On the Complexity of Client-Waiter and Waiter-Client Games
Abstract
Positional games were introduced by Hales and Jewett in 1963, and their study became more popular when Erdős and Selfridge showed their connection to Ramsey theory and hypergraph coloring in 1973. Several conventions of these games exist, and the most popular one, Maker-Breaker was proved to be -complete by Schaefer in 1978. The study of their complexity then stopped for decades, until 2017 when Bonnet, Jamain, and Saffidine proved that Maker-Breaker is -complete when parameterized by the number of moves. The study was then intensified when Rahman and Watson improved Schaefer’s result in 2021 by proving that the -hardness holds for -uniform hypergraphs. More recently, Galliot, Gravier, and Sivignon proved that computing the winner on rank hypergraphs is in , and Keopke proved that the -hardness also holds for -uniform hypergraphs.
We focus here on the Client-Waiter and the Waiter-Client conventions. Both were proved to be -hard by Csernenszky, Martin, and Pluhár in 2011, but neither completeness nor positive results were known. In this paper, we complete the study of these conventions by proving that the former is -complete, even restricted to -uniform hypergraphs, and by providing an -algorithm for the latter, parameterized by the size of its largest edge. In particular, the winner of Waiter-Client can be computed in polynomial time in rank hypergraphs for any fixed integer . Finally, in search of the exact location of the complexity gap in the Client-Waiter convention, we focus on rank 3 hypergraphs. We provide an algorithm that runs in polynomial time with an oracle in .
Keywords and phrases:
Complexity, positional games, Maker-Breaker, Client-Waiter, Waiter-Client, PSPACE-complete, FPTCategory:
Track A: Algorithms, Complexity and GamesCopyright and License:
![[Uncaptioned image]](x1.png)
2012 ACM Subject Classification:
Mathematics of computing Combinatorial algorithms ; Mathematics of computing Hypergraphs ; Theory of computation Problems, reductions and completenessFunding:
This research was partly supported by the ANR project P-GASE (ANR-21-CE48-0001-01).Editors:
Keren Censor-Hillel, Fabrizio Grandoni, Joël Ouaknine, and Gabriele PuppisSeries and Publisher:

1 Introduction
Positional games were introduced by Hales and Jewett [24] as a generalization of the Tic-Tac-Toe game. These games are played on a hypergraph , on which the players alternately claim an unclaimed vertex of . Tic-Tac-Toe corresponds to the convention Maker-Maker: the first player who claims all vertices of an edge of wins. However, since the second player cannot win in Maker-Maker games, and since Maker-Maker games are not hereditary for the outcome, the study quickly switched to another convention: Maker-Breaker.
In the most studied convention, Maker-Breaker, Maker (one player) tries to claim all the vertices of an edge, while Breaker (the other player) aims to prevent her from doing so. If not explicitly specified, we assume here that Maker plays the first move. The study became more popular in 1973 when Erdős and Selfridge [15] obtained the following criterion.
Theorem 1 (Erdős-Selfridge criterion).
Let be a hypergraph. If
then the position is a win for Breaker.
In the biased version of Maker-Breaker (where at each turn Maker selects one vertex, and then Breaker selects vertices), one can ask, for different families of hypergraphs, what is the threshold for which changes the outcome of the game. Surprisingly, for many hypergraph families (but not all), it has been found that the threshold for the random game (where both players play randomly) is essentially the same as the threshold for optimal play. This phenomenon, known as the “probabilistic intuition” was first demonstrated by Chvátal and Erdős [11], and then further developed in numerous papers, in particular by Beck (see for example the books [3] and [25]).
Schaefer proved in 1978 that it is -complete to determine the winner of a Maker-Breaker game [35] (the problem appears there under the name ). More precisely, he shows -hardness even for hypergraphs of rank , that is when edges have size at most . A simplified proof can be found in [10]. The result was improved in 2021 by Rahman and Watson [33] and very recently by Koepke [28]: the problem is -complete for -uniform hypergraphs with (i.e., hypergraphs where all edges have size exactly ). Byskov [10] also notices that deciding the winner in a Maker-Breaker game can be reduced to the same problem in the Maker-Maker convention (up to increasing by the maximal size of its edges). In particular, deciding who wins in a Maker-Maker game is -complete for -uniform hypergraphs as soon as .
On the positive side, Kutz [29] proved in 2005 that the problem is tractable for -uniform hypergraphs and -uniform linear ones111Hypergraphs whose intersection of each pair of edges is of size at most one.. It was improved recently by Galliot, Gravier, and Sivignon [19, 20]: deciding the winner is tractable for rank hypergraphs.
