Positional games on random graphs - Semantic Scholar
Positional games on random graphs Miloˇs Stojakovi´c
∗†
Tibor Szab´o∗
December 8, 2003
Abstract We introduce and study Maker/Breaker-type positional games on random graphs. Our main concern is to determine the threshold probability pF for the existence of Maker’s strategy to claim a member of F in the unbiased game played on the edges of random graph G(n, p), for various target families F of winning sets. More generally, for each probability above this threshold we study the smallest bias b such that Maker wins the (1: b) biased game. We investigate these functions for a number of basic games, like the connectivity game, the perfect matching game, the clique game and the Hamiltonian cycle game.
1
Introduction
(Un)biased positional games. Let X be a finite nonempty set and F ⊆ 2X . The pair (X, F) is a positional game on X. The game is played by two players Maker and Breaker, where in each move Maker claims one previously unclaimed element of X and then Breaker claims one previously unclaimed element of X. Maker wins if he claims all the elements of some set in F, otherwise Breaker wins. Unless otherwise stated, we assume that Maker starts the game. Note, however, that the asymptotic statements discussed are not influenced by which player makes the first move. The set X will be referred to as the board, and the set F as the set of winning sets. Whenever there is no confusion about what the board is, we may refer to the game (X, F) as just F. The set of all positional games could be partitioned into two classes. The game (X, F) is called a Maker’s win if Maker has a winning strategy, that is, playing against an arbitrary opponent Maker can occupy a member of F. Clearly, if (X, F) is not a Maker’s win, then Breaker is able to prevent an arbitrary opponent from occupying a winning set. These games are called Breaker’s win. Typical, well-studied examples of such positional games are played on the edges of a complete graph, i.e. X = E(Kn ). Maker’s goal usually is to build a ∗ Institute of Theoretical Computer Science, ETH Zurich, CH-8092 Switzerland. Email addresses: {smilos, szabo}@inf.ethz.ch † Supported by the joint Berlin/Zurich graduate program Combinatorics, Geometry and Computation, financed by ETH Zurich and the German Science Foundation (DFG).
1
graph theoretic structure–like a spanning tree, a perfect matching, a Hamiltonian cycle, or a clique of fixed size. It turns out that all these games are won easily by Maker, so in order to make things more fair (if such thing exists; actually no game of perfect information is fair as the winner–in theory–is known in the beginning of the game) one could give Breaker extra power by allowing him to claim more than 1 edge in each move. If X is a finite nonempty set, F ⊆ 2X and a, b are positive integers (possibly functions of the board size), then the 4-tuple (X, F, a, b) is a biased (a: b) game. In a biased (a: b) game, Maker claims a elements (instead of 1) and Breaker claims b elements (instead of 1) in each move. For a family F the smallest integer bF is sought (and sometimes found; see [1, 2, 3, 4, 5, 7]) for which Breaker wins the (1 : bF ) game. In the connectivity game Maker’s goal is to build a connected spanning subgraph; i.e. in this game the family of winning sets is the family T = Tn of all spanning trees on n vertices. Chv´atal and Erd˝os proved [7] that bT = Θ( logn n ). Beck [1] established bH = Θ( logn n ), where H = Hn is the family of all Hamiltonian cycles on n vertices. For the family Kk = Kk,n of all k-cliques on n vertices, Bednarska and 2
Luczak [4] showed that bKk = Θ(n k+1 ). More generally, they proved that in the game in which Maker’s goal is to claim an arbitrary fixed graph G, the ′ e(H)−1 over all threshold bias is Θ(n1/m (G) ). (Here m′ (G) is the maximum of v(H)−2 subgraphs H of G with at least 3 vertices.) Playing on a random board. In the present paper we introduce another approach to even out the advantage Maker has in a (1: 1) game, by randomly reducing the board size and keeping only those winning sets, which survive this thinning intact. Definition 1 Let (X, F, a, b) be a biased game. Random game (Xp , Fp , a, b) is a probability space of games where each x ∈ X is independently included in Xp with probability p, and Fp = {W ∈ F : W ⊆ Xp }.
Only if p = 0 or 1, or if ∅ ∈ F or F = ∅, we can be sure who (Maker or Breaker) wins the random game Fp . In all other cases the best we can conclude is that Maker (or Breaker) wins a.s. (almost surely), i.e. the probability that Maker (Breaker) wins tends to 1 if the board size tends to infinity. (So we actually talk about an infinite family of probability spaces of games. . . ) Excluding the trivial case ∅ ∈ F, Breaker always wins when p = 0. The graph games we consider are (easy) Maker’s wins, when p = 1. For a particular sequence of games (X, F), the first natural question to ask is: What is the threshold probability pF at which an almost sure Breaker’s win turns into an almost sure Maker’s win. More precisely we would like to determine pF for which • Pr[(Xp , Fp , 1, 1) is a Breaker’s win] → 1 for p ≪ pF , and • Pr[(Xp , Fp , 1, 1) is a Maker’s win] → 1 for p ≫ pF . 2
Such a threshold pF exists [6], since being a Maker’s win is an increasing property. The main goal of this paper is to establish a connection between the natural threshold values, bF and pF , corresponding to the two different weakenings of Maker’s power: bias and random thinning, respectively. We find that there is an intriguing reciprocal connection between these two thresholds in a number of well-studied games on graphs. Recall the notations T , H, and Kk , and let us denote by M the set of all perfect matchings on the graph Kn . Theorem 1 For positional games, played on E(Kn ), we have (i) pT = (ii) pM = (iii)
log n n
(iv) n
log n n ,
≤ pH ≤
1 − a(k)
2
log n n ,
log √ n, n 2
≤ pKk ≤ n− k+1 , where a(k) =
k+1 2
−
10 , 2k/2
5
(v) n− 3 ≤ pK3 ≤ n− 9 . For the connectivity game T an even more precise statement is true. In Corollary 19 we observe that Maker starts to win a.s. at the very moment when the last vertex of a random graph process picks up its second incident edge. More generally, for every p we would like to find the smallest bias bpF s.t. Breaker wins the random game (Xp , Fp , 1, bpF ) a.s. Note that by definition bF = b1F . Another trivial observation is that bpF = 0 provided p is less than the threshold for the appearance of the first element of F in the random graph. We obtain the following. Theorem 2 (i) bpT = Θ (pbT ) = Θ p logn n , provided p = Ω b1T , 1 (ii) bpM = Θ (pbM ) = Θ p logn n , provided p = Ω bM ,
√ √n , (iii) Ω p lognn ≤ bpH ≤ O p logn n , provided p = Ω log n c 2 k (iv) bpKk = Θ (pbKk ) = Θ pn k+1 , provided p = Ω logbK n . k
One can see that bpF is of order p/pF = pbF for the connectivity game and the perfect matching game, provided p ≥ pF . In particular for these games pF = Θ(1/bF ). In part (iv) of Theorem 2, generalizing the arguments of Bednarska and Luczak [4] we show that one can estimate bpKk up to a constant 3
factor, for all probabilities down to a polylogarithmic factor away from 2 critical probability 1/bKk = n− k+1 . On the other hand Theorem 1 part shows that we cannot get arbitrarily close to probability 1/bKk , since in triangle game Maker can win even for probabilities below 1/bK3 = n−1/2 . Nevertheless we think the Hamiltonian cycle game behaves “nicely”, i.e. same way as the connectivity game and the perfect matching game.
the (v) the the
Conjecture 1 Let H be the set of Hamiltonian cycles in Kn . n p , provided p = Ω logn n . bH = Θ p log n In particular,
log n . n Observe that the validity of the conjecture would mean that in a random graph with edge probability p ≫ logn n Maker could build a Hamiltonian cycle. So P´osa’s Theorem (which only proves the existence of a Hamiltonian cycle) would be true constructively even if an adversary is playing against us. The paper is organized as follows. In Section 2 we prove a general criterion for Breaker’s win in a different, auxiliary random game. In Section 3, the analysis of four biased random games is presented. In particular, in Subsections 3.1, 3.2, 3.3 and 3.4 we look at the connectivity game, the Hamiltonian cycle game, the perfect matching game and the clique game, respectively. In Section 4 we analyze more precisely a couple of (1: 1) games – the connectivity game (Subsection 4.1) and the clique game (Subsection 4.2). Finally, in Section 5 we give a collection of open questions and conjectures. pH =
Notation. For a graph G, e(G) and v(G) denote the number of edges and vertices (respectively) of G, and E(G) and V (G) denote the sets of edges and vertices (respectively). If C ⊆ V (G) and v ∈ V (G), then NC (v) denotes the set of neighbors of v in C. The logarithm log n in this paper is always of natural base.
2
A criterion
One of few general, but still very applicable results to decide the winner of biased positional games is the biased version of the Erd˝os–Selfridge Theorem [9, 2]. It provides a criterion for Breaker’s win, applicable on any game. Theorem 3 (Beck, [2]) If X
(1 + b)−|A|/a
0 are chosen so that p > δ)pb, 1) is Breaker’s win a.s.
