Optimally reconnecting weighted graphs against an edge-destroying adversary
aa r X i v : . [ m a t h . C O ] F e b OPTIMALLY RECONNECTING WEIGHTED GRAPHS AGAINST AN EDGE-DESTROYING ADVERSARYDaniel C. McDonald
Wolfram Research Inc., Champaign, Illinois [email protected]
Abstract
We introduce a model involving two adversaries Buster and Fixer taking turns modifying a connected graph, where each roundconsists of Buster deleting a subset of edges and Fixer responding by adding edges from a reserve set of weighted edges to leave thegraph connected. With the weights representing the cost for Fixer to use specific reserve edges to reconnect the graph, we provide areasonable definition for what should constitute an optimal strategy for Fixer to keep the graph connected for as long as possible ascheaply as possible, and prove that a greedy strategy for Fixer satisfies our conditions for optimality.
1. Introduction
Suppose a network must stay connected in the face of some adversary that periodically destroys subsets of its edges, with the networkbeing reconnected after each attack by adding replacement edges, each of which has its own cost. Beyond the requirement that eachindividual selection of replacement edges must reconnect the network, ideally taken together these selections should, in some sense,keep the network connected for as long as possible as cheaply as possible. In Subsection 1.1 we introduce a graph-theoretic frameworkto formally model this situation, including a definition of optimality for selection of replacement edges, and in Subsection 1.2 we givean example of the model in action. In Subsection 1.3 we state our main theorem, that a greedy strategy for selecting replacementedges is in fact optimal, and in Subsection 1.4 we compare and contrast our model to the well-studied Maker-Breaker game as wellas outline some directions for future research. Using tools developed in Section 2, we prove our main theorem in Section 3.
To initialize an instance of our model, we are given a finite multigraph G (for this paper, all vertex sets of multigraphs will beunambiguous, so we view multigraphs simply as multisets of edges) and a finite multiset R of weighted ”reserve” edges betweenvertices of G . Each edge r ∈ R has some nonnegative, finite weight w ( r ); for R ′ ⊆ R , define w ( R ′ ) = P r ∈ R ′ w ( r ).There are two parties in this model: a positive actor Fixer and an antagonist Buster. A multigraph is to be first ”busted” byBuster and then ”fixed” by Fixer in each of a series of rounds. A given series will be given a name such as S . The k th round of series S begins with a finite multigraph G S k on the vertex set V and a multiset R S k of weighted ”reserve” edges in the complete graph on V .Start with G S = G and R S = R .Buster begins the k th round by removing some nonempty multiset B S k of edges from the current multigraph G S k . If adding all ofthe remaining multiset R S k of reserve edges to G S k − B S k does not result in a connected multigraph, then we say Buster wins in the k thround , and for convenience set F S k = ∅ (Fixer cannot reconnect the graph no matter how much she spends on reserve edges, so shemight as well not spend anything at all). Otherwise, Fixer responds by creating a connected multigraph from G S k − B S k by adding a(potentially empty) multiset F S k of edges from the remaining multiset R S k of reserve edges. In either case, set G S k +1 = ( G S k − B S k ) ∪ F S k and R S k +1 = R S k − F S k . If Buster does not win in the k th round, he has the option to quit (for instance, in a real-life scenario, externalfactors could prevent Buster from continuing), in which case we say Fixer wins in the k th round . If either Fixer or Buster wins inthe k th round, denote it by | S | = k .If S is a series satisfying | S | ≥ k , then a Fixer strategy to continue S after the k th round is a set φ of series satisfying the following:1. The series consisting solely of the first k rounds of S belongs to φ .2. If T ∈ φ , Fixer wins T in the j th round, and B ⊆ G T j is nonempty, then there exists exactly one series U ∈ φ such that B U i = B T i and F U i = F T i for 1 ≤ i ≤ j , B U j +1 = B , and | U | = j + 1.3. All series in φ are identical to S through the first k rounds.4. If T and U are in φ , with B T i = B U i for 1 ≤ i ≤ j and F T i = F U i for 1 ≤ i < j , then F T j = F U j .Equivalently, φ can be defined as the set of series represented by directed paths originating from the root vertex in some decision tree T with the following structure. The root vertex r of T represents the first k rounds of S , and a non-root vertex v of T at distance j from r represents the ( k + j )th round of a series whose first k rounds are represented by r , and whose ( k + i )th round for 1 ≤ i < j is represented by the vertex on the path from r to v at distance i from r ; the path P in T from r to v represents the series T whose k + j rounds are represented by the vertices along P . Given a vertex v in T and a series T whose first k + j rounds are identicalto the series represented by the path from r to v , for each possible Buster move B T k + j +1 in the ( k + j + 1)st round of T (i.e. eachnonempty subset of a connected G T k + j +1 ), there is a child of v representing the ( k + j + 1)st round of T ; that is, each possible Bustermove B T k + j +1 is assigned a single Fixer response F T k + j +1 . Note that each path in T from r to a leaf represents a series won by Buster,while each path in T from r to an interior vertex (including the path consisting solely of the root r , if T has multiple vertices, i.e.Buster doesn’t win S in the k th round) represents a series won by Fixer.Let S and S ′ be series such that G S = G S ′ and R S = R S ′ . Say S is Fixer-superior to S ′ if each of the following holds:1. Fixer wins S or Buster wins S ′ .2. P | S | j =1 | B S j | ≥ P | S ′ | j =1 | B S ′ j | P | S | j =1 w ( F S j ) ≤ P | S ′ | j =1 w ( F S ′ j )The first requirement means that in terms of winning or losing, Fixer does at least as well in S as in S ′ . The second requirementmeans that Buster works at least as hard deleting edges in S as in S ′ . The third requirement means that Fixer spends at leastas much on reserve edges in S ′ as in S . In establishing the second and third requirements, it is often helpful to observe that P | T | j =1 | B T j | = | G T | + | R T | − | G T | T | +1 | − | R T | T | +1 | and P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T | T | +1 ) for any series T . Note that every move isFixer-superior to itself.Fixer move F S k is optimal if there exists a Fixer strategy φ to continue S after the k th round, such that for any series S ′ identical to S though Buster’s move of the k th round and any Fixer strategy φ ′ to continue S ′ after the k th round, for any T ∈ φ there exists T ′ ∈ φ ′ for which T is Fixer-superior to T ′ . That is, for a series T identical to S through the first k rounds, against any Buster moves Fixer canalways continue play in T with B T k , F T k , B T k +1 , F T k +1 , . . . , B T | T | , F T | T | , in such a way that for any series T ′ identical to S through Buster’smove in the k th round, against any Fixer moves Buster can always continue play in T ′ with B T ′ k , F T ′ k , B T ′ k +1 , F T ′ k +1 , . . . , B T ′ | T ′ | , F T ′ | T ′ | , insuch a way that T is Fixer-superior to T ′ . It is not immediately obvious that an optimal move need exist. We consider a series S starting with G S = { e , e , e } , where these edges form a triangle, and R S = { e , e } , where e has the sameendpoints as e and satisfies w ( e ) = 1, while e has the same endpoints as e and satisfies and w ( e ) = 2; see Figure 1. e e e e e Figure 1: The solid edges form the graph G S , while the dashed edges form the reserve set R S , with w ( e ) = 1 and w ( e ) = 2.Suppose Buster’s first move in S is to play B S = { e , e } , and Fixer responds with F S = { e } . We detail a Fixer strategy φ = { T , . . . , T } to continue S after the first round below. A series T i in the same row as the j th round means that the rounds of T i are given by the first j rows of the given box (ignoring the “Winner” column until the j th round). T j G T j R T j B T j F T j P | B T i | P w ( F T i ) Winner T { e , e , e } { e , e } { e , e } { e } T { e , e } { e } { e } { e } T / T / T { e , e } {} { e } / { e } / { e , e } {} T { e , e , e } { e , e } { e , e } { e } T { e , e } { e } { e } { e } T / T / T { e , e } {} { e } / { e } / { e , e } {} T { e , e , e } { e , e } { e , e } { e } T { e , e } { e } { e , e } {} S ′ identical to S up through Buster’s removal of a set of edges in the first round, but supposeFixer responds in S ′ with F S ′ = { e } . We detail a Fixer strategy φ ′ = { T ′ , . . . , T ′ } to continue S ′ after the first round below. T ′ j G T ′ j R T ′ j B T ′ j F T ′ j P | B T ′ i | P w ( F T ′ i ) Winner T ′ { e , e , e } { e , e } { e , e } { e } T ′ { e , e } { e } { e } { e } T ′ / T ′ / T ′ { e , e } {} { e } / { e } / { e , e } {} T ′ { e , e , e } { e , e } { e , e } { e } T ′ { e , e } { e } { e } { e } T ′ / T ′ / T ′ { e , e } {} { e } / { e } / { e , e } {} T ′ { e , e , e } { e , e } { e , e } { e } T ′ { e , e } { e } { e , e } {} φ ′ is, in fact, the only Fixer strategy to continue S ′ after the first round. Indeed, G S ′ = { e , e } , which is a pathgraph, and R S ′ = { e } , where e spans the enpdoints of that path. Thus Buster deleting any single edge from G S ′ necessitates Fixerreconnecting the graph using e , from which point on any further deletions by Buster cannot be countered by Fixer, and Busterinitially deleting both edges from G S ′ also cannot be countered by Fixer. Furthermore, see that for 1 ≤ i ≤
10, the series T i in φ isFixer-superior to the series T ′ i in φ ′ . Hence φ is a Fixer strategy to continue S after the first round, such that for any Fixer strategy φ ′ to continue S ′ after the first round, for any T ∈ φ there exists T ′ ∈ φ ′ for which T is Fixer-superior to T ′ .Now consider another alternate series S ′′ identical to S up through Buster’s removal of a set of edges in the first round, but supposeFixer responds in S ′′ with F S ′′ = { e , e } . We detail a Fixer strategy φ ′′ = { T ′′ , . . . , T ′′ } to continue S ′′ after the first round below. T ′′ j G T ′′ j R T ′′ j B T ′′ j F T ′′ j P | B T ′′ i | P w ( F T ′′ i ) Winner T ′′ { e , e , e } { e , e } { e , e } { e , e } T ′′ { e , e , e } {} { e } {} T ′′ / T ′′ / T ′′ { e , e } {} { e } / { e } / { e , e } {} T ′′ { e , e , e } { e , e } { e , e } { e , e } T ′′ { e , e , e } {} { e } {} T ′′ / T ′′ / T ′′ { e , e } {} { e } / { e } / { e , e } {} T ′′ { e , e , e } { e , e } { e , e } { e , e } T ′′ { e , e , e } {} { e } {} T ′′ / T ′′ / T ′′ { e , e } {} { e } / { e } / { e , e } {} T ′′ { e , e , e } { e , e } { e , e } { e , e } T ′′ / T ′′ / T ′′ / T ′′ { e , e , e } {} { e , e } / { e , e } / { e , e } / { e , e , e } {} φ ′′ is the only Fixer strategy to continue S ′′ after the first round since R S ′′ is empty, leaving Fixer with nooptions besides playing the empty set when possible. See that for 1 ≤ i ≤
9, the series T i in φ is Fixer-superior to the series T ′′ i in φ ′′ ,while T is Fixer-superior to T ′′ . Hence φ is a Fixer strategy to continue S after the first round, such that for any Fixer strategy φ ′′ to continue S ′′ after the first round, for any T ∈ φ there exists T ′′ ∈ φ ′ for which T is Fixer-superior to T ′′ .Since F S , F S ′ , and F S ′′ are Fixer’s only choices for responding to Buster in the first round, we have shown that φ is a Fixerstrategy to continue S after the first round, such that for any Fixer strategy φ ′ or φ ′′ to continue any alternate move F S ′ or F S ′′ afterthe first round, for any T ∈ φ there exists T ′ ∈ φ ′ and T ′′ ∈ φ ′′ for which T is Fixer-superior to T ′ and T ′′ . Hence F S is optimal. Notethat because T ′ and T ′′ are part of φ ′ and φ ′′ , respectively (the only Fixer strategies to continue S ′ and S ′′ after the first round), butare not Fixer-superior to any series in φ , by definition F S ′ and F S ′′ are not optimal. Fixer move F S k ⊆ R S k to create a connected multigraph G S k +1 by adding F S k to G S k − B S k is greedy if, for any other series S ′ identical to S through Buster’s move of the k th round, Fixer move F S ′ k ⊆ R S ′ k to create a connected multigraph G S ′ k +1 by adding F S ′ k to G S ′ k − B S ′ k satisfies w ( F S k ) ≤ w ( F S ′ k ). That is, Fixer plays greedily in response to a move by Buster by adding no reserve edge if the graphremains connected, and otherwise adding some cheapest set of reserve edges that connects the graph.Note that every optimal Fixer move is necessarily greedy. Indeed, if S and S ′ are identical through Buster’s move of the k th round,but w ( F S k ) > w ( F S ′ k ), then for any Fixer strategy φ to continue S after the k th round, the series T ∈ φ identical to S through the k thround but ending with | T | = k cannot be Fixer-superior to any series T ′ in a Fixer strategy φ ′ to continue S ′ after the k th round.This is because if | T ′ | = k then P | T | j =1 w ( F T j ) > P | T ′ | j =1 w ( F T ′ j ), while if | T ′ | > k then P | T | j =1 | B T j | < P | T ′ | j =1 | B T ′ j | . Our main theoremstates that greediness is also a sufficient condition for optimality. Theorem 1.1.
