Graph sharing games: complexity and connectivity ^{†}^{†}thanks: The research was supported by the project CEITI (GACR P2020/12/G061) of the Czech Science Foundation and by the grant SVV2010261313 (Discrete Methods and Algorithms). Viola Mészáros was also partially supported by OTKA Grant K76099 and by the grant no. MSM0021620838 of the Ministry of Education of the Czech Republic. Josef Cibulka and Rudolf Stolař were also supported by the Czech Science Foundation under the contract no. 201/09/H057. Rudolf Stolař was also supported by the Grant Agency of the Charles University, GAUK project number 66010.
Abstract
We study the following combinatorial game played by two players, Alice and Bob, which generalizes the Pizza game considered by Brown, Winkler and others. Given a connected graph with nonnegative weights assigned to its vertices, the players alternately take one vertex of in each turn. The first turn is Alice’s. The vertices are to be taken according to one (or both) of the following two rules: (T) the subgraph of induced by the taken vertices is connected during the whole game, (R) the subgraph of induced by the remaining vertices is connected during the whole game. We show that if rules (T) and/or (R) are required then for every and for every there is a connected graph for which Bob has a strategy to obtain of the total weight of the vertices. This contrasts with the original Pizza game played on a cycle, where Alice is known to have a strategy to obtain of the total weight.
We show that the problem of deciding whether Alice has a winning strategy (i.e., a strategy to obtain more than half of the total weight) is PSPACEcomplete if condition (R) or both conditions (T) and (R) are required. We also consider a game played on connected graphs (without weights) where the first player who violates condition (T) or (R) loses the game. We show that deciding who has the winning strategy is PSPACEcomplete.
Department of Applied Mathematics,
Charles University, Faculty of Mathematics and Physics,
Malostranské nám. 25, 118 00 Praha 1, Czech Republic;
cibulka@kam.mff.cuni.cz, ruda@kam.mff.cuni.cz
Department of Applied Mathematics and Institute for Theoretical Computer Science,
Charles University, Faculty of Mathematics and Physics,
Malostranské nám. 25, 118 00 Praha 1, Czech Republic;
kyncl@kam.mff.cuni.cz
Bolyai Institute, University of Szeged,
Aradi vértanúk tere 1, 6720 Szeged, Hungary;
viola@math.uszeged.hu
1 Introduction
The pizza problem.
Dan Brown devised the following pizza puzzle in 1996 which we formulate using graph notation. The pizza with slices of not necessarily equal size, can be considered as a cycle with nonnegative weights on the vertices. Two players, Bob and Alice, are sharing it by taking turns alternately. In every turn one vertex is taken. The first turn is Alice’s. Afterwards, a player can take a vertex only if the subgraph induced by the remaining vertices is connected. Dan Brown asked if Alice can always obtain at least half of the sum of the weights. Peter Winkler and others solved this puzzle by constructing a weighted cycle where Bob had a strategy to get at least of the total weight [1]. Peter Winkler conjectured that Alice can always gain at least of the total weight. This conjecture has been proved independently by two groups of authors [1, 5].
The games T, R and TR.
In this paper we investigate a generalized setting where the cycle is replaced with an arbitrary connected graph . Consider the following two conditions:

the subgraph of induced by the taken vertices is connected during the whole game,

the subgraph of induced by the remaining vertices is connected during the whole game.
Observe that the two conditions are equivalent if is a cycle or a clique.
The generalized game is called game T, game R, or game TR if we require condition (T), condition (R), or both conditions (T) and (R), respectively. Note that in games T and R, regardless of the strategies of the players, there is always an available vertex for the player on turn until all vertices are taken as the graph is connected. This need not be the case in game TR, however. Therefore we consider game TR played only on connected graphs, where the game never ends prematurely; see Lemma 5. In Proposition 7 we give a complete characterization of the graphs where the game TR ends with all vertices taken for all strategies of the players.
Playing on connected graphs.
In games T, R and TR, there are graphs where Bob can ensure (almost) the whole weight to himself. See Fig. 1 for examples in games T and R where the graph is a tree. For any , examples of connected graphs exist in any of the three variants of the game. See Fig. 2 for examples of such connected graphs.
Theorem 1.
For each of the games T, R and TR, for every and for every there is a connected graph for which Bob has a strategy to obtain at least of the total weight of the vertices.