4 log 2 δ2 b
(2)
holds, then the game (Xp , Fp∩ , (1−
Proof. For each A ∈ F and its corresponding set A′ ∈ Fp∩ we have E[|A′ |] = p|A|. If all winning sets A′ ∈ Fp∩ have size at least (1 − δ)p|A|, then X
′
|A | − (1−δ)pb
2
A′ ∈Fp∩
≤
X
− (1−δ)p|A| (1−δ)pb
2
=
A∈F
X
2−
|A| b
< 1.
A∈F
Using the Erd˝os–Selfridge theorem we obtain that Breaker wins the (Xp , Fp∩ , (1− δ)pb, 1) game, provided |A′ | ≥ (1 − δ)p|A| for all A′ ∈ Fp∩ . Next we check that this condition holds almost surely. Using a Chernoff bound, we obtain that X − δ2 p|A| 2 . Pr ∃A ∈ F : |A′ | ≤ (1 − δ)p|A| ≤ e A∈F
If we denote minA∈F X
A∈F
e−
δ 2 p|A| 2
≤
|A| b
X
A∈F
by mn , then we have 2−2
|A| b
≤
X
A∈F
2−mn 2−
|A| b
< 2−mn → 0,
and therefore all winning sets A′ ∈ Fp∩ have size at least (1 − δ)p|A| a.s. 5
2
3 3.1
Games Connectivity game
The first game we study is a random version of the biased connectivity game (E(Kn ), T , 1, b) on a complete graph on n vertices Kn . Maker’s goal is to build a spanning, connected subgraph, i.e. T is the set of all spanning trees on n vertices. It is obvious that pT = Ω( logn n ), since for lower probabilities the random graph is a.s. not connected, and Breaker wins even if he does not claim any edges. First we generalize this for arbitrary probability by providing Breaker with a strategy to isolate a vertex. Theorem 6 There exists K0 > 0 so that for arbitrary p ∈ [0, 1] and b ≥ K0 p logn n Breaker, playing the (1 : b) game on the edges of random graph G(n, p), can achieve that Maker’s graph has an isolated vertex a.s. Proof. Let us fix b = ⌊K0 pn/ log n⌋, where K0 is a constant to be determined later. Note that we can assume p > log n/2n, since otherwise the random graph does have an isolated vertex a.s., thus Breaker achieves his goal without making any moves. We present a strategy for Breaker to claim all the edges incident to some vertex of G(n, p). If successful, this strategy of course prevents Maker from building a connected subgraph. Similar strategy was introduced by Chv´atal and Erd˝os [7] for solving the problem on the complete graph. Breaker plays in two stages. First, he claims all edges of G(n, p) within a set C of n/ log n vertices such that Maker at the end of the first stage has not yet claimed any edge with an endpoint in C. Then, in the second stage Breaker claims all the remaining edges of G(n, p) incident to one of the vertices in C thus preventing Maker from connecting that vertex to any of the other vertices. In the first stage Breaker maintains a set Ct of t vertices, such that he claimed all edges of the random graph within Ct and Maker has not claimed an edge with an endpoint in Ct . Breaker will increase the size of this set by 1 in each move, so eventually he ends up with the desired C in n/ log n moves. Before each of his moves he spots two vertices v1 , v2 6∈ Ct for which dCt (v1 )+ dCt (v2 ) is minimized, s.t. no edge incident with either of them has been claimed by Maker. Then, in his (t + 1)st move Breaker claims edge v1 v2 and all edges connecting v1 or v2 with vertices in Ct . That way Ct got two new vertices and Maker in the following move claiming one edge can eliminate at most one vertex out of it. Therefore, after (t + 1)st round Breaker has the appropriate subset Ct+1 of size t + 1. For this strategy to work, Breaker should not be forced to play more then b edges per move. We are going to show that this conflict does not take place a.s. After the tth move of Breaker, t ≤ n/ log n, let St be the set of at most 3n/ log n vertices, containing Ct and the vertices incident to Maker’s edges. We prove that for any set S of size at most 3n/ log n there is vertex a ∈ / S, b−1 such that |NS (a)| ≤ 2 . Obviously it is enough to show this for sets of size 6
3n 3n equal to ⌊ log n ⌋. Let S be a fixed set with |S| = ⌊ log n ⌋. Then ≤ Pr ∀v ∈ / S: dS (v) > b−1 2 b−1 3n ≤ Pr e(S, V − S) ≥ n − log n 2
pn2
Pr[e(S, V − S) ≥ 2|S||V − S|p] ≤ e−3|S||V −S|p/8 ≤ e− log n , provided K0 is large enough. So the probability that there exists S so that for all v ∈ / S we have dS (v) > b−1 2 is at most
n 3n log n
2
pn − log n
e
≤
e log n 3 2
3n log n
2
pn − log n
e
log n pn + 3n log − log n log n
≤ e
1
≤ e−n( 2 −
3 log log n ) log n
→ 0.
n In the last inequality we used that p ≥ log 2n . In the second stage Breaker wants to claim all the edges connecting some vertex v ∈ C to the vertices that are not in C (thus preventing Maker from claiming any edge incident to v). We can model this situation with a simplified game: Breaker wants to occupy one of n/ log n disjoint winning sets, where each of them contains edges incident to a particular vertex of C into V − C. From the construction of C it follows that these sets are untouched by Maker. To prove that Breaker can win this game on disjoint winning sets a.s. we use the following theorem by Chv´atal and Erd˝os.
Theorem 7 [7] In a biased (b: 1) game with k disjoint winning sets of size s first player wins if and only if sk ≤ f (k, b), where f is given recursively by f (1, b) = 0;
f (k, b) = ⌊k(f (k − 1, b) + b)/(k − 1)⌋.
It is not difficult to verify that (b − 1)k
k−1 X 1 i=1
i
≤ f (k, b) ≤ bk
k−1 X 1 i=1
i
.
Our winning sets are the k = n/ log n sets of edges incident to a vertex of C which go to C¯ = V \ C. To apply Theorem 7 it is enough then to show that from above, i.e. for the size dC¯ (v) of each winning set is appropriately P bounded 1 a.s. each v ∈ C we have dC¯ (v) ≤ K40 pn ≤ (b − 1) k−1 i=1 i Indeed, using a Chernoff bound and a large enough K0 , we obtain that for every v ∈ C K0 pn K0 K0 K0 Pr dC¯ (v) > pn ≤ Pr d(v) > pn ≤ e− 4 ≤ n− 8 . 4 4 Therefore we have K0 K0 pn ≤ n · n− 8 → 0, Pr ∃v ∈ C : dC¯ (v) > 4 7
provided K0 is large enough. Then Theorem 7 guarantees Breaker’s win a.s., and the proof of Theorem 6 is complete. 2 Next we give a winning strategy for Maker in the connectivity game, thus determining the threshold bias bpT up to a constant factor. Obviously, Breaker wins if and only if he claims all the edges of a cut, i.e. all the edges connecting some set of vertices with its complement. Thus, in order to win Maker has to claim one edge in each of the cuts. We will consider a different game in which winning sets are cuts and roles are exchanged – Breaker makes and Maker breaks. This new point of view enables us to give Maker a winning strategy using Theorem 5, which is a criterion for Breaker’s win. Observe, that in this “cutgame” Breaker only cares about occupying the existing edges of a cut, that’s why we are going to look at the family Fp∩ instead of Fp . n and b ≤ k0 p logn n Maker Theorem 8 There exists k0 > 0, so that for p > 32 log n wins the random connectivity game (E(Kn )p , Tp , 1, b) a.s.
Proof. For b0 = log2 2 · logn n we are going to prove that the conditions of Theorem 5 are satisfied if F is the set of all cuts in a complete graph with n vertices. First we prove that condition (1) holds. We have n/2 X n k=1
k
−
2
k(n−k) b0
64 logn n and b ≤ m0 p logn n Maker wins the random perfect matching game (E(Kn )p , Mp , 1, b) a.s. Proof. We can show that Maker can win in a slightly harder game. More precisely, if the set of vertices of Kn is partitioned into two sets A and B of equal size before the game starts, we are going to show that Maker can claim a perfect matching with edges going only between A and B. For disjoint sets X, Y ⊂ V (Kn ), we define E(X, Y ) to be the set of edges between X and Y . Let F be a family of sets of edges, F = {E(X, Y ) : ∅ = 6 X ⊂ A, ∅ = 6 Y ⊂ B, |X| + |Y | = 10
n + 1}. 2
Suppose that at the end of the game Maker has not claimed all edges of any perfect matching between A and B. Hall’s necessary and sufficient condition for existence of a perfect matching implies that there exist sets X0 ⊂ A and Y0 ⊂ B such that |X0 | > |Y0 | and all edges in E(Kn )p ∩ E(X0 , B \ Y0 ) were claimed by Breaker. Therefore, in order to win, Maker has to claim at least one edge in each of the sets from F, i.e. he should win as the second player in the game (E(Kn )p , Fp∩ , b, 1). To prove that second player (Maker) wins we are going to use Theorem 5. We set δ = 1/2 and b0 = log4 2 · logn n . First we show that condition (1) holds. We have n/2 X n/2 k=1
k
k(n/2−k+1) n/2 − b0 2 n/2 − k + 1
< 2
⌊n/4⌋
< 2
⌊n/4⌋
2
X
n/2 k
X
e2 log(n/2)−2 log n
X
1 4
k=1
− k(n/2−k+1) b
2
k=1
= 2
⌊n/4⌋ k=1
Since lim min
n→∞ A∈F
k
0
k
< 1.
|A| > lim log n = ∞, n→∞ b0
the condition (2) is also satisfied and we can apply Theorem 5 proving that second player wins the random game (E(Kn )p , Fp∩ , log8 2 p logn n , 1) a.s., provided p > 64 log n/n. This immediately implies that Maker wins (E(Kn )p , Mp , 1, b) a.s. 2 Theorem 6 ensures a win for Breaker in the perfect matching game, if b > K0 pn/ log n. This, together with the above Theorem 11 proves part (ii) of Theorems 1 and 2.
