aa r X i v : . [ m a t h . C O ] J un Maker Can Construct a Sparse Graph on a Small Board
Heidi Gebauer ∗ Abstract
We study Maker/Breaker games on the edges of sparse graphs. Maker and Breaker take turnsin claiming previously unclaimed edges of a given graph H . Maker aims to occupy a given targetgraph G and Breaker tries to prevent Maker from achieving his goal. We define a function f onthe integers and show that for every d -regular graph G on n vertices there is a graph H with atmost f ( d ) n edges such that Maker can occupy a copy of G in the game on H . We consider positional games played on edge-sets of graphs. Let P = P ( N ) ⊆ E ( K N ) be a graphproperty of N -vertex graphs, and let H be a graph on the vertex set V ( H ) = V ( K N ). The game( E ( H ) , P ) is played by two players, called Maker and Breaker, who take turns in claiming onepreviously unclaimed edge of H , with Maker going first. Following the standard notation we call H the base graph or the board . Maker aims to occupy a graph having property P and Breaker tries toprevent Maker from achieving his goal: Breaker wins if, after all edges of H were claimed, Maker’sgraph does not possess P . A round denotes a pair consisting of a Maker’s move and the consecutiveBreaker’s move.Let G be a fixed graph on n vertices. We consider the game where Maker’s goal is to occupya copy of G . Formally, let P G denote the property that a graph contains G as a subgraph and let H = ( V ( H ) , E ( H )) be the board. The G -game denotes the game ( E ( H ) , P G ). Note that Maker hasonly a chance to win if | V ( H ) | ≥ n .We show that if G has maximum degree at most d then there is a board H where | E ( H ) | is linearin n such that Maker has a strategy to win the G -game on H . Theorem 1.1.
Let G be a graph with maximum degree d on n vertices. Then there is a constant c = c ( d ) and a graph H with | E ( H ) | ≤ cn such that Maker has a strategy to occupy a copy of G inthe game on H . Feldheim and Krivelevich [2] showed that there are constants c, c ′ depending on d such that thefollowing holds: if the board H is the complete graph K N on N = cn vertices then Maker can occupy ∗ Institute of Theoretical Computer Science, ETH Zurich, CH-8092 Switzerland. Email: [email protected]. G in at most c ′ n rounds (actually, they proved this statement also for the more generalclass of d -degenarate graphs). In our proof we will adopt many of their ideas, constructions andstructures. Notation
Throughout this paper we will assume that Breaker starts the game. Otherwise Makercan start with an arbitrary move, then follow his strategy. If his strategy calls for something heoccupied before he takes an arbitrary edge; no extra move is disadvantegous for him. Accordingly,we slightly abuse notation and let a round denote a pair consisting of a Breaker’s move and theconsecutive Maker’s move.Let
U, U ′ ∈ V ( H ). With E H ( U, U ′ ) we denote the set of edges between U and U ′ in H . Let v ∈ V ( G ). The neighborhood N G ( v ) denotes the set of vertices which are adjacent to v in G . Let u, v ∈ V ( G ). The distance dist G ( u, v ) between u and v denotes the number of edges in a shortestpath in G connecting u and v . When there is no danger of confusion we sometimes omit the index G .Adapting the notation of Feldheim and Krivelevich, the board graph along with the sets of Maker’sand Breaker’s claimed vertices is called a Game Position . Let G be a fixed d -regular graph on n vertices. Let r = ed . We will define an appropriate labeling l which assigns to every vertex v ∈ V ( G ) an element of { , . . . , r } . The level of a vertex v denotes l ( v ). We first show that we can find a labeling with some particular properties. Lemma 2.1.
