Strong games played on random graphs

In a strong game played on the edge set of a graph G there are two players, Red and Blue, alternating turns in claiming previously unclaimed edges of G (with Red playing first). The winner is the first one to claim all the edges of some target structure (such as a clique, a perfect matching, a Hamilton cycle, etc.). It is well known that Red can always ensure at least a draw in any strong game, but finding explicit winning strategies is a difficult and a quite rare task. We consider strong games played on the edge set of a random graph G ~ G(n,p) on n vertices. We prove, for sufficiently large $n$ and a fixed constant 0<p<1, that Red can w.h.p win the perfect matching game on a random graph G ~ G(n,p).


Introduction
Strong games, as a specific type of Positional games, involve two players alternately claiming unoccupied elements of a set X, which is referred to as the board of the game. The two players are called Red (the first player) and Blue (the second player). The focus of Red's and Blue's attention is a given family H ⊆ 2 X of subsets of X, called the hypergraph of the game, or sometimes referred to as the winning sets of the game. The course of the game is that Red and Blue take turns in claiming previously unclaimed elements of X, exactly one element each time, with Red starting the game. The winner of such a strong game (X, H) is the first player to claim all elements of some winning set F ∈ H. If this has not happened until the end of the game, i.e. until all elements of X have been claimed by either Red or Blue, the game is declared a draw.
One classical example of a strong game is the child game Tic-Tac-Toe and its close relative n-in-a-row, where the target sets are horizontal, vertical and diagonal lines of a square grid. The player completing a whole line first wins. If at the end of the game no with Breaker going first. Each player claims exactly one element of X per turn. Again, the set X is called the board of the game and the members of H are referred to as the winning sets. Maker wins the game as soon as she occupies all elements of some winning set F ∈ H. If Maker does not fully occupy any winning set by the time every board element is claimed by some player, then Breaker wins the game. Note that being the first player is never a disadvantage in a Maker-Breaker game (see e.g. [1]). Hence, in order to prove that Maker can win some Maker-Breaker game as the first or second player, it suffices to prove that she can win this game as the second player.
Using fast strategies for Maker-Breaker games (see [8]), some (perhaps surprising) results about particular strong games played on the edge set of a complete graph K n have been obtained recently. The few examples of such strong games for which an explicit winning strategy, based on a fast Maker-Breaker strategy, is known, include the perfect matching game, the Hamilton cycle game and the k-vertex-connectivity game (see [4], [5]), where Red's aim is to build a perfect matching, a Hamilton cycle, and a k-vertexconnected spanning subgraph of a complete graph K n , respectively.
Since the problem of finding explicit winning strategies for Red is quite hard, and since there are no general tools for it, it is just natural to continue exploring such strategies on different type of boards. Hopefully, at some point in the near future a general tool will appear. A very natural candidate to begin with is the well known binomial random graph G ∼ G(n, p), where each edge of the complete graph K n is being kept with probability p, independently at random (for a very good survey on random graphs the reader is referred to the excellent book [3]).
In this paper we initiate the study of strong games played on the edge set of a typical G ∼ G(n, p). In particular, we analyse the perfect matching game played on G and provide Red with an explicit winning strategy (though a bit complicated). Here is our main result: Theorem 1. Let p ∈ (0, 1) be a fixed constant. Then, a graph G ∼ G(n, p) is w.h.p. (with high probability, meaning with probability going to 1 as n tends to infinity) such that Red has a winning strategy for the perfect matching game played on E(G).

Notation and terminology
Our graph-theoretic notation is standard and follows that of [11]. In particular, we use the following. For a graph G, let V (G) and E(G) denote its sets of vertices and edges respectively. Moreover, let e(G) := |E(G)| be the number of edges of G and, for any two disjoint subset S, T ⊂ V (G) let e(S, T ) be the number of edges with one endpoint in S and the other in T (this set is denoted by E(S, T )). For a set S ⊆ V (G), let G[S] denote the subgraph of G, induced on the vertices of S.
Assume that some strong game, played on the edge set of some graph G, is in progress. At any given moment during this game, we denote the graph spanned by Red's edges by R, and the graph spanned by Blue's edges by B. We furthermore denote by d R (v) and At any point during the game, the vertices of G \ (R ∪ B) are called free vertices and any edge not yet claimed is called a free edge. Moreover, any vertex v ∈ V with d R (v) = 0 and d B (v) > 0 is called distinct.

Proof outline
The main idea of our proof is to show the following two things: on the one hand, Red can build a perfect matching 'fast', i.e. by claiming not much more edges than needed for a perfect matching. On the other hand, we need to make sure that Blue cannot build a perfect matching faster than Red. These two things than assure that the strong perfect matching game on G ∼ G(n, p) is Red's win.
