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Ramsey numbers of cubes versus cliques
David Conlon ∗^ Jacob Fox †^ Choongbum Lee ‡^ Benny Sudakov §
Abstract The cube graph Qn is the skeleton of the n-dimensional cube. It is an n-regular graph on 2 n^ vertices. The Ramsey number r(Qn, Ks) is the minimum N such that every graph of order N contains the cube graph Qn or an independent set of order s. Burr and Erd˝os in 1983 asked whether the simple lower bound r(Qn, Ks) ≥ (s − 1)(2n^ − 1) + 1 is tight for s fixed and n sufficiently large. We make progress on this problem, obtaining the first upper bound which is within a constant factor of the lower bound.
1 Introduction
For graphs G and H, the Ramsey number r(G, H) is defined to be the smallest natural number N such that every red/blue edge-coloring of the complete graph KN on N vertices contains a red copy of G or a blue copy of H.
One obvious construction, noted by Chv´atal and Harary [16], which gives a lower bound for these numbers is to take χ(H) − 1 disjoint red cliques of size |G| − 1 and to connect every pair of vertices which are in different cliques by a blue edge. If G is connected, the resulting graph contains neither a red copy of G nor a blue copy of H, so that r(G, H) ≥ (|G| − 1)(χ(H) − 1) + 1. Burr [8] strengthened this bound by noting that if σ(H) is the smallest color class in any χ(H)-coloring of the vertices of H, we may add a further red clique of size σ(H) − 1, obtaining
r(G, H) ≥ (|G| − 1)(χ(H) − 1) + σ(H).
Following Burr and Erd˝os [8, 11], we say that a graph G is H-good if the Ramsey number r(G, H) is equal to this bound. If G is a family of graphs, we say that G is H-good if all sufficiently large graphs in G are H-good. When H = Ks, where σ(Ks) = 1, we simply say that G or G is s-good.
The classical result on Ramsey goodness, which predates the definition, is the theorem of Chv´atal [15] showing that all trees are s-good for any s. On the other hand, the family of trees is not H-good for every graph H. For example [13], a construction of K 2 , 2 -free graphs due to Brown [7] allows one to show that there is a constant c < 12 such that
r(K 1 ,t, K 2 , 2 ) ≥ t +
t − tc ∗Mathematical Institute, Oxford OX1 3LB, United Kingdom. Email: david.conlon@maths.ox.ac.uk. Research supported by a Royal Society University Research Fellowship. †Department of Mathematics, MIT, Cambridge, MA 02139-4307. Email: fox@math.mit.edu. Research supported by a Simons Fellowship, an MIT NEC Corp. award and NSF grant DMS-1069197. ‡Department of Mathematics, UCLA, Los Angeles, CA, 90095. Email: choongbum.lee@gmail.com. Research sup- ported in part by a Samsung Scholarship. §Department of Mathematics, UCLA, Los Angeles, CA 90095. Email: bsudakov@math.ucla.edu. Research sup- ported in part by NSF grant DMS-1101185, by AFOSR MURI grant FA9550-10-1-0569 and by a USA-Israel BSF grant.
for t sufficiently large. This is clearly larger than (|K 1 ,t| − 1)(χ(K 2 , 2 ) − 1) + σ(K 2 , 2 ) = t + 2.
In an effort to determine what properties contribute to being Ramsey good, Burr and Erd˝os [9, 11] conjectured that if ∆ was fixed then the family of graphs with bounded maximum degree ∆ should be s-good for any s (and perhaps even H-good for all H). This conjecture holds for bipartite graphs H [12] but is false in general, as shown by Brandt [6]. He proved that for ∆ ≥ ∆ 0 almost every ∆- regular graph on a sufficiently large number of vertices is not even 3-good. His result (and a similar result in [28]) actually prove something stronger, namely that if a graph G has strong expansion properties then it cannot be 3-good.
On the other hand, it has been shown if a family of graphs exhibits poor expansion properties then it will tend to be good [1, 28]. To state the relevant results, we define the bandwidth of a graph G to be the smallest number for which there exists an ordering v 1 ,... , vn of the vertices of G such that every edge vivj satisfies |i − j| ≤. This parameter is known to be intimately linked to the expansion properties of the graph. In particular, any bounded-degree graph with poor expansion properties will have sublinear bandwidth [4].