There is a labeling l of the vertices such that for every u, v ∈ V ( G ) with l ( u ) = l ( v ) we have dist ( u, v ) ≥ .Proof: Suppose that we assign to each vertex v ∈ V ( G ) a level in { , . . . , r } uniformly andindependently at random. Note that for any two vertices u, v ∈ V ( G ), Pr [ l ( u ) = l ( v )] = r . We willapply the symmetric version of the famous Lov´asz Local Lemma. Theorem 2.2. (Symmetric version of the Lov´asz Local Lemma.) Let A , A , . . . , A n be events inan arbitrary probability space. Suppose that each event A i is mutually independent of a set of all theother events A j but at most k , and that Pr [ A i ] ≤ p for all ≤ i ≤ n . If ep ( k + 1) ≤ then Pr [ V ni =1 ¯ A i ] > . For every vertex v ∈ V ( G ) let A v denote the event that l ( v ) = l ( w ) for some vertex w with dist ( v, w ) ≤
2. Since G has maximum degree d there are at most d + d vertices at distance at2ost 2 from v . So the probability p = Pr [ A v ] is at most d + d r . Let { u , . . . , u s } denote the setof vertices u i with d ( v, u i ) ≤
2. Note that the event A v is completely determined by the values l ( v ) , l ( u ) , l ( u ) , . . . , l ( u s ). Hence A v is independent of { A w : dist ( v, w ) ≥ } . Applying the LocalLemma with k = d + d + d + d yields that ep ( k + 1) ≤ e ( d + d )(1+ d + d + d + d ) r ≤ ed d r = 1,which concludes our proof.For every vertex v let d ≤ ( v ) denote the number of vertices at distance at most 2 from v in G .Suppose that l is the labeling from Lemma 2.1. Let u, v be two vertices with l ( u ) < l ( v ). We saythat u blocks v if (i) ( u, v ) ∈ E ( G ), or (ii) there is a vertex w with l ( w ) < l ( u ) < l ( v ) such that( w, u ) , ( w, v ) ∈ E ( G ). We construct the directed graph D on the vertex set V ( G ) such that we drawan arc from v to u if and only if u blocks v . A vertex u is called a descendant of v if there is adirected path from v to u in D . Observation 2.3.
Let v ∈ V ( G ) . For every arc ( v, u ) we have l ( v ) > l ( u ) . Moreover, every vertex v has out-degree at most d ≤ ( v ) ≤ d + d ( d −
1) = d in D . Hence there are at most ( d ) + ( d ) +( d ) + . . . + ( d ) r − ≤ ( d ) r descendants of v in D . We first need some more notation. The lower-level neighborhood N − G ( v ) of a vertex v denotes theset { u ∈ N G ( v ) : l ( u ) < l ( v ) } . Accordingly, the upper-level neighborhood N + G ( v ) denotes the set { u ∈ N G ( v ) : l ( u ) > l ( v ) } . Construction of the board H For every vertex v ∈ V ( G ) we let P ( v ) denote the set of de-scendants of v in D . Let s d := d d +4 . For every vertex v ∈ V ( G ) we provide a set S v of d ( s d ) | P ( v ) | + s d vertices in H . So V ( H ) := ∪ v ∈ V ( G ) S v . Moreover, for every ( u, v ) ∈ E ( G ) weadd an edge between every a ∈ S u and b ∈ S v in H . In other words, E ( H ) := { ( a, b ) : a ∈ S u , b ∈ S v such that ( u, v ) ∈ E ( G ) } . Note that by Observation 2.3, | P ( v ) | ≤ d r and thus | E ( H ) | ≤ | E ( G ) | ( d ( s d ) d r + s d ) ≤ dn d ed d +8 + d d +4 ) Note that | E ( H ) | is linear in n . As in [2], to distinguish between vertices of G and vertices of H wemark the vertices of H with a star.During the game we will define for every vertex v ∈ V ( G ) a subset B v ⊆ S v with | B v | = s d .We adopt the concepts of a candidate vertex and a candidate scheme from [2], and state modifiedversions of Definition 2.1 - 2.3 (in [2]). Definition 2.4. (Vertex candidate with respect to a specific edge) Let H ⋆ be a position in ( E ( H ) , P G ) ,let ( u, v ) ∈ E ( G ) with l ( u ) < l ( v ) and let { u , . . . , u t − } = { w ∈ N + G ( u ) : l ( w ) < l ( v ) } . A vertex x ⋆ ∈ S v is called a candidate with respect to the edge ( u, v ) , if i) B u , B u , B u , . . . , B u t − are already determined, and(ii) for every choice of vertices b ⋆ ∈ B u , b ⋆ ∈ B u , . . . , b ⋆t − ∈ B u t − we have |{ b ⋆ ∈ B u : Maker claimed ( b ⋆ , b ⋆ ) , ( b ⋆ , b ⋆ ) , . . . , ( b ⋆ , b ⋆t − ) , ( b ⋆ , x ⋆ ) in H ⋆ }|| B u | ≥ t t Definition 2.5. (Vertex candidate) Let H ⋆ be a position in ( E ( H ) , P G ) , let v ∈ V ( G ) and let x ⋆ ∈ B v . We call x ⋆ a candidate if for every u ∈ N − G ( v ) x ⋆ is a candidate with respect to ( u, v ) . Let v , v , . . . , v n be an ordering of the vertices in V ( G ) where l ( v ) ≤ l ( v ) ≤ . . . , ≤ l ( v n ). Definition 2.6.
