Fast Strategies in Waiter-Client Games on $K_n$

Waiter-Client games are played on some hypergraph $(X,\mathcal{F})$, where $\mathcal{F}$ denotes the family of winning sets. For some bias $b$, during each round of such a game Waiter offers to Client $b+1$ elements of $X$, of which Client claims one for himself while the rest go to Waiter. Proceeding like this Waiter wins the game if she forces Client to claim all the elements of any winning set from $\mathcal{F}$. In this paper we study fast strategies for several Waiter-Client games played on the edge set of the complete graph, i.e. $X=E(K_n)$, in which the winning sets are perfect matchings, Hamilton cycles, pancyclic graphs, fixed spanning trees or factors of a given graph.


Positional games.
Positional games belong to the family of perfect information games between two players, and they have become a field of intense studies throughout the last decades (for an introductory overview see e.g. [18] [24]). Starting with the seminal papers by Hales and Jewett [16] as well as Erdős and Selfridge [11], many beautiful results have been proven since then. Some of these even provide intriguing connections between positional games and other branches of combinatorics such as random graph theory (see e.g. [2], [4], [5], [14], [23], [25]), extremal combinatorics and Ramsey theory (see e.g. [2], [16], [28]).
Related to the so-called strong games, among which Tic-Tac-Toe and Hex are probably the most famous examples (see e.g. [2]), Maker-Breaker games were introduced in [11] and have since become the most studied type of positional games. In a more general form, the latter are played as follows. Given a hypergraph pX, F q and a positive integer b, two players, Maker and Breaker, alternate in turns. In each round Maker is allowed to claim (up to) one element from the board X, which has not been claimed before in the game, while Breaker is allowed to claim (up to) b such elements in each round. If, until the end of the game, Maker succeeds in fully claiming all the elements of any winning set F P F , she wins the game. Otherwise, Breaker has claimed at least one element of each of the winning sets, and is declared the winner of the game. No draw is possible. Here, the integer b is called the bias of Breaker. In the case when b " 1 holds, the game is called unbiased, and in all other cases the game is called b-biased or biased for short. Theorem 1.3 (Theorem 1.1 and Theorem 1.5(i) in [27]). There exist constants δ, c, C ą 0 such that for every large enough n and for every b ď δ n ln n the following holds. n 2`c b ď τ M B pPM n , bq ď n 2`C b ln b . 3 Note that in the above statement the second order term is determined up to a logarithmic factor.
For the Waiter-Client version of that game, we can get rid of this extra factor in the upper bound while also allowing the bias to be linear in n.
Theorem 1.4. There exist constants δ, C ą 0 such that for every large enough even n and for every b ď δn the following holds.
τ W C pPM n , bq ď n 2`C b .
Hamiltonicity game. Another game which is easily won by Maker when played on the edges of a complete graph K n is the Hamiltonicity game. This time, let H n denote the family of all Hamilton cycles of K n , and for a moment consider n to be sufficiently large. Already in their early paper, Chvátal and Erdős [7] could show that Maker wins this game in at most 2n rounds. This was later improved to τ M B pH n , 1q ď n`2 by Hefetz, Krivelevich, Stojaković and Szabó [17] through ad-hoc arguments coupled with the Pósa rotation technique [29]. Finally, Hefetz and Stich [20] were fighting for the exact result and proved that τ M B pH n , 1q " n`1 by providing a rather technical (13 pages long) proof involving multiple case distinctions. We will show that Waiter can win the Waiter-Client version of the unbiased Hamiltonicity game in the same number of rounds.
Theorem 1.5. For every large enough integer n the following holds: τ W C pH n , 1q " n`1 .
Considering the biased Hamiltonicity game, the following has been shown. Theorem 1.6 (Theorem 1.3 and Theorem 1.5(ii) in [27]). There exist constants δ, c, C ą 0 such that for every large enough n and for every b ď δ`n ln 5 n˘1 2 the following holds.
n`cb ď τ M B pH n , bq ď n`Cb 2 ln 5 b .
For the Waiter-Client version of that game, we are able to show a better upper bound on the number of rounds, while again allowing the bias to be of linear size. Theorem 1.7. There exist constants δ, C ą 0 such that for every large enough n and for every b ď δn the following holds.
τ W C pH n , bq ď n`Cb .
Pancyclicity game. Quite recently (in [4] and [14]) it was suggested to generalise the Hamiltonicity game even further by choosing the winning sets to be all the pancyclic subgraphs of K n -subgraphs containing the cycles of all possible lengths between 3 and n. Denote with PC n the 4 family of all such subgraphs. Ferber, Krivelevich and Naves [14] proved that for b " op ? nq the b-biased Maker-Breaker pancyclicity game on K n is won by Maker, while it was already known before that for b ě 2 ? n Breaker wins, since he can block all triangles [7]. In contrast to this result, it was shown by Bednarska-Bzdȩga, Hefetz, Krivelevich and Łuczak [4] that the threshold bias in the corresponding Waiter-Client game is linear in n. Apart from that, not so much is known for games with PC n being the family of winning sets. In particular, no tight results on the number of rounds has been given before. In this paper we prove the following. Theorem 1.8. In the unbiased Waiter-Client pancyclicity game the following holds: τ W C pPC n , 1q " n`p1`op1qq log 2 n.
Note that this means that Waiter wins almost perfectly fast, as every spanning pancyclic subgraph of K n has at least n`p1´op1qq log 2 n edges [6]. Moreover, the second order term in the above theorem will be made even more precise later on (see the remark at the end of Section 4).
Connectivity and fixed spanning trees. Another game easily won by Maker in its unbiased version is the so-called connectivity game on K n , introduced by Chvatál and Erdős [7] in which Maker's goal is to claim any spanning subgraph of K n . Indeed, we already discussed that for large enough n, Maker has a strategy to create a Hamilton cycle asymptotically fast. Moreover, since there is no reason for Maker to close cycles in the connectivity game, there needs to be a strategy which succeeds within n´1 rounds. In fact, following the result of Lehman [26] Maker can win the game when K n is replaced by any graph consisting of two edge-disjoint spanning trees.
Due to the simplicity of the aforementioned game, Ferber, Hefetz and Krivelevich [13] introduced a variant of the connectivity game in which Maker aims to occupy a copy of some given spanning tree T . Obviously, in order to have a winning strategy for Maker, we cannot choose T arbitrarily now as Breaker can block large stars. Thus, it is natural to put some degree constraints on the desired tree T . Let F T denote the family of all copies of T in K n . The following result has been proven first. Theorem 1.9 (Theorem 1.1 in [13]). Let ε ą 0. Then for every large enough integer n the following holds. Let T be any spanning tree on n vertices, with maximum degree ∆pT q ď n 0.05´ε , and let b ď n 0.005´ε be any positive integer. Then τ M B pF T , bq " n`opnq .
Thus, even when the maximum degree and Breaker's bias are increasing with n, Maker has a strategy to win the fixed spanning tree game asymptotically fast. Naturally, one may wonder whether the error term in the above theorem can be improved when we put stronger constraints on ∆pT q and b. This was answered in [8] as follows. 5 Theorem 1.10 (Theorem 1.1 and Theorem 1.4 in [8]). Let ∆ be a positive integer, then for every large enough integer n the following holds. For every spanning tree T on n vertices with maximum degree ∆pT q ď ∆, we have n´1 ď τ M B pF T , 1q ď n`1 .
Moreover, if T is a tree chosen uniformly at random among all labeled trees on n vertices (not necessarily having a constant bound on the maximum degree), then with high probability That is, for most choices of T , Maker wins the tree embedding game perfectly fast.
In our paper we will show that in the unbiased Waiter-Client game, Waiter has a fast winning strategy in the game pEpK n q, F T q which creates at most one wasted edge. Moreover, in contrast to the above theorems we may also allow the maximum degree to grow much faster with n. Theorem 1.11. There exists a constant ε ą 0 such that the following holds for every large enough integer n. Let T be spanning tree of K n with ∆pT q ď ε ? n, then Moreover, the lower and the upper bounds are tight, surprisingly also for ∆pT q ď 3.
H-factor game. Using our methods from the fixed spanning tree game, we are also able to describe fast winning strategies for games in which Waiter aims to create a factor of a fixed constant size tree. Note that the same kind of question was studied in the Maker-Breaker setting, but only in the case when the fixed tree is either a path or a star [9].
More precisely, for a fixed graph H and an integer n satisfying vpHq|n, an H-factor of K n is defined to be the vertex disjoint union of copies of H covering all vertices of K n . Let F n,H´f ac be the family of all such subgraphs. We prove the following result: Theorem 1.12. Let k ě 2 be a positive integer and let T be any fixed tree on k vertices. Provided that n is a large enough integer with k|n, the following holds: Moreover, the lower and the upper bound are tight.
Observe that in all the games considered so far, Waiter can always win at least asymptotically fast. We finish the paper by giving an example where this is not the case and challenge the reader to improve our bounds. We consider the triangle factor game, whose Maker-Breaker version has been discussed in [1] and [13].
6 Theorem 1.13. For every large enough integer n such that 3|n the following holds: 13 12 n ď τ W C pF n,K 3´f ac , 1q ď 7 6 n`opnq.
We note that the above game again provides an example where Waiter can win (asymptotically) at least as fast as Maker in the corresponding Maker-Breaker version. Indeed, as was observed by Szabó (a proof is contained in [13]), Maker cannot win the unbiased triangle factor game on K n within less than 7 6 n rounds.
Organisation of the paper. In Section 2 and Section 3 we focus on Theorem 1.2 and Theorem 1.5, proving slightly stronger statements so that we can use those as tools for the proofs in later sections. In Section 4 we prove Theorem 1.8, in Section 5 we show Theorem 1.11 and Theorem 1.12, in Section 6 we continue with the proof of Theorem 1.13, and in Section 7 we prove Theorem 1.4 as well as Theorem 1.7. Whenever we use results or methods from other papers, we will introduce the necessary concepts in the relevant sections. Finally, we finish the paper with a few concluding remarks and open problems in Section 8.
Notation. The graph-theoretic notation that we use is standard and closely follows the notation from [31]. We write rns :" tk P N : 1 ď k ď nu for every positive integer n.
Given any graph G, we let V pGq and EpGq denote the vertex set and the edge set of G, respectively, and set vpGq " |V pGq| and epGq " |EpGq|. If tv, wu P EpGq is an edge, we denote it with vw for short, and we call w a neighbour of v. We set N G pvq " tw P V pGq : vw P EpGqu to be the neighbourhood of v in G and call d G pvq " |N G pvq| the degree of v in G. Moreover, ∆pGq " max vPV pGq d G pvq denotes the maximum degree of G and δpGq " min vPV pGq d G pvq denotes the minimum degree of G. Given any two subsets A, B Ă V pGq and any vertex v P V pGq we write N G pv, Aq " N G pvq X A, d G pv, Aq " |N G pv, Aq|, N G pAq :" Ť vPA N G pvq, E G pAq :" tvw P EpGq : v, w P Au, e G pAq :" |E G pAq|, E G pA, Bq :" tvw P EpGq : v P A, w P Bu and e G pA, Bq " |E G pA, Bq|.
For any two graphs H and G we write H Ă G if both V pHq Ă V pGq and EpHq Ă EpGq hold, and call H a subgraph of G in this case. We also set GzH " pV pGq, EpGqzEpHqq in this case.
Given any A Ă V pGq, we call GrAs " pA, E G pAqq the subgraph of G induced by A.
Two graphs H and G are called isomorphic, denoted with H -G, if there exists a bijection f : V pHq Ñ V pGq such that vw is an edge of H if and only if f pvqf pwq is an edge of G. If the latter is the case, we also say that H forms a copy of G.
If we represent a path P by a sequence pv 1 , v 2 , . . . , v k q, this means that V pP q " tv 1 , v 2 , . . . , v k u and EpP q " tv i v i`1 : 1 ď i ď k´1u hold. Similarly, if a cycle C is represented by a sequence pv 1 , v 2 , . . . , v k q, we mean that V pCq " tv 1 , v 2 , . . . , v k u and EpCq " tv i v i`1 : 1 ď i ď k´1u Y tv k v 1 u. The length of both a path and a cycle is always the number of its edges. 7 Assume that some Waiter-Client game is in progress. We let W and C denote the graphs consisting of Waiter's edges and Client's edges, respectively. Any edge belonging to C Y W is said to be claimed, while all the other edges in play are called free.