The first such result, shown by Burr and Erd˝os [11], states that for any fixed the family of connected graphs with bandwidth at most is s-good for any s. This result was recently extended by Allen, Brightwell and Skokan [1], who showed that the set of connected graphs with bandwidth at most is H-good for every H. Their result even allows the bandwidth to grow at a reasonable rate with the size of the graph G. If G is known to have bounded maximum degree, their results are particularly strong, saying that for any ∆ and any fixed graph H there exists a constant c such that if G is a graph on n vertices with maximum degree ∆ and bandwidth at most cn then G is H-good.
Many of the original problems of Burr and Erd˝os [11] have now been resolved [28], but one that remains open is to determine whether the family of hypercubes is s-good for every s. The hypercube Qn is the graph on vertex set { 0 , 1 }n^ where two vertices are connected by an edge if and only if they differ in exactly one coordinate. This family of graphs has sublinear bandwidth but does not have bounded degree, so the result of Allen, Brightwell and Skokan does not apply.
To get a first bound for r(Qn, K 3 ), note that a simple greedy embedding implies that any graph with maximum degree d and at least dn + 2n^ vertices has a copy of Qn in its complement. Suppose now that the edges of a complete graph have been 2-colored in red and blue and there is neither a blue triangle nor a red copy of Qn. Then, since the blue neighborhood of any vertex forms a red clique, the maximum degree in blue is at most 2n^ − 1. Hence, the graph must have at most (2n^ − 1)n + 2n^ < 2 n(n + 1) vertices. We may therefore conclude that r(Qn, K 3 ) ≤ 2 n(n + 1).
It is not hard to extend this argument to show that for any s there exists a constant cs such that r(Qn, Ks) ≤ cs 2 nns−^2. This is essentially the best known bound. Here we improve this bound, obtaining the first upper bound which is within a constant factor of the lower bound.
Theorem 1.1. For any natural number s ≥ 3 , there exists a constant cs such that
r(Qn, Ks) ≤ cs 2 n.
The original question of Burr and Erd˝os [11] relates to s-goodness but it is natural to also ask whether the family of cubes is H-good for any H. For bipartite H, this follows directly from a result of Burr, Erd˝os, Faudree, Rousseau and Schelp [12]. Our result clearly implies that for any H, there is a constant cH such that r(Qn, H) ≤ cH 2 n.
For triangles, the rough idea of the proof is to show that if a red/blue edge-coloring of KN does
We will begin, in Section 2, by studying the Ramsey number of cubes versus triangles. We will then show, in Section 3, how our arguments extend to K 4 before treating the general case in Section 4. In Section 5, we will prove Theorem 1.2. We will conclude with some further remarks. All logarithms are base 2 unless otherwise indicated. For the sake of clarity of presentation, we systematically omit floor and ceiling signs whenever they are not crucial. We also do not make any serious attempt to optimize absolute constants in our statements and proofs.
2 Triangle versus cube
The argument works for n ≥ 6. Consider a coloring of the edges of the complete graph KN on the vertex set [N ] = { 1 , 2 , · · · , N } for N ≥ 7000 · 2 n^ with two colors, red and blue, and assume that there are no blue triangles. We will prove that this coloring contains a red Qn.
2.1 Preprocessing the coloring
For each d = 0, 1 , 2 , · · · , log n + 3, we use the following procedure to construct a family S of subsets of [N ] (note that log n + 3 ≤ n for n ≥ 6):
If there exists a set S which induces a red clique of order exactly 4 · 2 n−d, then arbitrarily choose one, add it to the family S, and remove the vertices of the clique from [N ]. We define the codimension d(S) of such a set as d(S) = d. When there are no more such red cliques, continue to the next value of d. In the end, if we have
S∈S |S| ≥^
N 2 , then we let S be our family. Otherwise, if
S∈S |S|^ <^
N 2 , we add the set of remaining vertices to^ S, and declare it to have codimension zero (note that this set has size at least N 2 ≥ 4 · 2 n).
In either of the cases, we have
S∈S |S| ≥^
N
- Note that we also have the following properties. Proposition 2.1. (i) For an integer i ≥ 1 , let X =
S∈S,d(S)≥i S. Then each vertex^ v^ ∈^ [N^ ]^ has at most 2 n−i+3^ blue neighbors in X. (ii) For every set S ∈ S, the subgraph induced by S has maximum blue degree at most 2
n−d(S) 2 n.