Every greedy move by Fixer is optimal.
After establishing some facts about spanning trees, Fixer-superiority, and optimality in Section 2, we use them to prove Theorem1.1 in Section 3. Our proof will be by induction, split into cases by the number c of components of G S − B S . We shall see that thecase c = 1 is mostly trivial, the case c = 2 is the most difficult and requires case analysis of each move to verify that certain invariantsare maintained, and the case c ≥ For a family F of subgraphs of of the complete graph K n , the unbiased Maker-Breaker game on F consists of players Maker andBreaker taking turns claiming edges of K n (see [5] and [1] for some notable early results, and [3] for a more recent survey). Makerwins by claiming all edges of some graph in F , while Breaker wins if all edges are claimed before Maker wins (equivalently, Breakerwins by claiming an edge from each minimal member of F ). The family F most relevant for comparison of the Maker-Breaker gameon F with our game is the family of connected spanning subgraphs of K n . The gameplay of Maker-Breaker differs from that ofBuster-Fixer in several obvious ways, including the following:1. Maker only needs to end up with a connected graph, while Fixer must maintain connectedness after each turn.2. Maker cannot replace edges claimed by Breaker, whereas the reserve edges Fixer may select could include edges with the sameendpoints as the edges deleted by Buster.3. The Maker-Breaker game does not typically include weighted edges, which are is a consideration in the Buster-Fixer model.4. Maker’s objective is to directly beat Breaker (and vice-versa), but the relationship between Fixer and Buster is less symmetric;Buster is more of an agent of chaos than a goal-oriented player (e.g. Buster can simply choose to stop participating at anypoint), and Fixer’s objective is to do her best to keep the graph connected for as long and cheaply as possible based on Buster’sactions, regardless of how well Buster’s edge deletions actually do to disconnect the graph in expensive ways.We believe the last difference listed is the most important to take note of, as it provides a contrast in the fundamental structureof the models, which further dictates how results are stated for each model. A standard Maker-Breaker result (similar to manytypical results in positional game theory) would be a statement of conditions on n and F for Maker or Breaker to win, most likelyconstructively proven via an explicit strategy for Maker or Breaker. Our Theorem 1.1, that every greedy move by Fixer is optimal, isof a different flavor though. Since Fixer can’t even “win” if Buster plays for long enough, and has no way of forcing Buster to quit,we must compare Fixer strategies against each other, rather than use a single Fixer strategy to prove Fixer can achieve a certaingoal. Hence the optimal strategy for Fixer is the statement, proven by showing its superiority to all other Fixer strategies.Many avenues exist for future research into variations on our Buster-Fixer model. In particular, we wonder about optimal Fixerstrategies for alternative games, where the condition that Fixer must maintain on the graph through each round is changed frommaintaining connectedness to one of the following conditions:1. Two given vertices s and t must stay in the same component.2. The graph must stay k -connected for a given k > s to (or from) all other vertices.(c) The directed graph must have a directed path from a given vertex s to a given vertex t .(d) The directed graph must have directed paths in both directions between given vertices s and t .
2. Preliminaries2.1. Spanning Trees and Prim’s Algorithm A bridge in a multigraph M is an edge e such M − { e } has one more component than M ; equivalently, e is part of no cycle in M .A spanning tree of a connected multigraph M is a subgraph T of M such that T is a tree (i.e. connected and acyclic) whose vertexset matches that of M . A minimum spanning tree of an edge-weighted multigraph M is a spanning tree of M minimizing the totalweight of the edges. Minimum spanning trees are of interest to us because F S k is greedy if and only if it is a minimum spanning treeof the multigraph whose vertices are the components of G S k − B S k and whose edges are the edges of R S k (identifying each endpoint ofthe edges in R S k with the component of G S k − B S k within which it lies). Prim’s Algorithm (first discovered by Jarnik [4] and later by Prim [6] and Dijkstra [2]) finds a minimum spanning tree T of aweighted connected multigraph M one edge at a time by the following construction: with T initialized as any vertex, iteratively addto T any cheapest edge of M joining a vertex in T to one not yet in T , until all vertices of M are in T .We require not just the fact that Prim’s Algorithm successfully produces a minimum spanning tree, but also the fact that anyminimum spanning tree can be constructed via Prim’s Algorithm. Proposition 2.1.
A spanning tree of a weighted connected multigraph is a minimum spanning tree if and only if it can be constructedvia Prim’s Algorithm.Proof.
Let M be a weighted connected multigraph, let P be a subgraph of M constructed by applying Prim’s Algorithm, and let T be a minimum spanning tree of M . We complete the proof by showing that P is in fact a minimum spanning tree of M , and T canbe constructed via Prim’s Algorithm.If P is constructed via Prim’s Algorithm, then see that P is a spanning tree of M :1. P is connected because P is initialized as a single component (its single starting vertex), and the connectedness of P is maintainedas each new vertex is added as the endpoint of an edge whose other endpoint was already in P .2. P spans M because M is connected, so if some vertex of M is not yet in P , then some new vertex can always be added to P .3. P is acyclic because every edge added is a bridge in P .If P = T then P is a minimum spanning tree. Otherwise, let e be the first edge added during the construction of P that is not in T , let V be the set of vertices connected by the edges added before adding e , and let f be an edge in the path through T betweenthe endpoints of e such that one endpoint of f is in V but the other is not. Let T ′ be the spanning tree of M constructed from T byreplacing f with e .Since e and f are each edges with exactly one endpoint in V and e was added to P by Prim’s Algorithm, e cannot weigh morethan f . Hence T ′ cannot weigh more than T , so T ′ must be a minimum spanning tree of M as well. This process of constructingminimum spanning trees of M each with one more edge in common with P than the last can be continued until P is the minimumspanning tree constructed.Since T and T ′ are both minimum spanning trees of M , they must weigh the same. Since T ′ was constructed from T by replacing e with f , e and f must weigh the same. Hence f could have also been added by Prim’s Algorithm to extend the construction of aminimum spanning tree of M . This process of growing by an edge the subtree of T that can be shown to have been created accordingto Prim’s Algorithm can be continued until all of T is shown to have been created by Prim’s Algorithm. We first verify that Fixer-superiority is transitive.
Proposition 2.2.
Suppose S , S ′ , and S ′′ are series such that G S = G S ′ = G S ′′ and R S = R S ′ = R S ′′ . If S is Fixer-superior to S ′ ,and S ′ is Fixer-superior to S ′′ , then S is Fixer-superior to S ′′ .Proof. We have1. Fixer wins S , or Buster wins S ′ (since S is Fixer-superior to S ′ ), in which case Buster wins S ′′ (since S ′ is Fixer-superior to S ′′ ).2. P | S | j =1 | B S j | ≥ P | S ′ | j =1 | B S ′ j | ≥ P | S ′′ | j =1 | B S ′′ j | P | S | j =1 w ( F S j ) ≤ P | S ′ | j =1 w ( F S ′ j ) ≤ P | S ′′ | j =1 w ( F S ′′ j )so S is Fixer-superior to S ′′ by definition.We next show that Fixer playing only optimal moves past some round leads to a series that is Fixer-superior to certain otherseries, further justifying our definition of “optimal”. Lemma 2.3.
Let S be identical to S ′ up through Buster’s removal of a set of edges in the k th round, and let φ ′ be a strategy for Fixerto continue S ′ after the k th round. If F S j is optimal for j ≥ k , then there exists T ′ ∈ φ ′ such that S is Fixer-superior to T ′ .Proof. For k ≤ j ≤ | S | , since F S j is optimal there exists a strategy φ j for Fixer to continue S after the j th round, such that for anyseries S j identical to S up through Buster’s move in the j th round, every series in φ j is Fixer-superior to some series in any strategyfor Fixer to continue S j after the j th round. For each k ≤ j < | S | , set ˆ φ j as the subset of φ j consisting of its series which either endby the j th round or have Buster’s move in the ( j + 1)st round match B S j +1 , and set ˆ φ k − = φ ′ and ˆ φ | S | = φ | S | . Note that S ∈ ˆ φ | S | ,and for k ≤ j ≤ | S | every series in ˆ φ j is Fixer-superior to some series in ˆ φ j − (since ˆ φ j ⊆ φ j and for a series S j identical to S upthrough Buster’s move in the j th round, ˆ φ j − is a strategy for Fixer to continue S j after the j th round).We iteratively construct a sequence T | S | , T | S |− , . . . , T k − of series, with T j ∈ ˆ φ j for each j . First set T | S | = S ∈ ˆ φ | S | . Then, havingalready constructed T | S | , T | S |− , . . . , T j for some j ≥ k , select T j − ∈ ˆ φ j − so that T j is Fixer-superior to T j − . Hence by Proposition2.2, T | S | is Fixer-superior to T k − . Since S = T | S | and T k − ∈ ˆ φ k − = φ ′ , the proof is complete.Finally, we show that to verify the optimality of some Fixer move, we need only compare it to alternate Fixer moves consistingsolely of bridges. Proposition 2.4.