Unbiased perfect matching game
In the section we will prove Theorem 1.2 by showing a slightly stronger statement which will also be applied later in the discussion of the tree embedding game (Section 5).
Theorem 2.1. For large enough n, the following holds: Let H Ă K n,n be any subgraph with epHq ď n 2 . Then, in the unbiased Waiter-Client game on K n,n zH, Waiter has a strategy to force a perfect matching of K n,n within n`1 rounds.
Proof. Let V " A Y B be the bipartition of K n,n . Throughout the game, we denote with R the set of vertices which are isolated in Client's graph. For n´4 rounds (Stage I), Waiter's strategy will be to force a large matching in Client's graph greedily, while making sure that e W YH pRq decreases with every round as long as this value is still positive. Within 5 further rounds (Stage II), Waiter will then complete this matching to a perfect matching.
If at any point during the game, Waiter is unable to follow her strategy, she forfeits the game.
(We will later see that this does not happen.) The set R is dynamically updated after every turn. Waiter's strategy consists of the following two stages: Stage I: This stage lasts n´4 rounds, in which Waiter forces a matching of size n´4 in Client's graph between A and B. Each round is played as follows: Let u P A X R be an arbitrary vertex maximizing d W YH pu, B X Rq. Then Waiter offers two free edges ub 1 , ub 2 with b 1 , b 2 P B X R. By symmetry, assume that Client claims ub 1 . Then, vertices u and b 1 are removed from R.
Stage II: When Waiter enters Stage II, Client's graph is a matching M 1 of size n´4. Let S " AzV pM 1 q and T " BzV pM 1 q at this point. Within 5 rounds, Waiter forces a matching of size 4 between S and T . The details are given later in the strategy discussion.
It is evident that, if Waiter can follow this strategy without forfeiting the game, she creates a perfect matching of K n,n zH in Client's graph within n`1 rounds. It thus remains to check that she does not forfeit the game.

Strategy Discussion:
Stage I: By induction on the number of rounds it follows that Waiter can follow the strategy of Stage I while she additionally maintains that Indeed, the above inequality holds at the beginning of the game, since at that point |R| " 2n and epH Y W q ď n 2 . Now consider any round r in Stage I, and assume that Waiter so far could follow the strategy and maintain Inequality 2.1. According to the strategy, she then picks a vertex u P A X R such that d W YH pu, B X Rq is maximal. By induction we have that the number of vertices b P B X R with ub being free is at least Hence, there exist at least two vertices b 1

Unbiased Hamiltonicity game
In the section we will prove Theorem 1.5 by showing a slightly stronger statement which will also be applied later in the discussions of the pancyclicity game (Section 4) and the tree embedding game (Section 5). (1) @v P V pK n q : d W pvq ă 10.
(2) Let e C 1 be the first edge Client claims in the game, then e C 1 P EpHq. (3) There exists a path P Ă C of length 1 5 n s.t. E W pV pP qq " H.
Proof. At the beginning of the game, let Waiter fix an arbitrary subset A 1 Ă V of size 4.
Waiter's strategy will be as follows: At first she forces four vertex disjoint paths P i (with i P r4s) in Client's graph, each having an endpoint in A 1 , such that these paths cover the whole vertex set V " V pK n q. Afterwards, she makes Client connect the mentioned paths to a Hamilton cycle such that the prescribed properties are satisfied.
Let A 1 " ta i : i P r4su and initially, for every i P r4s, let P i be the path consisting only of the vertex a i . Waiter will force Client to extend the path P i for every i P r4s, such that a i remains one of its endpoints, until the union of these four paths covers V . At any point of the game, we let P denote the collection of these four paths. We set V pPq " Ť iPr4s V pP i q and R " V zV pPq. Moreover, we always denote with a 1 i the other endpoint of P i different from a i (except when vpP i q " 1 where we set a 1 i " a i ), and we set A 2 " ta 1 i : i P r4su. During most of the game, Waiter's strategy is to consider the paths in pairs. She takes two turns to extend either P 1 and P 2 or P 3 and P 4 and does so alternately. In order to keep our notation short, we define π to be the permutation on r4s with cycles p1 2q and p3 4q, and we sometimes denote P 4 with P 0 when we consider indices modulo 4.
In the following, we will present a strategy for Waiter and then prove that this strategy allows her to force a Hamilton cycle within n`1 rounds such that all the prescribed properties are ensured. If at any point during the game, she is unable to follow her strategy, she forfeits the game. (We will later see that this does not happen.) The sets A 1 , A 2 , C, W, R and P are updated at the end of every turn. Waiter's strategy consists of the following three stages: Stage I: This stage lasts exactly n´4 rounds. During this stage Waiter extends the four paths P i until R " H. She does this by alternating between two types of moves: Type A: Let this be the i th round, and x P R be a vertex maximizing d W px, V pPqq (breaking ties arbitrarily). Then Waiter offers the edges xa 1 i and xa 1 i`1 (with indices taken mod 4). After Client has chosen one of these edges and thus extended one of the paths P i or P i`1 (indices taken mod 4), the sets A 2 , C, W, R and P are updated in the obvious way.
Type B: Let this be the i th round, and let x, y P R be picked arbitrarily. Moreover, let P t be the path which was extended in the previous round. Then Waiter offers the edges xa 1 πptq and ya 1 πptq . After Client has chosen one of these edges and thus extended the path P πptq , the sets A 2 , C, W, R and P are updated in the obvious way. 10 As long as |R| ě 2 holds, Waiter alternates between these two types of moves, with Type A being considered in odd rounds and Type B being considered in even rounds. Once |R| " 1 holds, Waiter plays one more round according to Type A. Afterwards, she proceeds with Stage II.
Stage II: This stage lasts exactly 2 rounds, in which Waiter forces Client to connect the paths from P. As long as |P| ą 2, Waiter connects two paths in Client's graph as follows: She fixes a vertex v P A 2 such that d W pv, A 2 q is maximal and offers vx, vy where x, y P A 1 are picked such that they do not belong to the same path as v. W.l.o.g. assume that Client claims vx and thus connects two paths P j 1 , P j 2 P P. Then update A 1 and A 2 by removing v and x respectively, and update P by removing P j 1 and P j 2 , while adding the path induced by EpP j 1 q Y EpP j 2 q Y tvxu.
Stage III: Within 3 rounds Waiter forces a Hamilton cycle as desired. The details of how she can do this can be found in the strategy discussion.
It is evident that, if Waiter can follow the strategy without forfeiting the game, she creates a Hamilton cycle H within n`1 rounds. It thus remains to check that she does not forfeit the game and that H fulfills the properties (1) -(3) from Theorem 3.1.