3.4
Clique game
Here we look at the random version of the (1: b) biased clique game (E(Kn ), Kk , 1, b) on a complete graph Kn , where Kk is the set of all cliques of constant size k. Maker’s goal is to occupy all edges of a clique of size k while Breaker wants to prevent that. The deterministic clique game was extensively studied by Bednarska and Luczak in [4]. They proved a more general result by determining the order of the threshold bias for the whole family of games in which Maker’s goal is to claim an arbitrary fixed graph H. In this section, we will largely rely on the constructions and ideas from their paper. If {F1 , . . . , Ft } is a family of k-cliques having two common vertices, and ei ∈ E(Fi ), i = 1, . . . , t are distinct edges, then we call the graph ∪ti=1 Fi a t-2-cluster and the graph ∪ti=1 (Fi − ei ) a t-fan. If furthermore the k-cliques have three vertices in common, then a t-2-cluster is called a t-3-cluster and a
11
t-fan is called a t-flower. A t-fan or a t-2-cluster is said to be simple, if the pairwise intersections (of any two k-cliques) have size exactly 2. In order to prevent Maker to occupy a clique Kk , Breaker will play two auxiliary games. In the first one he prevents Maker from occupying a 3-cluster of constant size. Lemma 12 There exists t = t(k), so that for ε = b > pn
2(1−ε) k+1
1 2(k+2) ,
2
p ≫ n− k+1 and
Breaker wins the game (E(Kn )p , t-3-clusters, 1, b) a.s.
Proof. For l < t k2 , let cl be the number of t-3-clusters C contained in Kn , with e(C) = l, and let Xl be the random variable counting the number of t-3-clusters C contained in G(n, p), with e(C) = l. Using the first moment method we get t k2 −→ 0. Pr[∃l : Xl ≥ E[Xl ] log n] ≤ log n It remains to check the condition of Theorem 3. We a.s. have
X
t(k2 ) X
(1 + b)−e(C) =
X
(1 + b)−l
l=1 t-3-cluster C in G(n, p)
t-3-cluster C in G(n, p)
e(C)=l k 2
≤ log n
t( ) X
E[Xl ] (1 + b)−l
l=1
t(k2)
= log n
X l=1
= log n
cl
p 1+b
l
X
t-3-cluster C in Kn
p 1+b
e(C)
.
To estimate the value of the last sum, we follow the counting arguments from [4]. Namely, for each t-3-cluster C we can order its t cliques F1 , . . . , Ft so that for each i ≤ t1 , V (Fi ) 6⊆ ∪i−1 j=1 V (Fj ), where t1 is as large as possible. If we denote |V (C)| by v, we have log n
X
t-3-cluster C in Kn
tk v X X · v=kt1/k t1 =1
X
H⊆Kk 2 0 so that for p ≥ n− k+1 log6k+12 n and b ≥ 2 C0 pn k+1 Breaker wins the random clique game (E(Kn )p , (Kk )p , 1, b) a.s. Proof. Breaker will use b/2 of his moves to defend “immediate threats”, i.e. to claim the remaining edge in all k-cliques in which Maker occupied all but one edge. In order to be able to do this Maker must ensure that he never has to block more than b/2 immediate threats, that is, there is no dangerous b/2-fan. He will use his other b/2 moves to prevent Maker from creating a dangerous (b/2)-fan. We need the following statement. Lemma 16 [4] For every 0 < ε < 1 there exists b0 so that every graph with b independent sets of b > b0 vertices and at most b2−ε edges has at least 12 bε/3 size bε/3 . From Corollary 13 we get that Breaker can prevent Maker from claiming a 2(1−ε)
1 dangerous f -flower (where f = tpn k+1 , ε = 2(k+2) and t is a positive constant) using less then b/4 edges per move. On the other hand, from Lemma 14 we have that if C0 is large enough Breaker can prevent Maker from claiming 21 b/2 s simple s-fans using b/4 edges per move, where s = (b/2)ε/3 . Suppose that Maker managed to claim a dangerous (b/2)-fan. We define an auxiliary graph G′ with the vertex set being the set of all b/2 k-cliques of this dangerous fan, and two k-cliques being connected with an edge if they have at least 3 vertices in common. Since there is no dangerous f -flower in Maker’s graph, the degree of each of the vertices of the graph G′ is at most f k and 2−ε . On the other hand, number of independent therefore e(G′ ) < bf2k ≤ 2b ′ sets in G of size s cannot be more then 12 b/2 s .
14
Since the last two facts are obviously in contradiction with Lemma 16, Maker cannot claim a dangerous b/2-fan and the statement of the theorem is proved. 2 To prove the theorem for Maker’s win, we need the following result. G(n, M ) denotes the graph obtained by choosing a graph on n vertices with M edges uniformly at random. Lemma 17 [4] There exist δ > 0 and n0 such that for n ≥ n0 and M = 2⌊n2−2/(k+1) ⌋ with probability at least 2/3 each subgraph of G(n, M ) with ⌊(1 − δ)M ⌋ edges contains a copy of Kk . Using the last lemma we can prove a theorem for Maker’s win in the random clique game. 2
2
Theorem 18 There exists c0 > 0 so that for p > c10 n− k+1 and b ≤ c0 pn k+1 Maker wins the random clique game (E(Kn )p , (Kk )p , 1, b) a.s. Proof. We will follow the analysis of the random Maker’s strategy proposed in [4]. In each of his moves Maker chooses one of the edges of G(n, p) that was not previously claimed by him, uniformly at random. If the edge is free he claims it and we call that a successful Maker’s move. If the edge was already claimed by Breaker, then Maker skips his move (e.g. claims an arbitrary free edge, and that edge we will not encounter for the future analysis). Let δ > 0 and n0 be chosen so that the conditions of Lemma 17 are satisfied. We look at the course of game after M = 2⌊n2−2/(k+1) ⌋ moves. Since the probability p is bounded from below it is easy to see that graph G(n, p) has at least 12 p n2 edges a.s. If c0 < δ/24, we have M
≤ ≤
δ ⌊n2−2/(k+1) ⌋ 12c0 δ 1 1 n p . 2b+12 2
That means that a.s. only δ/2 fraction of the total number of elements of the board E(Kn )p is claimed (by both players) after move M . Therefore, the probability that the edge randomly chosen in Maker’s mth move (m < M ) is already claimed by Breaker is bounded from above by δ/2. That means that with probability tending to 1 Maker has at least (1 − δ)M successful moves. Since in each of his moves Maker has chosen edges uniformly at random (without repetition) from E(Kn )p , the graph containing edges chosen by Maker in first M moves in both his successful and unsuccessful moves actually is a random graph G(n, M ). Applying Lemma 17, we get that graph containing edges claimed by Maker in his successful moves contains a clique of size k with positive probability a.s., which means that there exists a non-randomized winning strategy for Maker. 2 Combining the statements of Theorem 15 and Theorem 18 we obtain part (iv) of Theorem 2. 15
4
Unbiased games
4.1
Connectivity one-on-one
A theorem of Lehman enables us to determine the threshold probability pT with extraordinary precision. Namely, Lehman [10] proved that the unbiased connectivity game is won by Maker (now as a second player!) if and only if the underlying graph contains two edge-disjoint spanning trees. The threshold for the appearance of two edge-disjoint spanning trees was determined exactly by Palmer and Spencer [11]. To formulate the consequence of these two results we need the concept of graph process. Let e1 , . . . em be the edges of Kn , where m = n2 . Choose a permutation π ∈ Sm uniformly at random and define an increasing sequence of subgraphs (Gi ) where V (Gi ) = V (Kn ) and E(Gi ) = {eπ(1) , . . . , eπ(i) }. It is clear that Gi is an n-vertex graph with i edges, selected uniformly at random from all n-vertex graphs with i edges. Given a particular graph process (Gi ) and a graph property P possessed by Kn , the hitting time τ (P) = τ (P, (Gi )) is the minimal i for which Gi has property P. The consequence of the theorems of Lehman, and Palmer and Spencer is that the very moment the last vertex receives its second adjacent edge, the unbiased connectivity game is won by Maker a.s. More precisely, the following is true. Corollary 19 For the unbiased connectivity game we have that a.s. τ (Maker wins T ) = τ (∃ two edge-disjoint spanning trees) = τ (δ(G) ≥ 2). In particular, for edge-probability p = (log n + log log n + ω(n))/n, where ω(n) tends to infinity arbitrarily slowly, Maker wins the unbiased connectivity game a.s., while if ω(n) → −∞, then Breaker wins a.s.