For the first goal, the basic tool of our proof is to partition the graph G ∼ G(n, p) into small, disjoint cliques of even size, and then play on each of these 'subboards' separately. Since the strong perfect matching game is a fast win for Red on a clique (see [4]), Red will eventually build a perfect matching on all subboards and therewith also on the entire graph G. In general, Red tries to build a perfect matching in each subboard consecutively. If, theoretically, Blue would play along and always play in the same subboard as Red, the strategy to win the strong perfect matching game on K n from [4] would suffice to ensure Red's win. Unfortunately, Blue can claim whatever edges she likes, i.e. Blue can decide to not play in the same subboard as Red. Thus a main part of our proof is to slightly alter the strategy from [4] and adapt it to our needs, hence producing different strategies Red can use on different subboards. For each subboard Red will choose her strategy separately, her choice depending only on the blue edges present in the subboard right before Red starts claiming edges in it. Red skips her 'subboard by subboard approach only if Blue tries to isolate a vertex or claims to many edges in a subboard Red did not yet play on. In this case Red immediately reacts on Blue's last move and answers in the subboard Blue is 'attacking'.
The main idea to achieve the second goal is that Red needs to control the number of 'wasted moves' which, roughly speaking, is the number of edges claimed by a player which do not bring her closer to a perfect matching. For example, if at some point Blue claims an edge touching an already touched vertex, then clearly she cannot finish her matching in n/2 moves, where n is the number of vertices of the graph G. Thus, if Red can make sure at any point during the game that she did not 'waste more moves' than Blue, she for sure wins the game, as every perfect matching is of the same size and since she started the game. Hence, the difficulty of Red's strategy will be to maintain this 'advantage' on the number of 'wasted moves'. To this end, note that Blue can essentially force Red to 'waste' a move only once on each subboard, namely by blocking the last edge (say uv) Red needs to build a perfect matching. To circumvent such attacks, Red will create a 'trap' for Blue on every subboard. For every subboard it holds that, if Blue touches at some point a vertex v which is isolated in Red's graph, then Red can keep this vertex v isolated until she needs to claim only one edge uv to build a perfect matching (on this subboard). Then, as every time that Blue claims an edge touching this vertex v again, the electronic journal of combinatorics 24(1) (2017), #P1.35 Blue postpones her win by another step. This is true especially whenever Blue blocks Red's last edge uv, and thus in this case Red gets some spare moves to continue and eventually build her perfect matching on this subboard.
If however Blue did not create such a 'trap vertex' and nevertheless blocks the last edge uv of Red's perfect matching on a certain subboard, Red applies the following trick: she 'imports' two vertices x and y from the next subboard, hence giving her the opportunity to claim another free and independent edge in this subboard, e.g. ux, and in the same time ensuring that Blue created a 'trap vertex', namely v. So that Red is always able to do this 'importing', we need to refine our partitioning of the vertex set and make sure that such two vertices x and y always exist. This is achieved by showing that we can partition the vertex set not only into disjoint cliques of small, even size, but also arrange them in a ordering such that all edges between any two consecutive cliques are present as well.
The rest of the paper is structured as follows: the partitioning of the graph G ∼ G(n, p) into small, disjoint subboards of even size is derived in Theorem 2 and Lemma 3 in Section 2.1. In Section 2.2 we describe a slightly altered strategy to play the strong perfect matching game on a subclique of a graph (see Lemma 4). This strategy is then the key ingredient used to describe the different substrategies Red will choose from to play on each subboard of the partitioned graph. These substrategies are described in detail in Section 4. The proof of Theorem 1 can be found in Section 3, where we describe Red's winning strategy (using the aforementioned substrategies) and prove that Red can indeed follow it.

Preliminaries and tools
In this section we introduce some tools used in the proof of Theorem 1.

Partitioning of G ∼ G(n, p)
We first show the following auxiliary theorem and a partitioning lemma for a graph G ∼ G(n, p) which will allow Red to partition the board E(G) into suitable subboards.
Theorem 2. Let n be a sufficiently large integer and let 0 < p 1 be a fixed constant. Then, w.h.p., a graph G ∼ G(n, p) is such that the following holds: There exists a partition V (G) = V 1 ∪ · · · ∪ V t of G into disjoint subsets such that for all 1 i t we have: The proof of this theorem closely follows a nice argument of Krivelevich and Patkós from the proof of Theorem 1.2 in [10]. Proof. We follow the following greedy algorithm: Let k = n/ ln 1 3 n . If n k is even, we define r := n k and otherwise, we define r := n k − 1 (so that r is even in any case). Now, partition the vertex set V (G) = U 1 ∪· · ·∪U r−2 ∪W such that |U 1 | = |U 2 | = · · · = |U r−2 | = k and W = V \ r−2 j=1 U j . Note that k |W | 3k and that r ln 1 3 n. We build the k cliques in r rounds by starting with cliques of size 1 and, for all 2 i r − 2, adding in the ith round one vertex of U i to each clique. In the last two rounds, we add the vertices of W 'smartly' to ensure that all but one of the cliques are of even size. We denote the k cliques obtained after the ith round of the algorithm by C 1 i , . . . , C k i . In the first (r − 2) rounds the algorithm works as follows: For i = 1 we simply define {C 1 1 , . . . , C k 1 } := U 1 , and hence C 1 1 , . . . , C k 1 is a collection of k cliques, each of size 1. For 2 i r − 2, in the i th round we expose all edges between U i and i−1 j=1 U j . To find the extension of the cliques, we define an auxiliary random bipartite graph That is, A i represents the already formed cliques of size (i − 1) and the other part stands for the new vertices we want to add to those cliques. For C ∈ A i and x ∈ U i , we add the edge Cx to E(B i ) if and only if x is connected (in G) to all the vertices of C. Hence, any perfect matching of B i corresponds to an extension of the cliques C 1 i−1 , . . . , C k i−1 by one vertex each. Note that the auxiliary graph B i has edge probability p i = p i−1 . It is shown below that w.h.p. there exists a perfect matching in B i for all 1 i r − 2.