Suppose φ is a Fixer strategy to continue S after the k th round such that for every series S ′ identical to S throughBuster’s move of the k th round for which every edge of F S ′ k is a bridge in G S ′ k +1 , for any T ∈ φ and any Fixer strategy φ ′ to continue S ′ after the k th round, there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ . Then F S k is optimal.Proof. Let T ∈ φ , let S ′ and S ′′ be series identical to S through Buster’s move of the k th round such that F S ′ k ⊆ F S ′′ k and every edgeof F S ′ k is bridge in G S ′ k +1 , and let φ ′′ be any Fixer strategy to continue S ′′ after the k th round. To complete the proof, we constructa Fixer strategy φ ′ to continue S ′ after the k th round such that for every T ′ ∈ φ ′ there exists T ′′ ∈ φ ′′ for which T ′ is Fixer-superiorto T ′′ . Indeed, by the hypothesis of this proposition there would exist T ′ ∈ φ ′ such that T is Fixer-superior to T ′ , and if there exists T ′′ ∈ φ ′′ such that T ′ is Fixer-superior to T ′′ , then T would be Fixer-superior to T ′′ , by Proposition 2.2; hence F S k would be optimalby definition since T ∈ φ , F S ′′ k , and φ ′′ were arbitrary. To construct φ ′ , we define an arbitrary series T ′ ∈ φ ′ ; that is, we let T ′ beidentical to S ′ through the k th round, and for arbitrary plays B T ′ j from Buster in the j th round for j > k , we assign Fixer responses F T ′ j derived from φ ′′ in such a way that T ′ is Fixer-superior to some T ′′ ∈ φ ′′ .First suppose | T ′ | = k , in which case let T ′′ be the lone series in φ ′′ satisfying | T ′′ | = k (i.e. T ′′ consists of the first k rounds of S ′′ ).Note that1. Either Fixer wins T ′ , or Buster wins T ′ , in which case ( G T ′ k − B T ′ k ) ∪ R T ′ k is disconnected, leaving ( G T ′′ k − B T ′′ k ) ∪ R T ′′ k alsodisconnected since it’s the same graph, meaning Buster also wins T ′′ P | T ′ | j =1 | B T ′ j | = P kj =1 | B S j | = P | T ′′ | j =1 | B T ′′ j | P | T ′ | j =1 w ( F T ′ j ) = P k − j =1 w ( F S j ) + w ( F T ′ k ) ≤ P k − j =1 w ( F S j ) + w ( F T ′′ k ) = P | T ′′ | j =1 w ( F T ′′ j )so T ′ is Fixer-superior to T ′′ .Now suppose | T ′ | > k . Note that F S ′ k ⊆ F S ′′ k implies for any T ′′ ∈ φ ′′ both that R T ′′ k +1 ⊆ R T ′ k +1 (since R T ′ k +1 and R T ′′ k +1 wereconstructed from R S k by removing F S ′ k and F S ′′ k , respectively) as well as that G T ′ k +1 is a subgraph of G T ′′ k +1 (since G T ′ k +1 and G T ′′ k +1 wereconstructed from G S k − B S k by adding F S ′ k and F S ′′ k , respectively). By the latter of these observations, there exists T ′′ ∈ φ ′′ such that B T ′′ k +1 = B T ′ k +1 ; we shall choose our T ′′ ∈ φ ′′ for which T ′ is Fixer-superior to T ′′ to satisfy B T ′′ k +1 = B T ′ k +1 . Note that for D = F S ′′ k − F S ′ k ,( G T ′ k +1 − B T ′ k +1 ) ∪ R T ′ k +1 = ((( G T ′ k − B T ′ k ) ∪ F T ′ k ) − B T ′ k +1 ) ∪ ( R T ′ k − F T ′ k )= ((( G T ′′ k − B T ′′ k ) ∪ ( F T ′′ k − D )) − B T ′′ k +1 ) ∪ ( R T ′′ k − F T ′′ k ) ∪ D = ((( G T ′′ k − B T ′′ k ) ∪ F T ′′ k ) − B T ′′ k +1 ) ∪ ( R T ′′ k − F T ′′ k )= ( G T ′′ k +1 − B T ′′ k +1 ) ∪ R T ′′ k +1 so ( G T ′ k +1 − B T ′ k +1 ) ∪ R T ′ k +1 is connected if and only if ( G T ′′ k +1 − B T ′′ k +1 ) ∪ R T ′′ k +1 is connected.If Buster wins T ′ in the ( k + 1)st round, then1. Buster also wins T ′′ in the ( k + 1)st round, as ( G T ′′ k +1 − B T ′′ k +1 ) ∪ R T ′′ k +1 is disconnected since ( G T ′ k +1 − B T ′ k +1 ) ∪ R T ′ k +1 is disconnecteddue to Buster winning T ′ in the ( k + 1)st round2. P | T ′ | j =1 | B T ′ j | = P kj =1 | B S j | + | B T ′ k +1 | = P kj =1 | B S j | + | B T ′′ k +1 | = P | T ′′ | j =1 | B T ′′ j | P | T ′ | j =1 w ( F T ′ j ) = P k − j =1 w ( F S j ) + w ( F T ′ k ) + 0 ≤ P k − j =1 w ( F S j ) + w ( F T ′′ k ) + 0 = P | T ′′ | j =1 w ( F T ′′ j ) since F T ′ k +1 = F T ′′ k +1 = ∅ so T ′ is Fixer-superior to T ′′ .Thus we may suppose either Fixer wins T ′ in the ( k + 1)st round or | T ′ | ≥ k + 2. Then ( G T ′ k +1 − B T ′ k +1 ) ∪ R T ′ k +1 is connected, so( G T ′′ k +1 − B T ′′ k +1 ) ∪ R T ′′ k +1 is also connected, meaning either Fixer wins T ′′ in the ( k + 1)st round or | T ′′ | ≥ k + 2. Suppose according to φ ′′ that Fixer repairs G T ′′ k +1 − B T ′′ k +1 with the set F T ′′ k +1 ⊆ R T ′′ k +1 to create the connected graph G T ′′ k +2 . Define φ ′ so that Fixer repairs G T ′ k +1 − B T ′ k +1 with the set F T ′ k +1 = F T ′′ k +1 ∪ (( F T ′′ k − F T ′ k ) − B T ′ k +1 ) to create the graph G T ′ k +2 .First, note that F T ′ k +1 ⊆ R T ′ k +1 since F T ′′ k +1 ⊆ R T ′′ k +1 ⊆ R T ′ k +1 and F T ′′ k − F T ′ k = R T ′ k +1 − R T ′′ k +1 ⊆ R T ′ k +1 . Hence Fixer can play F T ′ k +1 aslong as it makes G T ′ k +2 connected, which is the case since G T ′′ k +2 is connected and G T ′ k +2 = G T ′′ k +2 : G T ′ k +2 = ( G T ′ k +1 − B T ′ k +1 ) ∪ F T ′ k +1 = ( G T ′′ k +1 − ( F T ′′ k − F T ′ k ) − B T ′ k +1 ) ∪ F T ′′ k +1 ∪ (( F T ′′ k − F T ′ k ) − B T ′ k +1 )= ( G T ′′ k +1 − B T ′ k +1 ) ∪ F T ′′ k +1 = ( G T ′′ k +1 − B T ′′ k +1 ) ∪ F T ′′ k +1 = G T ′′ k +2 Next, note that R T ′ k +2 = R T ′′ k +2 since R T ′ k +2 = R T ′ k − ( F T ′ k ∪ F T ′ k +1 )= R T ′′ k − ( F T ′ k ∪ F T ′′ k +1 ∪ (( F T ′′ k − F T ′ k ) − B T ′ k +1 ))= R T ′′ k − (( F T ′′ k − ( B T ′ k +1 − F T ′ k )) ∪ F T ′′ k +1 )= R T ′′ k − ( F T ′′ k ∪ F T ′′ k +1 )= R T ′′ k +2 (using the facts that R T ′ k = R T ′′ k , that F T ′ k +1 = F T ′′ k +1 ∪ (( F T ′′ k − F T ′ k ) − B T ′ k +1 ), and that F T ′′ k ∩ ( B T ′ k +1 − F T ′ k ) = ( F T ′′ k − F T ′ k ) ∩ B T ′ k +1 = ∅ since F T ′′ k − F T ′ k ⊆ R T ′ k +1 and R T ′ k +1 ∩ B T ′ k +1 = ∅ ).Finally, see that since G T ′ k +2 = G T ′′ k +2 and R T ′ k +2 = R T ′′ k +2 , φ ′ can continue to be defined by copying φ ′′ in the following way. Assuming T ′ has been defined up to the start of the j th round for some j ≥ k + 2 in such a way that G T ′ j = G T ′′ j and R T ′ j = R T ′′ j for some T ′′ ∈ φ ′′ , and Buster removes some set B T ′ j of edges from G T ′ j in T ′ , set B T ′′ j = B T ′ j and let F T ′′ j be Fixer’s response in T ′′ prescribedby φ ′′ . Then set F T ′ j = F T ′′ j , leaving G T ′ j +1 = G T ′′ j +1 and R T ′ j +1 = R T ′′ j +1 . Continuing this process up through the final round ℓ of T ′ ,which we also let be the final round of T ′′ (either automatically if Buster wins, or by letting Buster quit if Fixer wins), we see that T ′ is Fixer-superior to T ′′ because1. Either Fixer wins T ′ , or Buster wins T ′ , in which case ( G T ′ ℓ − B T ′ ℓ ) ∪ R T ′ ℓ is disconnected, leaving ( G T ′′ ℓ − B T ′′ ℓ ) ∪ R T ′′ ℓ alsodisconnected since it’s the same graph, meaning Buster also wins T ′′ P | T ′ | j =1 | B T ′ j | = | G S | + | R S | − | G T ′ ℓ +1 | − | R T ′ ℓ +1 | = | G S | + | R S | − | G T ′′ ℓ +1 | − | R T ′′ ℓ +1 | = P | T ′′ | j =1 | B T ′′ j | P | T ′ | j =1 w ( F T ′ j ) = w ( R S ) − w ( R T ′ ℓ +1 ) = w ( R S ) − w ( R T ′′ ℓ +1 ) = P | T ′′ | j =1 w ( F T ′′ j )and thus we have constructed φ ′ so that for every T ′ ∈ φ ′ there exists T ′′ ∈ φ ′′ for which T ′ is Fixer-superior to T ′′ . Hence F S k isoptimal.
3. Proof of Main Theorem
To prove our main theorem, that during any series S , any greedy Fixer move F S k is optimal, we perform induction on | G S k | + | R S k | .To help with the base case, we use the following proposition. Proposition 3.1.
If Buster wins S in the k th round, then F S k is greedy and optimal.Proof. If Buster wins S in the k th round, then ( G S k − B S k ) ∪ R S k is disconnected, and by convention F S k = ∅ , which is greedy. Clearlythe only series identical to S through Buster’s move in the k th round is S itself, and the only strategy for Fixer to continue S afterthe k th round is { S } , so F S k is optimal because S is Fixer-superior to itself.Let V be the vertex set of G S , with | V | = n , and without loss of generality assume k = 1. Note that | G S | ≥ n − G S isconnected, so for our base case we consider | G S | + | R S | = n −
1. In this case, | G S | = n − R S = ∅ , and B S ⊆ G S is nonempty;then ( G S − B S ) ∪ R S is disconnected because it has at most n − S during the first round, andProposition 3.1 applies.Hence we may suppose | G S | + | R S | ≥ n , and inductively assume during any series T such that G T is a connected graph on V and | G T k | + | R T k | < | G S | + | R S | , any greedy Fixer move F T k is optimal. Furthermore, by Proposition 3.1, we may assume that Buster doesnot win S during the first round. Let φ be a greedy Fixer strategy to continue S after Fixer’s greedy move F S of the first round; bythe inductive hypothesis, all Fixer moves in φ past the first round are optimal. Let S ′ be an arbitrary series identical to S throughBuster’s move of the first round such that every edge of F S ′ is a bridge in G S ′ , and let φ ′ be an arbitrary strategy for Fixer tocontinue S ′ after the first round. By Proposition 2.4, in order to show F S is optimal, it suffices to show that for any T ∈ φ thereexists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ . Note that if Fixer wins T in the first round then T is clearly Fixer-superior to theonly series T ′ ∈ φ ′ satisfying | T ′ | = 1, so we may assume | T | > φ to continue S after Fixer’s greedy move F S of the first round, fixany series T ∈ φ satisfying | T | >
1, and fix any Fixer strategy φ ′ to continue S ′ after the first round. Let c equal the number ofcomponents of G S − B S . Let M be the multigraph whose vertices are the components of G S − B S and whose edges are the edges of R S (identifying each endpoint of the edges in R S with the component of G S − B S within which it lies), so F S is a minimum spanningtree of M , and F S ′ is a spanning tree of M . We complete the proof by showing for each value of c that there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ , handling separately the cases c = 1, c = 2, and c ≥ c = 1If c = 1, then the only spanning tree of M is edgeless, so F S = F S ′ = ∅ . For our fixed series T ∈ φ , let T ′ be any series in φ ′ identicalto T through Buster’s move of the second round. Since F T j is optimal for j ≥
2, and the subset φ ′′ of φ ′ consisting of all series in φ ′ identical to T ′ through the second round forms a strategy for Fixer to continue T ′ after the second round, by Lemma 2.3 there exists T ′′ ∈ φ ′′ ⊆ φ ′ such that T is Fixer-superior to T ′′ . c = 2If c = 2, then the spanning trees of M are the individual edges in R S joining the two components of G S − B S . Hence for two suchedges s and s ′ , where no such edge is cheaper than s , we have F S = { s } and F S ′ = { s ′ } . We establish Lemmas 3.2 and 3.3 in orderto prove Proposition 3.4, which provides a strategy for proving that F S = { s } is optimal. Lemma 3.2.
