For each i ∈ [n] we write I_i for the indirect cost of vertex v_i, and moreover define its left indirect cost by I i left : = ∑ j = 1 i = 1 j Pr ( { e j } ) and its right indirect cost by I i right : = ∑ j = 1 n = 1 j Pr ( { e n − j } ). Then clearly I i = I i left + I i right. It suffices to show (4) for i < n 2. We have (in fact even for i ≤ n - 1) on the one hand: I i = I i left + ( n − i ) Pr ( { e i } ) + I i + 1 right = I i left + ( n − i ) Pr ( { e i } ) + I i + 1 − I i + 1 left = I i left + ( n − i ) Pr ( { e i } ) + I i + 1 − ( I i left + i Pr ( { e i } ) ) = ( n − 2 i ) Pr ( { e i } ) + I i + 1On the other hand, along the same lines we prove I_n−i+1 = (n − 2i) Pr({e_{n − i}}) + I_n − i. By anonymity, I_i = I_n-i+1 and I_i+1 = I_n-i. It follows (n − 2i) Pr({e_i}) = (n − 2i (Pr({e_n−i}), and since i < n 2, this means Pr({e_i}) = Pr({e_n-i}).

For anonymous disconnection cost, this proves existence of NE for all ranges of α provided that n ≥ 9, since then 2 − 1 n − 1 ≤ 2 ≤ 1 2 ⌊ n − 1 2 ⌋. In the range 2 − 1 n − 1 ≤ α ≤ 1 2 ⌊ n − 1 2 ⌋ two very different topologies—namely cycle and star—co-exist as NE.

For α ≤ 1 2 ⌊ n − 1 2 ⌋ the cycle is a NE as well as an optimum, and so the price of stability is 1. For α ≥ 2(n – 1) a star is a NE as well as an optimum, and so the price of stability is 1.

For 1 2 ⌊ n − 1 2 ⌋ ≤ α ≤ 2 ( n − 1 ), the star is a NE and the cycle is an optimum. The price of stability so is upper-bounded by ( n − 1 ) ( α + 1 ) n α ≤ 1 + 2 α ≤ 1 + 4 ⌊ n − 1 2 ⌋ ≤ 1 + 8 n − 2

We have the social cost of a path: m α + 1 m ∑ v ∈ V ∑ e ∈ E rel ( e , v ) = ( n − 1 ) α + 1 n − 1 ∑ e ∈ E ∑ v ∈ V rel ( e , v ) = ( n − 1 ) α + 1 n − 1 ∑ e ∈ E sep ( e ) = ( n − 1 ) α + 1 n − 1 ∑ e ∈ E 2 ν ( e ) ( n − ν ( e ) ) = ( n − 1 ) α + 2 n − 1 ∑ k = 1 n − 1 k ( n − k ) = ( n − 1 ) α + 2 n − 1 ( n ( n − 1 ) n 2 − ( n − 1 ) n ( 2 n − 1 ) 6 ) = ( n − 1 ) α + 1 3 n ( n + 1 ) = Θ ( n α + n 2 )

We repeat the counting argument from the proof of Lemma 5.2: R ( v ) = def ∑ e ∈ E rel ( e , v ) = ( 3 ) ∑ e ∈ E e is a bridge | { P ∈ P ( v ) ; e ∈ E ( P ) } | = ∑ P ∈ P ( v ) | { e ∈ E ( P ) ; e is a bridge } | ≤ ∑ P ∈ P ( v ) | E ( P ) | = ∑ w ∈ V dist ( v , w )This is maximal if G is a path with v at its end; then R ( v ) = n ( n − 1 ) 2.

Fix a player v. Let R ≔ R(v) and let R′ be the same quantity when an additional edge e is built by v. By the previous proposition, we have R , R ' ≤ n ( n − 1 ) 2. The benefit in disconnection cost of building this edge for player v is 1 m R − 1 m + 1 R '. Due to the change in denominators from “m” to “m + 1” this expression looks somewhat unhandy. Yet, we can give good bounds incorporating the change in relevances, ΔR ≔ R - R′ ≥ 0, with one denominator. We can do something similar for the case when the player sells an edge, where we put ΔR ≔ R′ - R ≥ 0.