Since the introduction of positional games, studies have focused on the Maker-Breaker convention. In order to have a better understanding of this convention, Beck [2] introduced the Client-Waiter and Waiter-Client conventions in 2002 under the names Chooser-Picker and Picker-Chooser. Their current names were suggested by Bednarska-Bzdȩga, Hefetz, Krivelevich and Łuczak [6] as they are less ambiguous. In both conventions, Waiter selects two vertices of the hypergraph and offers them to Client. Client then chooses one to claim, and the second one is given to Waiter. If the number of vertices is odd, the last vertex goes to Client. In the Client-Waiter convention, Client wins if she claims all vertices of an edge, otherwise Waiter wins. In the Waiter-Client convention, Waiter wins if he claims all vertices of an edge, otherwise Client wins. In the rest of the paper, Client is referred as female and Waiter as male.
Notice that these conventions can also be seen as variations of Avoider-Enforcer, the misère version of Maker-Breaker. In particular, the definition we give for Waiter-Client is the one which appears originally in Beck [2]. But since [26], Waiter-Client is often defined following the Avoider-Enforcer convention: Waiter wins if he can force Client to claim a whole edge (and if the number of vertices is odd, the last vertex goes to Waiter). One can notice that both definitions correspond in fact to the same game. Despite the fact that Maker-Breaker and Avoider-Enforcer are very different games, in this “I-cut-you’ll-choose way” paradigm, since Client picks a vertex from only two possibilities, it is symmetric to associate the chosen vertex to Client or to Waiter.
The study of Client-Waiter and Waiter-Client games was first motivated by their similarities with Maker-Breaker games. For example, Bednarska-Bzdȩga (improving previous results in [2, 12]) proved that Theorem 1 also holds in Waiter-Client convention [5]. Moreover, the “probabilistic intuition” still holds for these conventions [2]. More recently, Bednarska-Bzdȩga, Hefetz and Łuczak [7] conjectured that the value of the threshold bias in the biased Waiter-Client -game exactly matches, up to a multiplicative constant, the predictions given by a random heuristic. This conjecture has just been proved recently by Nenadov in 2023 [30]. These similarities led Beck [2] and Csernenszky, Mándity, and Pluhár [12] to conjecture that if Maker wins in a Maker-Breaker game on some hypergraph going second, then Waiter wins in Waiter-Client. Notice that up to considering the transversal of , the conjecture also implies that a win for Breaker as a second player implies a win for Waiter in the Client-Waiter game. But this conjecture was disproved by Knox [27] in 2012. Nowadays, Client-Waiter and Waiter-Client games are studied independently of Maker-Breaker games: Csernenszky [13] proved in 2010 that -in-a-row Waiter-Client is a Client win on the infinite grid, while this problem is still open in the Maker-Breaker convention, and Hefetz et al. [26] studied several classical games in Waiter-Client convention.
In terms of complexity, only few results were known about Waiter-Client and Client-Waiter games. In contrast to Maker-Breaker, Maker-Maker, Avoider-Avoider and Avoider-Enforcer, which are known to be -complete [35, 10, 17, 33, 23], the asymmetry between the players moves in Waiter-Client and Client-Waiter conventions makes it more difficult to obtain reductions. In fact, Waiter has more choices than Client in his moves, which makes most reduction techniques fail. Both problems have been conjectured -complete in [12]. In [14], the authors show that both problems are -hard. However, in the Waiter-Client convention, the hypergraph has an exponential number of edges but is given in a succinct way: it is the transversal of a given hypergraph. This leaves open the question of whether the problem of deciding who wins a Waiter-Client game is still -hard when the hypergraph is explicitly given via the list of its edges.
The parameterized complexity of combinatorial games emerged very early in the study of parameterized problems. Meanwhile, deep results in complexity theory were obtained using this approach. For instance, Abrahamson, Rodney, Downey, and Fellows [1] proved that , through the game Geography. Only few results are known on positional games with the parameterized complexity paradigm, and only related to three conventions: Maker-Breaker, Maker-Maker and Avoider-Enforcer. Their study started by Downey and Fellows, conjecturing that Short Generalized Hex was , but it was disproved (unless = [1]) by Bonnet, Jamain and Saffidine in 2016 [9], proving its [1]-hardness. Then, in 2017, Bonnet et al. [8] proved that general Maker-Breaker games are [1]-complete, Avoider-Enforcer games are co-[1]-complete and general Maker-Maker games are [*]-complete.
Notice that the parameterized results obtained on general positional games consider the number of moves as a parameter. This is mostly motivated by the fact that these problems are already -hard for bounded rank hypergraphs. However, this is not the case in the Waiter-Client convention, and the aim of this paper is to study the complexity of determining the winner of Waiter-Client games parameterized by the rank of the hypergraph. Note that the outcome of Client-Waiter games is hereditary (see Lemma 10), but contrary to what is true for Maker-Breaker and Avoider-Enforcer games, this property does not hold anymore when the number of moves is bounded. Indeed, since Waiter can control where Client plays, the addition of isolated vertices can be used by Waiter to waste turns. Therefore, the number of moves in this convention is not as relevant as in the others.