Remark. The assumption that Maker is the second player is just technical, for the sake of smooth applicability of Lehman’s Theorem. If Maker is the first player, then from the proof of Lehman’s Theorem one can infer that Maker wins if and only if the base graph contains a spanning tree and a spanning forest of two components, which are edge-disjoint. This property has the same sharp threshold as the presence of two edge-disjoint spanning trees, and the hitting time should be the same when the next to last vertex receives its second incident edge.
4.2
k-cliques one-on-one
Let us fix k and let (F1 , . . . , Fs ) be a sequence of k-cliques. Then F = ∪si=1 Fi is called an s-bunch if V (Fi ) \ (∪i−1 j=1 V (Fj )) 6= ∅ and |V (Fi ) ∩ (∪j v(A1 ) ≥ 2(2)−1 , implying v(A ) ≥ k2k/2 /5. 1
k
1
k
We take an arbitrary k-clique F1 from A1 , and build a large bunch recursively as follows. If we picked k-cliques F1 , . . . , Fi then we choose Fi+1 such that |V (Fi+1 ) ∩ (∪ij=1 V (Fj ))| ≥ 2 and V (Fi+1 ) \ (∪ij=1 V (Fj )) 6= ∅. Note that this means that ∪i+1 j=1 Fj is an (i + 1)-bunch. Since the auxiliary graph GA1 of 0 the collection is connected we can keep doing this until V (A1 ) = ∪ij=1 V (Fj ) for some i0 ≤ c1 . Since v(Fi ) = k for all i ≤ i0 , we have that k
22 v(A1 ) > . i0 ≥ k 5 So there exists a large i0 -bunch which is a subgraph of E(Kn )p . −
1
k This does not happen a.s. for p = O(n a(k) ), where a(k) = k+1 2 − (k−2)2k/2 /5+2 < m(Ci0 ), since we have that the first i0 -bunch that appears in the random graph is the one of the minimum maximum density, which, by the last lemma, is the simple i0 -2-cluster. Note here that there is a constant (depending on k) number of nonisomorphic (2k/2 /5)-bunches. Let us now deal with the case k = 3. We again consider the partition of the edge set E(Kn )p into the set N of edges which do not participate in a triangle, and the edgesets of the collections Ai . Breaker will play a game on each E(Ai ) separately and, if successful on each of them, win the whole game. If there is a collection of triangles on v vertices, then we can extract a (v − 2)-bunch from it as described previously. By the last two lemmas the first (v − 2)-bunch to appear is the simple (v − 2)-2-cluster with density 2 − v3 . For p ≪ n−2/3 , no (v − 2)-bunch will appear with v ≥ 6. That means there are no collections on more than 5 vertices, in particular there are only a constant number of different collections. Another observation is that if the board Ai on which Breaker plays is a 2-degenerate graph, then an easy pairing strategy will provide Breaker
19
with a win. Take the ordering v1 , . . . , vk of V (Ai ), such that |NVj (vj+1 )| ≤ 2 for j = 1, . . . , k − 1, where Vj = {v1 , . . . , vj }. Then Breaker’s strategy is the following: if Maker takes one of the two edges connecting vj+1 to Vj , then Breaker takes the other one. If there is only one such edge, then Breaker takes an arbitrary edge. Suppose for a contradiction that Maker managed to occupy a triangle vi vj vk against this strategy. This is impossible, since if i, j < k, then at the point when Maker claimed one of the edges vi vk and vj vk , Breaker would have taken the other one. So Breaker can win on any component which is a 2-degenerate graph. Now we just note that since p ≪ n−2/3 every component of finite order (and we showed that we only have order at most 5) does not contain a subgraph with minimum degree at least 3. 2 On the other hand, from Theorem 18 we get that Maker can win the game 2 (E(Kn )p , (Kk )p , 1, 1) for p = Θ(n− k+1 ) and thus we immediately get pKk = 2
O(n− k+1 ). For the triangle game K3 a stronger upper bound can be found. 5
Proposition 23 For p = ω(n− 9 ), Maker can win in the game (E(Kn )p , (K3 )p , 1, 1) a.s. Proof. It is easy to check that Maker can claim a triangle in the (1: 1) game if the board on which the game is played is K5 minus an edge. Therefore, as soon as the graph G(n, p) contains K5 − e a.s., the initial game can be won by Maker a.s. 2 Theorem 18, Proposition 22 and Proposition 23 together imply parts (iv) and (v) of Theorem 1.
5
Open questions
More sharp thresholds? We saw in the previous section that the connectivity game has a sharp threshold, and even more. We think that both the perfect matching game and the Hamiltonian cycle game have the same sharp threshold log n n , and maybe even more. . . It would be very interesting to decide whether the following conjectures are true. Conjecture 2 (i)
τ (Maker wins M) = τ (δ(G) ≥ 2),
(ii) τ (Maker wins H) = τ (δ(G) ≥ 4).
Clique game/H-game. The determination of the threshold pK3 for the triangle game remains outstanding. Note, however, that an approach similar to the one proving pK3 ≥ n−2/3 makes this a finite problem. Indeed, since the maximum density of an s-bunch of triangles for s > 12 is at least 9/5, we can conclude that in the random graph with edge probability o(n−5/9 ) a.s. there is no triangle collection on more then 14 vertices. Therefore, to check whether Breaker wins the triangle game in this setting it is enough to check whether he 20
can win on all triangle collections on at most 14 vertices. We conjecture that the answer is “yes” and our upper bound is optimal, i.e. pK3 = n−5/9 . We are convinced that the arguments of Bednarska and Luczak [4] could be extended to full generality to positional games on random graphs along the lines of Section 3.4. More precisely, the following should be true. Let KH be the family of subgraphs of Kn , isomorphic to H. Then for any fixed graph H there is a constant c(H), such that ′ bpKH = Θ (pbKH ) = Θ pn−1/m (H) , c(H) n . provided p ≥ Ω nlog 1/m′ (H) Concerning the one-on-one game, it would be desirable to extend the counterexample given in Theorem 1 part (v) for arbitrary H, i.e. we believe the following is true.
Conjecture 3 There exists a constant ǫ(H) > 0, such that the unbiased KH is ′ a.s. a Maker’s win if p = n−1/m (H)−ǫ(H) . Should the conjecture be true, it would make the determination of the threshold pKH a finite problem, in a way similar to the case H = K3 . Relationships between thresholds. It is an intriguing task to understand under what circumstances the following is true. Problem 1 Characterize those games (X, F) for which pF =
1 . bF
More generally, characterize the games for which bpF = Θ (pbF ) , for every p ≥ Ω
1 bF
.
This is not true in general as the triangle game shows. What is the reason it is true for the connectivity game and the perfect matching game? Is it because the appearance of these properties has a sharp threshold in G(n, p)? Or because the winning sets are not of constant size? Problem 2 Suppose pF = 1/bF . Is it true that for every p ≥ pF , bpF = Θ (pbF )? It would be very interesting to relate the thresholds bF and pF to some thresholds of the family F in the random graph G(n, p) (or, more generally, in the random set Xp ). It seems to us that if the family Fp is quite dense and well-distributed in X, then Maker still wins the (1: 1) game. Problem 3 Characterize those games (X, F) for which there exists a constant K, such that for any probability p with Pr minx∈Xp |{F ∈ Fp : x ∈ F }| > K −→ 1, we have pF = O(p) and/or bF = Ω(1/p). 21
References [1] J. Beck: Random graphs and positional games on the complete graph, Ann. Discrete Math. 28(1985), 7–13. [2] J. Beck: Remarks on positional games, Acta Math. Acad. Sci. Hungar. 40(1982), 65–71. [3] J. Beck: Positional games and the second moment method, Combinatorica 22(2002), 169–216. [4] M. Bednarska, T. Luczak: Biased positional games for which random strategies are nearly optimal, Combinatorica 20(2000), 477–488. [5] M. Bednarska, T. Luczak: Biased positional games and the phase transition, Random Structures & Algorithms 18(2001), 141–152. [6] B. Bollob´ as, A. Thomason: Threshold functions, Combinatorica 7(1987), 35–38. [7] V. Chv´atal, P. Erd˝os: Biased positional games, Annals of Discrete Math. 2(1978), 221–228. [8] V. Chv´atal, P. Erd˝os: A note on Hamiltonian circuits, Discrete Math. 2(1972), 111–113. [9] P. Erd˝os, J. Selfridge: On a combinatorial game, J. Combinatorial Theory 14(1973), 298–301. [10] A. Lehman, A solution of the Shannon switching game, J. Soc. Indust. Appl. Math. 12(1964) 687–725. [11] E.M. Palmer, J.J. Spencer, Hitting time for k edge-disjoint spanning trees in a random graph. Period. Math. Hungar. 31(1995), 235–240.
