Saturation Games for Odd Cycles

Given a family of graphs $\mathcal{F}$, we consider the $\mathcal{F}$-saturation game. In this game two players alternate adding edges to an initially empty graph on $n$ vertices, with the only constraint being that neither player can add an edge that creates a subgraph that lies in $\mathcal{F}$. The game ends when no more edges can be added to the graph. One of the players wishes to end the game as quickly as possible, while the other wishes to prolong the game. We let $sat_g(\mathcal{F};n)$ denote the number of edges that are in the final graph when both players play optimally. The $\{C_3\}$-saturation game was the first saturation game to be considered, but as of now the order of magnitude of $sat_g(\{C_3\},n)$ remains unknown. We consider a generalization of this game. Let $\mathcal{C}_{2k+1}:=\{C_3,\ C_5,\ldots,C_{2k+1}\}$. We prove that $sat_g(\mathcal{C}_{2k+1};n)\ge(\frac{1}{4}-\epsilon_k)n^2+o(n^2)$ for all $k\ge 2$ and that $sat_g(\mathcal{C}_{2k+1};n)\le (\frac{1}{4}-\epsilon'_k)n^2+o(n^2)$ for all $k\ge 4$, with $\epsilon_k<\frac{1}{4}$ and $\epsilon'_k>0$ constants tending to 0 as $k\to \infty$. In addition to this we prove $sat_g(\{C_{2k+1}\};n)\le \frac{4}{27}n^2+o(n^2)$ for all $k\ge 2$, and $sat_g(\mathcal{C}_\infty\setminus C_3;n)\le 2n-2$, where $\mathcal{C}_\infty$ denotes the set of all odd cycles.


Introduction
Hajnal proposed the following game. Initially G is an empty graph on n vertices. Two players alternate turns adding edges to G, with the only restriction being that neither player is allowed to add an edge that would create a triangle. The last player to add an edge wins the game, and the central question is which player wins this game as a function of n.
The answer to this problem is known only for small values of n, the most recent result being n = 16 by Gordinowicz and Pra lat [5]. A variation of this game was considered by Füredi, Reimer, and Seress [4]. In the modified version of the game, there are two players, R and B, who alternate turns adding edges to an initially empty graph on n vertices with the same rules as in Hajnal's original triangle-free game. The main difference is that once no more edges can be added to G, R receives a point for every edge in the graph and B loses a point for every edge in the graph, with both players trying maximize the number of points they receive at the end. The question is now to figure out how many edges are at the end of the game when both players play optimally.
This game can be generalized. For a family of graphs F , we say that a graph G is F -saturated if G contains no graph of F as a subgraph, but adding any edge to G would create a subgraph of F . The F -saturation game consists of two players, R and B, who alternate turns adding edges to an initially empty graph G on n vertices, with the only restriction being that G is never allowed to contain a subgraph that lies in F . The game ends when G is F -saturated. The payoff for R is the number of edges in G when the game ends, and B's payoff is the opposite of this. We let sat g (F ; n) denote the number of edges that the graph in the F -saturation game ends with when both players play optimally, and we call this quantity the game F -saturation number. We note that technically this game, and hence the value of sat g (F ; n), depends on whether R or B makes the first move. This choice will not affect our asymptotic results, but for concreteness we will assume that R makes the first move.
Let C k denote the cycle of length k. The {C 3 }-saturation game was the original saturation game studied in [4], where they proved what is still the best known lower bound of 1 2 n log n + o(n log n) for sat g ({C 3 }; n). Recently Biró, Horn, and Wildstrom [1] managed to prove the first non-trivial asymptotic upper bound of 26 121 n 2 + o(n 2 ) for sat g ({C 3 }; n). A number of other results have been obtained for specific choices of F , see for example [2], [3], and [8]. In addition to this, saturation games have recently been generalized to directed graphs [7], hypergraphs [9], and to avoiding more general graph properties such as k-connectivity [6].

Main Results.
Let C 2k+1 := {C 3 , C 5 , . . . , C 2k+1 }, and let C ∞ denote the set of all odd cycles. Most of this paper will be focused on studying the C 2k+1 -saturation games for k ≥ 2. The key idea with these games is that by forbidding either player from making C 5 's, both players can utilize a strategy that keeps the graph essentially bipartite throughout the game. This makes it significantly easier to analyze the correctness of our proposed strategies and to bound the number of edges that are in the final graph. Our main result is the following upper and lower bounds for sat g (C 2k+1 ; n) and most values of k. We can also obtain a quadratic lower bound for smaller values of k. We consider two more saturation games. The first is the game where only one odd cycle is forbidden. We also consider the "complement" of the {C 3 }-saturation game where every odd cycle except C 3 is forbidden. It turns out that in this setting the game saturation number is linear.
This result is in sharp contrast to the fact that sat g (C ∞ ; n) = ⌊ 1 4 n 2 ⌋, see [2]. Notation. Throughout the paper we let G t denote the graph in the relevant saturation game after t edges have been added, and we let e t denote the edge of G t that is not in G t−1 . We let N t (x) denote the neighborhood of x in G t , d t (x, y) the distance between x and y in G t , and so on. We let t = ∞ correspond to the point in time when the graph has become F -saturated. If X t is a real number depending on t, we let ∆(X t ) = X t − X t−2 .