High-level description of the results
We show that, similarly to the Maker-Breaker and Avoider-Enforcer conventions, deciding the winner of a positional game in the Client-Waiter convention is -complete. The result was already conjectured in 2009 [12]. It is an improvement on [14] where the problem is shown to be -hard. Moreover, we obtain the -completeness even for -uniform hypergraphs.
Theorem 2.
For , Client-Waiter games are -complete even restricted to -uniform hypergraphs.
The membership in directly follows from Lemma 2.2 in [35]. So the main point is the hardness part. To prove it, we reduce the problem of deciding the winner in Client-Waiter from the problem of computing the winner in the following game.
Definition 3 ().
Let be a -CNF Formula over a set of pairs of variables . The -game is played by two players, Satisfier and Falsifier as follows: while there is a variable that has not been assigned a valuation, Satisfier chooses a pair of variables that he has not chosen yet and gives a valuation, or , to . Then Falsifier gives a valuation to . When all variables are instantiated, Satisfier wins if and only if the valuation they have provided to the s and s satisfies .
The -game is a new variant of a -game where the play order is closer to the Client-Waiter convention: the first player chooses, at each turn, which variables she plays on and which variable the second player will have to play on.
Again, deciding who wins on this new game is -complete. It is proved in the full version of the paper by a reduction from the game -QBF (known to be -complete since Stockmeyer and Meyer seminal work [38]).
Theorem 4.
Deciding who is the winner of the -game is -complete.
Then Theorem 2 is obtained by reducing the -game to the Client-Waiter one. This is done by designing a gadget (given in Figure 2) which simulates a pair of variables of the -game. Notice that the reduction only creates a hypergraph of rank . But, similarly to the Maker-Breaker games [33, Corollary 4] and the Avoider-Enforcer ones [23, Lemma 7], Lemma 22 shows that the hypergraph can be turned into a -uniform one afterwards.
In the Maker-Breaker convention, as said before, it is known [33] that deciding if Maker has a winning strategy is -complete over hypergraphs of rank . On the other side, the problem is easy for hypergraphs of rank 2 since Maker wins if and only if there are two adjacent -edges or the graph contains a singleton edge (result already noticed in [29]). But only recently, after a series of results [29, 32], Galliot, Gravier, and Savignon [19, 20] showed that the problem is still polynomial for hypergraphs of rank . The same question arises in the Client-Waiter convention: despite being -complete over hypergraphs of rank , what can we say about the complexity of the problem for hypergraphs of low rank? For hypergraphs of rank , it is easily seen that the problem is polynomial (see Oijid’s thesis [31]). The question is already non-trivial for hypergraphs of rank .
We show that the problem of deciding if Client has a winning strategy in a rank hypergraph reduces to the problem of finding a specific structure (called a Tadpole) in a hypergraph. Tadpoles have been generalized to hypergraphs by Galliot, Gravier and Sivignon [20] to handle rank 3 Maker-Breaker games and their definitions are recalled in Definition 25. Intuitively, given and two vertices of , an -tadpole is given by the union of a path from to and a cycle containing such that the structure is simple (two edges intersect only if they are two consecutive edges of the path or the cycle, or if one is the last edge of the path and the other an edge of the cycle containing ) and linear (the intersection of two intersecting edges is of size exactly ). More precisely, we require this structure to be -uniform, simple and linear, which means that all edges have size , the intersection of two consecutive edges has always size one, and the same vertex can not happen at two different places of the structure. An tadpole is simply called tadpole. A tadpole is said rooted at , if it is an -tadpole for some vertex .
We consider the problem Tadpole: Given a vertex in a -uniform hypergraph , decide if there is in a tadpole rooted at .
It is easy to check that a given subhypergraph is a tadpole, so this problem is in . We do not know if this problem is tractable. We note however that it would be sufficient to loop for all vertex and all triples of edges of the form and check if there are two disjoint simple linear paths linking the two sources and to the targets and ( would be the contact between the path and the cycle). The complexity of the problem “Disjoint Connected Paths” for graphs has been a very fruitful research topic. In the case of two sources and two targets, the problem was shown to be tractable [37, 36]. In fact, in their well-known result, Robertson and Seymour [34] showed that the problem continues to be tractable for a constant number of sources and targets. The case where the number of sources and targets is unbounded is one of the first -complete problems in Karp’s list. For -uniform hypergraphs, it has been just proved recently [21] that the simple linear connectivity problem (one source and one target) is tractable. A corollary of the next theorem is that if the “Disjoint Connected Paths” problem for two sources and two targets is still tractable for -uniform hypergraphs, then deciding who is winning in a Client-Waiter game on a hypergraph of rank would also be tractable.
Theorem 5.
In Client-Waiter games, the problem of deciding if Client has a winning strategy for a given rank hypergraph can be solved by a polynomial time algorithm which uses the problem Tadpole as an oracle.