We have 1 m R − 1 m + 1 R ' = 1 m R − 1 m + 1 ( R + ( R ' − R ) ) = ( 1 m − 1 m + 1 ) R + 1 m + 1 Δ R = 1 m ( m + 1 ) R + 1 m + 1 Δ R { ≤ 1 2 + 1 m + 1 Δ R ≤ n 2 ≥ 1 m + 1 Δ R

It follows Δ R ≤ ∑ k = 1 ℓ − 1 ( n − k ) = ( ℓ − 1 ) n − ∑ k = 1 ℓ − 1 k = ( ℓ − 1 ) n ( ℓ − 1 ) ℓ 2 = ( ℓ − 1 ) ( n − ℓ 2 ). The statement follows with Proposition 8.4(i).

(i)By Proposition 8.5(i), the player suffers an impairment in her cost if α > 1 2 + 1 m + 1 ( ℓ − 1 ) ( n − ℓ 2 )Since m ≥ n – 1, this is the case if α > 1 2 + 1 n ( ℓ − 1 ) ( n − ℓ 2 ) , which is the same as n ( α + 1 2 ) > ℓ ( n − ℓ 2 + 1 2 ). Since ℓ ≥ 3 ≥ 1, this is the case if n ( α + 1 2 ) > ℓ n.

We show (ii) in almost exactly the same way, using Proposition 8.5(ii) and that m ≥ n, since the original graph contains a cycle.

It follows the structural result:

Selling a chord e = {v, w} from a cycle C = (v,…, w,…, v) does not increase the relevance of any edge for any player. To see this, let C′ be a cycle which contains e. Then C′ – e also forms a cycle with a part of C, say (v,…, w). Hence, if the graph is bridgeless, removing a chord would decrease the player's building cost without increasing the disconnection cost. Now let the graph contain a bridge e′. Due to the decrease in the denominator of the disconnection cost, removing a chord impairs the disconnection cost. However, the player owning the chord, say v, would rather remove the chord and instead build an edge to form a new cycle containing e′. The only case where this is impossible is when v is one endpoint of the bridge e′ = {v, u}, and u is a leaf vertex. Then, a double-edge between v and u would be needed, which is not allowed unless we use a multigraph.

We consider this case now and show that we in fact do not need a multigraph. By selling the chord, the disconnection cost for v increases by 1 m ( m − 1 ) R ( v ). If this increase is strictly smaller than α, we are done. Hence assume 1 m ( m − 1 ) R ( v ) ≥ α now. Edge {v, u} has relevance n – 1 for u. Due to the positions of v and u, we have R(u) = R(v) + (n - 1) - 1. If u builds an edge to any other vertex, save v, edge e′ is put on a cycle. The improvement in disconnection cost for u by building such an edge is at least 1 m R ( u ) − 1 m + 1 ( R ( u ) − ( n − 1 ) ) = 1 m ( R ( v ) + n − 2 ) − 1 m + 1 ( R ( v ) − 1 ) = ( 1 m − 1 m + 1 ) R ( v ) + n − 2 m + 1 m + 1 = 1 m ( m + 1 ) R ( v ) + n − 2 m + 1 m + 1 = ( 1 m ( m − 1 ) + 1 m ( m + 1 ) − 1 m ( m − 1 ) ) R ( v ) + n − 2 m + 1 m + 1 ≥ α − 1 m ( 1 m − 1 − 1 m + 1 ) R ( v ) + n − 2 m + 1 m + 1 ≥ α − 1 m ( 1 m − 1 − 1 m + 1 ) n ( n − 1 ) 2 + n − 2 m + 1 m + 1 ≥ α − ( 1 m − 1 − 1 m + 1 ) n − 1 2 + n − 2 m + 1 m + 1 = α − 2 ( m − 1 ) ( m + 1 ) n − 1 2 + n − 2 m + 1 m + 1 ≥ α − 1 m + 1 + n − 2 m + 1 m + 1 > αSo u has an incentive to buy an additional edge, a contradiction to NE.