Organization. In Section 2 we produce an algorithm for R that guarantees that the game ends with at least as many edges as stated in Theorem 1.2, which will work the same way for all k ≥ 2. In Section 3 we modify this algorithm to take into account the choice of k, and from this we obtain the lower bound of Theorem 1.1. In Section 4 we produce an algorithm for B that guarantees that the game ends with at most as many edges as the upper bound of Theorem 1.1. Theorem 1.3 is proven in Section 5. Theorem 1.5 is proven in Section 6. We end with some concluding remarks in Section 7.

The Setup.
Let uv be the edge of G 1 . Let 1 < γ ≤ 2 and δ = 1 γ−1 . We say that G t is γ-good if it satisfies the following four conditions.
(1*) G t contains exactly one non-trivial connected component, and this component is bipartite Define an analogous partition for V t .
We note that (2*) and (4*) are trivially satisfied if U t 1 = V t 1 = ∅. We prove Theorem 1.2 by first proving the following result.
Theorem 2.1. There exists a strategy for R in the C 2k+1 -saturation game when k ≥ 2 such that for all odd t, whenever G t−1 contains an isolated vertex, R can add an edge so that G t is 3 2 -good.

The Algorithm.
In this subsection we inductively construct the algorithm of Theorem 2.1 whenever G t−1 has an isolated vertex. We will assume throughout this subsection that t is odd and that G t−1 contains an isolated vertex z. Let e t−1 = xy. We will say that e t−1 is an Note that an AU move causes y to be added to U t−1 .
We note that if we assume that G t−2 satisfies (1*), any vertex not in U t−2 ∪ V t−2 must be isolated. When R plays it will always be obvious that (1*) is maintained so we ignore this case in our analysis. We state our results in terms of general γ whenever γ = 3 2 isn't required. Lemma 2.2. If G t−2 is γ-good and e t−1 is an I move, then R can play so that G t is γ-good.
, then R adds the edge u ′ v ′ , and it's not hard to see that in this case G t is γ-good. If no such pair of vertices exists, then U t−1 ∪ V t−1 is a complete bipartite graph with, say, |U t−1 | ≤ |V t−1 |, in which case R adds the edge zv. This gives ∆(|U t |) = 1 and ∆(|X t |) = 0 for every other set of interest. Since U t−1 ∪ V t−1 is a complete bipartite graph (and since R added no vertex to U t−1 so b t U ≤ 0 and (3*) holds, so G t is γ-good.
In this case, R adds the edge xv (otherwise R adds the edge xu), which leads to ∆(|U t |) = ∆(|V t |) = ∆(|V t 1 |) = 1, ∆(|U t 1 |) = 0. x and y satisfy (2*), so this continues to hold. We have ∆ The situation becomes more complex if e t−1 is an AU or AV move, in which case R makes his move depending on the overall "State" of the game. We first make an observation.
| is a non-negative integer, and thus |V t−2 The analysis for the other case is similar.
By Lemma 2.4, if we inductively assume that G t−2 is γ-good, then after doing an AU move the game must be in one of the following States.
Lemma 2.5. If G t−2 is γ-good and e t−1 is an AU move putting the game is State N, then R can play so that G t is γ-good.
Proof. R follows the same exact same strategy he did in response to B playing an I move, and the analysis remains the same. Lemma 2.6. If G t−2 is 3 2 -good and e t−1 is an AU move putting the game in State OU, then R can play so that G t is 3 2 -good.
Proof. First assume that U t−1 , then R adds the edge xv, otherwise R picks an arbitrary u ′ ∈ U t−1 1 and adds the edge u ′ v. After this we have ∆(|V t |) = 1, ∆(|U t |) = 0, ∆(|U t 1 |) = −1, ∆(|V t 1 |) ≤ 1. (2*) is maintained since we made sure that y's neighbor x was in U 0 . We have We automatically have b t V ≤ 0 by assumption of us not being in State OV and having The algorithm and analysis for AV moves is analogous. Proof. If v ′ , v ′′ ∈ V t , let u ′ , u ′′ ∈ U t 0 be neighbors of v ′ and v ′′ respectively, noting that such vertices exist by (2*).
Thus having e t+1 = v ′ v ′′ would create either a C 3 or a C 5 , which is forbidden in the C 2k+1 -saturation game for k ≥ 2. The analysis for U t is similar.
Proof of Theorem 2.1. G 1 is 3 2 -good (in fact, it's γ-good for any 1 < γ ≤ 2). Inductively assuming that G t−2 was 3 2 -good, e t−1 must be a move of type I, O, AU , or AV by Lemma 2.8. The lemmas of the previous subsection show that R can then play so that G t is 3 2 -good whenever G t−1 contains an isolated vertex. Proposition 2.9. Let 1 < γ ≤ 2. Assume that there exists a strategy for R in the C 2k+1 -saturation game when k ≥ 2 such that for all odd t, whenever G t−1 contains an isolated vertex, R can add an edge so that G t satisfies (1*), (2*), and (3*). Then Proof. Assume that R uses such a strategy until there are no isolated vertices left. Let T denote the smallest even number such that G T contains no isolated vertices. Let S denote the set of vertices of G T −1 that are isolated, noting that |S| ≤ 2. We claim that R can choose e T +1 so that G T +1 satisfies (1*) and (2*). Indeed, if G T −1 doesn't satisfy (1*) and (2*), then there must exist some x ∈ S such that G T doesn't contain either xu or xv. Since G T is bipartite by Lemma 2.8, R can choose e T +1 to be one of these edges, causing G T +1 to satisfy (1*) and (2*). Thus R can play so that G T +1 , and hence G ∞ , satisfies (1*) and (2*). After this R plays arbitrarily.