Strategy discussion:
Stage I: At any point of the game we call a vertex v P R bad if d W pv, V pPqq ě 1 holds. We observe first that there will never be more than one such vertex which at the same time helps Waiter to follow the strategy of Stage I. (b) if i is even, then e W pA 2 q " 0 and there is exactly one bad vertex z at the end of round i.
Moreover, (c) e W pA 2 q ď 2 at the end of round i " n´4.
Proof. The statement follows by induction on i. Waiter can obviously follow the strategy for round 1, where she offers two edges according to Type A. The edge claimed by Client extends P 1 or P 2 . After the update, the other edge belongs to E W pA 2 q and connects the endpoints of P 1 and P 2 , making sure that statement (a) holds for i " 1. Let i ą 1 then.
Consider first the case when i ď n´5 is even, and observe that |R| ě 2 before round i. In round i´1 Waiter played according to Type A and extended a path P t with t " i´1 or i (mod 4).
By induction, there was no bad vertex at the end of round i´1, but there was exactly one edge e W in E W pA 2 q, and e W connected endpoints of P i´1 and P i (indices taken mod 4). Now, in round i Waiter wants to play according to Type B and needs to offer two edges xa 1 πptq and ya 1 πptq with x, y P R. She can do this since |R| ě 2 and x, y cannot be bad. The edge claimed by Client extends P πptq . By this, e W is removed from E W pA 2 q after the update, leading to e W pA 2 q " 0.
The other edge goes to Waiter's graph and creates exactly one bad vertex z P tx, yu. Since i is even and thus πptq " i or i´1 (mod 4), it follows that Consider next the case when i ď n´5 is odd. By induction we know that e W pA 2 q " 0 and there was exactly one bad vertex z at the end of round i´1, but qq " 1 (indices taken mod 4). Now, in round i, Waiter wants to play according to Type A. She picks a vertex x P R maximizing d W px, V pPqq, hence setting x " z by the uniqueness of the bad vertex. She needs to offer the edges xa 1 i and xa 1 i`1 , which she can do since d W px, V pP i q Y V pP i`1 qq " 0 (indices taken mod 4). The edge claimed by Client extends a path in P by the vertex x, so that x is removed from R and there does not exist a bad vertex anymore. After the update of P in round i, the edge which goes to Waiter's graph connects the endpoints of P i and P i`1 (indices taken mod 4) belonging to A 2 , such that e W pA 2 q " 1 as claimed. Finally, consider the case when i " n´4. Then, Waiter can follow the strategy for round i analogously to the case when i ď n´5 is odd. By induction, using (a) or (b), it holds that e W pA 2 Y Rq " 1 at the end of round i´1. Since, during round i, the last vertex of R is moved to A 2 and since Waiter receives only one new edge, it is immediately clear that e W pA 2 q ď 2 afterwards.
Stage II: When Waiter enters Stage II, Client's graph is the disjoint union of four vertex disjoint paths covering V , with each path being of length roughly n 4 , since the pairs pP 1 , P 2 q and pP 3 , P 4 q were extended alternately during Stage I. Before we show that Waiter can follow Stage II of the proposed strategy, let us first observe how Waiter's edges are distributed at the end of Stage I.

Observation 3.3. Right at the moment when Waiter enters Stage II, the following holds:
Proof. For (a) notice that only in the first four rounds of Stage I Waiter offers edges incident to A 1 , and none of these edges is contained in A 1 . All the other endpoints of these edges are part of V pPq at the end of the 5 th round, since by property (a) of Observation 3.2, there do not exist bad vertices at that moment. But now, since all paths get extended further in Stage I by attaching edges to the vertices in A 2 and making appropriate updates, none of the mentioned endpoints belongs to A 2 later on. It thus follows that E W pA 1 , A 1 Y A 2 q " ∅ at the end of Stage I.
The inequality e W pA 2 q ď 2 is already given by property (c) in Observation 3.2.
For (b) observe that in Stage I, immediately after a vertex v is added to some path P i P P, it This degree may increase further by at most 2, when the pair of paths pP i , P πpiq q is considered for a further extension by a sequence of turns of Type A and Type B. But then, according to the strategy, both paths get extended, which ensures that from now on v is not an endpoint anymore and Waiter does not offer any further edges at v throughout Stage I.
For (c), let e W be any edge claimed by Waiter in Stage I. If this edge was offered by Type A, then after the corresponding round i, the edge e W belongs to E W pV pP i q, V pP i`1 qq (indices taken mod 4). Otherwise, if e W was offered by Type B, then after the corresponding round i, e W connects the unique bad vertex z with the endpoint of one of the paths P i´1 or P i . In the next round, playing according to Type A, Waiter makes sure that z is added to one of the paths P i`1 or P i`2 , leading to e W P E W pV pP r q, V pP s qq with r ‰ s. Now, having Observation 3.3 in hand, one can easily see that Waiter can follow Stage II of her strategy without forfeiting the game. Indeed, by property (a) from the observation, we know that all edges between A 1 and A 1 Y A 2 are free. Hence, she can offer edges vx and vy as required by her strategy. Moreover, we have e W pA 2 q ď 2 at the beginning of Stage II. Since in Stage II Waiter always picks v P A 2 such that d W pv, A 2 q is maximized and since v is removed from A 2 after the update, it follows that e W pA 1 Y A 2 q " 0 must hold at the end of Stage II.
Stage III: When Waiter enters Stage III, P consists of two paths, say P 1 and P 2 , such that all the four edges between their endpoints are still free. Moreover, it holds that d W pvq ď 6 for every v P V , since these degrees were bounded by 4 at the end of Stage I and since Stage II took only 2 rounds. Now, in Stage III, the first step for Waiter is to force a Hamilton path in Client's graph. To do so, she arbitrarily chooses an endpoint v of P 1 and offers the edges vx, vy with x, y being the endpoints of P 2 . Let P " pv 1 , v 2 , . . . , v n q be the Hamilton path that is created in Client's graph by this first move. Then by the conditions from the beginning of Stage III we know that v 1 v n is still unclaimed. Now, by using Pósa rotations [29], Waiter forces a Hamilton cycle in Client's graph. For her second move in Stage III Waiter then picks two vertices v i and v j with i, j R t1, nu such that they are not neighbours of each other, and such that e C 1 R tv i v i`1 , v j v j`1 , u where e C 1 denotes the edge claimed by Client in round 1, and such that the edges v 1 Such vertices must exist since Waiter's degree of all vertices is bounded by 7 at this moment. 13 She offers v i v n and v j v n to Client, who w.l.o.g. claims v i v n . In the last round Waiter then offers v 1 v i`1 and v 1 v n , and no matter which edge Client chooses, that edge closes a Hamilton cycle H. Hence, in order to finish our argument, it remains to prove that the properties of Theorem 3.1 hold.
Property (1) holds since for every v P V we had d W pvq ď 6 at the beginning of Stage III, while Stage III lasted exactly 3 rounds. Property (2) holds because Client's only edge which is not depending on which edge Client claimed in the last round, and v i was chosen in such a way that both edges differ from e C 1 . For Property (3) recall that, during Stage I, among the paths P 1 , . . . , P 4 there were no interior Waiter's edges, according to Observation 3.3, and each of these paths reached a length longer than 1 5 n. Also, when Waiter connects these paths during Stage II and Stage III no such interior edges are created, as Waiter only offers edges between endpoints. Moreover, when we remove the unique Client's edge which does not belong to H, at most one of these paths from Stage I can get destroyed, and hence there must remain at least three paths supporting Property (3).
Proof of Theorem 1.5. If Waiter would want to win the unbiased Hamiltonicity game on K n within n rounds, she would need a Hamilton path of length n´1 after n´1 rounds. However, since there is only one possible edge to extend this to a Hamilton cycle and Waiter needs to offer two edges, Client can easily prevent the Hamilton cycle in round n. Hence, τ W C pH n , 1q ě n`1 follows. For equality we just apply Theorem 3.1. l

Unbiased Pancyclicity game
Proof of Theorem 1.8. Set gpnq " rlog pkq 2 pnqs and f pnq " gpnq`100 for any positive integer k. In the following we will describe a strategy for Waiter in the unbiased Waiter-Client game on K n , and afterwards we will show that it is a strategy with which Waiter forces a pancyclic spanning subgraph of K n within at most n`log 2 pnq`Opmaxtf pnq, kuq rounds. Whenever Waiter is not able to follow the proposed strategy, she forfeits the game. (We will show later that this does not happen.) The strategy is split into the following five stages: Stage I: Within at most n`1 rounds, Waiter forces a Hamilton cycle H " pv 1 , v 2 , . . . , v n q such that the following holds right at the moment when the Hamilton cycle is completed: (H 1) d W pvq ă 10 for every v P V pK n q, The details can be found in the strategy discussion. Afterwards, Waiter proceeds with Stage II.
14 Stage II: This stage lasts two rounds. At first, Waiter offers the edges v i v f pnq`i with i P r2s. Among these edges, Client needs to claim one; denote it with w 1 w f pnq`1 , and afterwards let for every i P rns (with v n`1 :" v 1 ). In the second round, Waiter offers two free edges between w f pnq`1 and tw n´60 , . . . , w n´50 u, among which Client needs to choose one. Afterwards, Waiter proceeds with Stage III.

Stage III:
This stage lasts f pnq´2 rounds. In the i th round of Stage III, Waiter offers the edges w 1 w i`2 and w f pnq´i w f pnq`1 , among which Client always needs to claim one. Once all the f pnq´2 rounds of Stage III are played, Waiter proceeds with Stage IV.