Parity of the vertex set.
Micek and Walczak [6, 7] also studied, independently of us, generalizations of the pizza game. They investigated how the parity of the number of vertices affects Alice’s chances to gain a positive fraction of the total weight in games T and R, particularly when the game is played on a tree. They proved that Alice can gain at least of the total weight in game R on a tree with an even number of vertices and in game T on a tree with an odd number of vertices. For the opposite parities they constructed examples of trees (such as those in Figure 1) where Bob has a strategy to get almost all the weight. They conjectured that Alice can gain at least of the total weight in game R on a tree with an even number of vertices; this has been confirmed by Seacrest and Seacrest [11].
In our original proof of Theorem 1 the graphs for games T and TR had an even number of vertices and the graphs for game R an odd number of vertices. Micek and Walczak (personal communication) asked whether also connected graphs with opposite parities satisfying the conditions of Theorem 1 exist. In Subsection 2.2 we find examples of such graphs for all three variants of the game.
Micek and Walczak [7] also conjectured that there is a function such that in game T, Alice can gain at least of the weight of any graph with an odd number of vertices and with no minor. Gągol [3] showed that under the additional assumption that the input graph has only weights, Alice gan gain at least of the total weight. Gągol, Micek and Walczak [4] recently informed us about proving a stronger version of their conjecture: there is a function such that in game T, Alice can gain at least of the weight of any graph with an odd number of vertices and with no subdivision of . On the other hand, they show that the construction of odd graphs which are arbitrarily bad for Alice can be easily modified to yield graphs with bounded expansion.
Micek and Walczak [6] noted that all known examples of sequences of even graphs where Alice’s gain in game R tends to zero, contain arbitrarily large cliques as subgraphs. We show that the condition of containing arbitrarily large cliques is not necessary to force Alice’s gain to tend to zero. In fact, the graphs in our example have arbitrarily large girth and form a class with bounded local expansion.
Theorem 2.
For every there is a weighted even graph with arbitrarily large girth for which in game R, Bob has a strategy to obtain at least of the total weight of the vertices. Moreover,

there exists an infinite class of such graphs with maximum degree bounded by ; in particular, has bounded expansion.

There is a class with bounded local expansion of weighted even graphs for which the infimum of the fraction of Alice’s gain in game R is zero.
For definitions of the classes with bounded expansion and bounded local expansion, see [9].
Canonical game and misere game.
As game TR does not always end with all vertices taken, it is natural to consider the following variation of the game, which we call a canonical game TR. Given a connected graph , Alice and Bob take turns alternately. In each turn a player takes one vertex of . The first player who has no move satisfying conditions (T) and (R) loses the game. We also consider a misere game TR, where the first player who has no move satisfying conditions (T) and (R) wins the game.
Complexity results and open problems.
We determine the complexity of deciding who has the winning strategy in the canonical and the misere game.
Theorem 3.
It is PSPACEcomplete to decide who has the winning strategy in the canonical game TR and in the misere game TR.
We also consider the complexity of determining the winning strategy (i.e., gaining more than half of the total weight) for the original three types of the game. We show the following.
Theorem 4.
It is PSPACEcomplete to decide who has the winning strategy in game R and TR.
However, we are unable to determine the complexity of deciding the winner for game T.
Problem 1.
What is the complexity of deciding who has the winning strategy in game T?
Problem 2.
What is the complexity of deciding who has the winning strategy in game R and in game T when the input graph is a tree?
We are not able to show that the problem is polynomial even if the tree has only one branching vertex.
Since the weights used in our proof of Theorem 4 are growing exponentially, it is natural to ask the following question.
Problem 3.
What is the complexity of deciding who has the winning strategy in game R and TR, when the weights of the vertices of are only or ?
2 Proofs
2.1 Graphs where game TR ends with all vertices taken
Here we characterize the graphs where game TR always ends with all vertices taken, and also the graphs for which there is at least one sequence of turns in game TR such that the game ends with all vertices taken. First we prove an auxiliary observation implying that it is always “safe” to play game TR on a connected graph.
Lemma 5.
Let be a connected graph with vertices and let be a vertex of . Let be a set of vertices of such that and the two induced subgraphs and are connected. Then there is a vertex such that both induced subgraphs and are connected.
Proof.