Suppose U is a series identical to S through the first round, with F U j greedy for j > k for some k , which satisfies ≤ k ≤ | U | − if Buster wins U and ≤ k ≤ | U | − if Fixer wins U . Then there exists a series U ( k + ) identical to U through the k th round such that | B U ( k + ) k +1 | = 1 , F U ( k + ) j is greedy for j > k , and U is Fixer-superior to U ( k + ) .Proof. If | B U k +1 | = 1, set U ( k + ) = U , so U ( k + ) is identical to U through the k th round, | B U ( k + ) k +1 | = | B U k +1 | = 1, F U ( k + ) j = F U j is greedy for j > k , and U is Fixer-superior to U ( k + ) since every series is Fixer-superior to itself.Thus we may assume | B U k +1 | >
1. Let U ( k + ) be the series identical to U through the k th round, with the rest of U ( k + )constructed as follows. We show that there exists an edge b ∈ B U k +1 such that if Buster plays B U ( k + ) k +1 = { b } in U ( k + ), thenFixer can respond with some greedy F U ( k + ) k +1 ⊆ F U k +1 . If this is the case, then for the ( k + 2)nd round in U ( k + ) Buster could play B U ( k + ) k +2 = B U k +1 − { b } since B U ( k + ) k +2 = B U k +1 − { b }⊆ G U k +1 − { b } = G U ( k + ) k +1 − B U ( k + ) k +1 ⊆ ( G U ( k + ) k +1 − B U ( k + ) k +1 ) ∪ F U ( k + ) k +1 = G U ( k + ) k +2 and Fixer could respond with F U ( k + ) k +2 = F U k +1 − F U ( k + ) k +1 since F U ( k + ) k +2 = F U k +1 − F U ( k + ) k +1 ⊆ R U k +1 − F U ( k + ) k +1 = R U ( k + ) k +1 − F U ( k + ) k +1 = R U ( k + ) k +2 and G U ( k + ) k +3 = ( G U ( k + ) k +2 − B U ( k + ) k +2 ) ∪ F U ( k + ) k +2 = ((( G U ( k + ) k +1 − { b } ) ∪ F U ( k + ) k +1 ) − ( B U k +1 − { b } )) ∪ ( F U k +1 − F U ( k + ) k +1 )= ((( G U k +1 − { b } ) ∪ F U ( k + ) k +1 ) − ( B U k +1 − { b } )) ∪ ( F U k +1 − F U ( k + ) k +1 )= ( G U k +1 − B U k +1 ) ∪ F U k +1 = G U k +2 which is connected since Buster doesn’t win U in the ( k + 1)st round. Furthermore, Fixer’s move F U ( k + ) k +2 in U ( k + ) would begreedy, since otherwise Fixer’s move F U k +1 = F U ( k + ) k +1 ∪ F U ( k + ) k +2 in U would not have been greedy, contradicting the hypotheses ofthis lemma. Then G U ( k + ) k +3 = G U k +2 and R U ( k + ) k +3 = R U ( k + ) k +1 − ( F U ( k + ) k +1 ∪ F U ( k + ) k +2 ) = R U k +1 − F U k +1 = R U k +2 , so setting B U ( k + ) j +1 = B U j and F U ( k + ) j +1 = F U j for j ≥ k + 2 would be valid plays by Buster and Fixer in U ( k + ), with Fixer’s moves being greedy because theywere greedy in U . Thus we’d have1. Fixer wins U , or Buster wins U and thus also wins U ( k + ).2. P | U | j =1 | B U j | = P | U ( k + ) | j =1 | B U ( k + ) j | P | U | j =1 w ( F U j ) = P | U ( k + ) | j =1 w ( F U ( k + ) j )so U would be Fixer-superior to U ( k + ). We complete the proof by showing there exists b ∈ B U k +1 such that Fixer can respond to B U ( k + ) k +1 = { b } with some greedy F U ( k + ) k +1 ⊆ F U k +1 .If there exists b ∈ B U k +1 such that G U k +1 − { b } is connected, let Buster play B U ( k + ) k +1 = { b } in U ( k + ). Since G U ( k + ) k +1 − B U ( k + ) k +1 = G U k +1 − { b } and is therefore connected, Fixer can respond greedily in U ( k + ) with F U ( k + ) k +1 = ∅ ⊆ F U k +1 .Finally, suppose G U k +1 − { b } is disconnected for every b ∈ B U k +1 . Since F U k +1 is greedy, it is therefore a minimum spanning treeof the multigraph H whose vertices are the components of G U k +1 − B U k +1 and whose edges are the edges of R U k +1 (identifying eachendpoint of the edges in R U k +1 with the component of G U k +1 − B U k +1 within which it lies). By Proposition 2.1 there exists an orderingof F U k +1 where edges appear in the order they were added by Prim’s Algorithm; let f be the first edge in this ordering. Let b be anedge in B U k +1 such that ( G U k +1 − { b } ) ∪ { f } is connected; note that such a b exists because otherwise f would be a loop in H andtherefore couldn’t be part of any minimum spanning tree. Let Buster play B U ( k + ) k +1 = { b } in U ( k + ), and have Fixer respond with F U ( k + ) k +1 = { f } . Fixer’s move is greedy because if there were some edge r ∈ R U ( k + ) k +1 such that w ( r ) < w ( f ) and ( G U k +1 − { b } ) ∪{ r } wasconnected, then r would’ve been chosen before f by Prim’s Algorithm in constructing a minimum spanning tree of H , contradicting f being the first edge chosen. Lemma 3.3.
Suppose U ( k ) is a series identical to S through the first round, with F U ( k ) j greedy for j ≥ k for some < k < | U ( k ) | .If F is a subset of R U ( k ) k such that ( G U ( k ) k − B U ( k ) k ) ∪ F is connected, then there exists a series U ( k + ) identical to U ( k ) throughBuster’s move in the k th round such that F U ( k + ) k = F , | B U ( k + ) k +1 | = 1 if Buster doesn’t win U ( k + ) in the ( k + 1) st round, F U ( k + ) j is greedy for j > k , and U ( k ) is Fixer-superior to U ( k + ) .Proof. By the inductive hypothesis of this section, for j ≥ k , F U ( k ) j is optimal since F U ( k ) j is greedy. Let ˆ φ be a greedy strategy forFixer to continue the series ˆ U after the k th round, where ˆ U is identical to U ( k ) through Buster’s move in the k th round, and F ˆ U k = F .Since U ( k ) is part of some Fixer strategy to continue U ( k ) after the k th round where Fixer only makes optimal moves after the k thround, and ˆ U is a series identical to U ( k ) through Buster’s move in the k th round, by Lemma 2.3 U ( k ) is Fixer-superior to someseries U ∈ ˆ φ . Note that k < | U | , since otherwise | U ( k ) | X j =1 | B U ( k ) j | > k X j =1 | B U ( k ) j | = | U | X j =1 | B U j | contradicting U ( k ) being Fixer-superior to U . If Buster wins U in the ( k + 1)st round, then F U k +1 is empty and greedy by convention,so we can set U ( k + ) = U . If Buster doesn’t win U in the ( k + 1)st round, then by Lemma 3.2 there exists a series U ( k + ) identicalto U through the k th round such that | B U ( k + ) k +1 | = 1, F U ( k + ) j is greedy for j > k , and U is Fixer-superior to U ( k + ). Since U ( k ) isFixer-superior to U , and U is Fixer-superior to U ( k + ), by Proposition 2.2 U ( k ) is Fixer-superior to U ( k + ).0Recall that in order to show F S is optimal, we fixed a series T ∈ φ that we must show is Fixer-superior to some T ′ ∈ φ ′ , where φ is a greedy Fixer strategy to continue S after the first round, and φ ′ is any Fixer strategy to continue S ′ after the first round. Proposition 3.4.
Suppose there exists a Fixer strategy to continue S after the first round such that for its subset φ consisting ofeach of its series T satisfying | B T j | = 1 for < j < | T | , and also | B T j | = 1 for j = | T | if Fixer wins T (i.e. Buster is restricted toremoving singletons after the first round, except for the final round if Buster wins), for every T ∈ φ there exists a series T ′ g identicalto S ′ through the first round such that F T ′ g j is greedy for j > and T is Fixer-superior to T ′ g . Then there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ .Proof. We first show that there exists a series T ′ g identical to S ′ through the first round such that F T ′ g j is greedy for j > T isFixer-superior to T ′ g .If Buster wins T in the second round, then T ∈ φ , so by hypothesis there exists a series T ′ g identical to S ′ through the first roundsuch that F T ′ g j is greedy for j > T is Fixer-superior to T ′ g .If Buster doesn’t win T in the second round, then by Lemma 3.2, there exists a series U ( ) such that U ( ) is identical to T throughthe first round, | B U ( )2 | = 1, F U ( ) j is greedy for j >
1, and T is Fixer-superior to U ( ). We iteratively apply Lemma 3.3 to constructa sequence U ( ) , U ( ) , . . . , U ( ℓ ) such that for 1 < k < ℓ , U ( k + ) is a series identical to U ( k ) through Buster’s move in the k th roundsuch that F U ( k + ) k is the set F prescribed by φ , | B U ( k + ) k +1 | = 1 if Buster doesn’t win U ( k + ) in the ( k + 1)st round, F U ( k ) j is greedyfor j > k , and U ( k ) is Fixer-superior to U ( k + ); the sequence terminates after we reach a series U ( ℓ ) in φ (that is, either Busterwins U ( ℓ ) in round ℓ , or | B U ( ℓ ) ℓ | = 1 and Fixer wins U ( ℓ ) in round ℓ , or | B U ( ℓ ) ℓ | = 1 and Buster wins U ( ℓ ) in round ℓ + 1). Since U ( ℓ ) ∈ φ , by the hypothesis of this proposition there exists a series T ′ g identical to S ′ through the first round such that F T ′ g j is greedyfor j > U ( ℓ ) is Fixer-superior to T ′ g . Since T is Fixer-superior to U ( ), U ( k ) is Fixer-superior to U ( k + ) for 1 < k < ℓ , and U ( ℓ ) is Fixer-superior to T ′ g , by Proposition 2.2 T is Fixer-superior to T ′ g .Thus regardless of whether Buster wins T in the second round, there exists a series T ′ g identical to S ′ through the first round suchthat F T ′ g j is greedy for j > T is Fixer-superior to T ′ g . We now show that T ′ g is Fixer-superior to some T ′ ∈ φ ′ . Since all Fixermoves after the first round of T ′ g are greedy, by the inductive hypothesis of the section they are optimal. Let φ ′′ be the subset of φ ′ consisting of its series for which Buster’s move in the second round matches B T ′ g , and let T ′′ be any element of φ ′′ , so φ ′′ is a strategyfor Fixer to continue T ′′ after the second round. Since F T ′ g j is optimal for j ≥
2, by Lemma 2.3 there exists T ′ ∈ φ ′′ ⊆ φ ′ such that T ′ g is Fixer-superior to T ′ .Hence T is Fixer-superior to T ′ g , which is Fixer-superior to T ′ . By Proposition 2.2, T is Fixer-superior to T ′ , as desired since T ′ ∈ φ ′ .We use Proposition 3.4 to complete the proof for the case c = 2 by showing there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ .We define a subset φ of a Fixer strategy to continue S after the first round consisting of each of its series T satisfying | B T k | = 1 for1 < k < | T | , and also | B T k | = 1 for k = | T | if Fixer wins T , by constructing an arbitrary member T of φ . T will be constructedsimultaneously alongside some series T ′ g identical to S ′ through the first round such that F T ′ g k is greedy for k >
1, in the followingway. Let T be identical to S through the first round, and let T ′ g be identical to S ′ through the first round. For a given round k > B T k of edges from G T k in T (unless Buster wins T in the k th round, in which casethe singleton requirement is dropped for B T k ), then based on that move in T Buster will remove a set B T ′ g k of edges from G T ′ g k in T ′ g (or, in a particular case, have T ′ g skip a round with respect to T , only to make it up later). Fixer will then respond in T ′ g with somegreedy set F T ′ g k of edges from R T ′ g k to connect G T ′ g k − B T ′ g k , then based on that response in T ′ g Fixer will add a set F T k of edges from R T k to connect G T k − B T k in T . Since each Buster move in T is an arbitrary singleton after the first round and before the finalround, and in the final round is an arbitrary singleton if Fixer wins and an arbitrary set if Buster wins, T is an arbitrary memberof φ , so our procedure for defining T fully defines φ . Since T is an arbitary member of φ , if we show T is Fixer-superior to T ′ g ,then by Proposition 3.4 there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ .In order to analyze T and T ′ g , we categorize the corresponding rounds of each series into Scenarios 3.2.1, 3.2.2, and 3.2.3. Eachscenario will include a list of conditions that must be satisfied by T and T ′ g , plus round-by-round instructions for both Buster tomake moves in T ′ g based on his moves in T as well as for Fixer to respond in T based on her responses in T ′ g . After each roundwe shall show either that T and T ′ g are complete, with T Fixer-superior to T ′ g , or that T and T ′ g still satisfy the conditions of thecurrent scenario, or that T and T ′ g have advanced to a new scenario. s in T and Fixer has not used s in T ′ g This scenario involves T and T ′ g each starting the k th round with the following properties:1. s ∈ G T k , s ′ ∈ G T ′ g k , and G T k − { s } = G T ′ g k − { s ′ } (i.e. the only difference between graphs is s in G T k being replaced by s ′ in G T ′ g k )12. s ′ ∈ R T k , s ∈ R T ′ g k , and R T k − { s ′ } = R T ′ g k − { s } (i.e. the only difference between reserve sets is s ′ in R T k being replaced by s in R T ′ g k )3. s and s ′ are bridges in G T k and G T ′ g k , respectively, between the same subgraphs X k and Y k , but perhaps in different spots (i.e.removing both edges from their respective graphs leaves the same graphs, each with two components); see Figure 24. for every r ∈ R T k ∪ R T ′ g k such that r joins X k to Y k , w ( r ) ≥ w ( s ) (i.e. in either series, no reserve edge going between subgraphs X k and Y k can be cheaper than s ) sX k Y k (a) G T k s ′ X k Y k (b) G T ′ g k Figure 2: Graphs G T k and G T ′ g k from Scenario 3.2.1.We divide our analysis of this scenario in the following way. Proposition 3.5 deals with the case that Fixer wins T in the ( k − k th round of T ). Propositions 3.7 and 3.8 deal with the case that Buster wins T inthe k th round, each dealing with a subcase of whether s ∈ B T k . The remaining propositions in our analysis of this scenario deal withthe remaining case that Buster makes a move in the k th round, and Fixer is able to reconnect the graph in response. Proposition 3.9deals with the subcase where G T k − B T k is connected, while Proposition 3.10 deals with the subcase that B T k = { s } (which wouldresult in G T k − B T k being disconnected, since s is a bridge in G T k ). Propositions 3.11, 3.12, and 3.13 deal with the remaining subcasesin a manner described later on. Proposition 3.5.