Let H ⋆ be a position in ( E ( H ) , P G ) and suppose that B v , B v , . . . , B v n are alldetermined. We say that ( B v , B v , . . . , B v n ) form a candidate scheme if every x ⋆ ∈ B v ∪ B v ∪ . . . ∪ B v n is a candidate. The next lemma is a slight adaptation of Lemma 2.1 in [2].
Lemma 2.7. (Feldheim, Krivelevich) Let H ⋆ be a position in ( E ( H ) , P G ) and let ( B v , B v , . . . , B v n ) be a candidate scheme. Then Maker’s graph contains a copy of G . A proof of this lemma is given in [2]. G -Game into Subgames We first need some more notation. Let v ∈ V ( G ) and let x ⋆ ∈ S v . By a slight abuse of notation wedefine the level l ( x ⋆ ) of x ⋆ to be l ( v ). Let H ⋆ be a position in the game ( E ( H ) , P G ). We call a vertex x ⋆ ∈ S v touched in H ⋆ if Maker or Breaker claimed an edge of the form ( y ⋆ , x ⋆ ) with l ( y ⋆ ) < l ( x ⋆ ).Accordingly, we call x ⋆ untouched in H ⋆ if x ⋆ is not touched in H ⋆ . We say that a vertex is v ∈ V ( G ) completed in H ⋆ if every vertex in B v is a candidate. We say that v is ready if every vertex u with( v, u ) ∈ E ( D ) is completed. When there is no danger of confusion we will sometimes omit mentioning H ⋆ explicitly.At the beginning of the game we set B v := S v for every vertex v ∈ V ( G ) with out-degree zero in D . Hence v is already completed (and thus also ready).We now consider some smaller games separately and then show how Maker can divide the G -gameinto a combination of those smaller games. Definition 2.8.
Let H ⋆ be a position in ( E ( H ) , P G ) , let v ∈ V ( G ) be ready in H ⋆ and let x ⋆ ∈ S v .For every u ∈ N − ( v ) the game G u,x ⋆ is defined as follows. The board consists of the set of edges E H ( B u , { x ⋆ } ) ; Maker’s goal is to achieve that x ⋆ becomes a candidate with respect to the edge ( u, v ) . We now extend this game to a game where not only x ⋆ but several vertices of S v should becomea candidate. 4 efinition 2.9. Let H ⋆ be a position in ( E ( H ) , P G ) , let v ∈ V ( G ) be ready in H ⋆ and suppose that B v has been determined. The game G v is defined as follows. The board is ∪ u ∈ N − G ( v ) E H ( B u , B v ) ;Maker’s goal is to win G u,x ⋆ for every u ∈ N − G ( v ) , x ⋆ ∈ B v . The next lemma is an adaptation of Lemma 2.2 in [2].
Lemma 2.10.