22
∗†
Tibor Szab´o∗
December 8, 2003
Abstract We introduce and study Maker/Breaker-type positional games on random graphs. Our main concern is to determine the threshold probability pF for the existence of Maker’s strategy to claim a member of F in the unbiased game played on the edges of random graph G(n, p), for various target families F of winning sets. More generally, for each probability above this threshold we study the smallest bias b such that Maker wins the (1: b) biased game. We investigate these functions for a number of basic games, like the connectivity game, the perfect matching game, the clique game and the Hamiltonian cycle game.
1
Introduction
(Un)biased positional games. Let X be a finite nonempty set and F ⊆ 2X . The pair (X, F) is a positional game on X. The game is played by two players Maker and Breaker, where in each move Maker claims one previously unclaimed element of X and then Breaker claims one previously unclaimed element of X. Maker wins if he claims all the elements of some set in F, otherwise Breaker wins. Unless otherwise stated, we assume that Maker starts the game. Note, however, that the asymptotic statements discussed are not influenced by which player makes the first move. The set X will be referred to as the board, and the set F as the set of winning sets. Whenever there is no confusion about what the board is, we may refer to the game (X, F) as just F. The set of all positional games could be partitioned into two classes. The game (X, F) is called a Maker’s win if Maker has a winning strategy, that is, playing against an arbitrary opponent Maker can occupy a member of F. Clearly, if (X, F) is not a Maker’s win, then Breaker is able to prevent an arbitrary opponent from occupying a winning set. These games are called Breaker’s win. Typical, well-studied examples of such positional games are played on the edges of a complete graph, i.e. X = E(Kn ). Maker’s goal usually is to build a ∗ Institute of Theoretical Computer Science, ETH Zurich, CH-8092 Switzerland. Email addresses: {smilos, szabo}@inf.ethz.ch † Supported by the joint Berlin/Zurich graduate program Combinatorics, Geometry and Computation, financed by ETH Zurich and the German Science Foundation (DFG).
1
graph theoretic structure–like a spanning tree, a perfect matching, a Hamiltonian cycle, or a clique of fixed size. It turns out that all these games are won easily by Maker, so in order to make things more fair (if such thing exists; actually no game of perfect information is fair as the winner–in theory–is known in the beginning of the game) one could give Breaker extra power by allowing him to claim more than 1 edge in each move. If X is a finite nonempty set, F ⊆ 2X and a, b are positive integers (possibly functions of the board size), then the 4-tuple (X, F, a, b) is a biased (a: b) game. In a biased (a: b) game, Maker claims a elements (instead of 1) and Breaker claims b elements (instead of 1) in each move. For a family F the smallest integer bF is sought (and sometimes found; see [1, 2, 3, 4, 5, 7]) for which Breaker wins the (1 : bF ) game. In the connectivity game Maker’s goal is to build a connected spanning subgraph; i.e. in this game the family of winning sets is the family T = Tn of all spanning trees on n vertices. Chv´atal and Erd˝os proved [7] that bT = Θ( logn n ). Beck [1] established bH = Θ( logn n ), where H = Hn is the family of all Hamiltonian cycles on n vertices. For the family Kk = Kk,n of all k-cliques on n vertices, Bednarska and 2
Luczak [4] showed that bKk = Θ(n k+1 ). More generally, they proved that in the game in which Maker’s goal is to claim an arbitrary fixed graph G, the ′ e(H)−1 over all threshold bias is Θ(n1/m (G) ). (Here m′ (G) is the maximum of v(H)−2 subgraphs H of G with at least 3 vertices.) Playing on a random board. In the present paper we introduce another approach to even out the advantage Maker has in a (1: 1) game, by randomly reducing the board size and keeping only those winning sets, which survive this thinning intact. Definition 1 Let (X, F, a, b) be a biased game. Random game (Xp , Fp , a, b) is a probability space of games where each x ∈ X is independently included in Xp with probability p, and Fp = {W ∈ F : W ⊆ Xp }.
Only if p = 0 or 1, or if ∅ ∈ F or F = ∅, we can be sure who (Maker or Breaker) wins the random game Fp . In all other cases the best we can conclude is that Maker (or Breaker) wins a.s. (almost surely), i.e. the probability that Maker (Breaker) wins tends to 1 if the board size tends to infinity. (So we actually talk about an infinite family of probability spaces of games. . . ) Excluding the trivial case ∅ ∈ F, Breaker always wins when p = 0. The graph games we consider are (easy) Maker’s wins, when p = 1. For a particular sequence of games (X, F), the first natural question to ask is: What is the threshold probability pF at which an almost sure Breaker’s win turns into an almost sure Maker’s win. More precisely we would like to determine pF for which • Pr[(Xp , Fp , 1, 1) is a Breaker’s win] → 1 for p ≪ pF , and • Pr[(Xp , Fp , 1, 1) is a Maker’s win] → 1 for p ≫ pF . 2
Such a threshold pF exists [6], since being a Maker’s win is an increasing property. The main goal of this paper is to establish a connection between the natural threshold values, bF and pF , corresponding to the two different weakenings of Maker’s power: bias and random thinning, respectively. We find that there is an intriguing reciprocal connection between these two thresholds in a number of well-studied games on graphs. Recall the notations T , H, and Kk , and let us denote by M the set of all perfect matchings on the graph Kn . Theorem 1 For positional games, played on E(Kn ), we have (i) pT = (ii) pM = (iii)
log n n
(iv) n
log n n ,
≤ pH ≤
1 − a(k)
2
log n n ,
log √ n, n 2
≤ pKk ≤ n− k+1 , where a(k) =
k+1 2
−
10 , 2k/2
5
(v) n− 3 ≤ pK3 ≤ n− 9 . For the connectivity game T an even more precise statement is true. In Corollary 19 we observe that Maker starts to win a.s. at the very moment when the last vertex of a random graph process picks up its second incident edge. More generally, for every p we would like to find the smallest bias bpF s.t. Breaker wins the random game (Xp , Fp , 1, bpF ) a.s. Note that by definition bF = b1F . Another trivial observation is that bpF = 0 provided p is less than the threshold for the appearance of the first element of F in the random graph. We obtain the following. Theorem 2 (i) bpT = Θ (pbT ) = Θ p logn n , provided p = Ω b1T , 1 (ii) bpM = Θ (pbM ) = Θ p logn n , provided p = Ω bM ,
√ √n , (iii) Ω p lognn ≤ bpH ≤ O p logn n , provided p = Ω log n c 2 k (iv) bpKk = Θ (pbKk ) = Θ pn k+1 , provided p = Ω logbK n . k
One can see that bpF is of order p/pF = pbF for the connectivity game and the perfect matching game, provided p ≥ pF . In particular for these games pF = Θ(1/bF ). In part (iv) of Theorem 2, generalizing the arguments of Bednarska and Luczak [4] we show that one can estimate bpKk up to a constant 3
factor, for all probabilities down to a polylogarithmic factor away from 2 critical probability 1/bKk = n− k+1 . On the other hand Theorem 1 part shows that we cannot get arbitrarily close to probability 1/bKk , since in triangle game Maker can win even for probabilities below 1/bK3 = n−1/2 . Nevertheless we think the Hamiltonian cycle game behaves “nicely”, i.e. same way as the connectivity game and the perfect matching game.
the (v) the the
Conjecture 1 Let H be the set of Hamiltonian cycles in Kn . n p , provided p = Ω logn n . bH = Θ p log n In particular,
log n . n Observe that the validity of the conjecture would mean that in a random graph with edge probability p ≫ logn n Maker could build a Hamiltonian cycle. So P´osa’s Theorem (which only proves the existence of a Hamiltonian cycle) would be true constructively even if an adversary is playing against us. The paper is organized as follows. In Section 2 we prove a general criterion for Breaker’s win in a different, auxiliary random game. In Section 3, the analysis of four biased random games is presented. In particular, in Subsections 3.1, 3.2, 3.3 and 3.4 we look at the connectivity game, the Hamiltonian cycle game, the perfect matching game and the clique game, respectively. In Section 4 we analyze more precisely a couple of (1: 1) games – the connectivity game (Subsection 4.1) and the clique game (Subsection 4.2). Finally, in Section 5 we give a collection of open questions and conjectures. pH =
Notation. For a graph G, e(G) and v(G) denote the number of edges and vertices (respectively) of G, and E(G) and V (G) denote the sets of edges and vertices (respectively). If C ⊆ V (G) and v ∈ V (G), then NC (v) denotes the set of neighbors of v in C. The logarithm log n in this paper is always of natural base.
2
A criterion
One of few general, but still very applicable results to decide the winner of biased positional games is the biased version of the Erd˝os–Selfridge Theorem [9, 2]. It provides a criterion for Breaker’s win, applicable on any game. Theorem 3 (Beck, [2]) If X
(1 + b)−|A|/a
0 are chosen so that p > δ)pb, 1) is Breaker’s win a.s.