In particular, the problem lies in the class .
In fact, the proof of this theorem furthermore shows that the two problems are equivalent under polynomial-time Turing reductions. The problem of detecting a Tadpole can conversely be reduced to deciding on a winning position in a Client-Waiter game.
Proposition 6.
The problem Tadpole is polynomial-time many-one reducible to the problem of deciding if Client has a winning strategy in a Client-Waiter game played on a rank hypergraph.
We also focus on the second convention Waiter-Client. Surprisingly, the complexity of deciding if a position is winning is very different in this convention.
Theorem 7.
Waiter-Client is on rank hypergraphs, parameterized by .
More precisely deciding if a hypergraph of rank is winning for Waiter can be decided in time where is a computable function, i.e., in linear time when is fixed.
This result is obtained from a structural analysis of rank hypergraphs, using the famous sunflower lemma from Erdős and Rado [16]. This is a very natural approach since, intuitively, a huge sunflower (edges pairwise intersecting on the same center set) should be “reducible” in the sense that one could replace the sunflower by a unique edge given by the center set. Indeed, if Waiter can obtain the center and the sunflower is large enough, then he can make sure to get a petal and so a set of the sunflower. This intuition should be valid, but the main difficulty is to exactly state what is “huge”, as all other potentially useful sunflowers interact. We were unable to find a simple strategy along these lines. Instead, we describe a kernelization type algorithm which produces a subset of vertices (kernel) of the hypergraph such that Waiter wins on if and only if Waiter wins on the trace of on the kernel. The argument is based on a process which ultimately reaches a fixed point, the major drawback being that the kernel size is enormously large. This kernelization provides an algorithm for the Waiter-Client game on rank hypergraphs. This also proves that if Waiter can win, he wins using only moves for some function . However, the gap between the very large upper bound and the best known lower bound ( moves, required to win on pairwise disjoint edges of size ) indicates how little we understand strategies in Waiter-Client games.
Nevertheless, the fact that Waiter-Client is FPT for rank hypergraphs could indicate that this convention is simpler to analyse than Maker-Breaker. It would be interesting to revisit the classical topics in which the Maker-Breaker game was tried as a tool (for instance 2-colorability of hypergraphs or the Local Lemma, see [4, 22]) to see if Waiter-Client could provide more insight.
Complexity results for the various conventions are summarized in the table below.
Rank r | 2 | 3 | 4 | 5 | 6 | 7+ |
---|---|---|---|---|---|---|
Maker | P[Folklore] | P[19] | Open | -c | -c | -c |
-Breaker | [28] | [33] | [33, 35] | |||
Maker | P[Folklore] | Open | Open | Open | -c | -c |
-Maker | [28, 10] | [33, 10] | ||||
Avoider | -c | -c | -c | -c | -c | -c |
-Avoider | [17] | [17] | [17] | [17] | [17] | [17] |
Avoider | P [19] | Open | Open | Open | -c | -c |
-Enforcer | [23] | [23] | ||||
Client | P Prop [31] | Open | Open | -c | -c | |
-Waiter | Cor 5 | Thm 2 | Thm 2 | |||
Waiter | P Prop 12 | P Thm 7 | P Thm 7 | P Thm 7 | P Thm 7 | FPT w.r.t. |
-Client | Thm 7 |
Organization
We start with some preliminary results and definitions in Section 2. We first tackle the Waiter-Client convention in Section 3, where we prove that these games are parameterized by the rank. We then analyze the Client-Waiter convention. We prove in Section 4 that Client-Waiter games are -complete, even restricted to -uniform hypergraphs. In Section 5, we focus on rank 3 hypergraphs in Client-Waiter convention and we prove that the decision problem of determining the outcome of the game is in . For reasons of space, some of the proofs in Sections 4 and 5 are given in the full version of the paper.
2 Preliminaries
In this section, we first introduce the context of the game by providing some definitions, then we present some useful lemmas to handle Client-Waiter and Waiter-Client games.
Definition 8.
Let be a hypergraph and let be an integer. is said to have rank if all its edges have size at most . is said to be -uniform if all its edges have size exactly .
Definition 9.
The hypergraph is a subhypergraph of if and . If , the induced subhypergraph is the subhypergraph . The trace of on is the hypergraph .
We now present some general results about Client-Waiter and Waiter-Client games. The monotonicity of Client-Waiter and Waiter-Client conventions is immediate and folkloric (it directly follows a simple strategy stealing argument): if the “Maker” player (i.e. Client in Client-Waiter and Waiter in Waiter-Client) has a winning strategy on a subhypergraph, then they have one on the general hypergraph. A proof of this lemma is available in Oijid’s thesis [31].
Lemma 10.