For the last rounds of the algorithm, let us assume that after (r − 2) rounds we have k cliques C 1 r−2 , . . . , C k r−2 of even size (r − 2). Now, we need to define the algorithm to extend these cliques with the vertices of W such that all but at most one of the cliques C 1 r , . . . , C k r are of even size. To this end we act as follows: four sets of size |W |/4 or |W |/4 + 1 such that |X 1 | = |Y 1 | and ||X 2 | − |Y 2 || 1. Then, for i = 1, 2 expose all the edges within W i := X i ∪ Y i . Assume without lose of generality that |X 2 | |Y 2 | = k 2 and let |X 1 | = |Y 1 | = k 1 . Note that in any case we have that k/4 k 1 k 2 k. Define the auxiliary random bipartite graph B W i = X i ∪ Y i with edge probability p W i = p (note that we forget about all the edges exposed inside X i and Y i ). Let Z 1 := {(x 1 , y 1 ), . . . , (x k 1 , y k 1 )} be the vertices of X 1 and Y 1 which are paired up by a perfect matching in B W 1 and let (x 1 , y 1 ), . . . , (x k 2 −1 , y k 2 −1 ) be the vertices which are paired up by a perfect matching in B W 2 . We then define the set Z 2 = {z 1 , . . . , z k 2 } by z i := (x i , y i ) for all 1 i k 2 − 1 and Again, it is shown below that there exists a perfect matching in B W 1 and B W 2 .
In the second last round, the algorithm exposes all edges between W 1 and V \ W . We define the auxiliary random bipartite graph B r−1 = A r−1 ∪ Z 1 on 2k 1 vertices, with A r−1 = {C 1 r−2 , . . . , C k 1 r−2 }. An edge between z i and C j r−2 is present in E(B r−1 ) if and only if all edges between x i and C j r−2 as well as all edges between y i and C j r−2 are present. Hence we obtain an edge probability p r−1 = p 2(r−2) . If there exists a perfect matching in B r−1 , the algorithm extends the cliques C 1 r−2 , . . . , C k r−2 by the corresponding vertex-pair in Z 1 and therewith obtains k cliques C 1 r−1 , . . . , C k r−1 of even size. Then, in the last round, the algorithm exposes all edges between W 2 and V \ W 2 . We define the auxiliary random bipartite graph B r = A r ∪ Z 2 on 2k 2 vertices, with . . , C k 2 r−1 }. An edge between z i and C j r−1 is present in E(B r ) if and only if all edges between x i and C j r−1 as well as all edges between y i and C j r−1 are present. Hence we obtain an edge probability p r = p 2r (we achieve equality in the edge probability by flipping additional coins with the 'missing' success-probability q e for all edges e incident to C j r−1 with k 1 < j k and, if z k 2 = y k 2 , for all edges incident to z k 2 ). If there exists a perfect matching in B r , the algorithm extends the cliques C 1 r−1 , . . . , C k 2 r−1 by the corresponding vertex-pair in Z 2 and therewith obtains k cliques C 1 r , . . . , C k r of which all but at most one (namely C k 2 r ) are of even size. Furthermore, . Thus, after reordering, the algorithm outputs a partition V (G) = V 1 ∪ · · · ∪ V k into k disjoint subsets of size Θ(ln It remains to prove that the algorithm succeeds w.h.p., i.e. that the algorithm can find a perfect matching in all the auxiliary bipartite graphs. Note that all the edge probabilities of these auxiliary graphs are at least p 2r . Since 0 < p < 1 is a constant, we have that p 2r p 2 ln 1 3 n = Ω(n α−1 ) for some 0 < α < 1. By Remark 4.3 in Chapter 4 in [9], we know that the probability that there is no perfect matching in a random bipartite graph G ∼ G(n, n, p) on two sets of size n with edge probability p is O(ne −np ). Therefore, the probability that the Algorithm fails is upper bounded by . All in all, the algorithm succeeds with high probability and outputs t cliques with the desired properties.
Using the above theorem, we prove the following lemma which ensures us a partitioning of a random graph G ∼ G(n, p) into disjoint complete subgraphs which can be cyclically ordered in such a way that the union of any two consecutive cliques is a clique as well.