Proof. (i) Given i ≥ 1 and a set X defined as above, note that no subset of X induces a red clique of size at least 4 · 2 n−i+1^ = 2n−i+3, since such a set would have been added to S in the previous round. On the other hand, since there are no blue triangles, the blue neighborhood of every vertex induces a red clique. Therefore, for every v ∈ [N ], v has at most 2n−i+3^ blue neighbors in X.
(ii) The claim is trivially true for d > 0, since every set of codimension at least 1 is a red clique. For d = 0, if S is not a red clique, then it is the set of remaining vertices in our procedure. Thus, there are no subsets of S which induce a red clique of size at least 4 · 2 n−(log^ n+3), since such a set would have been added earlier to the family S. On the other hand, since there are no blue triangles, the neighborhood of every vertex induces a red clique. Therefore, every vertex of S has at most 4 · 2 n−(log^ n+3)^ ≤ 2 n 2 n blue neighbors in^ S.
2.2 Tiling the cube
Our strategy is to first decompose the cube Qn into smaller cubes, and then to embed it piece by piece into KN , placing one of the subcubes in each of the subsets from S. We represent the subcubes
of Qn by using vectors in { 0 , 1 , ∗}n. For example, the vector (0, 1 , ∗, · · · , ∗) will represent the subcube {(0, 1 , x 1 , · · · , xn− 2 ) : (x 1 , · · · , xn− 2 ) ∈ { 0 , 1 }n−^2 }.
We seek a tiling of Qn, which we define as a collection of vertex-disjoint cubes that covers all the vertices of Qn and uses only cubes that have all their fixed coordinates at the start of the vector. That is, they are of the form C = (a 1 , a 2 , · · · , ad, ∗, ∗, · · · , ∗) for some d ≥ 0 and a 1 , a 2 , · · · , ad ∈ { 0 , 1 }. We will call such a cube special and let d be the codimension d(C) of C. We say that two disjoint cubes C and C′^ are adjacent, if there are vertices x ∈ C and x′^ ∈ C′^ which are adjacent in Qn. Note that the following list of properties holds for such cubes and a tiling composed from them.
Proposition 2.2. (i) If two special cubes C and C′^ intersect, then we have C ⊂ C′^ or C′^ ⊂ C. (ii) Two disjoint cubes C = (a 1 , · · · , ad, ∗, · · · , ∗) and C′^ = (b 1 , · · · , bd′^ , ∗, · · · , ∗) for d′^ ≤ d are adjacent if and only if (a 1 , · · · , ad′ ) and (b 1 , · · · , bd′ ) differ in exactly one coordinate. (iii) For a tiling C and a special cube C ∈ C of codimension d, there are at most d other special cubes C′^ ∈ C of codimension d(C′) ≤ d which are adjacent to C.
Proof. (i) Let d = d(C), d′^ = d(C′) and, without loss of generality, suppose that d′^ ≤ d. Suppose that (a 1 , · · · , an) ∈ C ∩ C′. Then, since we are only considering special cubes, we must have C = (a 1 , · · · , ad, ∗, · · · , ∗) and C′^ = (a 1 , · · · , ad′^ , ∗, · · · , ∗). However, we then have C ⊂ C′.
(ii) If (a 1 , · · · , ad′^ ) = (b 1 , · · · , bd′^ ), then we have C ⊂ C′, and thus we may assume that this is not the case. Cubes C and C′^ are adjacent if and only if there are two vectors v = (a 1 , · · · , ad, xd+1, · · · , xn) and w = (b 1 , · · · , bd′^ , yd′+1, · · · , yn) in { 0 , 1 }n^ which differ in exactly one coordinate. However, since the two vectors restricted to the first d′^ coordinates are already different, this can happen only if (a 1 , · · · , ad′^ ) and (b 1 , · · · , bd′^ ) differ in exactly one coordinate. Moreover, one can see that if this is the case, then there is an assignment of values to xi and yj so that v and w indeed differ in exactly one coordinate.
(iii) Since C has codimension d, there are only d coordinates that one can ‘flip’ from C to obtain a cube adjacent to C.