4 log 2 δ2 b
(2)
holds, then the game (Xp , Fp∩ , (1−
Proof. For each A ∈ F and its corresponding set A′ ∈ Fp∩ we have E[|A′ |] = p|A|. If all winning sets A′ ∈ Fp∩ have size at least (1 − δ)p|A|, then X
′
|A | − (1−δ)pb
2
A′ ∈Fp∩
≤
X
− (1−δ)p|A| (1−δ)pb
2
=
A∈F
X
2−
|A| b
< 1.
A∈F
Using the Erd˝os–Selfridge theorem we obtain that Breaker wins the (Xp , Fp∩ , (1− δ)pb, 1) game, provided |A′ | ≥ (1 − δ)p|A| for all A′ ∈ Fp∩ . Next we check that this condition holds almost surely. Using a Chernoff bound, we obtain that X − δ2 p|A| 2 . Pr ∃A ∈ F : |A′ | ≤ (1 − δ)p|A| ≤ e A∈F
If we denote minA∈F X
A∈F
e−
δ 2 p|A| 2
≤
|A| b
X
A∈F
by mn , then we have 2−2
|A| b
≤
X
A∈F
2−mn 2−
|A| b
< 2−mn → 0,
and therefore all winning sets A′ ∈ Fp∩ have size at least (1 − δ)p|A| a.s. 5
2
3 3.1
Games Connectivity game
The first game we study is a random version of the biased connectivity game (E(Kn ), T , 1, b) on a complete graph on n vertices Kn . Maker’s goal is to build a spanning, connected subgraph, i.e. T is the set of all spanning trees on n vertices. It is obvious that pT = Ω( logn n ), since for lower probabilities the random graph is a.s. not connected, and Breaker wins even if he does not claim any edges. First we generalize this for arbitrary probability by providing Breaker with a strategy to isolate a vertex. Theorem 6 There exists K0 > 0 so that for arbitrary p ∈ [0, 1] and b ≥ K0 p logn n Breaker, playing the (1 : b) game on the edges of random graph G(n, p), can achieve that Maker’s graph has an isolated vertex a.s. Proof. Let us fix b = ⌊K0 pn/ log n⌋, where K0 is a constant to be determined later. Note that we can assume p > log n/2n, since otherwise the random graph does have an isolated vertex a.s., thus Breaker achieves his goal without making any moves. We present a strategy for Breaker to claim all the edges incident to some vertex of G(n, p). If successful, this strategy of course prevents Maker from building a connected subgraph. Similar strategy was introduced by Chv´atal and Erd˝os [7] for solving the problem on the complete graph. Breaker plays in two stages. First, he claims all edges of G(n, p) within a set C of n/ log n vertices such that Maker at the end of the first stage has not yet claimed any edge with an endpoint in C. Then, in the second stage Breaker claims all the remaining edges of G(n, p) incident to one of the vertices in C thus preventing Maker from connecting that vertex to any of the other vertices. In the first stage Breaker maintains a set Ct of t vertices, such that he claimed all edges of the random graph within Ct and Maker has not claimed an edge with an endpoint in Ct . Breaker will increase the size of this set by 1 in each move, so eventually he ends up with the desired C in n/ log n moves. Before each of his moves he spots two vertices v1 , v2 6∈ Ct for which dCt (v1 )+ dCt (v2 ) is minimized, s.t. no edge incident with either of them has been claimed by Maker. Then, in his (t + 1)st move Breaker claims edge v1 v2 and all edges connecting v1 or v2 with vertices in Ct . That way Ct got two new vertices and Maker in the following move claiming one edge can eliminate at most one vertex out of it. Therefore, after (t + 1)st round Breaker has the appropriate subset Ct+1 of size t + 1. For this strategy to work, Breaker should not be forced to play more then b edges per move. We are going to show that this conflict does not take place a.s. After the tth move of Breaker, t ≤ n/ log n, let St be the set of at most 3n/ log n vertices, containing Ct and the vertices incident to Maker’s edges. We prove that for any set S of size at most 3n/ log n there is vertex a ∈ / S, b−1 such that |NS (a)| ≤ 2 . Obviously it is enough to show this for sets of size 6
3n 3n equal to ⌊ log n ⌋. Let S be a fixed set with |S| = ⌊ log n ⌋. Then ≤ Pr ∀v ∈ / S: dS (v) > b−1 2 b−1 3n ≤ Pr e(S, V − S) ≥ n − log n 2
pn2
Pr[e(S, V − S) ≥ 2|S||V − S|p] ≤ e−3|S||V −S|p/8 ≤ e− log n , provided K0 is large enough. So the probability that there exists S so that for all v ∈ / S we have dS (v) > b−1 2 is at most
n 3n log n
2
pn − log n
e
≤
e log n 3 2
3n log n
2
pn − log n
e
log n pn + 3n log − log n log n
≤ e
1
≤ e−n( 2 −
3 log log n ) log n
→ 0.
n In the last inequality we used that p ≥ log 2n . In the second stage Breaker wants to claim all the edges connecting some vertex v ∈ C to the vertices that are not in C (thus preventing Maker from claiming any edge incident to v). We can model this situation with a simplified game: Breaker wants to occupy one of n/ log n disjoint winning sets, where each of them contains edges incident to a particular vertex of C into V − C. From the construction of C it follows that these sets are untouched by Maker. To prove that Breaker can win this game on disjoint winning sets a.s. we use the following theorem by Chv´atal and Erd˝os.
Theorem 7 [7] In a biased (b: 1) game with k disjoint winning sets of size s first player wins if and only if sk ≤ f (k, b), where f is given recursively by f (1, b) = 0;
f (k, b) = ⌊k(f (k − 1, b) + b)/(k − 1)⌋.
It is not difficult to verify that (b − 1)k
k−1 X 1 i=1
i
≤ f (k, b) ≤ bk
k−1 X 1 i=1
i
.
Our winning sets are the k = n/ log n sets of edges incident to a vertex of C which go to C¯ = V \ C. To apply Theorem 7 it is enough then to show that from above, i.e. for the size dC¯ (v) of each winning set is appropriately P bounded 1 a.s. each v ∈ C we have dC¯ (v) ≤ K40 pn ≤ (b − 1) k−1 i=1 i Indeed, using a Chernoff bound and a large enough K0 , we obtain that for every v ∈ C K0 pn K0 K0 K0 Pr dC¯ (v) > pn ≤ Pr d(v) > pn ≤ e− 4 ≤ n− 8 . 4 4 Therefore we have K0 K0 pn ≤ n · n− 8 → 0, Pr ∃v ∈ C : dC¯ (v) > 4 7
provided K0 is large enough. Then Theorem 7 guarantees Breaker’s win a.s., and the proof of Theorem 6 is complete. 2 Next we give a winning strategy for Maker in the connectivity game, thus determining the threshold bias bpT up to a constant factor. Obviously, Breaker wins if and only if he claims all the edges of a cut, i.e. all the edges connecting some set of vertices with its complement. Thus, in order to win Maker has to claim one edge in each of the cuts. We will consider a different game in which winning sets are cuts and roles are exchanged – Breaker makes and Maker breaks. This new point of view enables us to give Maker a winning strategy using Theorem 5, which is a criterion for Breaker’s win. Observe, that in this “cutgame” Breaker only cares about occupying the existing edges of a cut, that’s why we are going to look at the family Fp∩ instead of Fp . n and b ≤ k0 p logn n Maker Theorem 8 There exists k0 > 0, so that for p > 32 log n wins the random connectivity game (E(Kn )p , Tp , 1, b) a.s.
Proof. For b0 = log2 2 · logn n we are going to prove that the conditions of Theorem 5 are satisfied if F is the set of all cuts in a complete graph with n vertices. First we prove that condition (1) holds. We have n/2 X n k=1
k
−
2
k(n−k) b0
64 logn n and b ≤ m0 p logn n Maker wins the random perfect matching game (E(Kn )p , Mp , 1, b) a.s. Proof. We can show that Maker can win in a slightly harder game. More precisely, if the set of vertices of Kn is partitioned into two sets A and B of equal size before the game starts, we are going to show that Maker can claim a perfect matching with edges going only between A and B. For disjoint sets X, Y ⊂ V (Kn ), we define E(X, Y ) to be the set of edges between X and Y . Let F be a family of sets of edges, F = {E(X, Y ) : ∅ = 6 X ⊂ A, ∅ = 6 Y ⊂ B, |X| + |Y | = 10
n + 1}. 2
Suppose that at the end of the game Maker has not claimed all edges of any perfect matching between A and B. Hall’s necessary and sufficient condition for existence of a perfect matching implies that there exist sets X0 ⊂ A and Y0 ⊂ B such that |X0 | > |Y0 | and all edges in E(Kn )p ∩ E(X0 , B \ Y0 ) were claimed by Breaker. Therefore, in order to win, Maker has to claim at least one edge in each of the sets from F, i.e. he should win as the second player in the game (E(Kn )p , Fp∩ , b, 1). To prove that second player (Maker) wins we are going to use Theorem 5. We set δ = 1/2 and b0 = log4 2 · logn n . First we show that condition (1) holds. We have n/2 X n/2 k=1
k
k(n/2−k+1) n/2 − b0 2 n/2 − k + 1
< 2
⌊n/4⌋
< 2
⌊n/4⌋
2
X
n/2 k
X
e2 log(n/2)−2 log n
X
1 4
k=1
− k(n/2−k+1) b
2
k=1
= 2
⌊n/4⌋ k=1
Since lim min
n→∞ A∈F
k
0
k
< 1.
|A| > lim log n = ∞, n→∞ b0
the condition (2) is also satisfied and we can apply Theorem 5 proving that second player wins the random game (E(Kn )p , Fp∩ , log8 2 p logn n , 1) a.s., provided p > 64 log n/n. This immediately implies that Maker wins (E(Kn )p , Mp , 1, b) a.s. 2 Theorem 6 ensures a win for Breaker in the perfect matching game, if b > K0 pn/ log n. This, together with the above Theorem 11 proves part (ii) of Theorems 1 and 2.