If Fixer wins T in the ( k − st round, then Buster can quit after the ( k − st round of T ′ g , resulting in Fixerwinning T ′ g in the ( k − st round, and T being Fixer-superior to T ′ g .Proof. We have1. Fixer wins T P | T | j =1 | B T j | = | G T | + | R T | − | G T k | − | R T k | = | G T ′ g | + | R T ′ g | − | G T ′ g k | − | R T ′ g k | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k ) = w ( R T ′ g ) − ( w ( R T ′ g k ) − w ( s ) + w ( s ′ )) ≤ w ( R T ′ g ) − w ( R T ′ g k ) = P | T ′ g | j =1 w ( F T ′ g j )and thus T is Fixer-superior to T ′ g . Lemma 3.6.
If Buster wins T in the k th round, and there exists ℓ such that T ′ g satisfies either S ℓj = k B T ′ g j ⊆ B T k with ( G T ′ g ℓ − B T ′ g ℓ ) ∪ R T ′ g ℓ disconnected, or S ℓj = k B T ′ g j = B T k , then Buster wins T ′ g in the ℓ th round and T is Fixer-superior to T ′ g .Proof. To begin, note that G T k ∪ R T k = ( G T ′ g k − { s ′ } ) ∪ { s } ∪ ( R T ′ g k − { s } ) ∪ { s ′ } = G T ′ g k ∪ R T ′ g k . If ( G T ′ g ℓ − B T ′ g ℓ ) ∪ R T ′ g ℓ is disconnected,then Buster wins T ′ g in the ℓ th round by definition, so to show Buster wins T ′ g in the ℓ th round in the case S ℓj = k B T ′ g j = B T k we show( G T ′ g ℓ − B T ′ g ℓ ) ∪ R T ′ g ℓ is a spanning subgraph of a disconnected graph and thus disconnected itself. Indeed( G T ′ g ℓ − B T ′ g ℓ ) ∪ R T ′ g ℓ = (( G T ′ g k ∪ ( ℓ − [ j = k F T ′ g j )) − ℓ [ j = k B T ′ g j ) ∪ ( R T ′ g k − ℓ − [ j = k F T ′ g j )= ( G T ′ g k − B T k ) ∪ ( R T ′ g k − B T k )= ( G T ′ g k ∪ R T ′ g k ) − B T k = ( G T k ∪ R T k ) − B T k = ( G T k − B T k ) ∪ R T k which is disconnected since Buster wins T in the k th round. Furthermore, noting that the convention F T k = ∅ implies R T k +1 = R T k ,we have21. Buster wins T ′ g in the ℓ th round because ( G T ′ g ℓ − B T ′ g ℓ ) ∪ R T ′ g ℓ is disconnected2. P | T | j =1 | B T j | = | G T | + | R T | − | G T k | − | R T k | + | B T k | ≥ | G T ′ g | + | R T ′ g | − | G T ′ g k | − | R T ′ g k | + P ℓj = k | B T ′ g j | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k ) = w ( R T ′ g ) − ( w ( R T ′ g k ) − w ( s ) + w ( s ′ )) ≤ w ( R T ′ g ) − w ( R T ′ g k ) = P k − j =1 w ( F T ′ g j ) ≤ P | T ′ g | j =1 w ( F T ′ g j )so T is Fixer-superior to T ′ g , as desired. Proposition 3.7.
If Buster wins T in the k th round and s / ∈ B T k , then Buster can play B T ′ g k = B T k in T ′ g , resulting in Busterwinning T ′ g in the k th round and T Fixer-superior to T ′ g .Proof. Buster can play B T ′ g k = B T k in T ′ g because B T k ⊆ G T k − { s } ⊆ G T ′ g k , so Buster wins T ′ g in the k th round and T is Fixer-superiorto T ′ g by Lemma 3.6. Proposition 3.8.
If Buster wins T in the k th round and s ∈ B T k , then in T ′ g Buster can play B T ′ g k = B T k − { s } , as well as B T ′ g k +1 = { s } following any Fixer response if ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k is connected, resulting in Buster winning T ′ g in the k th or ( k + 1) stround and T Fixer-superior to T ′ g .Proof. Note that B T k = { s } , since otherwise setting F T k = { s ′ } would contradict Buster winning T in the k th round since G T k +1 =( G T k − { s } ) ∪ { s ′ } = G T ′ g k , which is connected. Hence ∅ 6 = B T k − { s } ⊆ G T k − { s } ⊆ G T ′ g k , so Buster can play B T ′ g k = B T k − { s } .If ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k is disconnected, then Buster wins T ′ g in the k th round and T is Fixer-superior to T ′ g by Lemma 3.6.If ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k is connected, then Fixer will respond with some greedy F T ′ g k to create a connected graph G T ′ g k +1 . Note that s ∈ F T ′ g k , since otherwise G T ′ g k +1 = ( G T ′ g k − B T ′ g k ) ∪ F T ′ g k ⊆ ( G T k − B T k ) ∪ R T k , meaning G T ′ g k +1 is a spanning subgraph of a disconnectedgraph and thus disconnected itself, a contradiction. Hence Buster can play B T ′ g k +1 = { s } , so B T ′ g k ∪ B T ′ g k +1 = B T k , so Buster wins T ′ g inthe ( k + 1)st round and T is Fixer-superior to T ′ g by Lemma 3.6. Proposition 3.9.
If Buster plays in the k th round of T and G T k − B T k is connected, then Buster can copy her move from T to T ′ g by playing B T ′ g k = B T k = { b } , and Fixer can respond greedily in both T ′ g and T with F T ′ g k = F T k = ∅ to maintain the conditions ofthis scenario.Proof. Note that b = s , since s is a bridge and G T k − { b } is connected. Hence Buster can play B T ′ g k = { b } ⊆ G T k − { s } ⊂ G T ′ g k . Both X k − { b } and Y k − { b } are connected (since the only edge in G T k with one endpoint in X k and the other in Y k is s , which is a bridge),and thus G T ′ g k − B T ′ g k is connected, as the only edge in G T ′ g k with one endpoint in X k and the other in Y k is s ′ , which is a bridge suchthat b = s ′ (since b ∈ G T k but s ′ / ∈ B T k ). Hence the only greedy response for Fixer is F T ′ g k = ∅ , which Fixer can copy in T with F T k = ∅ . Setting X k +1 = X k − { b } and Y k +1 = Y k − { b } , we have1. s ∈ G T k − { b } = G T k +1 , s ′ ∈ G T ′ g k − { b } = G T ′ g k +1 , and G T k +1 − { s } = G T k − { b, s } = G T ′ g k − { b, s ′ } = G T ′ g k +1 − { s ′ } s ′ ∈ R T k = R T k +1 , s ∈ R T ′ g k = R T ′ g k +1 , and R T k +1 − { s ′ } = R T k − { s ′ } = R T ′ g k − { s } = R T ′ g k +1 − { s } s and s ′ are bridges in G T k and G T ′ g k , respectively, between X k +1 and Y k +1
4. for every r ∈ R T k +1 ∪ R T ′ g k +1 such that r joins X k +1 to Y k +1 , w ( r ) ≥ w ( s ), since r ∈ R T k +1 ∪ R T ′ g k +1 = R T k ∪ R T ′ g k and r would havealso joined X k to Y k (because X k and X k +1 share the same set of vertices, as do Y k and Y k +1 )and thus the conditions of this scenario are maintained. Proposition 3.10. If B T k = { s } , then Fixer can play F T k = { s ′ } to advance to Scenario 3.2.2.Proof. Letting T ′ g fall a round behing T and setting X k +1 = X k and Y k +1 = Y k , we have1. s ′ ∈ G T k +1 = ( G T k − { s } ) ∪ { s ′ } = G T ′ g k R T k +1 = R T k − { s ′ } = R T ′ g k − { s } and s ∈ R T ′ g k s ′ and s are bridges in G T k +1 and ( G T ′ g k − { s ′ } ) ∪ { s } , respectively, between X k +1 and Y k +1
34. for every r ∈ R T ′ g k such that r joins X k +1 to Y k +1 , w ( r ) ≥ w ( s ), since r ∈ R T k ∪ R T ′ g k and r would have also joined X k to Y k and thus the conditions of Scenario 3.2.2 are satisfied.Now suppose Fixer doesn’t win T in the ( k − T in the k th round, G T k − B T k is disconnected, and B T k = { b } 6 = { s } . Since s bridges X k and Y k in G T k , either b ∈ X k and X k − { b } is disconnected with two components, or b ∈ Y k and Y k − { b } is disconnected with two components. Without loss of generality, assume b ∈ X k and X k − { b } is disconnected with twocomponents X k and X k , with s bridging X k and Y k ; see Figure 3a. Note that Buster can copy his move from T in T ′ g by playing B T ′ g k = B T k = { b } , since b ∈ G T k − { s } ⊂ G T ′ g k ; the two possibilities for G T ′ g k are shown in Figures 3b and 3c. Furthermore, if s / ∈ F T ′ g k and ( G T k − B T ′ g k ) ∪ F T ′ g k is connected, then Fixer can copy her move from T ′ g in T by playing F T k = F T ′ g k since F T ′ g k ⊆ R T ′ g k − { s } ⊆ R T k . sbX k X k Y k (a) G T k s ′ bX k X k Y k (b) One G T ′ g k possibility s ′ bX k X k Y k (c) The other G T ′ g k possibility Figure 3: G T k and the two possibilities for G T ′ g k We separate into the following cases that together comprise every remaining possibility. Proposition 3.11 deals with the case that s ′ bridges X k and Y k in G T ′ g k (see Figure 4a). Propositions 3.12 and 3.13 deal with the case that s ′ bridges X k and Y k in G T ′ g k , withthe former dealing with the subcase that the cheapest connecting edge f in R T ′ g k is between X k and X k (see Figure 4b) and the latterdealing with the subcase that the cheapest connecting edge f in R T ′ g k is between X k and Y k (see Figure 4c). s ′ X k X k Y k (a) s ′ bridges X k and Y k s ′ fX k X k Y k (b) s ′ bridges X k and Y k , and cheapestconnecting edge f ∈ R T ′ g k joins X k to X k s ′ fX k X k Y k (c) s ′ bridges X k and Y k , and cheapest con-necting edge f ∈ R T ′ g k joins X k to Y k Figure 4: The three remaining possibilities for G T ′ g k − B T ′ g k , where in the latter two f is the cheapest edge in R T ′ g k such that ( G T ′ g k − B T ′ g k ) ∪ { f } is connected. Proposition 3.11. If s ′ bridges X k and Y k in G T ′ g k , then for B T ′ g k = B T k = { b } Fixer can copy her greedy response from T ′ g in T byplaying F T k = F T ′ g k = { f } in order to maintain the conditions of this scenario.Proof. If f bridges X k and X k , set X k +1 = ( X k − { b } ) ∪ { f } and Y k +1 = Y k . Then s and s ′ are bridges in G T k +1 and G T ′ g k +1 ,respectively, between X k +1 and Y k +1 , as s and s ′ each bridged X k and Y k , and the only new edge f resides entirely inside X k +1 . Forevery r ∈ R T k +1 ∪ R T ′ g k +1 such that r joins X k +1 to Y k +1 , w ( r ) ≥ w ( s ), since r ∈ R T k +1 ∪ R T ′ g k +1 ⊂ R T k ∪ R T ′ g k and r would have alsojoined X k and Y k (because X k and X k +1 share the same set of vertices, as do Y k and Y k +1 ).If f bridges X k and Y k , set X k +1 = X k and Y k +1 = X k ∪ { f } ∪ Y k . Then s and s ′ are bridges in G T k +1 and G T ′ g k +1 , respectively,between X k +1 and Y k +1 , as s and s ′ each bridged X k and Y k , and the only new edge f resides entirely inside Y k +1 . For every r ∈ R T k +1 ∪ R T ′ g k +1 such that r joins X k +1 to Y k +1 , w ( r ) ≥ w ( s ), since r ∈ R T k +1 ∪ R T ′ g k +1 ⊂ R T k ∪ R T ′ g k and either r joined X k to Y k (so w ( r ) ≥ w ( s ) by hypothesis of this scenario), or r joined X k to X k (so w ( r ) ≥ w ( f ) because F T ′ g k = { f } was greedy, and w ( f ) ≥ w ( s )by hypothesis of this scenario, since f ∈ R T k ∪ R T ′ g k and f joined X k to Y k ).Thus41. s ∈ G T k − { b } ⊂ G T k +1 , s ′ ∈ G T ′ g k − { b } ⊂ G T ′ g k +1 , and G T k +1 − { s } = ( G T k − { b, s } ) ∪ { f } = ( G T ′ g k − { b, s ′ } ) ∪ { f } = G T ′ g k +1 − { s ′ } s ′ ∈ R T k − { f } = R T k +1 , s ∈ R T ′ g k − { f } = R T ′ g k +1 , and R T k +1 − { s ′ } = R T k − { f, s ′ } = R T ′ g k − { f, s } = R T ′ g k +1 − { s } s and s ′ are bridges in G T k +1 and G T ′ g k +1 , respectively, between X k +1 and Y k +1