As we have already mentioned, our tiling C of the cube Qn will be constructed in correspondence with the family S constructed in Section 2.1. We will construct the tiling C by finding cubes of the tiling one by one. We slightly abuse notation and use C also to denote the ‘partial’ tiling, where only part of the cube Qn is covered. At each step, we will find a subcube C which covers some non-covered part of Qn, and assign it to some set SC ∈ S. We say that such an assignment is proper if the following properties hold.
Proper assignment.
- If C has codimension d, then SC has codimension d.
- Suppose that C is adjacent to some cube C′^ already in the tiling, and that C′^ is assigned to SC′^. Then the bipartite graph induced by SC and SC′^ contains at most |SC ||SC′ | 16 δ^2 blue edges, where^ δ^ = max{d(S), d(S
We use SC to denote the set in S to which C is assigned. Our algorithm for finding the tiling C and the corresponding sets in S is as follows.
Tiling Algorithm. At each step, consider all possible special cubes C which
C to S is a proper assignment (thus we have (c)), since C 1 , · · · , Cj are the only cubes adjacent to C already in the tiling.
Suppose, for the sake of contradiction, that i < d and consider the time t immediately after we last embedded a cube of codimension at most i. At time t, since (a 1 , · · · , an) was not covered, the cubes in C are disjoint from C. Moreover, the set of cubes of codimension at most i which are adjacent to C is the same as at the current time. Therefore, C could have been added to the tiling at time t as well, and this contradicts the fact that we always choose a cube of minimum codimension. Thus we have i ≥ d as claimed.
Note that as an outcome of our algorithm, we obtain a tiling C for which every pair of adjacent cubes C, C′^ ∈ C have at most (^161) δ 2 |SC ||SC′ | blue edges between SC and SC′ , where δ = max{d(C), d(C′)}.
2.3 Imposing a maximum degree condition
Having constructed the tiling C and made the assignment of the cubes to sets in S, we now wish to impose certain maximum degree conditions between the sets SC for C ∈ C. For a set C ∈ C of codimension d = d(C), let C 1 , · · · , Cj be the cubes of codimension at most d which are adjacent to C. Note that we have j ≤ d by Proposition 2.2. Moreover, there are at most (^161) d 2 |SC ||SCa | blue edges between SC and SCa for all a ≤ j.
Now, for each a, remove all the vertices in SC which have at least (^81) d |SCa | blue neighbors in SCa , and let TC be the subset of SC left after these removals. Since there are at most (^161) d 2 |SC ||SCa | blue
edges between SC and SCa , for each index a, we remove at most |S 2 Cd^ |vertices from SC , and thus the
resulting set TC is of size at least |S 2 C |≥ 2 · 2 n−d. Note that all the vertices in TC have blue degree at most (^81) d |SCa | ≤ (^41) d |TCa | in the set TCa. That is, we have the following property.
Maximum degree condition. For each pair of adjacent cubes C, C′^ ∈ C with d(C) ≥ d(C′), every vertex in TC has at most (^4) d(^1 C) |TC′ | blue neighbors in the set TC′.
2.4 Embedding the cube
We now show how to embed Qn. Recall that Qn was tiled by cubes in C, each corresponding to a subset from the family S. We will greedily embed these cubes one by one into their assigned sets from the family S, in decreasing order of their codimensions. If there are several cubes of the same codimension, then we arbitrary choose the order between them.
For each C ∈ C, we will greedily embed the vertices of C into the set TC ⊆ SC. Let d = d(C). Suppose that we are about to embed x ∈ C and let f : Qn → [N ] denote the partial embedding of the cube Qn obtained so far. Note that x has at most d neighbors x 1 , x 2 , · · · , xj (for j ≤ d) which are already embedded and belong to a cube other than C. Since we have only embedded cubes of codimension at least d, the maximum degree condition imposed in Section 2.3 implies that all the vertices f (xi) have blue degree at most (^41) d |TC | in the set TC. Together, these neighbors forbid at most 14 |TC | vertices of TC from being the image of x.
In addition, x has at most n − d neighbors y 1 , · · · , yk (for k ≤ n − d) which are already embedded and belong to C. By Proposition 2.1, each vertex f (yi) has blue degree at most 2 n−d 2 n in the set^ TC^. Together, these neighbors forbid at most 2 n−d 2 vertices of^ TC^ from being the image of^ x.^ Finally,
there are at most 2n−d^ − 1 vertices in TC which are images of some other vertex of C that is already embedded. Therefore, the number of vertices in TC into which we cannot embed x is at most
1 4
|TC | +
· 2 n−d^ + (2n−d^ − 1),
which is less than |TC | since |TC | ≥ 2 · 2 n−d. Hence, there exists a vertex in TC which we can choose as an image of x to extend the current partial embedding of the cube. Repeating this procedure until we finish embedding the whole cube Qn completes the proof.