3. The Lower Bound of Theorem 1.1 Theorem 3.1. There exists a strategy for R in the C 2k+1 -saturation game when k ≥ 3 such that for all odd t, whenever G t−1 contains an isolated vertex, R can add an edge so that G t is γ ′ -good.
The algorithm of Theorem 3.1 is essentially the same as that of Theorem 2.1, but with new definitions for the relevant parameters. Throughout this section we use the same notation as in the previous section unless stated otherwise, and we always assume that k ≥ 3.
Order the vertices of U t in some way, say based on the order that they were added to the set. We will say that a vertex x ∈ U t is the representative for (2) x lies along a shortest path from u ′ to u.
(3) x is the minimal vertex (with respect to the ordering of U t ) satisfying these properties.
Redefine U t 1 to be the set of vertices that are representatives for some vertex ofŨ 1 We make the following observation.
with the ℓ − 2 term coming from the fact that d t (v ′ , v) < ℓ is even. Since u ′ ∈Ũ t 1 implies d t (u ′ , u) ≥ ℓ, the distance must be exactly ℓ. The analysis forṼ t 1 is similar.
An application of Lemma 3.2 can be used to show that Lemma 2.8 continues to hold when we define (2*) in terms of these new sets. We omit the details.
The algorithm for Theorem 3.1 is the same algorithm as that of Theorem 2.1, except we now use these new definitions for U t 0 , U t 1 , V t 0 , V t 1 , and we make a slight change to how R responds in State C. Namely, in the previous algorithm when B added in the isolated vertex y, we checked to see if its neighbor x was in, say, U t 1 , in which case we added the edge xv. We now instead check if x ∈Ũ t 1 , and if it is we add the edge zv where z is the representative for x. Adding this edge strictly decreases d t−1 (x, u), so by Lemma 3.2 we will have d t (x, u) < ℓ and y will have a neighbor in U t 0 , so (2*) will still hold. One can check that outside of this specific subcase of State C, all of the previous analysis we did continues to hold with these new definitions of U t 1 and V t 1 . It remains to address the two points in the algorithm where we required γ = 3 2 , namely Lemma 2.3 and Lemma 2.6. Lemma 3.3. If G t−2 is γ-good and e t−1 is an O move, then R can play so that G t is γ-good.
Proof. R reacts as he did in the previous algorithm. Observe that no vertices are added toŨ t−1 . Indeed, the new vertices are within distance 2 < ℓ of u and v, so they'll both be added to . In particular, ∆(|U t 1 |) = ∆(|V t 1 |) = 0, and the remaining analysis is straightforward.
Proof. For each x ∈ U t 1 , let u x denote a vertex that x is the representative for, and let P x denote the set of vertices that make up a shortest path from x to u x . We claim that P x and P y are disjoint if By using a symmetric argument we see that we must have d t (w, x) = d t (w, y). If we had, say, x < y in the ordering of U t , then y couldn't be the representative for u y since x is less than y and satisfies the relevant properties. A similar result occurs if x > y. We conclude that the only way P x ∩ P y could be non-empty is if x = y.
For each x ∈ U t 1 , we observe that the number of vertices in P and none of these vertices appear in any other P y for x = y ∈ U t 1 . Since we can associate to each x ∈ U t 1 a set of at least k/4 elements of U t without any element of U t appearing in more than one set, we must have |U t Proof. R reacts as he did in the previous algorithm. By definition of State OU, we have V t 1 = ∅ and |U t−1 | > γ ′ |V t−1 | + δ. The latter implies that Combining these observations with Lemma 3.4 gives , with the last equality coming from the fact that γ ′ is a root of −x 2 + 4k −1 x + 1. It's not too hard to see that the remaining value is at most −2.
Proof of Theorem 3.1. The proof is essentially the same as that of Theorem 2.1, except here we use Lemma 3.3 and Lemma 3.5 instead of Lemma 2.3 and Lemma 2.6.
Proof of the lower bound of Theorem 1.1. Note that The result is then a consequence of Theorem 3.1 and Proposition 2.9.

The Upper Bound of Theorem 1.1
Throughout this section we will consider the C 2k−1 -saturation game, so that the smallest odd cycle that can be made is a C 2k+1 , and we will always have k ≥ 5. We again let e 1 = uv.
The algorithm needed to prove the upper bound of Theorem 1.1 will take some time to develop. The main idea is that B will maintain a number of long, disjoint paths in G t , and then eventually either B will be able to join these paths together and create many odd cycles, or the graph will look like a bipartite graph with one side much larger than the other.

Paths.
We wish to define a special set of paths P t in G t , with each path having v as one of its endpoints. We start with P 1 = {uv}, and inductively we define P t based on the following procedure (regardless of the parity of t).
Step 1. Consider the case that B added the edge e t = xw where x is either an isolated vertex or a vertex in an isolated edge xy (i.e. d( Step Step 3. Let d t P (x, y) denote the distance between x and y in the graph induced by the vertices of P t . If w is a vertex of p ∈ P t with d t (w, v) < d t P (w, v), then P t := P t \ {p}.