3.4
Clique game
Here we look at the random version of the (1: b) biased clique game (E(Kn ), Kk , 1, b) on a complete graph Kn , where Kk is the set of all cliques of constant size k. Maker’s goal is to occupy all edges of a clique of size k while Breaker wants to prevent that. The deterministic clique game was extensively studied by Bednarska and Luczak in [4]. They proved a more general result by determining the order of the threshold bias for the whole family of games in which Maker’s goal is to claim an arbitrary fixed graph H. In this section, we will largely rely on the constructions and ideas from their paper. If {F1 , . . . , Ft } is a family of k-cliques having two common vertices, and ei ∈ E(Fi ), i = 1, . . . , t are distinct edges, then we call the graph ∪ti=1 Fi a t-2-cluster and the graph ∪ti=1 (Fi − ei ) a t-fan. If furthermore the k-cliques have three vertices in common, then a t-2-cluster is called a t-3-cluster and a
11
t-fan is called a t-flower. A t-fan or a t-2-cluster is said to be simple, if the pairwise intersections (of any two k-cliques) have size exactly 2. In order to prevent Maker to occupy a clique Kk , Breaker will play two auxiliary games. In the first one he prevents Maker from occupying a 3-cluster of constant size. Lemma 12 There exists t = t(k), so that for ε = b > pn
2(1−ε) k+1
1 2(k+2) ,
2
p ≫ n− k+1 and
Breaker wins the game (E(Kn )p , t-3-clusters, 1, b) a.s.
Proof. For l < t k2 , let cl be the number of t-3-clusters C contained in Kn , with e(C) = l, and let Xl be the random variable counting the number of t-3-clusters C contained in G(n, p), with e(C) = l. Using the first moment method we get t k2 −→ 0. Pr[∃l : Xl ≥ E[Xl ] log n] ≤ log n It remains to check the condition of Theorem 3. We a.s. have
X
t(k2 ) X
(1 + b)−e(C) =
X
(1 + b)−l
l=1 t-3-cluster C in G(n, p)
t-3-cluster C in G(n, p)
e(C)=l k 2
≤ log n
t( ) X
E[Xl ] (1 + b)−l
l=1
t(k2)
= log n
X l=1
= log n
cl
p 1+b
l
X
t-3-cluster C in Kn
p 1+b
e(C)
.
To estimate the value of the last sum, we follow the counting arguments from [4]. Namely, for each t-3-cluster C we can order its t cliques F1 , . . . , Ft so that for each i ≤ t1 , V (Fi ) 6⊆ ∪i−1 j=1 V (Fj ), where t1 is as large as possible. If we denote |V (C)| by v, we have log n
X
t-3-cluster C in Kn
tk v X X · v=kt1/k t1 =1
X
H⊆Kk 2 0 so that for p ≥ n− k+1 log6k+12 n and b ≥ 2 C0 pn k+1 Breaker wins the random clique game (E(Kn )p , (Kk )p , 1, b) a.s. Proof. Breaker will use b/2 of his moves to defend “immediate threats”, i.e. to claim the remaining edge in all k-cliques in which Maker occupied all but one edge. In order to be able to do this Maker must ensure that he never has to block more than b/2 immediate threats, that is, there is no dangerous b/2-fan. He will use his other b/2 moves to prevent Maker from creating a dangerous (b/2)-fan. We need the following statement. Lemma 16 [4] For every 0 < ε < 1 there exists b0 so that every graph with b independent sets of b > b0 vertices and at most b2−ε edges has at least 12 bε/3 size bε/3 . From Corollary 13 we get that Breaker can prevent Maker from claiming a 2(1−ε)
1 dangerous f -flower (where f = tpn k+1 , ε = 2(k+2) and t is a positive constant) using less then b/4 edges per move. On the other hand, from Lemma 14 we have that if C0 is large enough Breaker can prevent Maker from claiming 21 b/2 s simple s-fans using b/4 edges per move, where s = (b/2)ε/3 . Suppose that Maker managed to claim a dangerous (b/2)-fan. We define an auxiliary graph G′ with the vertex set being the set of all b/2 k-cliques of this dangerous fan, and two k-cliques being connected with an edge if they have at least 3 vertices in common. Since there is no dangerous f -flower in Maker’s graph, the degree of each of the vertices of the graph G′ is at most f k and 2−ε . On the other hand, number of independent therefore e(G′ ) < bf2k ≤ 2b ′ sets in G of size s cannot be more then 12 b/2 s .
14
Since the last two facts are obviously in contradiction with Lemma 16, Maker cannot claim a dangerous b/2-fan and the statement of the theorem is proved. 2 To prove the theorem for Maker’s win, we need the following result. G(n, M ) denotes the graph obtained by choosing a graph on n vertices with M edges uniformly at random. Lemma 17 [4] There exist δ > 0 and n0 such that for n ≥ n0 and M = 2⌊n2−2/(k+1) ⌋ with probability at least 2/3 each subgraph of G(n, M ) with ⌊(1 − δ)M ⌋ edges contains a copy of Kk . Using the last lemma we can prove a theorem for Maker’s win in the random clique game. 2
2
Theorem 18 There exists c0 > 0 so that for p > c10 n− k+1 and b ≤ c0 pn k+1 Maker wins the random clique game (E(Kn )p , (Kk )p , 1, b) a.s. Proof. We will follow the analysis of the random Maker’s strategy proposed in [4]. In each of his moves Maker chooses one of the edges of G(n, p) that was not previously claimed by him, uniformly at random. If the edge is free he claims it and we call that a successful Maker’s move. If the edge was already claimed by Breaker, then Maker skips his move (e.g. claims an arbitrary free edge, and that edge we will not encounter for the future analysis). Let δ > 0 and n0 be chosen so that the conditions of Lemma 17 are satisfied. We look at the course of game after M = 2⌊n2−2/(k+1) ⌋ moves. Since the probability p is bounded from below it is easy to see that graph G(n, p) has at least 12 p n2 edges a.s. If c0 < δ/24, we have M
≤ ≤
δ ⌊n2−2/(k+1) ⌋ 12c0 δ 1 1 n p . 2b+12 2
That means that a.s. only δ/2 fraction of the total number of elements of the board E(Kn )p is claimed (by both players) after move M . Therefore, the probability that the edge randomly chosen in Maker’s mth move (m < M ) is already claimed by Breaker is bounded from above by δ/2. That means that with probability tending to 1 Maker has at least (1 − δ)M successful moves. Since in each of his moves Maker has chosen edges uniformly at random (without repetition) from E(Kn )p , the graph containing edges chosen by Maker in first M moves in both his successful and unsuccessful moves actually is a random graph G(n, M ). Applying Lemma 17, we get that graph containing edges claimed by Maker in his successful moves contains a clique of size k with positive probability a.s., which means that there exists a non-randomized winning strategy for Maker. 2 Combining the statements of Theorem 15 and Theorem 18 we obtain part (iv) of Theorem 2. 15
4
Unbiased games
4.1
Connectivity one-on-one
A theorem of Lehman enables us to determine the threshold probability pT with extraordinary precision. Namely, Lehman [10] proved that the unbiased connectivity game is won by Maker (now as a second player!) if and only if the underlying graph contains two edge-disjoint spanning trees. The threshold for the appearance of two edge-disjoint spanning trees was determined exactly by Palmer and Spencer [11]. To formulate the consequence of these two results we need the concept of graph process. Let e1 , . . . em be the edges of Kn , where m = n2 . Choose a permutation π ∈ Sm uniformly at random and define an increasing sequence of subgraphs (Gi ) where V (Gi ) = V (Kn ) and E(Gi ) = {eπ(1) , . . . , eπ(i) }. It is clear that Gi is an n-vertex graph with i edges, selected uniformly at random from all n-vertex graphs with i edges. Given a particular graph process (Gi ) and a graph property P possessed by Kn , the hitting time τ (P) = τ (P, (Gi )) is the minimal i for which Gi has property P. The consequence of the theorems of Lehman, and Palmer and Spencer is that the very moment the last vertex receives its second adjacent edge, the unbiased connectivity game is won by Maker a.s. More precisely, the following is true. Corollary 19 For the unbiased connectivity game we have that a.s. τ (Maker wins T ) = τ (∃ two edge-disjoint spanning trees) = τ (δ(G) ≥ 2). In particular, for edge-probability p = (log n + log log n + ω(n))/n, where ω(n) tends to infinity arbitrarily slowly, Maker wins the unbiased connectivity game a.s., while if ω(n) → −∞, then Breaker wins a.s.