3 Cliques of order 4
The argument for general cliques is similar to that for triangles given in the previous section. However, there are several new concepts involved and, to slowly develop the necessary concepts, we first provide a proof of the next case, which is K 4 versus a cube.
Recall that in the triangle case, we started by iteratively finding sets which formed a red clique (see Section 2.1). Red cliques were a natural choice, since the blue neighborhood of every vertex formed a red clique. Either we were able to find large red cliques or we were able to restrict the maximum blue degree of vertices in some way (see Proposition 2.1). If one attempts to employ the same strategy for the K 4 case, then the natural choice of sets that we should take instead of red cliques are blue triangle-free sets, since the blue neighborhood of every vertex now forms a blue triangle-free set. Suppose that we found a family S 1 of blue triangle-free sets. Since blue K 3 -free sets are not as powerful as red cliques for embedding subgraphs, we repeat the whole argument within each blue triangle-free set S ∈ S 1 , to obtain red cliques which are subsets of S. By so doing, we obtain a second family of sets S 2 , consisting of red cliques. We refer to sets in Sas level sets, and for a set S ∈ S, we define its level as(S) = `.
In order to find an embedding of the cube Qn using a strategy similar to that used in the triangle case, we wish to find a tiling of Qn, and for each cube C in the tiling, a red clique SC ∈ S 2 so that for two adjacent cubes C and C′, the sets SC and SC′ stand in ‘good’ relation. However, directly finding such a tiling and an assignment is somewhat difficult since we do not have good control on the blue edges between the sets in S 2. To be more precise, suppose that we are given S 1 , S 1 ′ ∈ S 1 and subsets S 2 ⊂ S 1 , S′ 2 ⊂ S′ 1 in S 2. Then the control on the blue edges between S 2 and S′ 2 is ‘inherited’ from the control on the blue edges between S 1 and S 1 ′, and thus depends on the relative sizes of S 1 and S′ 1 , not on the relative sizes of S 2 and S 2 ′ as in the triangle case (unless S 1 = S′ 1 ). To circumvent this difficulty, we will need to maintain tight control on the edge density between different sets in S 1 as well as those in S 2.
We seek a double tiling, which is defined to be a pair C = C 1 ∪ C 2 of tilings satisfying the property that for every C 2 ∈ C 2 there exists a cube C 1 ∈ C 1 such that C 2 ⊂ C 1 (in other words, C 2 is a refined tiling of C 1 ). We refer to cubes in Cas level cubes, and for a cube C ∈ C, we define its level as(C) = . Our goal is to find, for each = 1, 2, an assignment of cubes in Cto sets in S. The following are the key new concepts involved in the K 4 case.
Definition 3.1. (i) For a cube C ∈ C 1 , we define its 1-codimension as d 1 (C) = d(C). For a cube C ∈ C 2 contained in a cube C 1 ∈ C 1 , we define its 1-codimension as d 1 (C) = d(C 1 ) and its 2-codimension as d 2 (C) = d(C) − d(C 1 ).
Proposition 3.1. (i)
S∈S 1 |S| ≥^
N 2 , and for every^ S^1 ∈ S^1 , we have^
S∈S 2 ,S⊂S 1 |S| ≥^
|S 1 |
(ii) For an integer i ≥ 1 , let X =
S∈S 1 ,d 1 (S)≥i S. Then each vertex^ v^ ∈^ [N^ ]^ has at most^2 (^19) · 2 n−i blue neighbors in X. (iii) For a set S 1 ∈ S 1 and an integer i ≥ 1 , let
X =
S∈S 2 ,S⊂S 1 ,d 2 (S)≥i
S.
Then each vertex v ∈ S 1 has at most 16 · 2 n−d^1 (S^1 )−i^ blue neighbors in X. (iv) For every set S ∈ S 2 , the subgraph induced by S has maximum blue degree at most 2
n−d(S) n.
Proof. Part (i) is clear and we omit the proof of (ii), since its proof is similar to that of part (i) of Proposition 2.1.