Lemma 3. Let 0 < p 1 be a fixed constant. Then, w.h.p., a graph G ∼ G(n, p) is such that the following holds: There exists a partition V (G) = V 1 ∪ · · · ∪ V t of G into disjoint subsets such that for all 1 i t we have the following: for more details about this 'multiple exposure' trick, see [3]). Let V (G 1 ) = U 1 ∪ · · · ∪ U l be the partition of G 1 into disjoint subsets obtained by applying Theorem 2. Hence we have that |U i | is even for all 1 i l − 1.
Further, partition each subset Hence, all subsets but R l are of even size and, for all 1 i l, we have that L i and R i are of size Θ(log Before exposing G 2 , define an auxiliary digraph D = (V, E) such that the set of vertices is defined by , which is present if and only if all edges between R i and L j appear in G 2 .
Note that Using the main result of [6], we know that the digraph D contains a directed Hamilton cycle. By relabelling we may assume that (

The perfect matching game on the complete graph K n
The main tools used in the proof of Theorem 1 are the following two strategies concerning the perfect matching game on the complete graph K n . One is the strategy described in the proof of Theorem 1.2 in [8] which ensures that Maker can win the weak perfect matching game on K n in at most n/2 + 1 moves if n is even, respectively in n/2 moves if n is odd. We will henceforth denote this strategy by S weak n . The second strategy is a slight alteration of the strategy which ensures that Red can win the strong perfect matching game on K n , as described in the proof of Theorem 1.3 in [4]. The basic idea of this strategy is that Red makes sure of two things: on one hand, she fully claims a perfect matching 'fast', and on the other hand, she ensures not to 'waste more moves' than Blue. Thus, since all perfect matchings are of the same size, Blue cannot fully occupy a perfect matching before Red does, and hence Red wins the game (note that this idea works in general for all strong games with winning sets of the same size). To , be the number of wasted moves of Blue, respectively Red, on the edge set of the induced subgraph G[V ].
A simple but crucial observation on the number of wasted moves is that it can be bounded from below by the degree of any vertex, since at any time in a perfect matching We now describe how Red achieves the following:  In the above Lemma, one can think of E(H) as the 'subboard' where Red's focus is momentarily fixed on. Assuming that Blue claimed at most one edge of E(H) before Red starts claiming her edges, it then holds that Red can build a perfect matching on H 'fast' (at most 2 edges more than necessary). Note that we do not claim that Red finishes her perfect matching on H before Blue does (which is not possible because Blue can already have claimed an edge in E(H)). But in terms of the number of wasted moves, (i) and (ii) ensure that Red builds a perfect matching on H 'as fast as' Blue, i.e. Blue wastes at least as many moves as Red on the 'subboard' E(H), respectively at the vertex u Thus, seen on the whole graph, Blue 'wasted' at least as many moves as Red). Thus one can think of the strategy as an 'almost' strong strategy.
In the strategy described below, the main idea of Red is to assure that there exists what we call a trap vertex, i.e. a vertex u ∈ V (H) with d R (u) = 0 and d B (u) 1. Then, whenever Red claimed a matching of size n/2 − 1 on H, which she can do in n/2 − 1 moves (details in the proof), and therefore misses only one edge to complete her perfect matching, it holds that this missing edge is incident to a vertex u with d B (u) 1. Hence, if Blue claims uv, therewith preventing Red from building a perfect matching in n/2 moves, it holds that w B d B (u) − 1 1. But this in turns allows Red to 'waste a move' herself and build a perfect matching in at most two additional moves (while keeping Blue 'forced' to claim edges incident to u).
The proof of Lemma 4 closely follows the lines of Theorem 1.3 in [4].

Proof of Lemma 4.
In what follows, we present a strategy for Red and then prove that, by following it, Red can build a perfect matching on H in at most n/2 + 2 moves ensuring that the statements of Lemma 4 are true.
Assume first that n is odd. Following Maker's strategy S weak n on E(H), Red can build a perfect matching on H in n/2 moves, which is optimal (note that a perfect matching on an odd number of vertices is a matching covering all but 1 vertex).
Else, if n is even, Red's strategy is divided into the following three stages: Stage I: In her first move, depending on whether Blue claimed an edge in E(H) or not, Red distinguishes between the following two cases: To describe Stage II, we need to introduce more notation.
Moreover, we denote by D j the number of V (H)-distinct vertices immediately after Red's jth move and by D j be the number of V (H)-distinct vertices immediately before Red's jth move. Note that whenever Blue creates a V (H)-distinct vertex, it may take over the role of the designated trap vertex (and thus the vertex u of Stage I, Case 2 will be 'reintegrated' into the game).
Stage II: For every 2 j n/4+2, in her jth move Red claims an edge e j ∈ E(G[U ]) which is independent of her previously claimed edges while making sure that D j 1. Red can even ensure that, if D k = 1 for some 1 k n/2 − 1, then D j = 1 for all k j n/2 − 1. Hence, let 2 k n/2 − 1 be the smallest integer such that D k = 1. Then Red updates U := V (H) in her kth move, since she does no longer need the trap vertex u ∈ V (H).