Observe that by these Steps, each vertex besides v is in at most one path of P t . Let D t ℓ denote the set of vertices that are the endpoint of a path of length ℓ in P t and that aren't v. Proof. Let p i denote the path for which w i is an endpoint for, noting that p 1 = p 2 since w 1 = w 2 and neither are equal to v. Since p 1 = p 2 , w 1 and w 2 lie in different components of G t − {v} by Step 2. Thus if the edge w 1 w 2 created a forbidden cycle it would have to involve the vertex v. Since Step 3, the smallest cycle that could be formed is a C 2k+1 , which is allowed.

Phases and Phase Transitions.
For the rest of the section we assume that t is even. We wish to describe each G t as belonging to a certain "Phase." To do this formally we'll need some definitions.
Set U 0 := ∅, V 0 := {v}. Let e t = xy. If x ∈ V t−1 and y is an isolated vertex, then U t := U t−1 ∪ {y}, V t := V t−1 . If x ∈ V t−1 and y is in an isolated edge yz in G t−1 , then U t := U t−1 ∪ {y}, V t := V t−1 ∪ {z}. If x ∈ U t−1 we define U t and V t analogously. For any other case of e t , U t := U t−1 , V t := V t−1 . Let i t denote the number of isolated vertices in G t . Let c 1 = 1 2 , c 2 = 1 9 , c k = c −1 = (1000k 2 ) −1 . We will say that G t is in Phase ℓ for some −1 ≤ ℓ ≤ k based on the following set of rules.
• G 0 is in Phase 0.
• If G t−2 is in Phase ℓ and if G t satisfies none of the above situations, then G t is in Phase ℓ.
We note that our exact choices of c 1 , c 2 , c k , and c −1 aren't important. The only thing that matters is that they satisfy the following set of inequalities for k ≥ 5 and n sufficiently large, which isn't too difficult to verify for the values we've chosen (in particular one can use that 4.3. The Beginning of the Game. We will say that a path in U t ∪ V t is alternating if the vertices in the path alternate being in U t and V t , and we define d t a (x, y) for x, y ∈ U t ∪ V t to be the length of the shortest alternating path Theorem 4.3. For n sufficiently large there exists a strategy for B in the C 2k−1 -saturation game for k ≥ 5 such that, for every even t with G t in Phase ℓ for 0 ≤ ℓ < k, the following conditions hold.
(1') G t contains exactly one non-trivial connected component whose vertices are U t ∪ V t .
Proof. To be concrete we'll describe the algorithm for ℓ even, where we'll have D t ℓ ⊆ V t and D t ℓ+1 ⊆ U t . The algorithm and analysis for ℓ odd is exactly the same except the roles of U t and V t are reversed throughout. To deal with the case ℓ = 0, we define D t 0 = {v}. It will always be clear that (1') is maintained, so we omit this from our analysis.
We classify e t−1 as being of type I, O, AU, or AV essentially as we did before, except now I moves are defined to be an edge between two vertices of U t ∪ V t which aren't necessarily from distinct parts. It follows that e t−1 must be one of these four types of moves. Since G t−2 satisfies (3'), we have i t−2 ≥ 3 and |D t−2 ℓ | ≥ 3 when ℓ = 0. Thus i t−1 ≥ 1 (the number of isolated vertices always decreases by at most 2) and |D t−1 ℓ | ≥ 1 by Lemma 4.1 (and we always have |D t−1 0 | = 1). Thus there exists If e t−1 is an I move, B plays xv ′ . Then d t a (x, v) = ℓ + 1 and no other distances increase, so (2') is maintained.
If e t−1 = yz is an O move (noting that yz is an isolated edge in G t−1 since G t−2 satisfies (1')), then B plays yv ′ . We have d t a (y, v) = ℓ + 1 and d t a (z, v) = ℓ + 2, so (2') is maintained.
Say e t−1 = yw ′ is an AU move with w ′ ∈ V t−2 and y / ∈ U t−2 ∪ V t−2 . If d t−1 a (w ′ , v) ≤ ℓ, then B plays xv, essentially skipping his turn and maintaining (2') (in the analogous situation with ℓ odd, B plays xu to skip his turn). If instead d t−1 a (w ′ , v) = ℓ + 2, let w ′ y ′ w ′′ · · · be a shortest alternating path from w ′ to v. Then B adds the edge yw ′′ , and if this is a legal move we have d t a (y, v) = ℓ + 1, maintaining (2'). The following lemma shows that this move is indeed allowed. Lemma 4.5. Let w ′ be such that d t−2 a (w ′ , v) ≥ 2, say with w ′ y ′ w ′′ · · · a shortest alternating path from w ′ to v, and let y be an isolated vertex in G t−2 . If e t−1 = yw ′ , then e t = yw ′′ is a legal move.