(iii) Suppose that S 1 is an exceptional set. Then S 1 is the unique set S satisfying S ∈ S 2 and S ⊂ S 1. Since d 2 (S 1 ) = 0, X is an empty set for every i ≥ 1. Thus the conclusion follows.
Now suppose that S 1 is not an exceptional set. Since S 1 induces a blue triangle-free set, a blue neighborhood in S 1 of a vertex v ∈ S 1 forms a red clique. Thus if a vertex v ∈ S 1 has more than 16 · 2 n−d^1 (S^1 )−i^ blue neighbors in X, then inside the neighborhood of v, we can find a set inducing a red clique of size at least 8 · 2 n−d^1 (S^1 )−(i−1). However, this set must have been added in the previous step. Therefore, there are no such vertices.
(iv) Suppose that S ⊂ S 1 with S 1 ∈ S 1. If both S and S 1 are not exceptional sets, then S induces a red clique, and thus the conclusion follows. Suppose that S is exceptional, and S 1 is not. Then as in (iii), we can see that there is no vertex in S which has at least 8 · 2 n−d^1 (S)−(log^ n+3)^ = 2 n−d(S) n blue neighbors in S. One can similarly handle the case when S 1 is exceptional, since we have S = S 1 in this case.
3.2 Tiling the cube
The following proposition is similar to Proposition 2.2 (we omit its proof).
Proposition 3.2. Let C = C 1 ∪ C 2 be a double tiling, and let C ∈ C be a special cube of codimension d.
(i) If C is a level 1 cube, then for each = 1, 2 , there are at most d 1 (C) special cubes of level and codimension at most d which are adjacent to C (and C has level 1 adjacency with all these cubes). (ii) If C is a level 2 cube, then there are at most d 2 (C) special cubes of codimension at most d which have level 2 adjacency with C (they necessarily are of level 2), and for = 1, 2 , at most d 1 (C) cubes of level and codimension at most d which have level 1 adjacency with C.
Our double tiling C of the cube Qn will be constructed in correspondence with the family S con- structed in Section 3.1. We will construct C by finding cubes of the tiling one by one. We slightly abuse notation and use C also to denote the ‘partial’ double tiling, where only part of the cube Qn is covered. Ideally, we would like to construct C by constructing the level 1 tiling C 1 first, and then the level 2 tiling C 2. However, as we will soon see, it turns out that constructing the tiling in increasing order of codimension is more effective than in increasing order of level. At each step, we find a
subcube C which covers some non-covered part of Qn, and assign it to some set SC ∈ S. We say that such an assignment is proper if the following properties hold.
Proper assignment.
(C) =(SC ) and, for =(C), we have d(C) = d(SC ).- If C 2 ⊂ C 1 with C 1 ∈ C 1 and C 2 ∈ C 2 , then SC 2 ⊂ SC 1.
- Suppose that C is adjacent to some cube C′^ which is already in the tiling, where C′^ is assigned to SC′^ and C and C′^ have level ρ adjacency. Then the number of blue edges in the bipartite graph induced by SC and SC′^ is at most |SC ||SC′^ | · (8δ)
(C)+(C ′)− 6 , where δ = max{dρ(C), dρ(C′)}.
We use SC to denote the set in S to which C is assigned. Our algorithm for finding the tiling C and the corresponding sets in S is as follows.
Tiling Algorithm. At each step, consider all possible special cubes C which
(a) can be added to C to extend the partial tiling, (b) have d(C) ≥ d(C′) for all C′^ ∈ C, and (c) for which there exists a set S ∈ S which has not yet been assigned and such that assigning C to S gives a proper assignment.
Take a cube C 0 of minimum codimension satisfying (a),(b),(c), and add it to the tiling. Assign C 0 to the set SC 0 ∈ S given by (c).
Condition (a) is equivalent to saying that either C is disjoint to all cubes in C 1 , or is contained in some cube in C 1 and is disjoint to all cubes in C 2. The following proposition shows that the algorithm will terminate successfully.
Proposition 3.3. If the tiling is not complete, then the algorithm always chooses a cube from a non-empty collection.
Proof. Suppose that in the previous step we embedded some cube of codimension d (let d = 0 for the first iteration of the algorithm). Since the tiling is not complete, there exists a vertex (a 1 , · · · , an) ∈ Qn which is not covered twice.
Case 1: (a 1 , · · · , an) is not covered by any of the cubes in C 1.