4. for every r ∈ R T k +1 ∪ R T ′ g k +1 such that r joins X k +1 to Y k +1 , w ( r ) ≥ w ( s )so the conditions of this scenario are maintained. Proposition 3.12. If s ′ bridges X k and Y k in G T ′ g k , and the cheapest edge f ∈ R T ′ g k such that ( G T ′ g k − { b } ) ∪ { f } is connected hasone endpoint in X k and the other in X k , then for B T ′ g k = B T k = { b } Fixer can copy her greedy response from T ′ g in T by playing F T k = F T ′ g k = { f } in order to maintain the invariants in this scenario.Proof. By hypothesis of this proposition, Fixer can greedily play F T ′ g k = { f } in T ′ g to create a connected graph G T ′ g k +1 . Note that f ∈ R T k , since f ∈ R T ′ g k and R T ′ g k − R T k = { s } 6 = { f } , as f has both endpoints in X k whereas s has one in Y k . Furthermore, G T k − B T k consists of two components, one X k and the other the featuring X k and Y k being bridged by s , so Fixer can also play F T k = { f } in T to create a connected graph G T k +1 . Set X k +1 = ( X k − { b } ) ∪ { f } and Y k +1 = Y k , so1. s ∈ G T k − { b } ⊂ G T k +1 , s ′ ∈ G T ′ g k − { b } ⊂ G T ′ g k +1 , and G T k +1 − { s } = ( G T k − { b, s } ) ∪ { f } = ( G T ′ g k − { b, s ′ } ) ∪ { f } = G T ′ g k +1 − { s ′ } s ′ ∈ R T k − { f } = R T k +1 , s ∈ R T ′ g k − { f } = R T ′ g k +1 , and R T k +1 − { s ′ } = R T k − { f, s ′ } = R T ′ g k − { f, s } = R T ′ g k +1 − { s } s and s ′ are bridges in G T k +1 and G T ′ g k +1 , respectively, between X k +1 and Y k +1
4. for every r ∈ R T k +1 ∪ R T ′ g k +1 such that r joins X k +1 to Y k +1 , w ( r ) ≥ w ( s ), since r ∈ R T k +1 ∪ R T ′ g k +1 ⊂ R T k ∪ R T ′ g k and r would havealso joined X k to Y k (because X k and X k +1 share the same set of vertices, as do Y k and Y k +1 )so the conditions of this scenario are maintained. Proposition 3.13. If s ′ bridges X k and Y k in G T ′ g k , and the cheapest edge f ∈ R T ′ g k such that ( G T ′ g k − { b } ) ∪ { f } is connected has oneendpoint in X k and the other Y k , then for B T ′ g k = B T k = { b } Fixer can greedily play F T k = { s } in T ′ g , as well as play F T k = { s ′ } in T , to advance to Scenario 3.2.3.Proof. Since the cheapest connecting reserve edge in T ′ g is between X k and Y k , s is such an edge by hypothesis of this scenario, soFixer can greedily play F T ′ g k = { s } in T ′ g . Since G T k − B T k consists of two components, one X k and the other the featuring X k and Y k being bridged by s , and s ′ ∈ R T k bridges X k and Y k , Fixer can play F T k = { s ′ } in T to leave T and T ′ g so that1. G T k +1 = ( G T k − { b } ) ∪ { s ′ } = ( G T ′ g k − { b } ) ∪ { s } = G T ′ g k +1 R T k +1 = R T k − { s ′ } = R T ′ g k − { s } = R T ′ g k +1 which are the conditions of Scenario 3.2.3. s in T but Buster has not used s in T ′ g , which is a round behind T This scenario involves T starting the k th round and T ′ g starting the ( k − s ′ ∈ G T k = G T ′ g k − (i.e. the graphs are identical and contain s ′ )2. R T k = R T ′ g k − − { s } and s ∈ R T ′ g k − (i.e. the only difference between reserve sets is s being in R T ′ g k − but not in R T k )3. s ′ and s are bridges in G T k and ( G T ′ g k − − { s ′ } ) ∪{ s } , respectively, between the same connected subgraphs X k and Y k , but perhapsin different spots (i.e. removing both edges from their respective graphs leaves the same graphs, each with two components)4. for every r ∈ R T ′ g k − such that r bridges X k and Y k , w ( r ) ≥ w ( s ) (i.e. in either series, no reserve edge bridging subgraphs X k and Y k can be cheaper than s )5Note that in this scenario Buster can always copy his move from T with B T ′ g k − = B T k since B T k ⊆ G T k = G T ′ g k − . Furthermore, ifBuster doesn’t win T in the k th round and B T ′ g k − = B T k , then Buster doesn’t win T ′ g in the ( k − G T k − B T k ) ∪ R T k is connected, then so must be ( G T ′ g k − − B T ′ g k − ) ∪ R T ′ g k − because G T k = G T ′ g k − , B T k = B T ′ g k − , and R T k ⊆ R T ′ g k − .We divide our analysis of this scenario in the following way. Proposition 3.14 deals with the case that Fixer wins T in the ( k − k th round of T ). Propositions 3.15 and 3.16 deal with the case that Buster wins T inthe k th round, each dealing with a subcase of whether ( G T k − B T k ) ∪ R T k ∪ { s } is connected. Propositions 3.17 and 3.18 deal withthe remaining case that Buster makes a move in the k th round, and Fixer is able to reconnect the graph in response; each deals witha subcase of whether Fixer can respond greedily in T ′ g to B T ′ g k − = B T k with F T ′ g k − containing s . Proposition 3.14.
Suppose Fixer wins T in the ( k − st round. Then for B T ′ g k − = { s ′ } Fixer can play F T ′ g k − = { s } as a greedyresponse in T ′ g , and if Buster subsequently quits then Fixer wins T ′ g in the ( k − st round and T is Fixer-superior to T ′ g .Proof. Since G T ′ g k − − { s ′ } is the graph consisting of the components X k and Y k , and s ∈ R T ′ g k − is a bridge between X k and Y k , F T ′ g k − = { s } is a valid move by Fixer in T ′ g . Furthermore, F T ′ g k − = { s } is a greedy move because for every r ∈ R T ′ g k − such that r bridges X k and Y k , w ( r ) ≥ w ( s ). Note that this leaves s ′ ∈ G T k , s ∈ G T ′ g k , and G T ′ g k − { s } = G T ′ g k − − { s ′ } = G T k − { s ′ } (i.e. the only differencebetween graphs is s ′ in G T k being replaced by s in G T ′ g k ) as well as R T ′ g k = R T ′ g k − − { s } = R T k . Hence1. Fixer wins T P | T | j =1 | B T j | = | G T | + | R T | − | G T k | − | R T k | = | G T ′ g | + | R T ′ g | − | G T ′ g k | − | R T ′ g k | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k ) = w ( R T ′ g ) − w ( R T ′ g k ) = P | T ′ g | j =1 w ( F T ′ g j )so T is Fixer-superior to T ′ g . Proposition 3.15.
Suppose Buster wins T in the k th round and ( G T k − B T k ) ∪ R T k ∪ { s } is disconnected. Then for B T ′ g k − = { s ′ } Fixer can play F T ′ g k − = { s } as a greedy response in T ′ g . Furthermore, Buster playing B T ′ g k = B T k in T ′ g if s ′ / ∈ B T k , or Buster playing B T ′ g k = ( B T k − { s ′ } ) ∪ { s } in T ′ g if s ′ ∈ B T k , both result in Buster winning T ′ g in the k th round and T Fixer-superior to T ′ g .Proof. Since G T ′ g k − − { s ′ } is the graph consisting of the components X k and Y k , and s ∈ R T ′ g k − is a bridge between X k and Y k , F T ′ g k − = { s } is a valid move by Fixer in T ′ g . Furthermore, F T ′ g k − = { s } is a greedy move because for every r ∈ R T ′ g k − such that r bridges X k and Y k , w ( r ) ≥ w ( s ). Note that this leaves s ′ ∈ G T k , s ∈ G T ′ g k , and G T ′ g k − { s } = G T ′ g k − − { s ′ } = G T k − { s ′ } (i.e. the only differencebetween graphs is s ′ in G T k being replaced by s in G T ′ g k ) as well as R T ′ g k = R T ′ g k − − { s } = R T k .Let Buster play B T ′ g k = B T k in T ′ g if s ′ / ∈ B T k , or play B T ′ g k = ( B T k − { s ′ } ) ∪ { s } in T ′ g if s ′ ∈ B T k . We first show in either case thatBuster wins T ′ g in the k th round by showing that ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k is a spanning subgraph of ( G T k − B T k ) ∪ R T k ∪ { s } and thusdisconnected as well. If s ′ / ∈ B T k , then B T ′ g k = B T k and( G T ′ g k − B T ′ g k ) ∪ R T ′ g k = ((( G T k − { s ′ } ) ∪ { s } ) − B T k ) ∪ R T k ⊆ ( G T k − B T k ) ∪ R T k ∪ { s } since s / ∈ B T k , as s / ∈ G T k and B T k ⊆ G T k . If s ′ ∈ B T k , then B T ′ g k = ( B T k − { s ′ } ) ∪ { s } and( G T ′ g k − B T ′ g k ) ∪ R T ′ g k = ((( G T k − { s ′ } ) ∪ { s } ) − (( B T k − { s ′ } ) ∪ { s } )) ∪ R T k = ( G T k − B T k ) ∪ R T k since s ′ ∈ G T k ∩ B T k and s / ∈ G T k ∪ B T k .In either case of final Buster moves, noting that the convention of F T ′ g k = F T k = ∅ implies | G T k +1 | = | G T k |−| B T k | = | G T ′ g k − |−| B T ′ g k | = | ( G T ′ g k − − { s ′ } ) ∪ { s }| − | B T ′ g k | = | G T ′ g k +1 | as well as R T k +1 = R T k = R T ′ g k = R T ′ g k +1 ,1. Buster wins T ′ g P | T | j =1 | B T j | = | G T | + | R T | − | G T k +1 | − | R T k +1 | = | G T ′ g | + | R T ′ g | − | G T ′ g k +1 | − | R T ′ g k +1 | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k +1 ) = w ( R T ′ g ) − w ( R T ′ g k +1 ) = P | T ′ g | j =1 w ( F T ′ g j )so T is Fixer-superior to T ′ g . Proposition 3.16.