Proof. LetG t denote G t−1 with the edge yw ′ , and assume this created a forbidden odd cycle C. yw ′ and yw ′′ must be two edges of C. If y ′ is not a vertex of C, then let C ′ be C after replacing the edges yw ′ and yw ′′ with y ′ w ′ and y ′ w ′′ . Then C ′ is a forbidden odd cycle in G t−1 , a contradiction. Thus y ′ must be a vertex of C, so C contains in G t−1 a path from w ′′ to y ′ and a path from y ′ to w ′ , with exactly one of these paths being of even length. Any path of even length from w ′′ to y ′ in G t−1 has length at least 2k, since otherwise this path together with the edge y ′ w ′′ would create a forbidden odd cycle in G t−1 . A similar observation holds for paths of even length from y ′ to w ′ . We conclude in this case that C has length at least 2k + 3, which is allowed.
During Phase ℓ define g t = |V t | − |U t | if ℓ is even and g t = |U t | − |V t | if ℓ is odd. It's not too difficult to observe the following by using Lemma 4.1. We omit the details. Lemma 4.6. Assume that G t−2 satisfies (') and is in Phase ℓ for 0 ≤ ℓ < k. If B uses the algorithm described in Lemma 4.4, then the following holds for ℓ even (where whenever we write ∆(|D t ℓ |) we assume that ℓ = 0).
Analogous results hold for ℓ odd by switching AU with AV .
Let t ℓ denote the smallest value such that G t ℓ is in Phase ℓ, with t ℓ = ∞ if no such value exists. Let s ℓ = min{t ℓ , t −1 }. Lemma 4.7. If n is sufficiently large and B uses the algorithm described in Lemma 4.4, then G t satisfies (3') for all even 0 ≤ t < s 1 .
Let r 1 denote the number of even t with 0 ≤ t < t ′ 0 such that e t+1 is of type I or AU , and similarly define r 2 for O and AV moves. Note that If r 1 > c −1 n, then G t ′ 0 is in Phase −1, and if r 1 ≤ c −1 n then |D t ′ 0 ℓ | ≥ c 2 n + 9c −1 n by (2). We conclude that G t ′ 0 is either in Phase −1 or Phase 1, i.e. s 1 ≤ t ′ 0 .
Lemma 4.8. Assume that t ℓ is finite with 1 ≤ ℓ < k and that n is sufficiently large. If G t ℓ −2 satisfies (') and if B uses the algorithm described in Lemma 4.4, then G t satisfies (3') for all even t ℓ ≤ t < s ℓ+1 .
Proof. Assume first that ℓ is even. Let r 1 denote the number of t ℓ ≤ t < s ℓ+1 such that G t satisfies (3') and such that e t+1 is an I or AU move, and similarly define r 2 for moves of type O and AV . Let t ′ ℓ = t ℓ + 2(r 1 + r 2 ). We claim that t ′ ℓ = s ℓ+1 , which would imply the desired result. We trivially have t ′ ℓ ≤ s ℓ+1 , so assume for contradiction that t ′ ℓ < s ℓ+1 . Since G t satisfies (3') for t ℓ ≤ t < t ′ ℓ , B can indeed use the algorithm up to and including the point where the game reaches G t ′ ℓ , and further, t ′ ℓ is the smallest value such that G t ′ ℓ is in Phase ℓ and doesn't satisfy (3'). By Lemma 4.6 we have g t ′ ℓ − g t ℓ ≤ −r 1 . Since g t ℓ < c −1 n and g t ′ ℓ > −c −1 n (otherwise G t ℓ or G t ′ ℓ would be in Phase −1), we can assume that r 1 ≤ 2c −1 n.
Proof of Theorem 4.3. G 0 satisfies ('). If G t−2 satisfies (') and B uses the algorithm described in Corollary 4.9. For n sufficiently large there exists a strategy for B in the C 2k−1 -saturation game for k ≥ 5 such that G t ′ is in Phase k or Phase −1 for some sufficiently large t ′ .
Proof. For some large t ′ , i t ′ < 3. By Theorem 4.3 we know that B can play so that i t ≥ 3 whenever G t is in Phase ℓ for 0 ≤ ℓ < k. Thus if B uses this strategy, G t ′ must be in Phase k or Phase −1.

Endgame.
Theorem 4.10. There exists a strategy for B in the C 2k−1 -saturation game for k ≥ 5 such that if the game eventually reaches Phase k or Phase −1, then G ∞ contains at most ( 1 4 − c 2 −1 )n 2 + o(n 2 ) edges.
We first describe B's strategy for Phase k, recalling that once the game enters this Phase it never leaves it (and similarly for Phase −1). In Phase k, B connects two vertices of D t−1 k as long as |D t−1 k | ≥ 2, which is a legal move by Lemma 4.2. If |D t−1 k | ≤ 1, B plays arbitrarily. Note that ∆(|D t k |) ≥ −4 by Lemma 4.1. Since |D t k k | ≥ 4c k n, B is able to create at least c k n − 1 C 2k+1 's which all share a common vertex v and with no other vertices shared between the cycles. We will call such a structure a (c k n − 1)-bouquet.
Lemma 4.11. Let G be a graph that contains no odd cycles smaller than C 2k+1 with k ≥ 2. If C is a C 2k+1 in G, then no vertex of G has more than two neighbors in C.

Proof.