The connectivity of implies that has at least two vertices that are neighbors of . Let be such a pair of vertices with the largest possible distance in . We claim that neither of the vertices separates and therefore we can choose . Suppose for contradiction that is a cut vertex of and let be a component of that does not contain . Since does not separate , the component contains a neighbor of . The shortest path in between and passes through , which contradicts the choice of the pair . ∎
Corollary 6.
Let be a connected graph with vertices and let and be two distinct vertices of . Then the vertices of can be ordered as so that for each , both induced graphs and are connected. ∎
Proposition 7.
(1) For game TR on a graph there is a sequence of turns to take all the vertices of if and only if each cut vertex of separates the graph into precisely two components and every connected component of contains at most two cut vertices of .
(2) Game TR on a graph will always end with all vertices taken if and only if each cut vertex of separates the graph into precisely two components and every connected component of with at least three vertices contains at most one cut vertex of .
Proof.
(1) First we show that the two conditions are necessary.
Suppose is a cut vertex separating into more than two components. Let be the component where Alice made her first turn, and let and be two other components. By rule (T), before is taken, only vertices of can be taken. Therefore by rule (R), the vertex never becomes available since it separates from in the remaining graph.
Let be a connected component of . Suppose for contradiction that contains at least three cut vertices of . Then has at least three components . Let be the only vertex of neighboring with (). By rule (T), before any vertex of is taken, the players can take vertices from at most one of the components , say . Before any vertex of is taken, one of the vertices has to be taken. But these vertices never become available by rule (R), since they both separate from .
If both conditions are satisfied, then the connected components of can be arranged into a sequence , where for each , the components and share a cut vertex . By applying Corollary 6 for each of the components and the cut vertices it contains, we obtain the following order in which the players can take the vertices:
where
(2) If game TR always ends with all vertices taken, then we may assume that the two conditions from part (1) are satisfied. Suppose that has a connected component with at least vertices and with two cut vertices of : a vertex separating from and separating from . If Alice starts with taking a vertex from , then neither of the vertices becomes available during game TR.
If both conditions are satisfied, then the graph is connected or it is a union of two connected subgraphs (including the degenerate cases when or has only one vertex) and a path of cut vertices where and . By rule (R), Alice has to start in or (or in or in the degenerate cases). Suppose without loss of generality that she starts with taking a vertex in . By rule (R), all vertices of have to be taken before . By Lemma 5 applied to the set and the vertex , all the vertices of will be indeed taken. After that the players have to take the vertices of the path sequentially from to . If has at least three vertices, then by Lemma 5, also all the vertices of will be taken before the game ends. ∎
2.2 Proof of Theorem 1
In games T and TR, Bob can choose the following connected graph with an even number of vertices (for any given ): Take a large even cycle and replace each vertex in it by a clique and each edge by a complete bipartite graph . Assign weight to one vertex in every other clique, and weight to all the other vertices of the graph. See Figure 3, left. Bob uses the following strategy.

Take an availabe vertex of weight .

If no vertex of weight is available, take an available vertex from one of the cliques where at least one vertex has already been taken.
It is easy to see that by this strategy Bob makes sure that Alice takes at most one vertex of weight , and only in her first or second turn.
In game R, Bob can choose the following connected (bipartite) graph with an odd number of vertices. The vertex set is a disjoint union of sets and where is an element set for some large , is the set of all element subsets of and is a set of or elements so that the total number of vertices is odd. The edge set of consists of all edges between and and all edges that connect a vertex with each of the vertices such that . Each vertex from has weight , all the other vertices have weight . See Figure 3, right.
Consider a position in the game. Let be a subgraph of induced by the remaining vertices. A vertex is available if it is not a cut vertex of (equivalently, can be taken in the next turn). A leaf is a vertex in of degree one.
Claim 8.
Bob can force Alice to get at most vertices of weight .
Proof.
Bob has the following strategy.

If possible, he takes an available vertex of .

If no vertex of is available, then if possible, he takes either a vertex of that is not a leaf or a leaf vertex of whose neighbor in neighbors at least one other leaf in .

If neither 1) nor 2) applies, he takes any vertex of .
There are three phases of the game. The th phase lasts as long as Bob acts only according to the rules with numbers at most .