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 [43] (Prop. 3.1.3) for the version for vertex-connectivity

Clearly, any graph that was constructed in this manner is connected and bridgeless. Now let G be connected and bridgeless and H a subgraph of G that is constructible in this manner, chosen such that it has a maximum number of edges among all such subgraphs. Since G contains a cycle, H is not empty. Also, H is an induced subgraph since H + e is also constructible for any edge e. If H ≠ G, then since G is connected, there is an edge e = {v, w} with v ∉ V(H) and w ∈ V(H). Since G is bridgeless, this edge is on a cycle C = (w, e, v₁ = v₁,…, v_k = w). Let v_i be the first vertex with v_i ∈ V(H). Then P ≔ (w,…, v_i) is a path or cycle of the form used in the construction, and so H + P is constructible and has more edges than H, a contradiction.

Let G be a chord-free graph, w.l.o.g. being connected. We first consider the case that G is bridgeless. By the previous proposition, G can be constructed from a cycle on, say, N₀ vertices, by successively adding paths of the form (u, e₁, v₁,…, v_k, e_k+1, w), where u, w are vertices of the already constructed graph and v₁,…, v_k, k ∈ ℕ₀, are zero or more new vertices. For any two vertices u, w in the already constructed graph, there is a cycle C with u,w ∈V(C). Since G is chord-free, we may not add a path (u,e₁, w). Hence k ≥ 1 in each step, i.e., at least one new vertex is added. It follows that there are at most t ≤ n – N₀≕ N₁ steps in this construction. Let n_i and m_i be the number of new vertices and edges, respectively, inserted in step i. Then m_i = n_i + 1 for each i ∈ [t] and so we add ∑ i = 1 t m i = ∑ i = 1 t ( n i + 1 ) = N 1 + t ≤ 2 N 1 edges to the initial cycle. It follows that G has at most N₀ + 2N₁ ≤ 2N₀ + 2N₁ = 2n edges.

If G is not bridgeless, we consider each of its BCCs; these correspond to vertices of the bridge tree. Altogether, they cannot contribute more than 2n edges. In addition, there are at most n—1 edges, namely bridges of G. So we have a bound of 2n + n — 1 ≤ 3n.

Follows from Prop. 8.8 and 8.10.

Now we know that the total building cost in a NE is O(nα), hence it is of the same order as the optimal social cost. In order to bound the price of anarchy, we are left with bounding the disconnection cost. To this end, we make use of the bridge tree. The following is a corollary to Lemma 5.2.

We have by Lemma 5.2: 1 m ∑ v ∈ V ∑ e ∈ E rel ( e , v ) = 1 m ∑ v ∈ V R ( v ) ≤ 1 m ∑ v ∈ V ( n − 1 ) diam ( G ˜ ) = n m ( n − 1 ) diam ( G ˜ ) ≤ n diam ( G ˜ )A bound on the diameter of the bridge tree holding for all NE will hence yield a bound on the price of anarchy. This is accomplished by the following Lemma.

Let G be a NE. Let P˜ = (v₀, e₁, v₁,…, e_ℓ, v_ℓ) be a path in the bridge tree G˜ connecting two leaves v₀ and v_e. Let ℓ ¯ : = ⌈ ℓ 2 ⌉ ≥ 1. Then at least one of the following is true (recall the convention on page 313 regarding vertex-counting in the bridge tree): –

At least ⌈ n 2 ⌉ vertices lie beyond e_ℓ̄ from the view of v₀.