Let v be a vertex with neighbors v 1 , v 2 , v 3 ∈ C. Let d C (x, y) denote the length of the shortest path from x to y using only edges of C. First assume v ∈ C, and without loss of generality that In this case G contains cycles of length ℓ + 1 and 2k + 1 − ℓ + 1, one of which must be an odd number that is at most 2k − 1 since k ≥ 1, a contradiction. Now assume v / ∈ C. Then G contains cycles of length 2 + d C (v i , v j ) and 2k + 1 − d C (v i , v j ) + 2 for each i = j. The only way these values can both be either even or at least 2k + 1 is if d C (v i , v j ) = 2 for all i = j. This is impossible since C is not a C 6 . Lemma 4.12. Let G be an n-vertex graph that contains no odd cycles smaller than C 2k+1 with k > 1.
Note that in the statement and proof of this lemma we implicitly assume that 2k(c k n − 1) + 1 ≤ n, which certainly holds for our choice of c k .
Proof. For simplicity assume that G contains a (c k n, C 2k+1 )-bouquet and that c k n is an integer (this won't affect the asymptotics). The number of edges involving some vertex of a given cycle of the bouquet is at most 2n by Lemma 4.11, so the number of edges involving some vertex of the bouquet is at most 2c k n 2 . The number of edges involving vertices that are not part of the bouquet is at most Adding these two values gives the desired result.
We now describe B's strategy when the game reaches Phase −1 from Phase ℓ < k. B will play arbitrarily if G t−1 is connected. Otherwise B plays the same strategy he did for Phase ℓ, except instead of having v ′ ∈ D t−1 ℓ we take v ′ = v if ℓ is even and v ′ = u if ℓ is odd. If one goes back through the algorithm one can verify that with this strategy, for all even t ≥ t −1 , G t satisfies (1'), (2') for ℓ, and that ||U t | − |V t || ≥ c −1 n − 1.
Let D ′ ℓ denote the set of vertices that were in D t ℓ for some t.
No vertex of U ′ is adjacent to any vertex of D ′ 2 , and no vertex of V ′ is adjacent to any vertex of D ′ 1 . (2'). Since k + 5 < 2k + 1 by assumption of k ≥ 5, this is a forbidden odd cycle, a contradiction. The proof for V ′ is analogous.
Proof of Theorem 4.10. If B uses the above algorithm and the game is in Phase k, then G ∞ will contain a (c k − 1)-bouquet, and hence it will contain at most ( 1 4 − c 2 −1 )n 2 + o(n 2 ) edges by Lemma 4.12 and (5). Now assume that B uses the above algorithm and the game is in Phase −1. Further assume that G ∞ is not bipartite, or equivalently that U ′ ∪ V ′ is non-empty. Since d ∞ a (w, v) ≥ k − 1 ≥ 4 for any w ∈ U ′ ∪ V ′ by Lemma 4.13, U ′ ∪ V ′ will be empty unless the game enter Phase −1 from Phase ℓ ≥ 2 since B maintains that d ∞ a (x, y) ≤ ℓ + 2. In particular, at some point the game reaches Phase 2 with |D t2 2 | ≥ c 2 n. By Lemma 4.13, (1'), and (2') for ℓ = 1, every vertex of U ′ ∪ V ′ must have been isolated during all of Phase 1 in order to have d ∞ a (w, v) ≥ 4. Hence s := |U ′ ∪ V ′ | ≤ c 1 n + 1 − c 2 n (Phase 1 starts with at most c 1 n + 1 isolated vertices, and at least c 2 n vertices are in D t2 2 ). Let G ′ be the complete bipartite graph with bipartition U ∞ ∪ V ∞ (which will be G ∞ if s = 0). The only edges of G ∞ that aren't in G ′ are those contained in U ′ ∪ V ′ , and there are at most 1 4 s 2 + 1 such edges by Mantel's theorem. However, G ′ contains all of the edges from D ′ 2 to U ′ and D ′ 1 to V ′ , and none of these edges are in G ∞ by Lemma 4.13. There are at least |D ′ 2 ||U ′ | + |D ′ 1 ||V ′ | ≥ c 2 ns edges of this type, so in total G ′ contains at least c 2 ns − 1 4 s 2 − 1 more edges than G ∞ does. This number is non-negative if s = 0 by (6). It's thus enough to give an upper bound for the number of edges of G ′ , which is exactly Proof of the upper bound of Theorem 1.1. Corollary 4.9 and Theorem 4.10 show that for k ≥ 4, B has a strategy for the C 2k+1 -saturation game guaranteeing G ∞ contains at most ( 1 4 − c 2 −1 )n 2 + o(n 2 ) edges, where c −1 = (1000(k + 1)) −2 (recalling that Theorem 4.3, Theorem 4.10, and c −1 were defined in this section in terms of the C 2k−1 -saturation game).

Proof of Theorem 1.3
Lemma 5.1. Let k ≥ 2 and ℓ = max(3, ⌊ √ 2k⌋). There exists a constant t 0 such that, for n sufficiently large, B can play in the {C 2k+1 }-saturation game such that G t0 contains a clique on the vertex set U = {u 1 , . . . , u ℓ }, and such that there exists ℓ vertex disjoint paths of length k − 2, each with a distinct u i as its endpoint.