During the first phase Bob takes only vertices from . Let and be the numbers of vertices from taken in the first phase by Alice and by Bob, respectively. Since the first phase ends by Alice’s turn, Alice takes exactly vertices of in the first phase. Due to rules (R) and 1), all the vertices of whose all neighbors are taken in the first phase must be taken by Alice in the first phase. There are such vertices. It follows that . Equivalently,
If then the righthand side equals ; otherwise it is smaller than . Thus, . It means that Alice takes at most vertices of weight in the first phase. We further show that Alice takes only vertices of weight in the other two phases.
At the beginning of the second phase each remaining neighbors at least one leaf in . During the second phase this property is preserved because in each turn at most one vertex from may become available. Note that only Alice can make a vertex in available, so Bob will take such in the following turn. Thus Alice is forced to take vertices only from during the second phase.
As has at least two more vertices than , at some point the third phase has to start. At the beginning of the third phase the remaining vertex set is a union of three sets and that are subsets of and , respectively. There is a complete bipartite graph on and there is a matching between and . The first turn in the third phase is Bob’s. He takes a vertex from (there is an available vertex in since the size of is positive and even). Alice may take vertices from . Whenever Alice takes a vertex of , then in the consecutive turn Bob will do the same. Whenever Alice takes a vertex from , then in the consecutive turn Bob will take the available vertex from . This implies that either the entire is taken in the previously described way or the graph transforms into a star centered in the only remaining vertex of with each edge subdivided by a vertex in , and with each leaf in . As the original graph had odd number of vertices, the next turn is Alice’s and consequently Bob collects all the remaining vertices of . The claim follows. ∎
Now we show the constructions of graphs with the parities opposite to those in the proof above.
For game T and game TR and for every , we can construct a connected graph with an odd number of vertices starting from the graph described by Micek and Walczak [7, Example 2.2], replacing each vertex of weight by vertices of weight forming a clique, and replacing each original edge by a complete bipartite graph. The graph consists of vertices of weight , vertices of weight , and a vertex of weight for every nonempty subset . Each is joined by an edge to and each is joined to all such that . For the graph Bob has a strategy to take all but one vertex of weight , analogous to the strategy for [7].
For game R, for every we construct a connected weighted graph with vertices (we may assume that for some positive integer so that the total number of vertices is even). The construction generalizes the graph [7, Example 5.2] consisting of a clique of vertices of weight , with a leaf of weight attached to each vertex of the clique. The graph consists of a clique on vertices of weight and a vertex of weight for each element subset . The vertex is connected by an edge to all vertices such that .
Alice can collect at most vertices of weight if Bob plays as follows.

If possible, Bob takes a vertex of weight .

Otherwise he takes a vertex of weight which is not a unique leaf neighbor of some vertex (in the graph induced by the remaining vertices).
Bob can always play according to one of these two rules because the number of remaining vertices before his turn is odd. It follows that Bob plays in such a way that no vertex of weight becomes available after his turn, except the turn after which only two vertices remain. By the same argument as in the proof of Theorem 1, Alice takes at most vertices of weight at the beginning of the game, while Bob plays only by rule 1). Then she takes at most one more vertex of weight in her last turn.
2.3 Proof of Theorem 2
Let be fixed. We construct a weighted even graph as follows. The graph consists of a graph with all vertices of weight 1, and one leaf of weight 0 connected to each vertex of . In a similar construction of the graph by Micek and Walczak [7, Example 5.2], was a complete graph. Here we take as a much sparser graph on vertices, which is still a “good expander” in the following sense: the complement of contains no complete bipartite subgraph . In addition, is a graph with girth and we may also require to have a bounded maximum degree. We show that such a graph exists using probabilistic method.
In the rest of this section we omit the explicit rounding in the expressions involving , to keep the notation simple.
Lemma 9.
Let , and . There exists such that for every there exists a connected graph with vertices of girth at least and of maximum degree at most , whose complement contains no copy of .
Proof.
Let be a random graph from the probability space . That is, is a subgraph of where every edge is taken independently with probability . First we show that the expected number of cycles of length at most in is less than . The number of cycles of length in the complete graph with vertices is . A cycle of length appears in with probability . Hence, the expected number of cycles of length is at most . Therefore, the expected number of cycles of length at most in is at most
The expected number of subgraphs in the complement of is at most
The expected average degree of is smaller than but the expected maximum degree of is unbounded. However, the number of vertices of large degree is small. The probability that a given vertex has degree larger than is at most
(This is also a direct consequence of Markov’s inequality). The expected number of vertices of degree larger than is thus at most .