Stage IV:
This stage lasts at most rlog 2 pnqs rounds. Waiter makes sure that at the end of the i th round of Stage IV, there exist vertices w t 0 , w t 1 , . . . , w t i such that the following holds: In order to do so, in the i th round Waiter offers two free edges of the form w t i´1 w j with mint2t i´12 i, nu´20 ď j ď mint2t i´1´2 i, nu. For the edge w t i´1 w j chosen by Client, Waiter then sets Once there is a round s such that n´20 ď t s ď n holds, Waiter stops with Stage IV and proceeds with Stage V.

Stage V:
This stage lasts at most k´1 rounds. For her i th move, Waiter aims to make Client claim an edge w 1 w ℓ with 2 log piq 2 pnq ď t j ď ℓ ď t j`2 0 ď 10 log piq 2 pnq for some j ď s. In case Client does not already possess such an edge, Waiter just offers two free edges of the mentioned kind. Otherwise, Waiter just skips that move and proceeds to her next move.
In the following discussion, we need to check two properties for the given strategy: (1) Waiter can always follow the proposed strategy without forfeiting the game, and (2) when Stage V is over, Client's graph is pancyclic. Just note that then a pancyclic graph will be forced within at most n`log 2 pnq`f pnq`k " n`p1`op1qq log 2 pnq rounds, according to the bounds on the number of rounds given in the descriptions of the stages.

Strategy discussion:
(1) -Following the strategy: Waiter can follow Stage I because of Theorem 3.1. According to that theorem, Waiter can force a Hamilton cycle H within n`1 rounds such that Property (H 1) holds immediately after H is created. Moreover, she can make sure that right at this moment there is a path P Ă H of length n 5 such that E W pV pP qq " ∅ holds. Split P into two subpaths Q 1 and Q 2 of length n 10 each. Since epCzHq " 1 holds at the end of round n`1, we know that there must be some i P r2s with E C pV pQ i qqzEpQ i q " ∅. Labelling the vertices of H in such a way that V pQ i q " tv 1 , . . . , v n 10 u holds, we obtain Property (H 2). Afterwards, in Stage II and in Stage III, Waiter needs to offer several edges contained in Eptv i : i ď f pnq`2uqzEpHq, which are still free by Property (H 2) and since f pnq`2 ă n 5 . She also needs to offer two edges between w f pnq`1 and tw n´60 , . . . , w n´50 u which is possible since d W pw f pnq`1 q ă 10 at the end of Stage I. Next consider Stage IV and observe the following: if Waiter can follow this part of her strategy and as long as t i ă n´20 holds, we have t i ě 2t i´1´2 i´20 and t 0 ě f pnq ą 100, leading to at the end of Stage II. Since afterwards (in Stage III-IV) until the current round, each offered edge was incident to some w ℓ , ℓ ă t i´1 ă mint2t i´1´2 i, nu´20, there need to be at least two free edges as required by the strategy description. Once Client has claimed one of these edges, it is obvious that the Properties (W 1)-(W 3) hold again for i.
Finally, consider the i th round of Stage V for i P rk´1s. Since t 0 " f pnq`1, t s ě n´20 and since t j`1 ă 2t j for all j ď s, it follows that there must be some j P rss with 2 log piq 2 pnq ď t j ď 5 log piq 2 pnq. Having such t j fixed, it is enough to find two free edges w 1 w ℓ with t j ď ℓ ď t j`2 0. This is possible, because d W pw 1 q ă 10 at the end of Stage I and since in Stage II-IV no such edge was offered.
(2) -Finding pancyclicity: Let H " pw 1 , w 2 , . . . , w n q be the Hamilton cycle from Stage I. It is the edge disjoint union of two paths P 1 " pw 1 , w 2 . . . , w f pnq`1 q and P 2 " pw f pnq`1 , . . . , w n´1 , w n , w 1 q of lengths f pnq and n´f pnq, respectively. Both paths are closed to cycles in Client's graph by the edge w 1 w f pnq`1 which was claimed in Stage II. We next observe that after Stage III the following holds: 1 has length f pnq´t, (iii) w 1 and w f pnq`1 are the endpoints of P t 1 . 16 Proof. For t " 0 and t " f pnq´1 we let P f pnq´1 1 consist of the edge w 1 w f pnq`1 and P 0 1 " P 1 . For every 1 ď t ď f pnq´2 Client claimed either w 1 w t`2 or w f pnq´t w f pnq`1 in round t of Stage III, and thus we can choose either P t 1 " pw 1 , w t`2 , w t`3 , . . . , w f pnq , w f pnq`1 q or P t 1 " pw 1 , w 2 , . . . , w f pnq´t´1 , w f pnq´t , w f pnq`1 q.
Let w f pnq`1 w n´p be the edge claimed by Client in the second round of Stage II, and observe that 50 ď p ă f pnq. By closing the above paths P t 1 into cycles, either using the edge w 1 w f pnq`1 or the path pw f pnq`1 , w n´p , w n´p`1 , . . . , w n , w 1 q, we obtain cycles of all lengths between 3 ď ℓ ď f pnq`p.
Hence, it remains to find cycles of all the lengths larger than f pnq`p ě f pnq`50. In order to do so, we will fix 0 ď m ď k´1 from now on and we will explain how we find cycles of all lengths between log pm`1q 2 pnq`50 and mint2 log Having m fixed, set w km " w n if m " 0 and otherwise let w km be the vertex w ℓ from the m th move in Stage V. By Stage IV (in case m " 0) or Stage V (in case m ‰ 0q there is some index j m ď s such that k m´2 0 ď t jm ď k m . Moreover, set a i :" t i´ti´1´1 to be the number of vertices between w t i´1 and w t i on P 2 , for every i ď s. Then, at the end of Stage V the following holds: Proof. If we extend the subpath pw f pnq`1 , . . . , w km q from P 2 by Client's edge w km w 1 , we obtain a path P 1 2 of length k m´f pnq. By replacing any subpath pw t i´1 , . . . , w t i q, where i ď j m , with the edge w t i´1 w t i which was claimed in Stage IV, the path P 1 s can be shortened by exactly a i edges. As this can be done for any i P S, we can shorten P 1 2 to a path of length k m´f pnq´ř iPS a i . This proves the observation. Now, by joining the path P t 1 with the path P S 2 , for any 0 ď t ď f pnq´1 and any S Ă rj m s, we obtain a cycle of length k m´p t`ř iPS a i q. We will see in the following that this will indeed give us cycles of all lengths between log pm`1q 2 pnq`50 and mint2 log pmq 2 pnq, nu. We start with the following observation. Proof. Inductively, one may show that for every 0 ď j ď j m (S) every integer in rf pnq´1`ř iPrjs a i s can be written as a sum t`ř iPS a i with 0 ď t ď f pnq´1 and S Ă rjs .
The beginning of the induction (j " 0) should be obvious. So, let j`1 ą 0, and assume (S) to be true for j. By (W 3), the definition of a j and since t 0 " f pnq`1, it follows that As, by induction, the integers up to the last sum can be written as t`ř iPS a i with 0 ď t ď f pnq´1 and S Ă rjs, adding a j`1 to the latter creates all integers in Note that the last set contains all the remaining integers for completing the induction step. This shows (S) for j`1 and finishes the proof of the observation.
Finally, observe that f pnq´1`ř iPrjms a i " t jm´jm´2 and hence, by Observation 4.3 and by the argument immediately after Observation 4.2, we see that we can find cycles of all lengths between k m´p t jm´jm´2 q and k m . Now, note that k m ě t jm ě k m´2 0 by the choice of k m , and k m ě mint2 log pmq 2 pnq, nu since t jm ě 2 log pmq 2 pnq by Stage V (in case when m ‰ 0). Moreover, using that t jm ď 10 log Remark: In the above argument, we need that the interval r3, log pkq 2 pnq`50s and all the intervals rlog pm`1q 2 pnq`50, mint2 log pmq 2 pnq, nus with 0 ď m ď k´1 cover all integers from 3 to n. Hence we only need to ensure that 2 log pmq 2 pnq ě log pmq 2 pnq`50 holds for all m ď k´1, i.e. log pmq 2 pnq ě 50, which is given when log pk`2q 2 pnq ě 2. Thus, if we choose Hpnq to be the smallest integer t such that log ptq 2 pnq ă 2 holds, then the above proof gives us that the game is won within n`log 2 pnq`Hpnq`Op1q rounds. This is only an additive constant away from the best known general upper bound on the minimal size of pancyclic graphs as mentioned in [6].