Let us assume the first; the other case can be treated alike. Let v ≔ v₀ and w ≔ v_l and recall that we may treat vertices of the bridge tree G˜ as single players with respect to building of new links. Then e₁,…, e_l̄ for v have relevance at least ⌈ n 2 ⌉ each. So ∑ i = 1 ℓ ¯ rel ( e i , v ) ≥ ℓ ¯ n 2 ≥ ℓ 2 n 2 = Ω ( ℓ n ). By building {v,w}, player v would have a benefit in disconnection cost of at least 1 m + 1 ℓ n 4 ≥ 1 3 n + 1 ℓ n 4 = Ω ( ℓ ), using the bound m ≤ 3n from Corollary 8.11. Since the edge is not built, α is larger than this benefit, so ℓ ≤ ( 3 n + 1 ) 4 n α = O ( α ).

Follows from Corollary 8.12 and Lemma 8.13.

The building cost and the disconnection cost in a NE are both O(nα) by Corollary 8.11 and 8.14. The theorem follows with Proposition 7.1, which states that the optimum social cost is Θ(nα).

A closer look at Lemma 8.13 and its proof reveals that there exists a constant c > 0 such that if m = O(n) then there are players who can improve their disconnection cost by c through the building of new links, as long as the graph contains bridges. It follows that for α < c, all NE are bridgeless, i.e., they have disconnection cost 0 and the adversary cannot harm them. This does not rule out, however, that they may contain an unnecessarily high number of links, compared to an optimum. On the other hand, the ratio cannot be more than O(1) by Theorem 8.15.

Building cost of a NE is bounded by 3nα by Corollary 8.11. Disconnection cost of a NE is bounded by (12n + 4) α by Corollary 8.14. In total, social cost of a NE is bounded by (15n + 4) α. Using Remark 7.2 with c₁≔15 and c_o≔4, and finally n ≥ 9 proves the claim.

Let sep_max > 0. For any two distinct bridges e and e′, one component of G — e is strictly contained in one component of G — e′. Therefore, with multiple critical edges, ν ( e ) < n 2 for all e ∈ E_max, and so also for all other bridges (since they have smaller ν(·) value). In other words, there is always a small and a large component of G – e, with e being a bridge.

Let P = (v₀, e₁, v₁,…, v_ℓ−1,e_l,v_l) be a path in the bridge tree G˜ with e₁ and e_l being distinct critical edges. First assume that v_l is in the larger component of G — e_l. Then v₀ is in the smaller component of G – e₁. Then the smaller component of G — e₂ cannot contain v₀, since otherwise ν(e₁) < ν(e₂), and e₁ would not be critical. So the component of G — e₂ containing v₀ is the larger one, and then the same holds for the component of G — e₁ containing v₀. This contradicts that v_l is in the larger component of G — e_l. We can carry out the same argument with v₀ and e₁. Summarizing, now we know that the smaller component of G — e_l is located “before” P and that the smaller component of G – e_l is located “beyond” P.

If ℓ ≥ 3, then there is an edge f between e₁ and e_l on P. The smaller component of G – f strictly contains either the smaller component of G — e₁ or G — e_l. Since ν(e₁) = ν(e₂), we have thus in particular, ν(f) > ν(e₁), a contradiction that e₁ is critical. Hence there is no such edge f, and so ℓ = 2. Since this holds for all pairs (e₁, e_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:

The social cost of the path is ( n − 1 ) α + ⌊ n 2 ⌋ + ⌈ n 2 ⌉ = Θ ( n α + n 2 ). The adversary removes the one middle edge if n is even, or one of the two middle edges if n is odd. The disconnection cost for each player is n 2 ≤ α if n is even and at most 1 2 ( ⌊ n 2 ⌋ + ⌈ n 2 ⌉ ) = n 2 ≤ α if n is odd. Hence, there is no incentive for any player to build more edges than she currently owns, even after exchanging the currently built edges for others.

Now consider that a player v sells one (or two) of her edges and buys one (or two) different ones instead. First consider that one edge is exchanged. Since all edges point outwards, the part of the path that becomes disconnected from v does not contain the critical edge(s). So, after reconnecting it with v via a new edge, there are as many vertices on both sides of the formerly critical edge(s) as before the exchange. No separation value increases. Hence the formerly critical edges remain critical. They also maintain their relevance for v. With the same argument, if the exchanged edge itself was critical, the new one will be critical as well, also with the same relevance for v.