It follows that there exists a graph with vertices such that by deleting some vertices, we obtain a graph with vertices, with maximum degree at most , with no cycle shorter than , and with no in the complement.
In case the graph is not connected, we add the necessary edges connecting different components of to make connected, in such a way that the maximum degree of does not exceed . ∎
Let be fixed. By taking a graph from Lemma 9 for every and attaching a leaf to each vertex of , we get an infinite class of graphs with bounded maximum degree, thus with bounded expansion. Note that to obtain a class of bounded expansion it is not necessary to delete vertices of high degree as in the proof of Lemma 9, since a.a.s. the random graphs form a class with bounded expansion [8].
The class is constructed as follows. For every , let and be as in Lemma 9. Let be such that and let be the graph given by Lemma 9. Let be the graph obtained by attaching a leaf to each vertex of . Then is a class of graphs satisfying and therefore has bounded local expansion [9].
Now we show that Alice’s gain in game R played on the graph is bounded by of the total weight. That is, Alice takes at most vertices of during the game. We need only the property that the complement of contains no and that is connected.
During the game, we call a remaining vertex of exposed if its neighboring leaf of weight has been taken. Let be the remaining subgraph of . Let be the set of exposed cut vertices of . Let . Call a component of the graph large if it has at least vertices and small otherwise. Observe that has at most one large component, which we denote by . Let be the set of exposed cut vertices adjacent to . Let . Observe that .
Bob’s strategy is the following.

If Alice took a vertex of weight 0 neighboring a vertex in the previous turn, Bob takes , if it is available (that is, if is not a cut vertex in ).

If Alice makes a vertex available (in which case does not belong to after her turn), Bob takes .

If neither of the previous rules apply, Bob takes any vertex such that , is not a vertex of weight neighboring a vertex of , and taking does not make any vertex of available. In case there is no large component , Bob may take any available vertex.
We claim that Bob’s strategy is complete, that is, he can always play by one of the rules. We show this in a series of observations.
Observation 10.
The large component contains at most one exposed vertex, and that can happen only after Alice’s turn.
Proof.
This is a direct consequence of Bob’s strategy: he is not allowed to expose a vertex of , and by rule 1) he immediately takes a vertex of exposed by Alice. ∎
Corollary 11.
Alice never takes a vertex of . ∎
Observation 12.
A vertex of can become available only after Alice’s turn and Bob takes it immediately in his next turn.
Proof.
By following rules 1) and 2) Bob takes a vertex from or and thus cannot make a vertex of available. Rule 3) explicitly forbids making a vertex of available. Alice can make a vertex of available only by taking its neighbor from . For Bob’s next turn only rule 2) applies and the observation follows. ∎
Corollary 13.
Alice never takes a vertex of . ∎
Observation 14.
If Bob has to follow rule 3), there is a vertex he can take.
Proof.
Suppose that Bob is forced to follow rule 3) and that has a large component . By Observation 10, none of the vertices of is exposed. In particular, the total number of vertices in and the remaining leaves of weight attached to is even. If taking every available vertex makes some vertex available, then is the only neighbor of outside . In particular, and hence the total number of vertices of is even, which is a contradiction with Bob being on turn. Therefore there is an available vertex satisfying the conditions of rule 3). ∎
It remains to estimate Alice’s gain. To this end, we need an upper bound on the size of the “latest” large component. Let , , be the large component after Bob’s th turn. The number is chosen as the largest possible.
Observation 15.
For , we have . There is no large component after or more Bob’s turns.
Proof.
The large component expands only when some vertex of becomes available (and thus is added to ). By Observation 12, is taken by Bob in the next turn and thus the large component remains the same as two turns before. Similar observation applies also for small components. ∎
The large component shrinks after Alice’s and Bob’s th turns (that is, ) if either some of its vertices is taken or if some of its cut vertices becomes exposed. In the first case the size of drops exactly by , in the second case the size of can drop by up to .
Now consider the last large component . If some of the vertices of is taken in the next Bob’s turn, then has precisely vertices, since the remaining subset of after Bob’s turn is a small component of .
Suppose that a cut vertex of is exposed in the next Alice’s turn. Then consists of small components only.