Proof. Implicitly we assume throughout the proof that G t−1 contains an isolated vertex, which will certainly be true if t is bounded by a constant and n is sufficiently large. First assume ℓ = ⌊ √ 2k⌋ ≥ 3. We claim that any choice of e t for t ≤ 2 ℓ 2 is a legal move. Indeed, for any t ≤ 2 ℓ 2 − 1, G t will contain at most 2 ℓ 2 − 1 ≤ ℓ 2 − 1 ≤ 2k − 1 edges, and hence any choice of e t+1 won't create a C 2k+1 in G t+1 . With this in mind, B spends his first ℓ 2 moves creating a K ℓ on U = {u 1 , . . . , u ℓ }. For his next move after forming this K ℓ , B adds the edge e t = u 1 x 1 where x 1 is an isolated vertex of G t−1 . For his ith move after adding the edge u 1 x 1 , B adds the edge e t = x i x i+1 with x i+1 an isolated vertex of G t−1 . B continues this up to i = k − 3, creating a path of length k − 2 with u 1 as an endpoint. The next edge B adds is e t = u 2 y 1 with y 1 an isolated vertex of G t−1 . As before B will extend this path until its length reaches k − 2. B repeats this process for each vertex of U , giving the desired subgraph by time t 0 = 2 ℓ 2 + 2ℓ(k − 2). If instead ℓ = 3, then it's not too difficult to see that B can create a K 3 in his first 3 moves for any k ≥ 2. After this he does essentially the same strategy as before and creates the desired subgraph by time t 0 = 6 + 6(k − 2). For all even t > t 0 , B uses the following strategy. If i t−2 < 2, B plays arbitrarily. Otherwise if R plays xy with x, y isolated vertices of G t−1 , B plays xv t (a legal move since this doesn't create a cycle). Otherwise B plays xv t with x an isolated vertex of G t−1 .
We wish to bound the number of edges of G ∞ when B uses this strategy. To this end, let P denote the vertices that belong to some p i (including u i and v i ), , let e(X, Y ) denote the number of edges in G ∞ where one vertex lies in X and the other in Y .
Proof. This is immediate from the fact that |P | = ℓ(k − 1).
Proof. If this were not the case, then by the Erdős-Gallai Theorem there would exist a path of length 2k in V i . Since v i is adjacent to the two endpoints of this path, this would imply that G ∞ contains a C 2k+1 , a contradiction.
Proof. Assume G ∞ contained the edge w i w j with w i ∈ V i , w j ∈ V j . Then for any r = i, j (and such an r exists since ℓ ≥ 3), G ∞ would contain the cycle w i p ′ i u r p j w j , but this is a C 2k+1 , a contradiction. Claim 5.5. For any w ∈ W , e({w}, V i ) = 0 for at most one i.
Proof. Assume G ∞ contained the edges ww i and ww j with w i ∈ V i , w j ∈ V j , i = j. Then G ∞ would contain the cycle ww i p ′ i p j w j , a C 2k+1 , a contradiction. Claim 5.6. e(W, V i ) ≤ |W | max(|V i |).
Proof. This is an immediate consequence of the previous claim. Proof. Note that i t0 ≥ n − 2t 0 and ∆(i t ) ≥ −2 for all even t ≥ t 0 + 2 by the way the algorithm was constructed. It follows that there are at least n/2 + O(1) values of t with i t−2 ≥ 2, and hence B adds an edge of the form xv t for at least this many values of t. Thus B ensures that each of the ℓ vertices v i have at least n 2ℓ + O(1) neighbors in G ∞ .
Proof. This is an immediate consequence of the fact that |V i | = n − j =i |V j | − |W | − |P | and the previous claim.
We conclude that the number of edges in G ∞ will be e(W, W ) + e(W, V i ) + o(n 2 ) ≤ This value is maximized when |W | = 1 6 (1 + 1 ℓ )n, giving an upper bound of as desired.
6. Proof of Theorem 1.5 We will say that a vertex v is good if all but at most one edge incident to v is contained in a triangle. We will say that a graph G is k-good if there exists a set of edges B(G) with |B(G)| ≤ k such that every vertex of G \ B(G) is good. Observe that if G is k-good and G ′ is G plus an edge, then G ′ is (k + 1)-good. Theorem 6.1. There exists a strategy for B in the (C ∞ \ C 3 )-saturation game such that for all even t, either G t−1 is (C ∞ \ C 3 )-saturated or G t is 1-good. Lemma 6.2. Let G be a 2-good graph that contains no cycle C 2k+1 for any k ≥ 2. Then G contains no C ℓ for any ℓ ≥ 5.
Proof. Let C denote a C 2k in G with k ≥ 3 on the vertex set {v 1 , . . . , v 2k }, and let C ′ = C \ B(G). Since k ≥ 3, there exists an i such that C ′ contains the edges v i−1 v i and v i v i+1 . Since these edges are in G \ B(G), at least one of these edges is in a triangle, would be a C 2k−1 in G, a contradiction. A similar result holds if j = i + 3. Observe that G contains the cycles v i v i+1 · · · v j and v i+1 v i+2 · · · v j . One of these cycles must have odd parity with length at least 5 since j = i+2, i+3, a contradiction. We conclude that G contains no C 2k with k ≥ 3, proving the result.