When two edges are exchanged, v is the center vertex, and in particular all critical edge(s) are among the exchanged ones, see Figure 3. This again means that the disconnected parts do not contain critical edges, and so the exchange cannot change that each of the two edges has ⌊ n 2 ⌋ vertices on the one and ⌈ n 2 ⌉ vertices on the other side, so they remain critical. Also, their relevance for v does not change. The MaxNE property is clear by Remark 2.1.

Removing a chord does not change the relevance of any edge, nor does it change sep_max, hence it does not change E_max. Selling a chord so is always beneficial.

With Proposition 8.10, it follows immediately:

An additional edge e only changes the ν (·) value of those edges which are put on a cycle by e, namely it reduces them to 0. Hence, none of the edges in {e₂,…, e_k} (or {e₃,…, e_k}) becomes less attractive for the adversary when e is added. Also no other edge becomes more attractive by the addition of e, since no ν(·) value increases.

Fix two critical edges e₁ and e₂, and set n₀ ≔ v(e₁). For each i ∈ {1, 2} fix a player v_i in the smaller component of G – e_i. Then for each i ∈ {1,2} we have rel(e_i, v_i) = n – n₀ and rel(e, v_i) = n₀ for all critical edges e ≠ e_i; recall that all critical edges have the same v(·) value. Building {v₁,v₂} puts e₁ and e₂ on a cycle and leaves the other m_max – 2 critical edges critical by Remark 9.5. For each i ∈ {1,2} [46], player v_i has her disconnection cost decreased by: 1 m max ∑ e ∈ E max rel ( e , v i ) − 1 m max − 2 ∑ e ∈ E max e ∉ { e 1 , e 2 } rel ( e , v i ) = 1 m max ( ( m max − 1 ) n 0 + n − n 0 ) − 1 m max − 2 ( m max − 2 ) n 0 = 1 m max ( ( m max − 2 ) n 0 + n ) − n 0 = 1 m max ( n − 2 n 0 )

Since we are in a NE, this is at most α. Since n ≥ m_maxn₀, we have n − 2n₀ ≥ (m_max − 2) n₀, and so it follows α ≥ ( 1 − 2 m max ) n 0 ≥ 1 3 n 0. Moreover, it follows m_maxα + n₀ ≥ n − n₀. With these two inequalities at hand, we can bound sep_max. We have sep max = 2 n 0 ( n − n 0 ) ≤ 2 n 0 ( m max α + n 0 ) ≤ 2 ( n 0 m max α + 9 α 2 ) ≤ 2 ( n α + 9 α 2 ) ≤ α ≤ c n 2 ( n α + 9 c n α ) = 2 ( 1 + 9 c ) n α

First we consider the case m_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 T₁ and T₂ that are connected to the rest by the two critical edges e₁ and e₂, respectively. They both have n₀ ≔ ν(e₁) = ν(e₂) vertices. There may be more subtrees T₃,…, T_N connected by e₃,…, e_N to the center vertex. To streamline notation, we often write T_k instead of V(T_k), k ∈ [N], when we refer to the set of vertices of a tree. Figure 4 on the next page depicts the situation. Figure 5 shows how a new edge would put e₁ and e₂ on a cycle.