For , let denote the graph after Bob’s th turn. Let . By Corollaries 11 and 13, the only vertices of weight Alice can take belong to . The following lemma thus provides an upper bound on Alice’s gain.
Lemma 16.
and .
Proof.
Observe that there is no edge between a vertex of and a vertex of . Indeed, every vertex of belongs to for some , and is nonadjacent to by the definition of . Since , the inequality follows, otherwise the complement of would contain a forbidden copy of .
The graph consists of small components only. If has at least vertices, consider a minimal subgraph of that has at least vertices and is a union of some components of . Then has less than vertices and has less than vertices, due to the forbidden in the complement of . Therefore has less than vertices. ∎
2.4 Proof of Theorem 3
First we consider the canonical game TR. We proceed by polynomial reduction from the standard PSPACEcomplete problem TQBF (also called QBF). An instance of the TQBF problem is a fully quantified boolean formula with variables and alternating quantifiers, and the question is whether is true. Without loss of generality we may assume that is even and that the formula starts with the existential quantifier:
We may also assume without loss of generality that in the previous expression is a 3SAT formula.
As the game always ends after polynomially many turns, one can search through all possible game states and determine who has the winning strategy in PSPACE. To show that the problem is also PSPACEhard, it suffices to prove that for every formula there is a graph constructible in polynomial time such that is true if and only if Alice has a strategy to win on in the canonical game TR.
2.4.1 The construction of
First we introduce a Vgadget that will be used many times in the construction of . The Vgadget is a path of length four. The middle vertex of is distinguished because will be identified with other vertices during the construction.
For every variable of we build a variable gadget. For the variable the gadget consists of a path of length two between the vertices and that represent the two possible values of , and we attach a Vgadget to the middle vertex of , see Figure 4. The variable gadgets are connected by edges , , and in for all , see Figure 5.
For every clause of we introduce a new vertex in . The vertex is connected to the three vertices in corresponding to the literals of the clause . In case a variable stands with negation in , the vertex is connected to in , otherwise is connected to . We attach a Vgadget to . Further we add one special vertex to and edges , , and for each , see Figure 5.
The order enforcer for a vertex is a gadget that prevents Alice from starting at . This property is proved in Observation 19 below. The special neighbors of are the neighbors of among the vertices , , , , , , , . The order enforcer for the vertex connects to a newly added vertex , adds a path of length two between and each special neighbor of , and adds a Vgadget to the middle vertex of each , see Figure 6. We attach an order enforcer simultaneously for each vertex .
2.4.2 The game
In the following we make some easy observations.
Observation 17.
In a game where Alice wins, no vertex of a Vgadget can be taken.
Proof.
Each Vgadget in is connected to other vertices only at its middle vertex , see Figures 4 and 5. By rule (R), the first vertex taken from the Vgadget is or . Consequently, by rule (T), it can be taken only in the first turn. If Alice takes vertex in the first turn, then Bob has to take vertex . By the rules there is no further vertex that Alice could take. So she loses the game. The case of is analogous. ∎
As a straightforward consequence of Observation 17 we get that no vertex corresponding to some clause can be taken from .
Observation 18.
For every , only one of and can be taken.
Proof.
Since has at least two Vgadgets, Observation 17 implies that taking both and would disconnect . ∎
Observation 19.
Let be a vertex in to which an order enforcer is attached. If in the first turn Alice takes or , then she loses the game.
Proof.
Suppose that Alice takes . Then Bob takes and there is no further vertex for Alice to take, as each special neighbor of would disconnect the order enforcer and every other neighbor of or is the middle vertex of a Vgadget. If Alice takes , then Bob takes . Similarly, Alice cannot take any further vertex. In both cases Alice loses the game. ∎
Using similar arguments as in the previous proof we also observe the following fact.
Observation 20.
If some neighbor of is taken, cannot be taken anymore. ∎
The game must proceed as follows. As a consequence of Observations 17, 18 and 19 Alice will take or in the first turn. By Observations 18 and 20, the only possible choices in the th turn for the player on turn are and for . In the st turn Alice has to take or she has no turn and loses the game.
If Alice cannot take , it means that some vertex corresponding to a clause would get disconnected from the part of that contains the remaining vertices and . This occurs if and only if previously all three vertices corresponding to the literals of were taken. If after taking the subgraph induced by the remaining vertices of is connected, then there is no further vertex to take as it would necessarily disconnect .