Let be a hypergraph and let be a subhypergraph of . If Client (resp. Waiter) wins the Client-Waiter (resp. Waiter-Client) game on , then Client (resp. Waiter) wins the Client-Waiter (resp. Waiter-Client) game on .
In the Client-Waiter convention, edges of size are forced moves for Waiter.
Lemma 11 (Proposition 9 from [12]).
Consider a game in the Client-Waiter convention on some hypergraph . Let be the set of vertices already claimed by Waiter and Client respectively. If there exists such that and , then an optimal move for Waiter is to propose the two unclaimed elements of with his next move.
3 Waiter-Client games are FPT on rank k hypergraphs
In Oijid’s thesis [31], it is proved that if Waiter can win a rank 2 Waiter-Client game, he has a winning strategy in at most three moves. This leads to the following result:
Proposition 12 (Theorem 1.70 from [31]).
Let be a rank hypergraph. The winner of a Waiter-Client game played on can be computed in polynomial time.
We here extend this result by proving Theorem 7, i.e., that for any , the winner of a Waiter-Client game on a rank hypergraph can be computed in time parameterized by . This result is a far-reaching generalization of the following easy fact: if a rank hypergraph has pairwise disjoint edges, it is Waiter’s win. In each phase, he reduces the size of the edges by one and divide their number by two, by proposing disjoint pairs of elements from different edges, preventing Client from playing in more than a half of the remaining edges.
Let be a rank hypergraph. We call -sunflower a set of edges of pairwise intersecting on a fixed set called center of . When the center is empty, the sunflower simply consists of pairwise disjoint edges. We call petal of every set where . We authorize multisets in the definition, in particular, any edge of can be considered as an -sunflower for every (the petals being the emptyset). Such a sunflower with empty petals is called trivial. The celebrated Sunflower Lemma from Erdős and Rado [16] asserts that every rank hypergraph with at least distinct edges contains a non trivial -sunflower. We first show that the number of inclusion-wise minimal centers is bounded in terms of and . This is the first step of the FPT algorithm for the Waiter-Client game.
We say that an -sunflower of is minimal (with respect to inclusion) if no -sunflower of has a center strictly included in the center of . Let be a subset of vertices of , we say that is outside if all petals of are disjoint from . In particular, every edge forms a trivial -sunflower outside for every subset .
Lemma 13.
There exists a function for which every rank hypergraph has at most distinct centers of minimal -sunflowers. Moreover, there exists a function such that whenever is a subset of vertices of size , has at most distinct centers of minimal -sunflowers outside .
Proof.
We focus on the existence of , and postpone the outside case to the end of the proof.
We just have to show that if are distinct centers of size of a family of minimal -sunflowers , then is bounded in terms of and . Indeed, this implies the first part of the lemma since will be at most times this bound. Let us fix . We denote by the petal of . If is large enough, we can extract a subfamily of sunflowers from which, up to reordering, is assumed to be with the following additional properties:
-
all (distinct) centers form a sunflower with center .
-
for all , all petals form a sunflower with center .
Indeed, we just have for this to iterate extractions using the Sunflower Lemma, one for the centers, and for the petals. We now extract, by a simple greedy argument, an -sunflower centered at in . This gives a contradiction to the minimality of centers .
Let us define and . With this notation, the edge of the sunflower is .
Our first edge of is the first edge of , that is . Observe that intersects at most of the (pairwise disjoint) subsets and also at most of the subsets . Since , there is an index such that is disjoint from (notice that is automatically distinct from since is disjoint of ). We pick the edge as the second edge of , that is . Since all edges of intersect on , the edge is disjoint from . In particular . By a similar argument, since , there is an index such that is disjoint from . We now pick the edge as the third edge of . Iterating the process leads to an -sunflower centered at , a contradiction.
The proof for sunflowers outside some set is similar, but we have to take care of the fact that the final sunflower must have its petals outside . This is granted for the subsets and , which are petals of the original -sunflowers outside . But the (pairwise disjoint) sets can intersect . Observe that this happens at most times. Hence, choosing leads to a contradiction.
We now turn to the key-definition. Given some integer , we say that a set is an -kernel of if for every edge , there exists and an -sunflower outside centered at . Note that we can assume that is minimal outside . Observe that the union of all edges of is an -kernel for every , and that if has pairwise disjoint edges, then is an -kernel. We now show that there is always a bounded size kernel.
Theorem 14.
There exists a function such that every rank hypergraph has an -kernel of size at most , for every integer .
Proof.
Let and be some integer. We denote by the set of all centers of minimal -sunflowers, whose size is bounded by by Lemma 13. We then set and start to define our sequence as follows: Once is defined, we denote by the set of all centers of minimal -sunflowers outside and set . Our goal is to show that there exists a step , bounded by a function on and , such that . Observe that when this step is reached, is an -kernel. Indeed, let be an edge. Note that by itself is a trivial -sunflower outside , hence there is a included in which is the center of a minimal -sunflower outside . By definition of , we have , and therefore , implying that is an -kernel. To reach our conclusion, we need to show that the size of is upper bounded by some fixed function (depending on and ). For this, we just have to bound the size of in terms of the size of . This is easily obtained by Lemma 13, since the total number of minimal -sunflowers outside is at most .