Unbiased games involving trees
In this section we will prove Theorem 1.11. Based on ideas from [8] and [13], we will split the given tree T into a subgraph T 1 and a nice behaving structure (large matching or long path).
In her strategy, Waiter will first force a copy of T 1 more or less greedily and without wasting any move, while additionally caring about the distribution of her edges. Afterwards, Waiter will force the appropriate nice behaving structure while wasting at most one round.
Let T be any tree. We denote by L " LpT q the set of leaves of T and by N T pLq the set of vertices which are in the neighbourhood of the leaves w.r.t. T . For every x P N T pLq let ℓpxq be the number of leaves which are neighbours of x in T and define ∆pN T pLqq " max xPN T pLq ℓpxq.
We start with the following lemma, similar to Lemma 2.1 in [22], which states that each of the trees T considered in Theorem 1.11 has a nice behaving structure as mentioned above: a large matching where every edge is incident to a leaf or a long bare path, i.e. a path such that all the inner vertices have degree 2 in T . Proof. Set ε " µ{3 and assume that |N T pLq| ă µ ? n. We will show now that T needs to contain a bare path of length at least µ ? n. By our assumption, we obtain |L| ď ∆pN T pLqq¨|N T pLq| ă εµn . Now, let T 1 " T´L then n 1 :" |V pT 1 q| ě n´µεn ą 2n 3 . Let S i " tv P V pT 1 q | d T 1 pvq " iu and S ěi " tv P V pT 1 q | d T 1 pvq ě iu for every i P rns, and observe that S 1 Ď N T pLq. Further, let P be the collection of maximal bare paths in T 1 and let T be the tree obtained from T 1 by contracting each path in P to an edge. Then |P| " epT q " vpT q´1 " |S 1 |`|S ě3 |´1.
Also, by the Handshake Lemma it holds that leading to |S ě3 | ă |S 1 | and hence By the Pigeonhole Principle and since each vertex from S 2 belongs to exactly one path in P, there exists a bare path of length at least where last inequality uses that µ ă 1 2 , n 1 ą 2 3 n and that n is large enough. Theorem 1.11 will follow from the next slightly stronger result which will be used later as well for the study of the tree-factor game.
Theorem 5.2. There exists ε ą 0 such that the following holds for every large enough integer n: Let T be a tree on n vertices and let v P V pT qzpL Y N T pLqq be such that the following holds: (1) d T pvq ď n 3 and , (2) ∆pT ztvuq ď ε ? n.
Moreover, let p P V pK n q. Then, in an unbiased Waiter-Client game on K n , Waiter has a strategy to force Client to claim a copy of T within n rounds such that (a) in Client's copy of T , the vertex p P V pK n q represents the vertex v P V pT q, and

(b) in each round of her strategy, Waiter offers either 2 edges or no edge incident to p.
Proof. Let µ " 1 3 and choose ε ă µ 20 according to Lemma 5.1. Then, given a tree T with the properties from the theorem above, there exists a bare path of length at least µ ? n in T (Case A) or we have |N T pLq| ě µ ? n (Case B). We provide a different strategy for Waiter for each case.
In order to describe Waiter's strategy, we use notation similar to that from [8] and [13]. Let S Ď V pT q be an arbitrary set, then an S-partial embedding of T in G is an injective mapping f : S Ñ V pGq such that f pxqf pyq is an edge in G whenever xy is an edge in T . Vertices in S are called embedded vertices. Let any subgraph T 1 Ă T be given. Then a vertex v P f pSq is called If at any point during the game, Waiter is unable to follow the strategy, she forfeits the game.
(We will see later that this does not happen.) Case A -Long bare path. Consider first the case when T contains a bare path of length at least µ ? n. Let P be such a path of length µ ? n, and denote its endpoints with u and w. Let u 1 and w 1 be the neighbours of u and w in P respectively. T zP is a forest with two tree 20 components, say T 1 and T 2 . We let T 1 Ă T be the forest induced by EpT 1 q Y EpT 2 q Y tuu 1 , ww 1 u.
W.l.o.g. we may assume that both v and u belong to T 1 .
In broad terms, Waiter's strategy is to first force a copy of T 1 (Stage I and II) and then to force a copy of the bare path P (Stage III) in such a way that a copy of T is created within n rounds.
Throughout the game, she maintains a set S and an S-partial embedding f of T into K n in order to represent the subgraph of T which currently is isomorphic to Client's graph. Initially, set S " tv, wu, f pvq " p and f pwq " q for arbitrary p, q P V pK n q. Waiter's strategy is split into the following stages: Stage I: This stage lasts for d T pvq rounds in which Waiter closes the vertex v w.r.t. T . Each round is played as follows: First, Waiter fixes an arbitrary vertex t P N T pvqzS. Waiter then offers two edges pa 1 , pa 2 such that both edges are free and a 1 , a 2 P A. By symmetry, assume that Client chooses the edge pa 1 .
Then Waiter updates A, S and f by removing a 1 from A, adding t to S and setting f ptq :" a 1 .
Stage II: This stage lasts epT 1 q´d T pvq rounds in which it is Waiter's goal to create a V pT 1 qpartial embedding. For each round, she plays as follows: If S " V pT 1 q holds, Waiter proceeds to Stage III. Otherwise, fix an arbitrary vertex x P f pSztu 1 , w 1 uq X O T and let t " f´1pxq. Since x is open, there exists a vertex z P pV pT 1 q Y V pT 2 q Y tu 1 , w 1 uqzS such that tz P EpT 1 q. Waiter then offers two free edges a 1 x and a 2 x such that a 1 , a 2 P A. By symmetry, we may assume that Client chooses a 1 x. Then Waiter updates A, S and f by removing a 1 from A, adding z to S and setting f pzq :" a 1 . Afterwards, she repeats Stage II.
Stage III: When Waiter enters Stage III, Client's graph is a copy of the subgraph T 1 . Within n´epT 1 q rounds, Waiter now forces a Hamilton path on V pK n qzf pV pT 1 qYV pT 2 qq with endpoints f pu 1 q and f pw 1 q. The details of how Waiter can do this are given later in the strategy discussion.
Strategy Discussion: It is obvious that if Waiter can follow the given strategy without forfeiting the game, she forces a copy of T within at most d T pvq`pepT 1 q´d T pvqq`pn´epT 1 qq " n rounds. Hence, it remains to show that Waiter can indeed do so. However, before we study each stage separately, let us observe the following:

Observation 5.3. Throughout Stages I and II, as long as Waiter can follow the proposed strategy, it holds that
n´2 and e CYW pAq " 0, (ii) d W px, Aq ď d C pxq for every x P f pSq.

Proof. Property (i) is immediately clear. The inequality |A| ě µ
? n´2 holds, since the strategy for the mentioned stages is to force a copy of T 1 without wasting any move and since epT 1 q " epT 1 Y T 2 q`|tuu 1 , ww 1 u| " n´µ ? n`2. The equation e CYW pAq " 0 holds, since Waiter always only offers edges that intersect f pSq. For Property (ii), observe that d W px, Aq may only increase, when x P f pSq (since e CYW pAq " 0) and Waiter offers an edge between x and A. However, when this happens in any of the Stages I and II, Waiter actually offers two edges between x and A, which increases d C pxq by 1 at the same time. Hence, d W px, Aq cannot become larger than d C pxq.
In the following we check that Waiter always can follow the strategy without forfeiting the game.
Stage I: According to the strategy, Waiter needs to offer 2d T pvq ď 2n 3 edges at v. She can easily do so, since there exists n´1 edges to choose from.

Stage II:
The vertex z, which is described in the strategy, exists because of our assumption that x is an open vertex. Moreover, by Observation 5.
n, which in turn means that at least |A|´ε ?
n edges between x and A are free. Hence there exist two free edges xa 1 and xa 2 as desired and Waiter can follow the proposed strategy.
Stage III: When Waiter enters Stage III, she has successfully managed to force a copy of T 1 . Let A 1 " V pK n qzf pV pT 1 q Y V pT 2 qq and observe that e CYW pA 1 q " 0 holds right at this moment. Indeed, according to Observation 5.3 we have e CYW pAq " 0. Moreover, e CYW ptf pu 1 q, f pw 1 qu, Aq " 0 holds, since in Stage II Waiter always chooses x different from f pu 1 q and f pw 1 q, which in turn ensures that, once these vertices are embedded, Waiter never offers edges incident to those again.
For Stage III, Waiter now plays as follows. At first she considers a fake round which is not played at all but where Waiter pretends that Client claimed the edge e C :" f pu 1 qf pw 1 q. Afterwards, she continues according to the strategy from Theorem 3.1 (with K n replaced by K n rA 1 s -K |A 1 | ), which ensures that within |A 1 |`1 rounds there is a Hamilton cycle in Client's graph on A 1 which contains the edge e C . Since the first round was a fake round, this actually means that, within |A 1 | " n´epT 1 q rounds, Waiter obtains a Hamilton path in A 1 as desired.
Case B -Many leaf neighbours. Consider next the case when |N T pLq| ě µ ? n holds. Then, there exists a matching M 0 of size at least µ ? n which consists of edges that are incident to leaves of T . Define the sets L 0 :" L X V pM 0 q, D i " tw P V pT q : dist T pv, wq " iu, D odd :" Ť i odd D i and D even :" Ť i‰0 even D i . By the Pigeonhole Principle we have that there is a set D good P tD odd , D even u such that D good X N T pL 0 q has size at least µ 1 ? n with µ 1 :" µ{3. Let M 1 " te P M 0 : e X D good ‰ ∅u, L 1 " L X V pM 1 q and T 1 " T´L 1 . By the choice of D good we have that |M 1 | ě µ 1 ? n, v R V pM 1 q and dist T 1 px, yq ě 2 for every x, y P N T pL 1 q.
In broad terms, Waiter's strategy now is to first force a copy of T 1 (Stage I and II) and then to force a copy of the matching M 1 (Stage III) in such a way that a copy of T is created within n rounds. Throughout the game, she again maintains a set S and an S-partial embedding f of T into K n in order to represent the subgraph of T which currently is isomorphic to Client's graph.
Initially, set S " tvu and f pvq " p for an arbitrary p P V pK n q. Additionally, at any moment in the game we define S 1 :" S X N T pL 1 q.
Stage I: This stage lasts for d T 1 pvq " d T pvq rounds in which Waiter closes the vertex v w.r.t. T 1 . Each round is played as follows: First, Waiter fixes an arbitrary vertex t P N T 1 pvqzS. Waiter then offers two edges pa 1 , pa 2 such that both edges are free and a 1 , a 2 P A. By symmetry, assume that Client chooses the edge pa 1 .
Then Waiter updates A, S and f by removing a 1 from A, adding t to S and setting f ptq :" a 1 .
Stage II: This stage lasts epT 1 q´d T 1 pvq rounds in which it is Waiter's goal to create a V pT 1 qpartial embedding, while also taking care of the distribution of Waiter's edges between the open and the available vertices. For each round, she plays as follows: If S " V pT 1 q then Waiter proceeds to Stage III. Otherwise, Waiter considers the following case distinction: Case 1. Let there be two vertices u 1 , u 2 P O T 1 , and let t 1 " f´1pu 1 q and t 2 " f´1pu 2 q. By assumption, there exist vertices z 1 , z 2 P V pT 1 qzS such that t 1 z 1 , t 2 z 2 P EpT 1 q. Waiter then picks any vertex a P A such that au 1 and au 2 are free, where she prefers vertices satisfying d W pa, f pS 1 qq ě 1, and offers au 1 and au 2 to Client. By symmetry we may assume that Client chooses au 1 . Then Waiter updates A, S and f by removing a from A, adding z 1 to S and setting f pz 1 q :" a .