Let S′′^ be the subfamily of S 1 consisting of sets which are already assigned to some cube in C 1 , and let S′^ = S 1 \ S′′. If S′′^ contains a set of codimension zero, then the corresponding cube is the whole cube Qn, and it contradicts the fact that (a 1 , · · · , an) is not covered. Thus we may assume that there is no set of codimension zero in S′′, from which it follows that |SC | = 2^18 |C| for all C ∈ C 1. Note that ∣∣ ∣
S∈S′
S
S∈S 1
S
S∈S′′
S
∣ ≥^
N
C∈C
SC
∣ ≥^245 ·^2 n^ −^218 ·^2 n^ >^244 ·^2 n.
to C 2 in each level, and thus |A(1)| ≤ 2 d 1. Let F = {S ∈ S 2 : S ⊂ SC 1 }. By Proposition 3.1, we
have
S∈F S
∣ ≥^
|SC 1 | 2.^ We say that a set^ S^ ∈ F^ is^ bad^ for a cube^ A^ ∈ A
(1) (^) if there are at least
(8δA)(S)+(SA)−^6 |S||SA| = (^) (8δ|AS||) 4 S−A`|(A) blue edges between S and SA (where δA = max{d 1 , d 1 (A)}). Otherwise, we say that S is good for A. For each fixed A, let FA be the subfamily of F consisting of sets which are bad for F. By the properness of the assignment up to this point, we know that there
are at most |SC 1 ||SA| (8δA)^5 −`(A)^ blue edges between^ SC^1 and^ SA^ for every^ A^ ∈ A
(1). Therefore, by counting
the number of blue edges between SC 1 and SA in two ways, we see that
∑
S∈FA
|S||SA|
(8δA)^4 −`(A)^
|SC 1 ||SA|
(8δA)^5 −`(A)^
from which we have
S∈FA |S| ≤^
1 8 δA |SC^1 | ≤^
1 8 d 1 |SC^1 |. Let F′′^ be the subfamily of F of sets which are already assigned to some cube in C 2. There are no sets of relative codimension zero in F′′^ since this implies that (a 1 , · · · , an) is covered by a cube in C 2. It thus follows that for every C ∈ C 2 such that C ⊂ C 1 , we have |SC | = 8|C|, and ∣∣ ∣
S∈F′′
S
∣ =^
S∈F′′
|S| ≤ 8
C⊂C 1
|C| ≤ 8 |C 1 | = 8 · 2 n−d^1.
Let S′^ = F \ (F′′^ ∪
A∈A(1)^ FA) be the subfamily of^ F^ =^ {S^ ∈ S^2 :^ S^ ⊂^ SC 1 }^ of sets which are not assigned to any cubes yet and are good for all the cubes in A(1). Since |SC 1 | ≥ 218 |C 1 | = 2^18 · 2 n−d^1 , we have ∣ ∣∣ ⋃ S∈S′
S
S∈F
S
S∈F′′
S
A∈A(1)
S∈FA
S
|SC 1 |
− 8 · 2 n−d^1 − 2 d 1 ·
|SC 1 |
8 d 1
215 · 2 n−d^1.
For each i ≥ 0, let S i′ = {S ∈ S′^ : d 2 (S) = i} = {S ∈ S′^ : d(S) = d 1 + i}. Suppose that |S i′| ≤ 128 i^3 for all i. Then, since we have |S| = 8 · 2 n−d^1 −i^ for all S ∈ S′ i,
∣∣ S′
i
S i′
i
128 i^3 · 8 · 2 n−d^1 −i^ < 215 · 2 n−d^1 ,
which is a contradiction. Therefore, there exists an index i for which |S i′| > 128 · i^3.
Let C = (a 1 , · · · , ad 1 +i, ∗, · · · , ∗) and consider it as a level 2 cube. By Proposition 3.2(ii), there are at most i cubes in C which have level 2 adjacency with C and have codimension at most d 1 + i. Let A(2)^ be the family consisting of these cubes. For a cube A ∈ A(2), we say that a set S ∈ S′ i is bad for A if there are at least (8i)(SA)+(S)−^6 |SA||S| = (^) (8^1 i) 2 |SA||S| blue edges between SA and S. Otherwise,
we say that S is good for A. We claim that there are at most 128i^2 sets in S i′ which are bad for each fixed A.