Suppose Buster wins T in the k th round and ( G T k − B T k ) ∪ R T k ∪ { s } is connected. Then for B T ′ g k − = B T k Busterdoes not win T ′ g in ( k − st round, and any valid Fixer response in T ′ g must satisfy s ∈ F T ′ g k − . Furthermore, this allows Buster toplay B T ′ g k = { s } in T ′ g , resulting in Buster winning T ′ g in the k th round and T Fixer-superior to T ′ g .Proof. First see that Buster does not win T ′ g in ( k − G T ′ g k − − B T ′ g k − ) ∪ R T ′ g k − = ( G T k − B T k ) ∪ R T k ∪ { s } , which isconnected by hypothesis of this proposition. Next, note that any valid Fixer response in T ′ g must satisfy s ∈ F T ′ g k − , since otherwise( G T ′ g k − − B T ′ g k − ) ∪ F T ′ g k − ⊆ ( G T k − B T k ) ∪ R T k , which is disconnected because Buster wins T in the k th round.With B T ′ g k = { s } , Buster wins T ′ g in the k th round since( G T ′ g k − B T ′ g k ) ∪ R T ′ g k = ((( G T ′ g k − − B T ′ g k − ) ∪ F T ′ g k − ) − { s } ) ∪ ( R T ′ g k − − F T ′ g k − )= ( G T k − B T k ) ∪ ( R T ′ g k − − { s } )= ( G T k − B T k ) ∪ R T k which is disconnected because Buster wins T in the k th round. Furthermore, noting that the convention of F T ′ g k = F T k = ∅ implies R T k +1 = R T k and R T ′ g k +1 = R T ′ g k ,1. Buster wins T ′ g P | T | j =1 | B T j | = | G T | + | R T | − | G T k +1 | − | R T k +1 | = | G T | + | R T | − | ( G T k − B T k ) ∪ R T k | = | G T ′ g | + | R T ′ g | − | ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k | = | G T ′ g | + | R T ′ g | − | G T ′ g k +1 | − | R T ′ g k +1 | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k ) = w ( R T ′ g ) − w ( R T ′ g k − − { s } ) ≤ w ( R T ′ g ) − w ( R T ′ g k − − F T ′ g k − ) = w ( R T ′ g ) − w ( R T ′ g k ) = P | T ′ g | j =1 w ( F T ′ g j )so T is Fixer-superior to T ′ g . Proposition 3.17.
Suppose Buster plays some set B T k = { e } that does not win T for Buster in the k th round, and when Bustercopies that move in T ′ g with B T ′ g k − = { e } , Fixer responds greedily in T ′ g with a set F T ′ g k − , but no possible greedy response for Fixer in T ′ g contains s . Then Fixer can copy that move in T with F T k = F T ′ g k − to stay in this scenario.Proof. Fixer can validly play F T k = F T ′ g k − in T because F T ′ g k − ⊆ R T ′ g k − − { s } = R T k and G T k +1 = ( G T k − B T k ) ∪ F T k = ( G T ′ g k − − B T ′ g k − ) ∪ F T ′ g k − = G T ′ g k which is connected. In addition to G T k +1 = G T ′ g k , we also have R T k +1 = R T k − F T k = ( R T ′ g k − − { s } ) − F T ′ g k − = R T ′ g k − { s } and s ∈ R T ′ g k − − F T ′ g k − = R T ′ g k .We show that e = s ′ by showing that if e = s ′ then Fixer could have contradicted the assumption that no possible greedyresponse in T ′ g contained s by playing F T ′ g k − = { s } . Indeed, F T ′ g k − = { s } would have been a valid Fixer response since s is a bridgein ( G T ′ g k − − { s ′ } ) ∪ { s } between connected subgraphs X k and Y k . Furthermore, it would have been a greedy response since s ′ beinga bridge between X k and Y k implies that any greedy response would have to be a single edge in R T ′ g k − bridging X k and Y k , and forevery r ∈ R T ′ g k − such that r bridges X k and Y k , w ( r ) ≥ w ( s ) by assumption of this scenario.7To complete the proof that Fixer playing F T k = F T ′ g k − in T maintains the conditions of this scenario, we show that G T k +1 − { s ′ } consists of two components X k +1 and Y k +1 such that s ′ and s are bridges in G T k +1 and ( G T ′ g k − { s ′ } ) ∪ { s } , respectively, between X k +1 and Y k +1 , with every r ∈ R T ′ g k bridging X k +1 and Y k +1 also satisfying w ( r ) ≥ w ( s ). If e is not a bridge in X k or Y k , thenboth X k − { e } and Y k − { e } are connected, and furthermore G T ′ g k − − { e } is connected because s ′ is a bridge between X k − { e } and Y k − { e } and e = s ′ ; hence F T k = F T ′ g k − = ∅ since that would be the only greedy move by Fixer in T ′ g , so we can set X k +1 = X k − { e } and Y k +1 = Y k − { e } . Thus without loss of generality we may assume e is a bridge in X k between the two components X k and X k of X k − { e } , and F T k = F T ′ g k − = { f } for some edge f ∈ R T k either bridging X k and X k , or bridging Y k and one of X k or X k . If f bridges X k and X k , then we may set X k +1 = ( X k − { e } ) ∪ { f } and Y k +1 = Y k . Thus without loss of generality we may assume f bridges X k and Y k , so w ( f ) ≥ w ( s ) by the assumptions of this scenario, and w ( f ) ≤ w ( r ) for any r ∈ R T ′ g k − bridging X k and X k ,since otherwise F T ′ g k − = { r } would have been a cheaper valid response for Fixer in T ′ g , contradicting F T ′ g k − = { f } being greedy. Hencewe can set X k +1 = X k and Y k +1 = Y k ∪ X k ∪ { f } , since for every r ∈ R T k +1 such that r bridges X x +1 and Y k +1 , either r bridged X k and Y k in which case w ( r ) ≥ w ( s ) by the assumptions of this scenario, or r bridged X k and X k , in which case we’ve already shown w ( r ) ≥ w ( f ) ≥ w ( s ). Proposition 3.18.
Suppose Buster plays some set B T k that does not win T for Buster in the k th round, and when Buster copiesthat move in T ′ g with B T ′ g k − = B T k , Fixer can respond greedily in T ′ g with a set F T ′ g k − containing s . Then Buster can play B T ′ g k = { s } and Fixer can create a connected graph G T ′ g k +1 with some greedy F T ′ g k in T ′ g , and Fixer can play F T k = ( F T ′ g k − − { s } ) ∪ F T ′ g k in T , totrigger Scenario 3.2.3 for the ( k + 1) st round.Proof. Buster can play B T ′ g k = { s } in T ′ g because s ∈ F T ′ g k − ⊆ G T ′ g k . Fixer can respond with some F T ′ g k ⊆ R T ′ g k such that G T ′ g k +1 =( G T ′ g k − B T ′ g k ) ∪ F T ′ g k is connected (so Buster doesn’t win T ′ g in the k th round), because Buster’s failure to win T in the k th roundimplies ( G T k − B T k ) ∪ R T k is connected, and( G T ′ g k − B T ′ g k ) ∪ R T ′ g k = ( G T ′ g k ∪ R T ′ g k ) − { s } = (( G T ′ g k − ∪ R T ′ g k − ) − B T ′ g k − ) − { s } = ( G T ′ g k − − B T ′ g k − ) ∪ ( R T ′ g k − − { s } )= ( G T k − B T k ) ∪ R T k so ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k is connected as well. Fixer can play F T k = ( F T ′ g k − − { s } ) ∪ F T ′ g k in T because F T k = ( F T ′ g k − − { s } ) ∪ F T ′ g k ⊆ R T ′ g k − − { s } = R T k and G T k +1 = ( G T k − B T k ) ∪ F T k = ( G T ′ g k − − B T ′ g k − ) ∪ ( F T ′ g k − − { s } ) ∪ F T ′ g k = (( G T ′ g k − − B T ′ g k − ) ∪ F T ′ g k − ) − { s } ) ∪ F T ′ g k = ( G T ′ g k − B T ′ g k ) ∪ F T ′ g k = G T ′ g k +1 which we already showed was connected because there existed a valid Fixer move F T ′ g k to prevent Buster from winning T ′ g in the k th8round. Since we just showed G T k +1 = G T ′ g k +1 , and R T k +1 = R T k − F T k = ( R T ′ g k − − { s } ) − (( F T ′ g k − − { s } ) ∪ F T ′ g k )= ( R T ′ g k − − F T ′ g k − ) − F T ′ g k = R T ′ g k − F T ′ g k = R T ′ g k +1 Scenario 3.2.3 is triggered for the ( k + 1)st round. T and T ′ g are in the same state This scenario involves T and T ′ g each starting the k th round with the following properties:1. G T k = G T ′ g k R T k = R T ′ g k Note that in this scenario, after Buster plays some B T k in T , Buster can copy that move in T ′ g with B T ′ g k = B T k since G T k = G T ′ g k .We divide our analysis of this scenario in the following way. Proposition 3.19 deals with the case that Fixer wins T in the ( k − k th round of T ). Proposition 3.20 deals with the case that Buster wins T in the k thround. Proposition 3.21 deals with the remaining case that Buster makes a move in the k th round, and Fixer is able to reconnectthe graph in response. Proposition 3.19.
If Fixer wins T in the ( k − st round, then Buster can quit after the ( k − st round of T ′ g , resulting in Fixerwinning T ′ g in the ( k − st round, and T being Fixer-superior to T ′ g .Proof. We have1. Fixer wins T P | T | j =1 | B T j | = | G T | + | R T | − | G T k | − | R T k | = | G T ′ g | + | R T ′ g | − | G T ′ g k | − | R T ′ g k | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k ) = w ( R T ′ g ) − w ( R T ′ g k ) = P | T ′ g | j =1 w ( F T ′ g j )and thus T is Fixer-superior to T ′ g . Proposition 3.20.
If Buster wins T in the k th round, then Buster playing B T ′ g k = B T k results in Buster winning T ′ g in the k th roundand T Fixer-superior to T ′ g .Proof. We have ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k = ( G T k − B T k ) ∪ R T k , which is disconnected since Buster wins T in the k th round, so Buster wins T ′ g in the k th round. Hence F T ′ g k = F T k = ∅ by convention, implying G T k +1 = G T k − B T k = G T ′ g k − B T ′ g k = G T ′ g k +1 and R T k +1 = R T k = R T ′ g k = R T ′ g k +1 , so1. Buster wins T ′ g P | T | j =1 | B T j | = | G T | + | R T | − | G T k +1 | − | R T k +1 | = | G T ′ g | + | R T ′ g | − | G T ′ g k +1 | − | R T ′ g k +1 | = P | T ′ g | j =1 | B T ′ g j | P | T | j =1 w ( F T j ) = w ( R T ) − w ( R T k +1 ) = w ( R T ′ g ) − w ( R T ′ g k +1 ) = P | T ′ g | j =1 w ( F T ′ g j )and thus T is Fixer-superior to T ′ g . Proposition 3.21.