Case 2.
Let there be only one vertex u P O T 1 , but assume that u R f pS 1 q. Let t " f´1puq. By assumption, there exists a vertex z P V pT 1 qzS such that tz P EpT 1 q. Waiter then picks any vertices a 1 , a 2 P A such that a 1 u and a 2 u are free, where she prefers vertices satisfying d W pa i , f pS 1 qq ě 1, and offers these two edges to Client. By symmetry we may assume that Client chooses a 1 u. Then Waiter updates A, S and f by removing a 1 from A, adding z to S and setting f pzq :" a 1 .

Case 3.
Let there be only one vertex u P O T 1 , and moreover assume that u P f pS 1 q. Let t " f´1puq. By assumption, there exists a vertex z P V pT 1 qzS such that tz P EpT 1 q, and among such vertices we choose z such that d T 1 pzq is maximal. Waiter then picks any vertices a 3 , a 4 P A such that a 3 u and a 4 u are free and such that d W pa 3 , f pS 1 qq " d W pa 4 , f pS 1 qq " 0, and offers these two edges to Client. By symmetry we may assume that Client chooses a 3 u. Then Waiter updates A, S and f by removing a 3 from A, adding z to S and setting f pzq :" a 3 .
Afterwards, Waiter repeats Stage II.
Stage III: When Waiter enters Stage III, Client's graph is a copy of the subgraph T 1 . Within epM 1 q`1 rounds, Waiter now forces a perfect matching between V pK n qzf pV pT 1 qq and f pN T pL 1 qq. The details of how Waiter can do this are given later in the strategy discussion.
Strategy Discussion: It is obvious that, if Waiter can follow the given strategy without forfeiting the game, she forces a copy of T within at most d T pvq`epT 1 q´d T pvq`epM 1 q`1 " epT q`1 " n rounds. Hence, it remains to show that Waiter can indeed do so. To this end, we define a vertex u to be a stopping vertex if u P f pSq and if for the vertex t " f´1puq the following holds: for every y P N T 1 ptqzS we have d T 1 pyq " 1. We first observe the following which will help us later to show that Waiter can follow the proposed strategy.  4 with u P f pSq, Client gets the edge ua 1 or ua 3 , while a i P A for every i P r4s, and then a 1 or a 3 is moved from the set A to the set f pSq (and hence maybe to the set f pS 1 q) by the update of that case. But then, in Case 2, since e W pa 1 , Aq " 0 holds by Property (i) for the previous round and since u R f pS 1 q by assumption of that case, we conclude that d W px, f pS 1 qq does not increase for any x which remains in the set A. Moreover, in Case 3, since e W pa 3 , Aq " 0 holds analogously and d W pa 4 , f pS 1 qq " 0 was true at the end of the previous round (by the choice of a 4 in that case), we conclude that after following Case 3 we have d W pa 4 , f pS 1 qq " 1 and d W px, f pS 1 qq does not increase for any x ‰ a 4 which remains in the set A. Hence, in either case, the degrees d W px, f pS 1 qq never exceed 1 for the vertices x P A.
It remains to verify Property (iv). By the discussion above for Property (iii) we see that e W pf pS 1 q, Aq can only increase in Case 3 of Stage II, and if it does then it increases by exactly 1. Hence, it is enough to show that if e W pf pS 1 q, Aq " ε ? n`1 holds at the end of any round r and if at this moment O T 1 still does not solely consist of a stopping vertex, then e W pf pS 1 q, Aq ď ε ? n will hold at the end of round r`1. So, assume the mentioned conditions hold. Then, round r was played according to Case 3 of Stage II. That is, at the beginning of round r there was only one vertex u P f pSq which was open w.r.t. T 1 , and moreover u P f pS 1 q. By assumption the vertex u was not a stopping vertex. That is, for t " f´1puq we could find a vertex y P N T 1 ptqzS such that d T 1 pyq ě 2. In round r, Waiter played according to Case 3 of Stage II, thus the vertex z (for the strategy described in Case 3) with tz P EpT 1 q was chosen such that d T 1 pzq ě 2. Waiter then fixed vertices a 3 , a 4 P A such that a 3 u and a 4 u were free and she offered these two edges to Client. By symmetry, we may assume that Client chose a 3 u and the other edge a 4 u was added to Waiter's graph. Then Waiter updated A, S and f by removing a 3 from A, adding z to S and setting f pzq " a 3 . Using that d T 1 pzq ě 2, we conclude that a 3 needs to be open w.r.t. T 1 at the end of round r. Moreover, using Property (i), we also get that d W pa 3 , Aq " 0 holds at this moment. For round r`1, two possible cases may occur now. The first case is that u is still an open vertex w.r.t. T 1 at the beginning of round r`1. Then this round is played according to Case 1 with tu 1 , u 2 u " tu, a 3 u. If Waiter can follow the strategy, we already know from the discussion of Property (iii) that d W px, f pS 1 qq stays unchanged for every x which remains in the set A. However, by the strategy of Case 1, it also happens that Waiter picks some vertex a P A with d W pa, f pS 1 qq ě 1 such that u 1 a and u 2 a are free. Note that such a vertex a exists since by Property (iii) and under assumption of e W pf pS 1 q, Aq " ε ? n`1 there exist ε ? n`1 vertices 25 a P A with d W pa, f pS 1 qq ě 1, while At the end of round r`1 the vertex a gets removed from A, and hence e W pA, f pS 1 qq gets reduced by d W pa, f pS 1 qq ě 1.
The second case is that u is not an open vertex w.r.t. T 1 at the beginning of round r`1, and hence a 3 is the only open vertex w.r.t. T 1 at that point. Since t " f´1puq P N T pL 1 q holds and tz P EpT 1 q holds for the vertex z " f´1pa 3 q we know that z R N T pL 1 q by the choice of M 1 and L 1 . Hence a 3 R f pS 1 q and thus, in round r`1, Waiter plays according to Case 2 (with u :" a 3 ). That is, Waiter then offers two edges a 1 a 3 and a 2 a 3 such that a 1 , a 2 P A and such that d W pa i , f pS 1 qq ě 1 holds for i P r2s (which is possible since d W pa 3 , Aq " 0 and since there exist ε ? n`1 vertices a P A with d W pa, f pS 1 qq ě 1). By symmetry we may assume that Client chooses a 1 a 3 . Then a 1 gets removed from A by the update of that case, making sure that e W pA, f pS 1 qq gets reduced by d W pa 1 , f pS 1 qq ě 1.
With Observation 5.4 in hand, we can check easily that Waiter can follow the proposed strategy without forfeiting the game.
Stage I: According to the strategy, Waiter needs to offer 2d T pvq ď 2n 3 edges at v. She can easily do so, since there exists n´1 edges to choose from.
Stage II: Assume Waiter needs to make a move according to Stage II, but she could follow her strategy in all the previous rounds. Further, let us assume first that O T 1 does not solely consist of a stopping vertex yet. In Case 1, when there exists u 1 , u 2 P O T 1 , we know that where the last inequality uses Property (i) and ε ă µ 20 ă µ 1 6 . Hence, Waiter can find a vertex a P A such that u 1 a and u 2 a are free edges, and hence she can follow her strategy in that case. In Case 2 and Case 3, when there exists a unique vertex u P O T 1 which is not a stopping vertex, we similarly obtain that d W pu, Aq`e W pA, f pS 1 qq ă |A|´2 by Properties (i)-(iv), and hence Waiter can find vertices a 1 , a 2 or a 3 , a 4 as required to follow the strategy. Now let us assume that at some point O T 1 solely exists of a stopping vertex u. Then, in order to finish with Stage II, only the vertex u needs to get closed w.r.t. T 1 . As this takes at most d T 1 pf´1puqq ď ε ? n rounds played according to Case 2 or Case 3, while the Properties (ii) and (iv) were true before that point, we know that until the end of Stage II, e W pf pS 1 q, Aq and d W pu, Aq cannot exceed 2ε ? n`1. But then, observing analogously that d W pu, Aq`e W pf pS 1 q, Aq ă 2p2ε ? n`1q ă |A|´2 , it follows that Waiter can find vertices a 1 , a 2 or a 3 , a 4 as desired by her strategy.
Stage III: When Waiter enters Stage III, Client's graph is a copy of the subgraph T 1 . The sets A :" V pK n qzf pV pT 1 qq and B :" f pS 1 q " f pN T pL 1 qq both have size at least µ 1 ? n " epM 1 q. Moreover e W pA, Bq ă 2ε ? n holds, as explained in the discussion of Stage II, and e C pA, Bq " 0. Following the strategy given for Theorem 2.1, Waiter can force a perfect matching between A and B within epM 1 q`1 rounds.
Having Theorem 5.2 in hands, we are now able to prove Theorem 1.11 and Theorem 1.12.
Proof of Theorem 1.11. Let n be large enough, and let T be any tree on n vertices and with maximum degree at most ε ? n. The upper bound τ W C pF T , 1q ď n follows from Theorem 5.2; the lower bound τ W C pF T , 1q ě n´1 trivially holds since epT q " n´1.
If T is a path on n vertices, then τ W C pF T , 1q " n´1. Indeed, in the strategy given for Theorem 3.1, Waiter forces a Hamilton path in the first round of Stage III, which is the pn´1q st round in the game. This shows that the lower bound in Theorem 1.11 is tight.
If T is a tree obtained from a path on n´4 vertices by connecting two further vertices to each of its endpoints, then τ W C pF T , 1q " n. Indeed, if Waiter would want to force a copy of T within n rounds, then for some edge e P EpT q she would need to force a copy of T´e within n´2 rounds. However, since a unique edge extends this copy of T´e to a copy of T , Client can easily prevent the copy of T in round n´1, because Waiter needs to offer two edges. This shows that the upper bound is tight. l Proof of Theorem 1.12. Let T be any tree on k vertices, and let H by a T -factor on n vertices with k|n. Then τ W C pF n,T´f ac , 1q ě epHq " k´1 k n. In order to prove the upper bound on τ W C pF n,T´f ac , 1q, let H 1 by an arbitrary tree on n 1 " n`1 vertices, obtained from H by adding one further vertex v 1 and adding exactly one edge between v 1 and each copy of T in H. Waiter then pretends to play on K n 1 Ą K n with V pK n 1 qzV pK n q " tp 1 u. She plays according to the strategy given for Theorem 5.2 (with v :" v 1 and p :" p 1 ), and whenever this strategy makes her offer two edges incident to p 1 , she only pretends to play that round. This way, she forces a copy of H " H 1´v1 in the game on K n , wasting at most one round, and hence she wins within k´1 k n`1 rounds. For the tightness of both bounds, we can use the same trees as in the proof of Theorem 1.11.
For large enough k, if T is a path on k vertices and H is a T -factor, Waiter can win within k´1 k n. Before the game starts, she just splits the vertex set into n k sets of size k, and then on each of these parts she forces a Hamilton path without wasting a move. On the other hand, if T is a tree obtained from a path on k´4 vertices by connecting two further vertices to each of its endpoints, then analogously to the proof of Theorem 1.11 we get τ W C pF n,T´f ac , 1q " k´1 k n`1. l