If ∆(B[V (H)]) > 1 holds immediately after Blue's (n/4 + 2)nd move, then Red skips to Stage M. Otherwise, for every n/4 + 3 j n/2 − 1, in her jth move Red claims an edge e j ∈ E(G[U ]) which is independent of her previously claimed edges while making sure that D j 1. Red then proceeds to Stage III.
Stage III: Red completes her perfect matching in E(H) by claiming at most 3 additional edges as follows: Let u, v ∈ V (H) be the two last vertices Red needs to connect to build a perfect matching on E(H) and w.l.o.g. let u be the designated trap vertex, resp. the V (H)distinct vertex. In her (n/2)nd move, Red claims uv and finishes her perfect matching in E(H). If this is not possible, let xy and wz be two red edges such that B[{u, v, w, z, x, y}] consists solely of the edge uv. In her (n/2)nd move Red claims the edge vy. In her (n/2 + 1)st move, Red claims the edge ux and thus finishes her perfect matching in E(H). If this is not possible, Red claims the edge uz. Since Blue cannot claim both xw and vw in her next move, Red claims one of them in her (n/2 + 2)nd move and thus finishes her perfect matching in E(H). It remains to prove that Red can indeed follow all parts of the strategy. We will now prove inductively that Red can follow Stage II of her strategy (either for n/4 + 2 or n/2 − 1 moves). Observe that for both cases in Stage I we have that D 1 1, and thus the induction basis is fulfilled. Additionally, note that, since Blue can create at most two V (H)-distinct vertices in one round, we have that D j+1 − D j 2. We distinguish now two cases:  Henceforth, the strategy described in the proof above will be denoted by S a.strong n . It will be the key ingredient of the proof of Theorem 1. Before we start the proof of Theorem 1, let us make a last remark, which follows directly from the second last paragraph of the proof.

Proof of Theorem 1
For the description of Red's strategy to win the strong perfect matching game on a graph G ∼ G(n, p), we will use the following notation and definitions: Assume that the graph G ∼ G(n, p) is partitioned according to Lemma 3. Hence we have a partition , the number of wasted moves of Blue, respectively Red, on the subboard E i .
A subboard E i is called inactive if e(R i ) = 0, it is called active if e(R i ) 1, but R i does not contain a perfect matching, and it is called safe if R i contains a perfect matching. Additionally, if a subboard E i is inactive and w B i turns 1, we call the subboard E i dangerous. Every dangerous subboard stays dangerous until Red completed her perfect matching on it (at which point it becomes safe).
The strategy which we describe below enables Red to build a perfect matching on G ∼ G(n, p) 'quickly' while ensuring that w R w B stays true at all times. The main idea is that Red (mostly) neglects all the edges in between the subboards and plays on each subboard separately, with the aim to complete a perfect matching on each board while ensuring that w R i w B i . Note that this condition can be interpreted and is sometimes referred to a as 'Red does not waste more moves than Blue' on the subboard E i . Whenever possible, Red will consider the subboards in cyclic order, meaning that she tries to complete her perfect matching on E i before claiming an edge on E i+1 . Observe, crucially, that Red ignores most of the blue edges in between the subboards (until the very end of the strategy) and counts the number of wasted moves of Blue 'locally', that is, on every subboard separately. Otherwise, since we will use the degree of some specified vertices to give a lower bound on the number of wasted moves of Blue, it could happen that an edge u i u j with u i ∈ V i and u j ∈ V j adds a 'plus 1' to the number of wasted moves in two subboards. Another important observation is that the partition of V (G) allows us to nullify the 'natural' attack point of Blue, which is blocking the last edge Red needs for a perfect matching on a subboard E i . Whenever Blue blocks such an edge, using the fact that the union of two consecutive subboards is a clique as well, Red 'imports' two suitable vertices from V i+1 to circumvent Blue's attack.
Red only interrupts this 'subboard by subboard' approach if Blue tries to block a vertex or claims too many edges on an inactive subboard. Thus, when a subboard E i becomes dangerous, Red gives this board a special attention and, whenever Blue claims an edge in E i , Red skips her 'subboard by subboard' approach and answers on E i instead.
The description of Red's strategy comes in two parts. First, we introduce the three substrategies called S dangerous , S trap and S empty . These substrategies give Red three alternatives of how to play on the subboards E i . For each subboard E i Red will decide to follow one of these three substrategies, depending on how Blue's graph B i looks like right before Red claims her first edge on E i .
After introducing these substrategies, we describe the so called overall strategy, and prove that Red can indeed follow this overall strategy. This strategy shows how Red actually plays the game on the edge set of the graph G, i.e. how Red chooses the substrategies for the subboards E i .
In the following, we introduce the three substrategies S trap , S dangerous and S empty . Here we only describe under which conditions Red chooses a substrategy and what she can achieve by following it. The detailed description of each substrategy and the proofs that Red can indeed play according to these substrategies are postponed to Section 4.
S trap : The strategy S trap is used on subboards E i where, right before Red claims her first edge on E i , it holds that e(B i ) 1 and w B i = 0. By following the strategy S trap , Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that w R i w B i at any time.