Remark. The assumption that Maker is the second player is just technical, for the sake of smooth applicability of Lehman’s Theorem. If Maker is the first player, then from the proof of Lehman’s Theorem one can infer that Maker wins if and only if the base graph contains a spanning tree and a spanning forest of two components, which are edge-disjoint. This property has the same sharp threshold as the presence of two edge-disjoint spanning trees, and the hitting time should be the same when the next to last vertex receives its second incident edge.
4.2
k-cliques one-on-one
Let us fix k and let (F1 , . . . , Fs ) be a sequence of k-cliques. Then F = ∪si=1 Fi is called an s-bunch if V (Fi ) \ (∪i−1 j=1 V (Fj )) 6= ∅ and |V (Fi ) ∩ (∪j v(A1 ) ≥ 2(2)−1 , implying v(A ) ≥ k2k/2 /5. 1
k
1
k
We take an arbitrary k-clique F1 from A1 , and build a large bunch recursively as follows. If we picked k-cliques F1 , . . . , Fi then we choose Fi+1 such that |V (Fi+1 ) ∩ (∪ij=1 V (Fj ))| ≥ 2 and V (Fi+1 ) \ (∪ij=1 V (Fj )) 6= ∅. Note that this means that ∪i+1 j=1 Fj is an (i + 1)-bunch. Since the auxiliary graph GA1 of 0 the collection is connected we can keep doing this until V (A1 ) = ∪ij=1 V (Fj ) for some i0 ≤ c1 . Since v(Fi ) = k for all i ≤ i0 , we have that k
22 v(A1 ) > . i0 ≥ k 5 So there exists a large i0 -bunch which is a subgraph of E(Kn )p . −
1
k This does not happen a.s. for p = O(n a(k) ), where a(k) = k+1 2 − (k−2)2k/2 /5+2 < m(Ci0 ), since we have that the first i0 -bunch that appears in the random graph is the one of the minimum maximum density, which, by the last lemma, is the simple i0 -2-cluster. Note here that there is a constant (depending on k) number of nonisomorphic (2k/2 /5)-bunches. Let us now deal with the case k = 3. We again consider the partition of the edge set E(Kn )p into the set N of edges which do not participate in a triangle, and the edgesets of the collections Ai . Breaker will play a game on each E(Ai ) separately and, if successful on each of them, win the whole game. If there is a collection of triangles on v vertices, then we can extract a (v − 2)-bunch from it as described previously. By the last two lemmas the first (v − 2)-bunch to appear is the simple (v − 2)-2-cluster with density 2 − v3 . For p ≪ n−2/3 , no (v − 2)-bunch will appear with v ≥ 6. That means there are no collections on more than 5 vertices, in particular there are only a constant number of different collections. Another observation is that if the board Ai on which Breaker plays is a 2-degenerate graph, then an easy pairing strategy will provide Breaker
19
with a win. Take the ordering v1 , . . . , vk of V (Ai ), such that |NVj (vj+1 )| ≤ 2 for j = 1, . . . , k − 1, where Vj = {v1 , . . . , vj }. Then Breaker’s strategy is the following: if Maker takes one of the two edges connecting vj+1 to Vj , then Breaker takes the other one. If there is only one such edge, then Breaker takes an arbitrary edge. Suppose for a contradiction that Maker managed to occupy a triangle vi vj vk against this strategy. This is impossible, since if i, j < k, then at the point when Maker claimed one of the edges vi vk and vj vk , Breaker would have taken the other one. So Breaker can win on any component which is a 2-degenerate graph. Now we just note that since p ≪ n−2/3 every component of finite order (and we showed that we only have order at most 5) does not contain a subgraph with minimum degree at least 3. 2 On the other hand, from Theorem 18 we get that Maker can win the game 2 (E(Kn )p , (Kk )p , 1, 1) for p = Θ(n− k+1 ) and thus we immediately get pKk = 2
O(n− k+1 ). For the triangle game K3 a stronger upper bound can be found. 5
Proposition 23 For p = ω(n− 9 ), Maker can win in the game (E(Kn )p , (K3 )p , 1, 1) a.s. Proof. It is easy to check that Maker can claim a triangle in the (1: 1) game if the board on which the game is played is K5 minus an edge. Therefore, as soon as the graph G(n, p) contains K5 − e a.s., the initial game can be won by Maker a.s. 2 Theorem 18, Proposition 22 and Proposition 23 together imply parts (iv) and (v) of Theorem 1.
5
Open questions
More sharp thresholds? We saw in the previous section that the connectivity game has a sharp threshold, and even more. We think that both the perfect matching game and the Hamiltonian cycle game have the same sharp threshold log n n , and maybe even more. . . It would be very interesting to decide whether the following conjectures are true. Conjecture 2 (i)
τ (Maker wins M) = τ (δ(G) ≥ 2),
(ii) τ (Maker wins H) = τ (δ(G) ≥ 4).
Clique game/H-game. The determination of the threshold pK3 for the triangle game remains outstanding. Note, however, that an approach similar to the one proving pK3 ≥ n−2/3 makes this a finite problem. Indeed, since the maximum density of an s-bunch of triangles for s > 12 is at least 9/5, we can conclude that in the random graph with edge probability o(n−5/9 ) a.s. there is no triangle collection on more then 14 vertices. Therefore, to check whether Breaker wins the triangle game in this setting it is enough to check whether he 20
can win on all triangle collections on at most 14 vertices. We conjecture that the answer is “yes” and our upper bound is optimal, i.e. pK3 = n−5/9 . We are convinced that the arguments of Bednarska and Luczak [4] could be extended to full generality to positional games on random graphs along the lines of Section 3.4. More precisely, the following should be true. Let KH be the family of subgraphs of Kn , isomorphic to H. Then for any fixed graph H there is a constant c(H), such that ′ bpKH = Θ (pbKH ) = Θ pn−1/m (H) , c(H) n . provided p ≥ Ω nlog 1/m′ (H) Concerning the one-on-one game, it would be desirable to extend the counterexample given in Theorem 1 part (v) for arbitrary H, i.e. we believe the following is true.
Conjecture 3 There exists a constant ǫ(H) > 0, such that the unbiased KH is ′ a.s. a Maker’s win if p = n−1/m (H)−ǫ(H) . Should the conjecture be true, it would make the determination of the threshold pKH a finite problem, in a way similar to the case H = K3 . Relationships between thresholds. It is an intriguing task to understand under what circumstances the following is true. Problem 1 Characterize those games (X, F) for which pF =
1 . bF
More generally, characterize the games for which bpF = Θ (pbF ) , for every p ≥ Ω
1 bF
.
This is not true in general as the triangle game shows. What is the reason it is true for the connectivity game and the perfect matching game? Is it because the appearance of these properties has a sharp threshold in G(n, p)? Or because the winning sets are not of constant size? Problem 2 Suppose pF = 1/bF . Is it true that for every p ≥ pF , bpF = Θ (pbF )? It would be very interesting to relate the thresholds bF and pF to some thresholds of the family F in the random graph G(n, p) (or, more generally, in the random set Xp ). It seems to us that if the family Fp is quite dense and well-distributed in X, then Maker still wins the (1: 1) game. Problem 3 Characterize those games (X, F) for which there exists a constant K, such that for any probability p with Pr minx∈Xp |{F ∈ Fp : x ∈ F }| > K −→ 1, we have pF = O(p) and/or bF = Ω(1/p). 21
References [1] J. Beck: Random graphs and positional games on the complete graph, Ann. Discrete Math. 28(1985), 7–13. [2] J. Beck: Remarks on positional games, Acta Math. Acad. Sci. Hungar. 40(1982), 65–71. [3] J. Beck: Positional games and the second moment method, Combinatorica 22(2002), 169–216. [4] M. Bednarska, T. Luczak: Biased positional games for which random strategies are nearly optimal, Combinatorica 20(2000), 477–488. [5] M. Bednarska, T. Luczak: Biased positional games and the phase transition, Random Structures & Algorithms 18(2001), 141–152. [6] B. Bollob´ as, A. Thomason: Threshold functions, Combinatorica 7(1987), 35–38. [7] V. Chv´atal, P. Erd˝os: Biased positional games, Annals of Discrete Math. 2(1978), 221–228. [8] V. Chv´atal, P. Erd˝os: A note on Hamiltonian circuits, Discrete Math. 2(1972), 111–113. [9] P. Erd˝os, J. Selfridge: On a combinatorial game, J. Combinatorial Theory 14(1973), 298–301. [10] A. Lehman, A solution of the Shannon switching game, J. Soc. Indust. Appl. Math. 12(1964) 687–725. [11] E.M. Palmer, J.J. Spencer, Hitting time for k edge-disjoint spanning trees in a random graph. Period. Math. Hungar. 31(1995), 235–240.
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