Let us now denote, for every , the sequence where denotes the number of distinct centers of size of minimal -sunflowers outside . If , we claim that is lexicographically not larger than . To see this, assume that is minimum such that and suppose for contradiction that . This means that there is a set of size which is the center of a minimal -sunflower outside , but not the center of a minimal -sunflower outside . In particular, strictly contains which is the center of a minimal -sunflower outside , but not the center of a minimal -sunflower outside (by minimality of ). Since , there is a set of size which is the center of a minimal -sunflower outside , but not the center of a minimal -sunflower outside . This process always finds a smaller center which is minimum outside but not outside . This is a contradiction.
Note that the previous argument moreover shows that if , then the centers of minimal -sunflowers outside and outside are the same, thus .
Therefore, we just have to show that the sequence is ultimately constant after some fixed number of steps. Let us examine for this how the sequence evolves from to . At each step, the minimum index coordinate which differs in and decreases. Meanwhile, the coordinates where can change in an arbitrary way, but no more than some fixed amount since the size of is bounded by (a crude upper bound for the increase of would be for instance ). In all, this process terminates before some fixed number of steps depending on and . In all, since the number of minimal centers is bounded in terms of and , an upper bound on the number of steps can be computed as follows: when a nonzero coordinate of decreases by , all the next ones increases by at most . Therefore, the total increase on each coordinate can be computed inductively, and depends only on , – which is upperbounded by – and the maximum values of the previous coordinates which were also upperbounded by a function of and by induction hypothesis. Since the number of coordinates is , this gives an upper bound on the total number of steps depending only on and .
Kernels are relevant for Waiter-Client games. Indeed, if is a -kernel of a rank hypergraph and is the trace hypergraph on vertex set and edge set , we have the following equivalence:
Lemma 15.
is Waiter win if and only if is Waiter win.
Proof.
Since is obtained by reducing the size of some edges of (and deleting isolated vertices), if Waiter has a winning strategy on , he has one on .
Now assume that is Waiter win. Waiter can play his strategy on in order to select a set of vertices which contains some for . Since is a kernel, there exists a -sunflower outside with center . Now Waiter just has to play outside to select one of the petals of and win the game.
We can now prove Theorem 7, asserting that Waiter-Client is FPT on rank hypergraphs.
Proof.
Let be some input rank hypergraph on vertices and edges. We first describe an algorithm running in time which decides if is Waiter-win.
The strategy is to compute a -kernel as in Theorem 14 which size only depends on . Then, by Lemma 15, we just have to (brute force) test in time if is a Waiter win. The only point to check is how efficiently one can compute the sequence as in the proof of Theorem 14.
As the argument is similar for getting from , we just describe how to compute , that is the set of centers of minimal -sunflowers. An easy algorithm consists in (bottom up) testing all subsets included in some edge of and determine which ones are centers of -sunflowers (this single test being done in linear time, see for example Lemma 9.7 in [18]). The computation of all centers can be achieved in quadratic time. Once computed, determining which centers are minimal can be done in linear time. The size of the centers and, by Lemma 13, the number of minimal centers are bounded by . Therefore by going through the list of centers (ordered by their cardinal) only a function of number of steps is required to verify if it is minimal (i.e., it suffices to check if it contains a previously computed minimal center). In all, each can be computed in quadratic time, and the number of steps is bounded in terms of .
To improve the complexity of the above algorithm to linear, we just need to show that we can output the list of the centers of the -sunflowers in linear time. First, we define a tabular of size at most . Informally, for each , we will store in the potential elements for a -sunflower of center . So, for each edge and each subset , is a potential petal of a sunflower of center , and we add to . Then, if , there is a sunflower of size . But if a sunflower of center contains an edge of , it can always be replaced by another edge of . So we can safely remove from one edge of . Thus, is always bounded by . Therefore, adding or removing an edge from can always be done in time that only depends on . After this step, we can check for each if it is the center of a -sunflower by only considering the sets stored in .
4 6-uniform Client-Waiter games are PSPACE-complete
This section is dedicated to the proof of Theorem 2. As explained in the introduction, we start (in Section 4.1) with hypergraphs of rank :
Proposition 16.
Computing the winner of a Client-Waiter game is -complete, even restricted to hypergraphs of rank .
We notice that the membership in follows from an argument of Schaefer [35].
Lemma 17.
Both Client-Waiter and Waiter-Client positional games are in .