For , let be TRUE if was taken by one of the players and FALSE if was taken. It follows that the formula is satisfied if and only if Alice can take at the end of the game, that is, if and only if she wins the game. As Alice’s turns in correspond to the variables with the existential quantifier in and Bob’s turns in to the variables with the universal quantifier, it follows that Alice wins if and only if is a true formula.
Obviously the construction of was carried out in polynomial time. This completes the proof of Theorem 3 for the canonical game TR.
2.4.3 The misere game TR
The proof of the PSPACEcompleteness for the misere game TR is very similar to the proof for the canonical game TR, with the following differences. Instead of even we consider odd, so that Alice chooses between and and Bob then takes or has no move in the st turn. The Vgadget now consists of a path of length attached by its middle vertex, and the order enforcer for a vertex is now doubled, with an additional edge . See Figures 7 and 8.
2.5 Proof of Theorem 4
As in the proof of Theorem 3, we show a polynomial reduction from the TQBF problem. Without loss of generality we may assume that the considered formula is of the form where is a NAESAT formula. That is, each clause of has three literals and it is satisfied if and only if at least one of the three literals is evaluated as TRUE and at least one as FALSE.
2.5.1 The construction of
Let be the number of clauses in . We construct a connected weighted graph of size in the following way (see Figure 9). For each variable we take a path with 4 vertices. The end vertices of are labeled and . If a variable occurs in the clause , we add a vertex connected by an edge to and a vertex connected by an edge to . The vertices and are connected by clause gadgets depicted in Figure 10. We add a set of vertices that are connected to all vertices and . Finally, we add two vertices and connected to all other vertices (including the edge between and ). Observe that the constructed graph is connected and has an odd number of vertices.
The weights are set as follows:
All other vertices (that is, the inner points on the paths and the vertices of ) have weight . The vertices of positive weight can be partitioned into the following groups: , and the groups , each consisting of vertices of the clause gadget corresponding to the clause . Note that the weights are chosen so that each vertex in a group has larger weight than the sum of weights of all the vertices from groups . Let denote the total weight of all vertices in .
In the rest of this section we prove the following statement.
Proposition 21.
Alice has a winning strategy in game TR played on if and only if is true. Alice has a winning strategy in game R played on if and only if is true.
2.5.2 Starting the game
We start with the following easy observations.
Observation 22.
Once one of the vertices or is taken, game TR reduces to game R as condition (T) is always satisfied further on. ∎
We will call the graph induced by the remaining vertices in some position in the game briefly the remaining graph.
Observation 23.
If in some position in game R played on the remaining graph has a cut vertex , then Bob has a strategy to get .
Proof.
Let be the remaining graph after some of Alice’s turns. If there is a component in containing at least two vertices, Bob takes a vertex from that component (at least one vertex from such component is always available). When no such component exists, the graph is a star with central vertex and odd number of leaves. Now Bob and Alice alternately take leaves of , until only two vertices are left. Then becomes available for Bob. ∎
Corollary 24.
If Bob is on turn in game R, he has a strategy to get any of the remaining vertices. ∎
In the analysis of the game we may without loss of generality assume that the players play optimally, that is, each of them follows a strategy maximizing his/her gain.
Now in a sequence of lemmas we show that the players do not have much freedom in choosing their strategy if they want to play optimally. Roughly speaking, the players will not deviate too much from a greedy strategy, which consists in taking an available vertex of largest weight.
Lemma 25.
Suppose that both players play optimally. Fix . Suppose further that Alice’s first turn is on , Bob’s first turn is on , and for each in the nd turn of the game either or is taken. Then Bob has a strategy to get all the remaining vertices from .
Proof.
For each , let be the remaining vertex of the pair and let . Every vertex of is a cut vertex in the remaining graph and is adjacent to all vertices from . Each has a neighbor in the path that is still not taken. Let . Bob’s strategy is to avoid taking vertices from as long as possible, and take a vertex of whenever it becomes available. When no such turn is possible, he takes an available vertex from .
First we observe that if the remaining graph after some of the Alice’s following turns is not covered by , then every component of contains a vertex that is not a cut vertex in . Indeed, since induces a connected subgraph of , we may take the end vertex of the longest path starting in and ending in .