S dangerous : The strategy S dangerous is used to play on subboards E i where, right before Red claims her first edge on E i , it holds that w B i = 1. By following the strategy S dangerous , Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that w R i < w B i at any time.
S empty : The strategy S empty is used to play on subboards E i where, right before Red claims her first edge on E i , it holds that e(B i ) = 0. By following the strategy S empty , Red achieves the following: Thus, assuming that these three substrategies exist, we can now describe the Overall Strategy and therewith prove Theorem 1.
Proof of Theorem 1. First we propose a strategy for Red and then prove that the proposed strategy is indeed a winning strategy for the perfect matching game played on a typical G ∼ G(n, p). From now on, we condition on G satisfying all the properties mentioned in the statements in Section 2.1. Hence let V (G) = V 1 ∪ . . . ∪ V t be the partition of G ∼ G(n, p) into t disjoint subsets as described in Lemma 3.
Overall Strategy: During the game, except for the first move, Red always reacts to Blue's last move. Hence, we consider one round as one move of Blue and a countermove of Red (except for round 0, which only consists of the first move of Red). At any point during the game, if Red is unable to follow the proposed strategy or if Red claims more than n/2 + 4t edges (where t is the number of subboards obtained by partitioning V (G)), then she forfeits the game. As long as Red's graph does not contain a perfect matching, Red does the following: In her first move of the game Red claims an edge ab ∈ E 1 obtained by following S empty on E 1 . In any other move, let e j be the edge which has just been claimed by Blue in her jth move. First, Red checks whether for some i t we have e j ∈ E i and E i is dangerous. If this is the case, Red answers according to the strategy S dangerous on E i . Otherwise, if there exists some active subboard, then Red chooses the smallest integer i t for which E i is active and plays on E i according to her chosen strategy S i (it is described below how Red chooses the strategy S i on each subboard E i ). Otherwise, there are no active subboards and Red's graph contains a perfect matching on the subboards E 1 , . . . , E k−1 for some k ∈ [t]. In this case Red wants to play on E k (which is not dangerous) according to the strategy S k , which she chooses in the following way: If e(B k ) 1, then Red defines S k = S trap . Else, e(B k ) = 0 and then Red defines S k = S empty . claims an edge xy ∈ E t in her first move on E t , Red skips to Stage II of S a.strong mt and finishes her perfect matching accordingly. Else, Red plays according to S weak mt−2 on E(G[V t \ {a, b}]). It remains to prove that Red can indeed follow this Overall Strategy and that, by following it, Red can build a perfect matching on E(G) in at most n/2+4t moves ensuring that w R w B at any time during the game.
The Overall Strategy considers all possible cases since either Blue has already claimed an edge in E i or not when Red needs to choose how to play on E i . Note that if w B i turns 1, Red immediately sets S i = S dangerous and answers on E i . Thus, whenever Red can choose the strategy to play on E i , it holds that w B i = 0. Therefore, whenever Red decides to play according to one of the three substrategies, the respective assumptions on the subboards are fulfilled.
Therewith, observe that for all 1 i t Red can follow the overall strategy and play accordingly on every subboard. Moreover, Red needs to claim at most m i /2 + 4 edges to finish her perfect matching on every E i . Hence, since m i is even for all 1 i t − 1 by Lemma 3, she finishes her perfect matching on E(G) in at most n/2 + 4t moves. Thus it remains to show that Blue cannot claim a perfect matching on E(G) before Red finishes her perfect matching.
Whenever Red plays according to Case 2.1 on E t , Red finishes the game in exactly m t /2 + 1 moves (as Red's graph contains a perfect matching of G[V \ V t ]). But since Blue's graph does not contain a perfect matching of G[V \ V t ] (and e(B t ) = 0), it follows that Blue cannot finish a perfect matching of G in less than m t /2 + 1 moves. Thus Red finishes her perfect matching before Blue does and hence wins the game.
Whenever Red plays according to Case 2.2 on E t , note that if Blue claims an edge xy ∈ E t after Red claimed ab ∈ E t , then by symmetry we may assume that d Bt (x) = 1 and d Rt (x) = 0. Thus, when Red plays according to S a.strong mt , we have that there exists a V t -distinct vertex, and therefore Remark 5 ensures that w Rt w Bt at any time. But this implies that Blue cannot build a perfect matching on E t faster than Red, thus Red wins the game. Else, if Blue does not claim an edge in E t after Red claimed ab ∈ E t , We know that Blue's graph does not contain a perfect matching of G[V \ (V t \ {a, b})] (as d B (a) = d B (b) = 0 right before Red claimed ab). Thus the same argument as for Case 2.1 shows that Red wins the game.