If Buster plays some set B T k that does not win T for Buster in the k th round, then after Buster plays B T ′ g k = B T k ,Fixer can copy her greedy move from T ′ g in T by playing F T k = F T ′ g k to stay in this scenario.Proof. Since Buster does not win T in the k th round and ( G T ′ g k − B T ′ g k ) ∪ R T ′ g k = ( G T k − B T k ) ∪ R T k , Buster also doesn’t win T ′ g in the k th round, so Fixer can respond greedily in T ′ g with some F T ′ g k . Since R T k = R T ′ g k , Fixer can copy that move in T with F T k = F T ′ g k .Hence G T k +1 = G T ′ g k +1 and R T k +1 = R T ′ g k +1 , leaving us in the same scenario.9 c ≥ c ≥
3, then F S and F S ′ each have multiple edges. Let S ′′ be a series such that G S = G S ′ = G S ′′ , R S = R S ′ = R S ′′ , and B S = B S ′ = B S ′′ . To show that for our fixed series T ∈ φ , there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ , we first show(via Proposition 3.23) that there exists a Fixer move F S ′′ and strategy φ ′′ for Fixer to continue S ′′ after the first round such that F S ′′ contains an edge e ∈ F S and for every T ′′ ∈ φ ′′ there exists T ′ ∈ φ ′ such that T ′′ is Fixer-superior to T ′ . We then show (viaProposition 3.24) that for every T ∈ φ there exists T ′′ ∈ φ ′′ such that T is Fixer-superior to T ′′ . Then for every T ∈ φ , we would have T ′′ ∈ φ ′′ and T ′ ∈ φ ′ such that T is Fixer-superior to T ′′ , and T ′′ is Fixer-superior to T ′ . Since Fixer-superiority is transitive, T wouldbe Fixer-superior to T ′ .In order to prove Propositions 3.23 and 3.24, we make use of the following observation. Suppose P and U are series such that forsome subset F of F P , the situation facing Fixer during her first move in U is the same situation she faced in P after having partiallyfixed G P − B P with F from R P (i.e. G U − B U = ( G P − B P ) ∪ F and R U = R P − F ). Further suppose Q is a series identical to P upthrough Fixer partially fixing each graph with F at the start of her first move, but Fixer finishes her first move in Q by copying herentire first move in U (i.e. G Q = G P , R Q = R P , B Q = B P , and F Q = F ∪ F U ). Finally, suppose F U is optimal, φ U is a strategy ofoptimal moves for Fixer to continue U after the first round, φ Q is the strategy for Fixer to continue Q after the first round constructedby replacing the first round of each series in φ U with the first round of Q , and φ P is a strategy for Fixer to continue P after the firstround. Then for every Q ′ ∈ φ Q , we should expect by dint of Fixer copying the first part of her move from P into Q ′ before (in a sense)finishing that move optimally, and playing all subsequent moves optimally, that there exists P ′ ∈ φ P such that Q ′ is Fixer-superior to P ′ . We formally verify below that this indeed holds true. Lemma 3.22.
Let P and U be series such that for some subset F of F P , G U − B U = ( G P − B P ) ∪ F and R U = R P − F . Let Q bea series such that G Q = G P , R Q = R P , B Q = B P , and F Q = F ∪ F U . If F U is optimal, φ U is a strategy of optimal moves for Fixerto continue U after the first round, φ Q is the strategy for Fixer to continue Q after the first round constructed by replacing the firstround of each series in φ U with the first round of Q , and φ P is a strategy for Fixer to continue P after the first round, then for every Q ′ ∈ φ Q there exists P ′ ∈ φ P such that Q ′ is Fixer-superior to P ′ .Proof. Fixer’s strategy φ Q against any Buster strategy will be a translation of Fixer’s strategy φ U against the same Buster strategy.Note that G Q = ( G Q − B Q ) ∪ F Q = ( G P − B P ) ∪ F ∪ F U = ( G U − B U ) ∪ F U = G U and R Q = R Q − F Q = R P − ( F ∪ F U )= ( R P − F ) − F U = R U − F U = R U so Q and U are equivalent starting in the second round.Since F U is optimal, and φ U is a strategy of optimal moves for Fixer to continue U after the first round, by Lemma 2.3 for anyseries V identical to U through Buster’s move of the first round, for any U ′ ∈ φ U and any strategy φ V for Fixer to continue V afterthe first round, there exists V ′ ∈ φ V such that U ′ is Fixer-superior to V ′ . Let V be identical to U through Buster’s move of the firstround, but set F V = F P − F . Note that G P = ( G P − B P ) ∪ F P = ( G P − B P ) ∪ F ∪ ( F P − F )= ( G U − B U ) ∪ ( F P − F )= ( G V − B V ) ∪ F V = G V R P = R P − F P = ( R P − F ) − ( F P − F )= R U − F V = R V − F V = R V so P and V are equivalent starting in the second round.Let φ Q be the strategy for Fixer to continue Q after the first round constructed by replacing the first round of each series in φ U with the first round of Q , let φ P be a strategy for Fixer to continue P after the first round, and let φ V be the strategy for Fixer tocontinue V after the first round constructed by replacing the first round of each series in φ P with the first round of V . Let Q ′ ∈ φ Q ,and let U ′ ∈ φ U be the series from which Q ′ was constructed by replacing the first round with the first round of Q . Let V ′ ∈ φ V be aseries such that U ′ is Fixer-superior to V ′ , and let P ′ ∈ φ P be the series from which V ′ was constructed by replacing the first roundof P ′ with the first round of V . Then1. Fixer wins Q ′ , or Buster wins Q ′ , in which case ( G Q ′ | Q ′ | − B Q ′ | Q ′ | ) ∪ R Q ′ | Q ′ | is disconnected, implying Buster wins U ′ since ( G U ′ | U ′ | − B U ′ | U ′ | ) ∪ R U ′ | U ′ | = ( G Q ′ | Q ′ | − B Q ′ | Q ′ | ) ∪ R Q ′ | Q ′ | , implying Buster wins V ′ since U ′ is Fixer-superior to V ′ , implying ( G V ′ | V ′ | − B V ′ | V ′ | ) ∪ R V ′ | V ′ | is disconnected, implying Buster wins P ′ since ( G P ′ | P ′ | − B P ′ | P ′ | ) ∪ R P ′ | P ′ | = ( G V ′ | V ′ | − B V ′ | V ′ | ) ∪ R V ′ | V ′ | P | Q ′ | j =1 | B Q ′ j | = | B Q | − | B U | + P | U ′ | j =1 | B U ′ j | ≥ | B P | − | B V | + P | V ′ | j =1 | B V ′ j | = P | P ′ | j =1 | B P ′ j | P | Q ′ | j =1 w ( F Q ′ j ) = w ( F ) + P | U ′ | j =1 w ( F U ′ j ) ≤ w ( F ) + P | V ′ | j =1 w ( F V ′ j ) = P | P ′ | j =1 w ( F P ′ j )so Q ′ is Fixer-superior to P ′ . Proposition 3.23.
There exists a move F S ′′ and strategy φ ′′ for Fixer to continue S ′′ after the first round such that F S ′′ containsan edge e ∈ F S and for every T ′′ ∈ φ ′′ there exists T ′ ∈ φ ′ such that T ′′ is Fixer-superior to T ′ .Proof. Since every series is Fixer-superior to itself, if F S ∩ F S ′ = ∅ , then we could set F S ′′ = F S ′ and φ ′′ = φ ′ . Hence we may assume F S ∩ F S ′ = ∅ .Since c ≥ F S has multiple edges. Let e be the cheapest edge of F S , and let e ′ be in the path through F S ′ between the endpointsof e ; see Figure 5a. Recalling that F S is a minimum spanning tree of the multigraph M whose vertices are the components of G S − B S and whose edges are the edges of R S (identifying each endpoint of the edges in R S with the component of G S − B S within whichit lies), note that no non-loop edge in M (i.e. edge in R S joining two components of G S − B S ) can be cheaper than e , or else byProposition 2.1 it would have been added to F S by Prim’s algorithm as the edge immediately after the first of its endpoints joinedthe tree.Consider the series U , initialized by the following constructions of G U and R U . Construct G U by adding F S ′ − { e ′ } to G S ′ andthen deleting from that graph all the edges in B S ′ except for an edge g joining the two components of G S ′ − e ′ ; see Figure 5b. Notethat g must exist, or else F S ′ would not be a tree. Construct R U by deleting F S ′ − { e ′ } from R S ′ .Note that G U is connected. Indeed, G S ′ is connected, and G U is G S ′ with e ′ replaced by g , where e ′ and g connect the same twocomponents of G S ′ − e ′ .Furthermore, | G U ∪ R U | < | G S ∪ R S | . The graph G U and reserve edge set R U are obtained from the graph G S and reserve edge set R S by transferring the edges F S ′ − { e ′ } from R S to G U , and then deleting a positive number of edges from the graph, since | B S | > c < B U of edges removed by Buster, any greedy choice of F U by Fixer is optimal.Set B U = { g } .We claim that F U = { e } is optimal. First, see that e ∈ R U : e ∈ F S − F S ′ ⊆ R S − F S ′ = R S ′ − F S ′ ⊂ R S ′ − ( F S ′ − { e ′ } )= R U Next, see that adding e would connect G U − B U , since G U − B U = G S ′ − { e ′ } , and G S ′ is connected, with e and e ′ both joiningthe two components of G S ′ − { e ′ } . Finally, see that no edge h in R U that would connect G U − B U can be cheaper than e , since1 R U ⊂ R S , and h ∈ M because G U − B U = G S ′ − { e ′ } and if h joins the two components of G S ′ − { e ′ } and e ′ ∈ F S ′ then h must jointwo components of G S ′ − F S ′ = G S ′ − B S ′ = G S − B S , so h being cheaper than e would contradict e being the cheapest edge in M .Hence F U = { e } is optimal, by the inductive hypothesis.Let φ U be a strategy of greedy moves for Fixer to continue U after the first round; by the inductive hypothesis, these greedymoves are optimal. Let S ′′ be a series such that G S ′′ = G S ′ , R S ′′ = R S ′ , B S ′′ = B S ′ , and F S ′′ = ( F S ′ − { e ′ } ) ∪ { e } , noting that for F = F S ′ − { e ′ } we have F ⊂ F S ′ , G U − B U = ( G S ′ ∪ F ) − ( B S ′ − { g } ) − { g } = ( G S ′ − B S ′ ) ∪ F (since F ⊂ F S ′ and F S ′ ∩ B S ′ = ∅ imply F ∩ B S ′ = ∅ ), R U = R S ′ − F , and F S ′′ = F ∪ F U . Let φ ′′ be the strategy for Fixer to continue S ′′ after the first round constructedby replacing the first round of each series in φ U with the first round of S ′′ . By Lemma 3.22, for every T ′′ ∈ φ ′′ there exists T ′ ∈ φ ′ such that T ′′ is Fixer-superior to T ′ . ee ′ (a) The dashed lines are the edges of F S , while the dotted linesare the edges of F S ′ . g (b) The graph G Q , where dotted lines are the edges of F S ′ −{ e ′ } . Figure 5: Two graphs from the proof of Proposition 3.23. The blobs are the components of G S − B S . Proposition 3.24. If F S ′′ contains an edge e ∈ F S and φ ′′ is a strategy for Fixer to continue S ′′ after the first round, then for every T ∈ φ there exists T ′′ ∈ φ ′′ such that T is Fixer-superior to T ′′ .Proof. Pick h ∈ B S that joins the two components of G S − { e } , and consider the series U satisfying G U = ( G S − { h } ) ∪ { e } , R U = R S − { e } , and B U = B S − { h } . Then G U − B U = ( G S ′′ − B S ′′ ) ∪ { e } and R U = R S ′′ − { e } . Since G U ∪ R U = ( G S ∪ R S ) − { h } where h ∈ B S ⊆ G S , by the inductive hypothesis any greedy play by Fixer is optimal, including F U = F S − { e } .Let φ U be the strategy for Fixer to continue U after the first round constructed by replacing the first round of each series in φ with the first round of U . Since all Fixer moves in φ are greedy, all Fixer moves in φ U are also greedy, so all Fixer moves in φ U are optimal by the inductive hypothesis. Furthermore, φ is the strategy for Fixer to continue S after the first round constructed byreplacing the first round of each series in φ U with the first round of S . Since S is a series such that G S = G S ′′ , R S = R S ′′ , B S = B S ′′ ,and F S = { e } ∪ F U , where { e } ⊆ F S ′′ satisfies G U − B U = ( G S ′′ − B S ′′ ) ∪ { e } and R U = R S ′′ − { e } , by Lemma 3.22 for every T ∈ φ there exists T ′′ ∈ φ ′′ such that T is Fixer-superior to T ′′ .Combining the previous two propositions with the transitivity of Fixer-superiority yields the following conclusion to this subsection. Corollary 3.25.
For every T ∈ φ , there exists T ′ ∈ φ ′ such that T is Fixer-superior to T ′ .Proof. By Proposition 3.23, there exists a move F S ′′ and strategy φ ′′ for Fixer to continue S ′′ after the first round such that F S ′′ contains an edge e ∈ F S and for every T ′′ ∈ φ ′′ there exists T ′ ∈ φ ′ such that T ′′ is Fixer-superior to T ′ . By Proposition 3.24, forevery T ∈ φ there exists T ′′ ∈ φ ′′ such that T is Fixer-superior to T ′′ . Hence for every T ∈ φ , there exists T ′′ ∈ φ ′′ and T ′ ∈ φ ′ suchthat T is Fixer-superior to T ′′ , and T ′′ is Fixer-superior to T ′ . By Proposition 2.2, T is Fixer-superior to T ′ . Acknowledgment.
The author thanks R.T. Solo for the initial problem statement.
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