Unbiased triangle factor game
In this section we give the proof of Theorem 1.13. However, before doing so, let us first prove the following lemma.
Lemma 6.1. Consider an unbiased Waiter-Client game on K 12 , and fix any two vertices u, v P V pK 12 q. Then Waiter has a strategy that, within 7 moves, forces Client to create 2 vertex disjoint triangles with the following additional properties: (1) Both u and v are in the two triangles.
(2) All edges within the set of 6 vertices, that are not in a triangle, and the edge uv have not been offered.
Proof. Throughout the proof we will often use the fact, that Waiter can offer two edges from For the lower bound on τ W C pF n,K 3´f ac , 1q we may provide a Client's strategy. Throughout the game Client will maintain a set of marked vertices M Ă V , which is initially empty. Moreover, we consider the sets X v :" txy : xv, vy P EpCqu for every v P V pK n q and the set X " Ť v X v which consists of those edges which would close a triangle in Client's graph. In the following we describe Client's strategy.
In any round of the game suppose that Waiter offers two edges x 1 y 1 and x 2 y 2 to Client. Then Client chooses his edge according to the following case distinction.
Case 1: Suppose at least one of the offered edges belongs to EpK n qzX. Then Client arbitrarily chooses such an edge.

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Case 2: Suppose otherwise that there exist z 1 , z 2 P V pK n q such that x 1 y 1 P X z 1 and x 2 y 2 P X z 2 . Client then considers three subcases: (a) If z 1 R M, Client chooses the edge x 2 y 2 and adds z 1 to the set M.
(b) Otherwise, if z 2 R M, Client chooses the edge x 1 y 1 and adds z 2 to the set M.
(c) Otherwise, if z 1 , z 2 P M, then Client chooses his edge arbitrarily.
It is obvious that Client can always follow the proposed strategy. Hence, it remains to show that it prevents a triangle factor for at least 13 12 n rounds. We start with the following observation. Suppose that a triangle factor T " tt 1 , . . . , t n 3 u is created in Client's graph. In order to conclude that at least 13 12 n rounds have been played, we consider two cases. Assume first that right at this moment |U| ě n 6 . Then where the first inequality follows from Observation 6.2(a) and since every vertex belongs to a triangle, and the second inequality follows from Observation 6.2(b). Assume |U| ă n 6 next, then where the first inequality follows from Observation 6.2(a). In any case, we obtain |EpCq| ě 13 12 n.
For the upper bound on τ W C pF n,K 3´f ac , 1q we may provide a Waiter's strategy. For this, let n 0 be an even integer such that Waiter has a strategy to force a copy of K 48 in the unbiased 29 Waiter-Client game on K n 0 . Such an integer exists by [2]. Now, playing on K n , fix any set of vertices W Ă V pK n q with |W | " n 0 . In the following we describe Waiter's strategy. If at any point during the game, Waiter is unable to follow the strategy, she forfeits the game. (We will later see that this does not happen.) Waiter's strategy consists of the following three stages: Stage I: Playing on K n rW s only, Waiter forces Client to create a clique of size 48.
Stage II: When Waiter enters Stage II, there exists a set K Ă W of size 48 such that CrKs -K 48 . Let S " W zK and T " V zW . Waiter will force Client to create a large family of vertex disjoint triangles and she will update S and T by always removing the vertices of these triangles. As long as |T | ě 12, she plays in sequences of at most 7 moves as follows: It is clear that, if Waiter can follow the proposed strategy without forfeiting the game, she forces a triangle factor within at most`n 0 2˘`7 6 n`2¨12 " 7 6 n`Op1q rounds. Therefore, it remains to verify that Waiter indeed can follow the proposed strategy.
Strategy discussion: The strategy in Stage I can be followed by the choice of n 0 . Note that when Stage I is over, no edges from E Kn pT, S Y T q have been offered yet.
For Stage II assume that, before Waiter plays a sequence of moves as described in the strategy, it is still true that all edges in E Kn pT, S Y T q are free. It then follows from Lemma 6.1 that Waiter can play her next moves according to the strategy of Stage II. Just note that in the case when S ‰ ∅ it may happen that uv P E Kn pSq has already been offered before; but this does not cause any problem, since for Lemma 6.1 Waiter does not need to offer uv at all. Finally also note that, when the two triangles are created, by (2) from Lemma 6.1 and by the update in Stage II it follows that E Kn pT, S Y T q again consists only of free edges. Hence, Waiter can repeatedly apply Lemma 6.1 and follow the strategy for Stage II.
Afterwards, when Waiter enters Stage III, it holds that S " ∅ and |T | ă 12. Since E Kn pK, T q consists solely of free edges at this point, Waiter can offer edges as desired. l 30

Biased games
In the proof of Theorem 1.4 and Theorem 1.7 we will make use of the following result due to Bednarska-Bzdȩga, Hefetz, Krivelevich and Łuczak [4].
Theorem 7.1 (Theorem 1.4(ii) in [4]). There exists a positive constant c P p0, 1q and an integer n 0 such that the following holds. If n ě n 0 and b ď cn, then playing a b-biased Waiter-Client game on EpK n q, Waiter has a strategy to force a spanning pancyclic graph.
We start with the proof of Theorem 1.7.
We let b ď δn from now on and, whenever necessary, we will assume n to be large enough.
In the following we will describe a strategy for Waiter in the b-biased Waiter-Client game on K n , and afterwards we will show that it is a strategy with which Waiter forces a Hamilton cycle within at most n`Cb rounds. Whenever Waiter is not able to follow the proposed strategy, she forfeits the game. (We will show later that this does not happen.) The strategy is split into five stages.
Stage I: Within n´C 0 b´1 rounds, Waiter forces a path P " pa 1 , . . . , a n´C 0 b q on n´C 0 b vertices according to the following rule: Initially set P " ta 1 u for an arbitrary vertex a 1 P V pK n q. Assume after i´1 rounds Waiter has already forced a path P " pa 1 , . . . , a i q on i vertices. Then in round i Waiter selects b`1 vertices x 1 , . . . , x b`1 from V zV pP q which have the smallest degree in her graph. Then she offers the edges {a i x j : j P rb`1su to Client, of which he needs to pick one. Waiter updates the path such that P " pa 1 , . . . , a i`1 q with a i`1 :" x j .
Once P has reached length n´C 0 b´1, Waiter proceeds with Stage II.
Stage II: Let P denote the path that Client claims at the end of Stage I. Let R " V zV pP q then. Playing only on K n rRs, within at most Cb rounds, Waiter forces a Hamilton cycle of K n rRs.
The details of how she can do this, can be found later in the strategy discussion. Afterwards, Waiter proceeds with Stage III.
Stage III: This stage lasts 1 round. Let P " pa 1 , . . . , a n´C 0 b q be the path from Stage I and let H be the Hamilton cycle from Stage II. Waiter now picks b`1 vertices x 1 , . . . , x b`1 P V pHq such that a 1 x j is free for every j P rb`1s. Then she offers all these b`1 edges.
Afterwards, Client needs to pick one of the edges a 1 x j . From then on, setx " x j and let x be one of the neighbours of x j on H. Next, Waiter proceeds with Stage IV.