The most classical -complete problem is 3-, the quantified version of -. Our hardness proof is a reduction from -, but since the roles of the players in Client-Waiter games are very different, we go through an intermediate problem which was defined in the introduction (Definition 3). This is a variant of - where, at each turn, Satisfier chooses an index and instantiates the variable , and then Falsifier instantiates the variable . The main idea of this game is to introduce a CNF-game mimicking the fact that one player chooses which variable the second player has to play on. We prove in the full version of the paper that solving this game is -complete by a reduction from -.
Theorem 18.
Determining the winner of the -game is -complete.
4.1 Reduction to Client-Waiter games
4.1.1 Blocks in Client-Waiter games
The main tool of several reductions of positional games is pairing strategies. However, this cannot be applied to Client-Waiter games, since only Waiter has choices about how to make the pairs. We present blocks-hypergraphs and block-strategies that will be used similarly to pairing strategies to ensure that Client can claim some vertices. A blocks-hypergraph is depicted in Figure 1. The idea of blocks was already used in the -hardness proof from [14], but we present here a more formal definition of them. Intuitively, Blocks are a generalization of Lemma 11, which are blocks of size .
Definition 19 (Blocks).
Let be a hypergraph. A block of size is a set of vertices such that for some , and any set of vertices of is an edge.
If the vertices of can be partitioned into blocks, we say that is a blocks-hypergraph.
Lemma 20.
Let be a hypergraph, and let be a block of . If Waiter has a winning strategy in , he has to offer the vertices of two by two.
Proof.
Suppose that Waiter has a winning strategy in which he does not offer all vertices of two by two. Let . The first time he presents a vertex with a vertex , Client can choose . Then, each time Waiter offers at least one vertex in , Client claims it. In the end, Client will claim at least vertices of and therefore wins. Thus, if Waiter has a winning strategy, he has to offer the vertices of two by two.
Corollary 21.
Let be a blocks-hypergraph. If Waiter has a winning strategy in , any pair he offers contains vertices belonging to the same block of .
4.1.2 Construction of the hypergraph
We show now the reduction. The main idea of the reduction is that we construct a blocks-hypergraph, such that each block corresponds to the valuation that will be given to a variable.
Let be an instance of where , and is a 3-CNF on the variables of . We build a hypergraph as follows.
Let us define the set of vertices. Let , with for , (gadget which encodes Satisfier’s choice for the variable ) and (gadget which encodes Falsifier’s choice for the variable ).
Now we focus on the construction of the edges.
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The block-edges , which make each and each a block:
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The pair-edges (see Figure 2):
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The clause-edges. Each clause is a set of three literals . We define first, for and , the set which encodes the property that the literal is instantiated to ,
We define now the set of edges:
with
For example, if , we have , and .
Finally, we have two edges to encode :
The gadget for the pair is depicted in Figure 2.
Intuitively, these edges are constructed in such a way that:
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The block-edges force Waiter to always propose two vertices in the same set.
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The pair-edges force Waiter to give to Client the choice of the value of after he has made his choice for .
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The clause-edges which represent the clauses of and make the equivalence between a win of Waiter and a valuation that satisfies .
Finally, the reduction associates to the instance of Paired SAT the hypergraph ) with of Client-Waiter. This reduction is polynomial as contains edges, contains edges, and contains at most edges (where is the number of clauses of ).
We define an underlying assignment of the variables, corresponding to the moves on the hypergraph as follows:
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If Client claims ,
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If Client claims ,
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If Client claims and one of , ,
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If Client claims ,
We prove in the full version of the paper that Waiter has a winning strategy on if and only if Satisfier has a winning strategy on . Then, we can obtain the proof of Proposition 16.
Proof of Proposition 16.
First, according to Lemma 17, Client-Waiter games are in . We prove the hardness by a reduction from .
Let be an instance of . Consider the hypergraph obtained from the reduction provided in Subsection 4.1.2. It has vertices and edges, which is polynomial. Satisfier wins in if and only if Waiter wins in (see full version for the details of this proof). Therefore, determining the winner of a Client-Waiter game is -complete.
Moreover, as any edge of has size at most , the problem is even -complete restricted to hypergraphs of rank .
4.2 Reduction to 6-uniform hypergraphs
As it is done by Rahman and Watson for Maker-Breaker games [33], or by Gledel and Oijid for Avoider-Enforcer games [23], Proposition 16 can be strengthened by transforming the hypergraph of rank into a -uniform one (with ). We achieved this with the following lemma (proved in the full version of the paper):
Lemma 22.
Let be a hypergraph of rank . Let . If , there exists a hypergraph of rank where , having and such that Client has a winning strategy in the Client-Waiter game on if and only if she has one in .
Proof.
5 Rank 3 Client-Waiter equivalent to the problem of detecting a tadpole
Similarly to Maker-Breaker, the -hardness of Client-Waiter games restricted to -uniform hypergraphs leads us to consider smaller rank hypergraphs. In this section, we prove that determining the winner for rank 3 hypergraphs is in .