Let H ⋆ be a position in ( E ( H ) , P G ) , let v ∈ V ( G ) be ready and let x ⋆ ∈ S v beuntouched in H ⋆ . Then, for every u ∈ N − G ( v ) Maker has a strategy to win the game G u,x ⋆ . Let N − G ( v ) = { u , . . . , u t } and note that if v is ready then every u i and every w ∈ N + ( u i ) with l ( w ) < l ( v ) is completed. Then Lemma 2.10 follows directly from the proof of Lemma 2.2 in [2]. Forcompleteness we restate the core of the proof. Proof of Lemma 2.10:
We need some notation. Let F = ( V ( F ) , E ( F )) be a hypergraph, i.e., E ( F ) is a subset of the power set 2 V ( F ) . In a positional game on F Maker and Breaker alternatelyclaim an unclaimed vertex of V ( F ) until all vertices are claimed. We will use the following result byAlon, Krivelevich, Spencer and Szab´o [1], extending a previous result by Sz´ekely [3]. Theorem 2.11. (Alon, Krivelevich, Spencer, Szab´o) Let F be a hypergraph with X hyperedges, whosesmallest hyperedge contains at least x vertices. In a positional game on F Maker has a strategy toclaim at least x − q x ln(2 X )2 vertices of each hyperedge. To prove Lemma 2.10 we fix a v ∈ V ( G ) and an untouched x ⋆ ∈ S v . If N − G ( v ) = ∅ then theclaim is clearly true. Otherwise let u ∈ N − G ( v ). Let { u , . . . , u t − } = { w ∈ N + G ( u ) : l ( w ) < l ( v ) } and recall that t ≤ d . Note that since v is ready we have that every u i is completed and therefore B u , . . . , B u t − are all candidates.If t = 1 then Maker can win the game G u,x ⋆ by connecting x ⋆ to half of the vertices of B u .Suppose that t ≥
2. We can express G u,x ⋆ as a positional game on a hypergraph F : Let V ( F ) bethe set of all edges in E H ( B u , x ⋆ ); and for every combination of b ⋆ ∈ B u , b ⋆ ∈ B u , . . . , b ⋆t − ∈ B u t − add to E ( F ) the hyperedge e b ⋆ ,b ⋆ ,...,b ⋆t − consisting of all edges of the form ( b ⋆ , x ⋆ ) where b ⋆ ∈ B u isconnected to b ⋆ , b ⋆ , . . . , b ⋆t − in Maker’s graph.We have | E ( F ) | ≤ ( s d ) t − .Note that | e b ⋆ ,b ⋆ ,...,b ⋆t − | = |{ b ⋆ ∈ B u : Maker claimed ( b ⋆ , b ⋆ ) , ( b ⋆ , b ⋆ ) , . . . , ( b ⋆ , b ⋆t − ) }| . Since B u , . . . , B u t − are all candidates we have | e b ⋆ ,b ⋆ ,...,b ⋆t − | ≥ | B u | t − ( t − ≥ s d t − ( t − s d t ( t − − r s d t ( t −
1) ln(2( s d ) t − ) (1)vertices in every hyperedge. A careful calculation (details can be found in [2]) yields that theexpression in (1) is at least s d t t . 5emma 2.10 allows to show the following corollary, a similar version of which has already beenstated in [2]. Corollary 2.12.
Let H ⋆ be a position in ( E ( H ) , P G ) and let v ∈ V ( G ) be ready in H ⋆ . Supposethat B v has been determined and that every vertex in B v is untouched in H ⋆ . Then Maker has astrategy to win the game G v .Proof: Consider the following strategy for Maker. Suppose that Breaker claims an edge ( b ⋆ , x ⋆ )with b ⋆ ∈ B u and x ⋆ ∈ B v for some u ∈ N − G ( v ) (note that due to the construction of the board everyclaimed edge is of this form). Then Maker responds in the game G u,x ⋆ Since the boards of the games G u,x ⋆ are pairwise disjoint Maker can treat each game G u,x ⋆ separately. Thus Lemma 2.10 yields a winning strategy for Maker in G v .The next observation shows that the game G v is finished after not too many rounds. Observation 2.13.