First assume that we can arrange v₁ ∈ T₁ and v₂ ∈ T₂ such that after building {v₁, v₂}, there are no critical edges in T₁ nor in T₂. If there are no subtrees except T₁ and T₂, i.e., if N = 2, this means that we can make the graph bridgeless by the additional edge. The improvement in disconnection cost for v₁ (and also for v₂) of building {v₁,v₂} is hence their original disconnection cost, i.e., 1 2 ( n − n 0 + n 0 ) = 1 2 n ≥ 1 6 n. If N > 3, then critical edges emerge in one or more of the T₃ + e₃,…, T_N + e_N after building. Fix k ∈ {3,…, N}. Since e_k is not critical without the new edge, we have |Tk| < n₀ or | T k | ≥ ⌈ n 2 ⌉.The latter can be excluded, since it would imply that the smaller (or equally sized) component of G – e_k includes T₁ and the center vertex, and so ν(e_k) > n₀ = ν(e₁), in which case e₁ would not be critical. Moreover, we have |T_k| ≤ n − 2n₀ < n − 2 |T_k|, so | T k | < 1 3 n. For a player in T₁ (or T₂), a critical edge in T_k + e_k can have relevance at most |T_k| and so no more than 1 3 n.The improvement in disconnection cost for v₁ (and also for v₂) gained by building {v₁, v₂} is hence at least the original disconnection cost minus 1 3 n, i.e., 1 2 ( n − n 0 + n 0 ) − 1 3 n = ( 1 2 − 1 3 ) n = 1 6 n.

Now consider that for all choices of v₁ ∈ T₁ and v₂ ∈ T₂, building { v₁; v₂} induces a critical edge in at least one of T₁ or T₂. For each i ∈ {1,2} we can do the following. Let u_i be the vertex where T_i is connected to the rest of the graph and consider T_i being rooted at u_i. Let P_i be a path starting at u_i and ending at one of the leaves of T_i, say w_i, such that the path always descends into a subtree that has a maximum number of vertices, as shown in Figure 6 on the following page. If we choose v_i ≔ w_i, i = 1; 2, then each P_i does not contain a critical edge when we build { v₁; v₂}, since these paths then both are located on a cycle. However, by assumption, there is a critical edge f in, say T₁. By construction of P₁, we have ν ( f ) ≤ n 0 2. So, player v₁ (and also v₂) can reduce her disconnection cost to no more than n 0 2. It follows that the improvement in disconnection cost is at least 1 2 ( n − n 0 + n 0 ) − 1 2 n 0 = 1 2 ( n − n 0 ), which is at most α, since we are in a NE. It follows sep_max = 2n₀ (n − n₀) ≤ 2n₀. 2α ≤ 4n α.

The case of m_max = 1 can be treated similarly. Let e₁ be the critical edge and T₁ the subtree with n₀ ≔ v(e₁) vertices. There are zero or more additional subtrees, say T₂,…, T_N. If there are zero such trees, define T₂ ≔ G˜ – T₁, which consists of just one vertex in the bridge tree then (but can consist of multiple vertices in G). Let the ordering be such that |T₂| ≥ |T_k| for all k ∈ {3,…, N}. Then we argue similar to before with T₁ and T₂ in the roles of the former subtrees of the same name. Assume first that we can find v₁ ∈ T₁ and v₂ ∈ T₂ such that building {v₁,v₂} does not induce any critical edges in T₁ nor T₂. If N ≤ 2, then we can make the graph bridgeless and this means an improvement for v₁ of at least n −n₀, and so sep_max = 2n₀ (n − n₀) ≤ 2n₀ α ≤ 4nα. If N ≥ 3, then fix k ∈{3,…, N}. We have |T_k| < n₀. Moreover, we have |T_k| ≤ n − (|T2| + n₀) ≤ n − 2 |T_k|, and so | T k | ≤ 1 3 n. Then building {v₁, v₂ } reduces the disconnection cost of v₁ to no more than 1 3 n. This means an improvement for v₁ of at least n − n 0 − 1 3 n ≥ 2 3 n − 1 2 n = 1 6 n.

If each choice of v₁ and v₂ induces a critical edge in T₁ or T₂, we can, as before, show that by a careful choice of these vertices, building {v₁, v₂} reduces the disconnection cost for v₁ (and v₂) to at most n 0 2. Player v₁ originally has disconnection cost n − n₀ ≥ n₀, so she experiences an improvement of at least n 0 2. (Player v₂ originally has disconnection cost n₀, so she as well experiences an improvement of at least n 0 2.) It follows n₀ ≤ 2α and so sep_max = 2n₀ (n — n₀) ≤ 4αn.