Proof of Theorem 6.1. G 0 is 1-good, so inductively assume that B has been able to play so that G t−2 is 1-good. If G t−1 is saturated then the game is over and B doesn't play anything, so assume this is not the case. If G t−1 is 0-good, then B plays e t arbitrarily and G t will be 1-good. Now assume that G t−1 isn't 0-good. That is, there exists edges v 1 x and v 2 x with v 1 = v 2 such that neither of these edges are contained in triangles. We claim that adding e t = v 1 v 2 is a legal move. If it were not, then there must exist a path P of length 2k with k ≥ 2 from v 1 to v 2 in G t−1 . If x is not a vertex of P , then G t−1 contains the cycle formed by taking P and adding the edges xv 1 and xv 2 , which is a C 2k+2 . Since inductively G t−2 is 1-good, G t−1 is 2-good, and hence doesn't contain such a C 2k+2 by Lemma 6.2. Thus x must be a vertex of P . Let P i denote the path from v i to x in P , and let k i denote the length of P i . G t−1 contains a C ki+1 , namely by taking P i together with the edge xv i . Thus k i ≤ 3 by Lemma 6.2. Also k i = 2, since this would contradict xv i not being contained in a triangle. Since k 1 + k 2 = 2k ≥ 4, we must have, say, k 1 = 3. Let C = v 1 abx be the 4-cycle formed from P 1 and xv 1 . If, say, ab were contained in a triangle abc, then we must have c = v 1 or c = x, as otherwise v 1 acbx defines a C 5 in G t−1 . But if c = v 1 or x, then v 1 x is contained in a triangle, a contradiction. A similar analysis shows that no edge of C is contained in a triangle. This is only possible if B(G t−1 ) consists of two edges of C that aren't both incident to x, as otherwise one of ab and v 1 a would be contained in a triangle. In particular, two of the edges {xv 1 , xv 2 , xb} are not in B(G t−1 ), and we conclude that at least one of these edges must be contained in a triangle. But we've assumed that none of these edges are in triangles, a contradiction. We conclude that v 1 v 2 is a legal move to play.
Note that at least one of the edges xv 1 and xv 2 must be in B(G t−1 ), as otherwise G t−1 \ B(G t−1 ) wouldn't have all good vertices (namely, x wouldn't be a good vertex). Since v 1 x, v 2 x, and the new edge v 1 v 2 are contained in a triangle of G t , the set B(G t ) := B(G t−1 ) \ {v 1 x, v 2 x} shows that G t is 1-good as desired.
Proof of Theorem 1.5. If B uses the strategy of Theorem 6.1, then G ∞ will be 2-good. Lemma 6.2 shows, in particular, that G ∞ can't contain two even cycles of the form C 2r and C 2r+2 for any r. The result then follows from Theorem 12 of [10] after taking k = 2.

Concluding Remarks
• Our analysis was far from sharp in various places, and one could certainly improve the bounds of Theorem 1.1 with a more careful analysis.
• We do not have any strong upper bounds for sat g (C 5 ; n) or sat g (C 7 ; n), though we can prove that both are strictly less than 1 4 n 2 . Specifically, we can prove the following bound for k ≥ 2: sat g (C 2k+1 ; n) ≤ 1 4 n 2 − k − 1 2 n + O(1).
In order to get this bound, B uses the algorithm in Section 4 with c k and c −1 scaled down by a factor of 25. All of the analysis we did before holds if G t is bipartite for all t, so we can assume that G ∞ contains an odd cycle. In this case, the number of edges involving a vertex from the smallest odd cycle of G will be at most 2n by Lemma 4.11, and there will be at most 1 4 (n − 2k − 3) 2 other edges in G. Adding these two values gives the desired bound. We note that this bound is non-trivial. In particular, the algorithm of Theorem 2.1 shows that for k ≥ 2, B has no strategy that guarantees the creation of any odd cycles in the C 2k+1 -saturation game.
• Theorem 1.1 shows that for all k ≥ 4, sat g (C 2k+1 ) ≤ ( 1 4 − c k )n 2 + o(n 2 ) for some c k > 0. We suspect that such a bound also holds for sat g (C 5 ; n) and sat g (C 7 ; n).
• As a consequence of the bounds of Theorem 1.1, we know that sat g (C 2k+1 ; n) ≤ sat g (C 2k ′ +1 ; n) when k ′ is sufficiently larger than k and n is sufficiently large. We conjecture that this remains true when k ′ = k + 1.
A positive answer to this conjecture would in particular provide a desirable upper bound for sat g (C 5 ; n) and sat g (C 7 ; n). Note that the bound sat g ({C 3 }; n) ≤ 26 121 n 2 + o(n 2 ) of [1] together with Theorem 1.2 shows that the conjecture is true for k = 2, and moreover that sat g (C 3 ; n) ≤ sat g (C 2k+1 ; n) for all k ≥ 2 and n sufficiently large.
• One can verify that the proof of Theorem 1.3 generalizes to bounding sat g (F ; n) where F is any set of odd cycles whose smallest cycle is a C 2k+1 with k ≥ 2. Similarly the lower bounds of Theorem 1.1 and Theorem 1.2 generalize to bounding sat g (F ; n) where F is any set of odd cycles containing C 2k+1 . This shows that sat g (C ∞ \ C 2k+1 ; n) is quadratic for any k ≥ 3 since the set contains C 2k−1 . Theorem 1.5 shows that this value is linear when k = 1, and it would be of interest to know whether sat g (C ∞ \ C 5 ; n) is quadratic or sub-quadratic.

Acknowledgments
The author would like to thank Jacques Verstraete for suggesting this research topic, as well as his assistance with the general structure of the paper.