We show that testing if a hypergraph of rank is winning for Client reduces to the problem of searching two structures (the -snakes and the -tadpoles) in . We start by defining them.
Definition 23.
Let be a hypergraph and .
An alternating sequence of vertices and edges of , , is an -path if , , and, for , and the edge contains both and . The number of edges is called the length of . An -cycle is an -path of length at least two. An -path is called linear if the size of the intersection of two consecutive edges is always one. An -cycle of length at least is linear if the -path is linear and if the intersection of the end edges is exactly . We continue to call linear an -cycle of length if . An -path is said simple if appears only in , appears only in , and whenever some edges and intersect, then . Similarly, an -cycle is simple if whenever and intersect, then or .
An -path is also called an -path or a path. A cycle is an -cycle for some in .
For the rest of this paper, if there is no ambiguity, we will denote paths and cycles by the set of edges in them.
Definition 24.
Let be a vertex of . An -snake is a simple linear -path with such that for all , has exactly size , and has size at most .
Definition 25.
Let be a hypergraph and . If , an -tadpole is a sequence of edges where:
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belongs to and no other edge;
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belongs to , , and no other edge;
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is a -uniform simple linear -path ;
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is a -uniform simple linear -cycle ;
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is the only vertex which appears both in and .
If , an -tadpole is just a -uniform simple linear -cycle. When is an -tadpole, we may simply say is an -tadpole, or even just a tadpole.
Notice that the first two bullets in the definition are in fact consequences of the following ones.
We consider the problem Tadpole: Given a vertex in a -uniform hypergraph , decide if there is a -tadpole in .
We denote by the outcome of an optimal Client-Waiter game on the hypergraph . We will write when Waiter has a winning strategy on , and otherwise.
Let be a vertex of , we consider the following family of subhypergraphs of :
Notice that -uniform simple linear -cycles are particular cases of -tadpoles, so such subhypergraphs are also in . In the absence of ambiguity over the considered hypergraph, we will simply write .
We notice that Waiter has a winning strategy in which starts by offering a pair if and only if Waiter has a winning strategy in both trace hypergraphs and . So to simplify notations, we will write for the trace and for the induced subhypergraph . In particular, we can just write and .
Lemma 26.
Let be a vertex of hypergraph of rank and let . Then .
Proof.
We do the proof by induction on the size (number of edges) of . By definition, each subhypergraph of contains at least one edge.
Assume that contains exactly one edge. It means that is a -snake of length , i.e., one edge of length at most two. Since is an edge of size at most one, it is a winning position for Client.
Assume otherwise that has at least two edges. Three cases can happen.
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is a -snake of length at least two. So is a sequence of edges of the form where has size at most and the other edges have size . Assume , then by Lemma 11, there is a winning strategy for Waiter which starts with the pair . However, if Client selects , then is a -snake of size smaller than , and so the position is winning for Client by induction hypothesis. Hence .
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is a -uniform simple linear -cycle. So is of the form where is a -uniform simple linear path of positive length and contains . The conclusion is similar to the previous case. Indeed, if in Waiter offers the pair and Client picks , we obtain a -snake of size smaller than .
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is a -tadpole which is not a cycle. So is a sequence of edges of the form where the edge contains a vertex such that is a -uniform simple linear -path, and is a -uniform simple linear -cycle. Assume , then by Lemma 11, there is a winning strategy for Waiter which starts with the pair . However, if Client selects , then is a -tadpole of size smaller than , and so the position is winning for Client by induction hypothesis. Hence .
We show in the full version of the paper that in fact snakes and tadpoles are the only winning patterns for Client: If there are two vertices and such that and are empty, then if Waiter has a winning strategy, he has one starting by the pair . Consequently, testing whether a hypergraph is winning for Waiter boils down to whether such pairs can be recursively found until the hypergraph is empty.
6 Open problems
The construction of the kernel for Waiter-Client implies that, whenever Waiter has a winning strategy in a game on a hypergraph of rank , he has one with a number of moves which is bounded by a function of . We can denote by the maximum number of moves needed for Waiter to win on a hypergraph of rank . Since Waiter needs moves to win on pairwise disjoints edges . Proposition 12 shows that there is equality in the case . However, even for the construction of our kernel would wield a value way larger than .
Problem 1: Given , what is the value of ?
Theorem 5 and Proposition 6, show that the problems Tadpole and Client-Waiter for 3-uniform hypergraphs are intrinsically linked. However, as stated before, we do not know yet if the problem Tadpole is tractable, which lead us to our second problem.
Problem 2: Is Tadpole in ?
Lastly, our study of Client-Waiter focused on hypergraphs of rank 6 and 3, showing that the first case is -complete while the second is not. This leaves open the cases of hypergraphs of rank 4 and 5.
Problem 3: What is the complexity of Client-Waiter on hypergraphs of rank 4 and 5 ?
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