For all other scenarios, we will show that w R w B at any point during the game, and thus Blue cannot claim a perfect matching before Red does. To this end, first consider the case when Red plays according to S dangerous or S trap on E t . Then, since Red plays according to one of the three substrategies on every subboard E i , we have that where the last inequality follows since E i and E j , E i and E(V j , V j+1 ), as well as Therewith, the only remaining case is when e(B t ) = 0 but there exists a trap vertex u ∈ V t with d B (u) 1, right before Red claims her first edge on E t . Note that e(B t ) = 0 the electronic journal of combinatorics 24(1) (2017), #P1.35 means that E t is not dangerous. Hence, the case P 2 from the strategy S empty cannot occur on E t−1 and therewith Vt] . Let us consider these two cases separately: if w R t−1 + 2 < w B[V t−1 ,Vt] , since w Rt 2 (Red plays according to S a.strong mt on E t ), it holds that w R t−1 + w Rt < w B[V t−1 ,Vt] (note that this is the only case where we need this 'strange looking' additional +2 in the property P 3). Thus at any time during the game.
Else, if w R t−1 w B t−1 , we have for the first t − 1 subboards at any time that Note that we did not consider blue edges incident to any vertex v ∈ V t to derive inequality (1) (as the second sum on the right side of the above inequality only sums up to t−2). Furthermore, as Red decides to play according to S a.strong mt on E t , Lemma 4 asserts us that either w Rt w Bt (in which case, together with inequality (1), we are done), or that w Rt = d Bt (u). In the latter case, let uv be an edge incident to u in Blue's graph right before Red claimed her first edge on E t . Hence, we have that Thus, since we did not consider blue edges in E(v, V t ) to derive inequality (1), it follows that w R w B , concluding the proof.

Substrategies
In this section, we describe the three substrategies S trap , S dangerous and S empty in detail and prove that Red can indeed follow them. Note that these substrategies rely on S weak n , which ensures that Maker can win the weak perfect matching game on K n in at most n/2 + 1 moves, and the strategy S a.strong n , as described in the proof of Lemma 4.

Substrategy S trap
S trap : The strategy S trap is used on subboards E i where, right before Red claims her first edge on E i , it holds that e(B i ) 1 and w B i = 0. Thus Blue's graph consists of 1 up to m i /2 independent edges. The main idea of Red is to 'dissect' all but at most one of Blue's edges and thus split the board E i into two 'halves', since then Red can use the strategy S a.strong n on both 'halves'. In what follows we propose a strategy and then prove that, by following it, Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that w R i w B i at any time.
The strategy S trap can be divided into the following two cases: be an edge such that u 1 ∈ U i,1 and u 2 ∈ U i,2 . Before her first move, Red chooses u 2 to be the designated trap vertex when playing according to the strategy S a.strong ). In her first move, Red then plays according to the strategy S a.strong be two edges such that u 1 , v 1 ∈ U i,1 and u 2 , v 2 ∈ U i,2 . Before her first move, Red chooses v 1 , resp. u 2 , to be the designated trap vertices when playing according to the strategy S a.strong It remains to prove that Red can indeed follow this strategy and, by following it, Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that w R i w B i .
First, for the partitioning of V i into U i,1 and U i,2 , recall that B i consists of independent edges only. Therefore, if m i is divisible by 4, Red can 'dissect' all edges of Blue and, by distributing free vertices equally, obtain a partition of V i into two equitable halves U i,1 and U i,2 of even size. If m i is however not divisible by 4 and there are no free vertices to distribute (i.e. if B i consists of m i /2 independent edges), Red might not be able to dissect all edges in order to keep |U i,1 | and |U i,2 | even. But then, by dissecting all but one of Blue's edges, the described partitioning of V i is possible.
For  2 ] , where u 2 is the designated trap vertex chosen right after the partitioning. Since there exists at least one edge u 1 u 2 with u 1 ∈ U i,1 , we have that  2 ] after Red finished her perfect matching on E i . These two equalities, together with the two different blue edges u 1 u 2 , v 1 v 2 ∈ E(U i,1 , U i,2 ) then give us that w R i w B i is true at any time.

Substrategy S dangerous
S dangerous : The strategy S dangerous is used to play on subboards E i where, right before Red claims her first edge on E i , it holds that w B i = 1. Note that in this case e(B i ) > 2 and B i consists of either one path of length 2 or one path of length 3 together with some independent edges. The main idea of Red is again to 'dissect' all but at most one of Blue's edges and thus split the board E i into two 'halves', since then Red can use the strategy S a.strong n . In what follows we propose a strategy and then prove that, by following it, Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that w R i < w B i at any time.
The strategy S dangerous can be divided into the following cases: Let xyz be the path of length 2 in Blue's graph. Red partitions the vertex set V i = U i,1 ∪ U i,2 into two subsets, as described in the strategy S trap , additionally making sure that x and z belong to U i,1 and y belongs to U i,2 . Then Red ignores the electronic journal of combinatorics 24(1) (2017), #P1.35 that the edge xy is coloured blue and plays as described in Case 2 of the strategy S trap .
Case 3: e(B i ) > 2 and Blue's graph contains a path of length 3. Let wxyz be the path of length 3 in Blue's graph. Red partitions the vertex set V i = U i,1 ∪ U i,2 into two subsets, as described in the strategy S trap , additionally making sure that x and z belong to U i,1 and w and y belong to U i,2 . Then Red ignores that the edge xy is coloured blue and plays as described in Case 2 of the strategy S trap .