As the vertices of are the only common neighbors of any pair of vertices from , at most one vertex of is made available after each Alice’s turn, unless Alice took the last remaining vertex of . But this will not happen as long as Bob is avoiding taking vertices from , since contains more than half of the vertices of .
It follows that if Bob has no available vertex outside , then all the vertices of the remaining graph belong to . Hence in the rest of the game, Bob’s strategy reduces to avoiding and taking a vertex of whenever Alice makes it available. When only one vertex remains, we have a similar situation as in the proof of Theorem 1: the remaining graph is a tree which is a union of paths , it has an odd number of vertices and so Alice is on turn. Consequently Bob can get all the remaining vertices . ∎
Lemma 26.
Suppose that both players play optimally. Suppose further that Alice’s first turn is on and Bob’s first turn is on . Then Bob has a strategy to get vertices of total weight at least .
Proof.
If Alice takes a vertex in her second turn, then Bob takes in his second turn. This makes unavailable for Alice’s third turn. By Corollary 24, Bob has a strategy to get in the rest of the game. Therefore he gets vertices of total weight at least .
Now suppose that Alice takes in her second turn (the case of is symmetric). Then Bob takes in his second turn. By Lemma 25, Bob has a strategy to take both and in the rest of the game. ∎
Lemma 27.
If both players play optimally in game R or game TR, then Alice’s first turn is on and Bob’s first turn is on .
Proof.
Suppose that Alice takes in her first turn and then Bob takes . If is not a cut vertex in , then Alice can take in her second turn and get vertices of total weight at least which implies that Bob’s first turn was not optimal. If is a cut vertex in (namely, one of the vertices ), then becomes the only cut vertex in the remaining graph, so it is the only unavailable vertex for Alice in her second turn. Thus Alice can take in her second turn. As still separates the remaining vertices, Bob cannot take in his next turn. It follows that Alice can take or in her third turn, depending on Bob’s choice in his second turn (we may assume without loss of generality that the formula has at least two variables). In this way Alice gained vertices of total weight at least . Therefore by Lemma 26, Bob’s first turn was not optimal.
Now suppose that Alice’s first turn is on . As both and are connected to all other vertices, Bob can play his first turn on . By Observation 22, we can consider only game R from now on. If Alice took in her second turn, then she is not playing optimally, as she can get to a position with the same remaining graph using a better strategy: she takes in her first turn, then by the previous paragraph Bob takes , and then Alice takes . Thus Alice takes in her second turn. Then by Corollary 24 Bob has a strategy to get , so he gains at least . Hence Alice’s only optimal first turn is on . ∎
2.5.3 Variable gadgets
A variable gadget for the variable is the path connecting vertices and , with only the end vertices connected to .
Lemma 28.
If both players play optimally, then for each , in the nd turn of the game either or is taken.
Proof.
Let be the smallest positive integer that violates the statement of the lemma.
Suppose first that the nd turn was Alice’s (thus is odd). Let be the vertex taken by Alice in the nd turn. If , then Bob can take the end vertex of adjacent to in the next turn. If , then Bob takes in the next turn.
We now argue that Bob has a strategy to get all the remaining vertices from . We would get the same remaining graph if Alice and Bob switched the vertices they took in the nd and the rd turn. This modified game satisfies conditions of Lemma 25 until the nd turn. Since Bob is now required to take in the rd turn, we cannot apply Lemma 25 directly. But since is large enough and taking does not make any remaining vertex of available to Alice, Bob may get all the remaining vertices from using the strategy from the proof of Lemma 25 from his next turn.
It follows that Bob has a strategy to get vertices of weight at least . Therefore the only optimal choice for Alice in the nd turn was or .
Now suppose that is even. If Bob takes a vertex in the nd turn, then either or is available for Alice in the next turn and she takes it. The total weight of the remaining vertices and the vertex is then smaller than . But by Lemma 25, Bob has a strategy to get at least that much of the weight if he takes or instead of in the nd turn. This finishes the proof of Lemma 28. ∎
Let be the sequence of the remaining vertices from the groups . The vertices determine a truth assignment of the variables: if and if . See Figure 11. For each , let be the neighbor of in the path .
2.5.4 Clause gadgets
The subgraph induced by the group acts as a clause gadget for the clause . Call the two vertices of the group of weight heavy and the six vertices of weight light.
Lemma 29.
Suppose that all vertices