Let v ∈ V ( G ) . The board size of the game G v is bounded by | N − G ( v ) | ( s d ) ≤ ds d ,hence G v lasts at most ds d rounds. We now describe a strategy S for Maker to obtain a candidate scheme. We choose S in such away that the following invariant I is maintained. Invariant I Suppose that t rounds have been played and let H ⋆ denote the corresponding position.Let v ∈ V ( G ) be a vertex which became ready in round t (i.e. v is ready after round t but was notready after round t − s d vertices in S v are untouched in H ⋆ .The invariant I clearly holds for t = 0. Indeed, every S v has cardinality at least s d and at thebeginning of the game every vertex is untouched.Let t > t − t −
1. For all vertices v which became ready in round t − S ⊆ S v of s d untouched vertices and sets B v := S . If all vertices of V ( G ) are completed after round t − B v , B v , . . . , B v n ) form a candidate scheme, which by Lemma 2.7 guarantees thatMaker’s graph contains a copy of G . In this case we are done. It remains to consider the case wherenot all vertices are completed.Suppose that in round t Breaker claims an edge ( x ⋆ , y ⋆ ) with x ⋆ ∈ S u , y ⋆ ∈ S v and l ( u ) < l ( v ).We distinguish three cases.Case 1 v is ready but not completed. Maker responds in the game G v .Case 2 v is not ready. Maker selects a w ∈ P ( v ) such that w is ready but not completed and acts asif Breaker claimed an edge in the board of G w (recall that P ( v ) denotes the set of vertices u for which there is a directed path from v to u in D ). Note that such a w always exist. Indeed,suppose otherwise and let w ′ be a non-ready vertex with minimum level among all vertices in P ( v ). By construction w ′ has out-degree at least one (otherwise w ′ would be ready since the6eginning of the game), and by assumption there is at least one out-neighbor w ′′ of w ′ whichis not completed. But then w ′′ is non-ready with l ( w ′′ ) < l ( w ′ ), contradicting the choice of w ′ .Case 3 v is completed. Maker selects a not yet completed vertex w ∈ V ( G ) and acts as if Breakerclaimed an edge in the board of G w .Note that since the boards G v are pairwise distinct Maker can treat each game G v separately.We now show that the invariant I is fulfilled after round t . Suppose that v became ready afterround t . We observe that for every vertex x ⋆ ∈ S v which was touched by Breaker in one of the first t rounds there is (at least) one vertex y = f ( x ⋆ ) ∈ V ( G ) such that Maker acted as if Breaker claimedan edge in the board of G y . Note that by construction, y ∈ P ( v ).By Observation 2.13, for every u ∈ P ( v ), G u lasted at most ds d rounds; thus there are at most ds d vertices x ⋆ ∈ S v with f ( x ⋆ ) = u . Hence there are at most ds d | P ( v ) | vertices x ⋆ ∈ S v where f ( x ⋆ ) ∈ P ( v ). Therefore at most ds d | P ( v ) | vertices of S v have been touched during the first rounds,which yields that at least s d vertices remain untouched, as claimed.Thus, if Maker follows S then by Corollary 2.12 he has a strategy to achieve that ( B v , B v , . . . , B v n )form a candidate scheme. Due to Lemma 2.7 this guarantees that Maker’s graph contains a copy of G . This concludes the proof of Theorem 1.1. References [1] N. Alon, M. Krivelevich, J. Spencer and T. Szab´o. Discrepency Games,
Electronic Journal ofCombinatorics , (1), (2005)[2] O. N. Feldheim and M. Krivelevich. Winning fast in sparse graph construction games. Combina-torics, Probability and Computing , (6), (2008) 781-791[3] L. A. Sz´ekely. On two concepts of discrepancy in a class of combinatorial games. Infinite andFinite Sets, Colloquia Mathematica Societatis Janos Bolyai ,37