It remains to prove that Red can indeed follow this strategy and, by following it, Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that w R i < w B i .
To this end, first observe that when B i consists only of the path xyz of length 2, Red can claim xz, hence ensuring that there exists exactly one V i -distinct vertex. Thus Red can play according to the strategy S a.strong m i −2 on V \ {x, z} and build a perfect matching in at most m i /2 + 2 moves. Additionally, since 0 = w R i < w B i = 1 and y is already V i -distinct in the beginning, it follows from Remark 5 that w R i < w B i is true at any time.
For the partition of V i note that 'dissecting' the edges immediately gives us the additional conditions on x, y, z and w, thus this partition can be obtained.
For Cases 2 and 3 note that by 'ignoring' the colouring of the edge xy (which lies in between U i,1 and U i,2 ), since e(B i ) > 2, we have the exact same conditions as in Case 2 of the strategy S trap (at least two independent edges in Blue's graph). Thus Red can play as described in Case 2 of the strategy S trap and build a perfect matching in at most m i /2 + 4 moves. Additionally, since Red does not 'waste more moves' than Blue when playing according to S trap , and again since w R i < w B i in the beginning, we have that w R i < w B i is true at any time.

Substrategy S empty
S empty : The strategy S empty is used to play on subboards E i where, right before Red claims her first edge on E i , it holds that e(B i ) = 0. It achieves the following: P 1: If there exists a V i -distinct vertex u ∈ V i after Red's (m i /2 − 1)st move on E i or if ∆(B i ) > 1 before Red's (m i /4 + 3)rd move on E i , then w R i w B i at any time.
P 2: Else, if E i+1 is (resp. was) dangerous, then w R i + w R i+1 w B i + w B i+1 at any time. The strategy S empty consists of the following stages: Stage I: Red claims an arbitrary edge ab ∈ E i and skips to Stage II. . Otherwise, we distinguish between the following two cases: Case 1: e(R i+1 ) 1.
Then Red claims uv. If this is not possible, Red plays as described in Stage III of the strategy S a.strong m i and finishes her perfect matching in at most three moves.
Case 2: e(R i+1 ) = 0. Then Red claims uv. If this is not possible, Red chooses two vertices x, y ∈ V i+1 such that (i) the edges ux and vy are still free, and (ii) the number of Blue edges yz with z ∈ V i is smaller than m i /4 + 1.
If this is not possible, Red uses Stage III of the strategy S a.strong m i to finish her perfect matching on the subboard E i in at most three moves. Otherwise, Red claims the edge ux. Before her next move, Red updates V i := V i ∪ {x, y} and V i+1 := V i+1 \ {x, y}. Then Red claims vy. If this is not possible, Red uses Stage III of the strategy S a.strong m i to finish her perfect matching on the updated subboard E i in at most three moves.
It remains to prove that Red can indeed follow this strategy and, by following it, Red builds a perfect matching on E i in at most m i /2 + 4 moves ensuring that the properties P 1, P 2 and P 3 are achieved.
Note that after Stage I there exists no V i -distinct vertex, hence Red can skip to Stage II of the strategy S a.strong m i . Thus, whenever Red skips to Stage M of the strategy S a.strong m i , meaning that ∆(B i ) > 1 before Red's (m i /4 + 3)rd move on E i , it follows that w R i w B i at any time (which gives us a first case of P 1). Moreover, Red then builds a perfect matching on E i in m i /2 + 1 moves.
Note that, while following Stages I and II of the strategy S a.strong m i , Red made sure there is at most one V i -distinct vertex after each of her moves, thus we can assume w.l.o.g. that that d B i (v) = 0 whenever Red reaches Stage III. Thus, If d B i (u) 1, i.e. there exists a V i -distinct vertex after Red's (m i /2 − 1)st move, Red essentially plays according to S a.strong m i on E i . Therefore, we have that Red finishes her perfect matching in at most three moves, and Remark 5 ensures that w R i w B i (which gives us another case of P 1).
Whenever Red plays according to Stage III, Case 1, note that e(R i+1 ) 1 means that E i+1 is, respectively was, a dangerous subboard. Therewith, w R i+1 < w B i+1 . Since there exists no V i -distinct vertex after Red's (m i /2 − 1)st move, no red edge is connected in Blue's graph to u or v. Thus, if Red cannot claim uv, she can indeed follow Stage III of the strategy S a.strong m i . In this case, Lemma 4 ensures only that d B i (u) = w R i , which is the same as saying w R i − 1 w B i . But then, as w R i+1 < w B i+1 , we have that w R i + w R i+1 w B i + w B i+1 (which gives us P 2). Moreover, Red builds a perfect matching on E i in at most m i /2 + 2 moves.
Whenever Red plays according to Stage III, Case 2, first note the following: if such two vertices x and y do not exist, then Blue claimed lots of edges between V i and V i+1 (remember that all edges between V i and V i+1 are present). Especially it means that the electronic journal of combinatorics 24(1) (2017), #P1.35