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Stage IV: This stage lasts exactly b rounds in which Waiter forces a few Hamilton paths on V pK n qzR using Pósa rotations [29]. More precisely, let P 0 " P " pa 1 , . . . , a n´C 0 b q be the path from Stage I. Set v 0 " a n´C 0 b . By playing on K n rV pP 0 qs only, for i P rbs Waiter will ensure that immediately after the i th round in Stage IV Client's graph contains a path P i such that the following properties hold: (P 1) V pP i q " V pP 0 q, (P 2) P i has endpoints a 1 and v i P V ztv 0 , . . . , v i´1 u, The details of how Waiter can do this can be found later in the strategy discussion. Afterwards, Waiter proceeds with Stage V.
Stage V: Within one round, Waiter forces Client's graph to contain a Hamilton cycle of K n . The details of how she can do this can be found later in the strategy discussion.
If Waiter can follow the proposed strategy without forfeiting the game, then it is obvious that she forces a Hamilton cycle within at most pn´C 0 b´1q`Cb`1`b`1 ă n`Cb rounds.
Therefore, it remains to prove that Waiter indeed can always follow the proposed strategy.

Strategy discussion:
Stage I: Consider the round i P rn´C 0 b´1s in Stage I. By then Waiter already forced a path P " pa 1 , . . . , a i q. Since in the previous rounds Waiter only offered edges which are incident to at least one of the vertices a j with j ă i, we know that before Waiter's i th turn all the edges between a i and V zV pP q are free. Moreover, since |V zV pP q| ě C 0 b, Waiter can easily find and offer b`1 edges as required by the strategy.
Stage II: Let P " pa 1 , . . . , a n´C 0 b q denote the path which Waiter has forced at the end of Stage I. Since so far she only offered edges which are incident to at least one of the vertices a j with j ă n´C 0 b, we know that at the beginning of Stage II all the edges inside R " V zV pP q are still free. Letñ :" |R| " C 0 b and, using (7.1), observe that b " C´1 0ñ ă cñ andñ ą n 0 .
According to Theorem 7.1, Waiter has a strategy for playing a pb : 1q game on K n rRs with bias b " cñ in such a way that Client is forced to get a pancyclic graph. Thus, following that strategy with bias b ăb (by pretending to addb´b extra edges to Waiter's graph in each round), Waiter can force a Hamilton cycle on R within at most ă Cb rounds, as promised in the strategy description.

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Stage III: When Waiter enters Stage III, it holds that d CYW pa 1 q " b`1, since only in the very first rounds she offered edges incident to a 1 . The number of available edges between a 1 and R is at least |R|´d CYW pa 1 q " C 0 b´pb`1q ą b`1 by the choice of C 0 . Hence, Waiter can offer edges as required for this stage of the proposed strategy.
Stage IV: Before we will show that Waiter can follow Stage IV, let us first observe that none of the vertices has a too large degree by now.
Observation 7.2. At the beginning of Stage IV it holds that d CYW pv, V pP 0 qq ă δ 0 n for every vertex v P V pK n q.
Proof. When Waiter forces the path P in Stage I, she always prefers to offer edges from the current endpoint a i to the vertices of smallest Waiter-degree in V zV pP q. This way she makes sure that the Waiter-degrees among the vertices in V zV pP q differ by at most 1 throughout Stage I. Now, Stage I lasts n´Cb´1 rounds, and thus epW q ď npb`1q holds throughout Stage I. In particular, all vertices v P V zV pP q then satisfy ă 0.5δ 0 n . Now consider the beginning of Stage IV. It holds that d CYW pa 1 q " 2pb`2q ă 3δn ă 0.1δ 0 n, since there were only two rounds in which Waiter offered edges at a 1 . For every vertex v P V pP qzta 1 u we then have d CYW pvq ă 0.5δ 0 n`pb`1q ă δ 0 n, since after v was added to P there was only one round in which Waiter offered edges incident to v. Moreover, for every remaining vertex v (i.e. v P R), we have d CYW pv, V pP 0 qq ă 0.5δ 0 n`1 ă δ 0 n since these vertices belong to V zV pP q at the end of Stage I and since afterwards, until the end of Stage III, the edge va 1 may be the only edge between v and V pP q that got offered. This proves the observation.
Having the observation in hands, we now show how Waiter can force the desired paths in Stage IV.
We will do it in such a way that the Properties (P 1)-(P 3) hold as well as the following property: (Q 1) d CYW pv, V pP i qq ă δ 0 n`i for every v P V pP i qztv 0 , . . . , v i´1 u.
We proceed by induction on i. For i " 0 the path P 0 " P from Stage I trivially satisfies (P 1) and (P 2). Property (Q 1) follows by Observation 7.2. Moreover, Property (P 3) holds by the following reason: In Stage I every offered edge is incident to at least one of the vertices in V pP 0 qztv 0 u, in Stage II every edge is disjoint from V pP 0 q and in Stage III every edge is incident to a 1 R tv 0 , xu. Thus, v 0 x has not been offered yet.
So, let i ą 0 then. Let P i´1 be the path given by induction and consider the i th move in Stage IV. For every vertex v P V pP i´1 qztv i´1 u denote with v`the unique neighbour of v in P i´1 33 with dist P i´1 pv`, v i´1 q " dist P i´1 pv, v i´1 q´1. Set B 1 :" y P V pP i´1 q : y`x P C Y W ( , B 2 :" y P V pP i´1 q : y`" v j for some j ă i ( , B 3 :" ty P V pP i´1 q : yv i´1 P C Y W u .
By Observation 7.2 and since in Stage IV Waiter only offers edges in V pP 0 q, we obtain |B 1 | ă δ 0 n.
Hence, also using (7.1), we conclude |V pP 0 qzpB 1 Y B 2 Y B 3 q| ě n´C 0 b´4δ 0 n ě n´C 0 δn´4δ 0 n ě n´0.01n´0.4n provided n is large enough. Waiter's strategy now is to offer b`1 edges of the form yv i´1 with y P V pP 0 qzpB 1 YB 2 YB 3 q, which is possible since y R B 3 implies that yv i´1 is free. Client then needs to claim one of these edges; by abuse of notation denote this edge with yv i´1 . Let v i " y`then, and let P i be the path induced by pEpP i´1q q Y tyv i´1 uqztyy`u. Property (P 1) is trivially satisfied.
Moreover, P i has endpoints a 1 and y`" v i and, since y R B 2 , Property (P 2) holds as well.
Property (P 3) is guaranteed as y R B 3 . For Property (Q 1) observe the following: In her i th move of Stage IV, Waiter only offers edges incident to v i´1 . Thus, the degrees d CYW pv, V pP i´1 qq " d CYW pv, V pP i qq can increase at most by 1 for every v ‰ v i´1 .

Stage V:
Let v 0 , . . . , v b be the distinct endpoints of the paths P 0 , . . . , P b from Stage IV. Each of the b`1 edges v i x with 0 ď i ď b is free, and hence Waiter can offer those in her next move. Let v j x be the edge claimed by Client afterwards, and let f " a 1x be the edge which Client claimed in Stage III. Then is a Hamilton cycle of K n which is fully claimed by Client.
Finally, we proceed with the proof of Theorem 1.4.
Proof of Theorem 1.4. Creating a perfect matching, under assumption of Theorem 7.1, is rather straightforward. Let δ 0 , δ, C 0 and C be defined as in the previous proof. For roughly 0.5pn´C 0 bq rounds, Waiter can force a large matching, by playing according to Stage I from the previous proof, where a long path was created, and just faking every second move (i.e. pretending to make the move, but not playing at all). Then, as in Stage II of the previous proof, she forces a Hamilton cycle on the remaining vertices within Cb rounds, thus a perfect matching is created. 34

Concluding remarks
Winning as fast as possible. As already mentioned in the introduction, there are quite a few games which Waiter win (almost) perfectly fast. Also, for almost every game that we considered in our paper, we were able to prove that Waiter can win it at least asymptotically fast. On the other hand, for the triangle factor game we know that it is not won asymptotically fast.
We believe that the upper bound in Theorem 1.13 is asymptotically tight and hence pose the following conjecture.
Another game to consider is the minimum degree k game D n,k played on K n , in which the winning sets consist of all spanning subgraphs H with δpHq ě k. In this paper we considered the unbiased and biased version of the perfect matching and Hamiltonicity game, which covers the minimum degree 1 and minimum degree 2 games. We also have an argument that would show that Waiter can win the unbiased game with winning sets D n,k within kn 2`O p1q rounds. We wonder whether this can be improved as follows: Problem 8.2. For k ą 2, show that τ W C pD n,k , 1q " t kn 2 u`1. Moreover, determine τ W C pD n,k , bq asymptotically when b ą 1.
Note that Maker can win the unbiased Maker-Breaker version of the game D n,k within at most t kn 2 u`1 rounds, as shown by Ferber and Hefetz in [12], therefore we are curious to know whether this holds in the Waiter-Client setup as well.
Finally, one could look at the k-clique factor game for k ě 3. The triangle factor game considered in this paper covers the case k " 3, but it is still unknown what happens for k ą 3. Problem 8.3. Find non-trivial upper and lower bounds for τ W C pF n,K k´f ac , 1q when k ą 3.
Lastly, all the games we considered in this paper are examples showing that Waiter can win (asymptotically) at least as fast as Maker. We wonder whether this is always the case. Further questions involving trees. In Section 5 we found a fast winning strategy for Waiter in the case where she wants to force a copy of a given spanning tree that fulfills some maximum degree condition. It would be interesting to see if it is possible to relax this condition when we do not intend to win fast. Problem 8.5. For trees T on n vertices, how large can ∆pT q be such that Waiter has a strategy in the game with winning sets F T on K n ? 35