Let Xi =
S∈S i′ S, and note that by Proposition 3.1, there are at most^ |SA| ·^16 ·^2 n−i (^) blue edges
between the sets SA and Xi. Each set S ∈ S i′ which is bad for A accounts for at least
1 (8i)^2
|SA||S| =
8 · 2 n−i 64 i^2
|SA| =
2 n−i 8 i^2
|SA|
such blue edges (note that |S| = 8 · 2 n−i). Therefore, in total, there are at most 128i^2 sets in S′ i which are bad for A, as claimed above. Since there are at most i sets in A(2)^ and |S i′| > 128 i^3 , there exists a set S ∈ S i′ which is good for all the cubes A ∈ A(2).
In order to show that C satisfies (a) and (b), it suffices to verify that d 1 + i ≥ d, since this implies the fact that C is disjoint from all the other cubes in C 2 (note that C ⊂ C 1 , and if C intersects some other cube of level 2, then that cube must contain C and therefore also contains (a 1 , · · · , an) by Proposition 2.2). Furthermore, if this is the case, assigning C to S is a proper assignment (thus we have (c)).
Now suppose, for the sake of contradiction, that d 1 + i < d and consider the time t immediately after we last embedded a cube of codimension at most d 1 + i. At time t, since d 1 + i ≥ d 1 = d(C 1 ), the cube C 1 was already embedded and, since there are no cubes of level 2 covering (a 1 , · · · , an), the cubes in C 2 are disjoint from C. A cube in the partial embedding at time t, which is of codimension at most d 1 + i, is 1-adjacent to C if and only if it is 1-adjacent to C 2. Hence, the family of cubes which are adjacent to C at time t is a subfamily of A(1)^ ∪ A(2). Therefore, C could have been added to the tiling at time t as well, and this contradicts the fact that we always choose a cube of minimum codimension. Thus we have d 1 + i ≥ d as claimed.
Note that as an outcome of our algorithm, we obtain a tiling C such that for every pair of adjacent cubes C, C′^ ∈ C, we have control on the number of blue edges between SC and SC′^ (as given in the definition of proper assignment).
3.3 Imposing a maximum degree condition
As in the triangle case, we now impose certain maximum degree conditions between the sets SC for C ∈ C. It suffices to impose maximum degree conditions between sets of level 2.
For a set C ∈ C 2 of codimension d = d(C), and relative codimensions d 1 = d 1 (C), d 2 = d 2 (C), recall that we have a set SC ∈ S such that |SC | ≥ 8 · 2 n−d. Let Aρ be the family of cubes in C 2 with codimension at most d which have level ρ adjacency with C. For each A ∈ Aρ, let δA,ρ = max{dρ, dρ(A)}. By Proposition 3.2(ii), we have |Aρ| ≤ dρ for each ρ = 1, 2. For each A ∈ Aρ, there are at most (^64) δ^12 A,ρ
|SC ||SA| blue edges between SC and SA.
Now for ρ = 1, 2, and each A ∈ Aρ, remove all the vertices in SC which have at least (^8) δ^1 A,ρ |SA| blue neighbors in SA, and let TC be the subset of SC left after these removals. Since there are at most 1 64 δ A,ρ^2 |SC^ ||SA|^ blue edges between^ SC^ and^ SA, we remove at most^
|SC | 8 δA,ρ vertices from^ SC^ , for each set A ∈ Aρ. Thus the resulting set TC is of size at least
|TC | ≥ |SC | − d 1 ·
|SC |
8 δA, 1 − d 2 ·
|SC |
8 δA, 2
|SC |
≥ 4 · 2 n−d.
For each A ∈ Aρ, all the vertices in TC have blue degree at most (^8) δ^1 A,ρ |SA| ≤ (^4) δ^1 A,ρ |TA| in the set TA. Thus we obtain the following property.
Maximum degree condition. Let C, C′^ ∈ C 2 be a pair of cubes having level ρ adjacency with d(C) ≥ d(C′). Then every vertex in TC has at most (^41) δρ |TC′ | blue neighbors in the set TC′^ (where δρ = max{dρ(C), dρ(C′)}).
and assigning each of them to some set SC ∈ S. Informally, this means that the subcube C of Qn will be found in the SC part of our graph. Note that the trivial level zero cube Qn gets assigned to the trivial level zero set [N ] and this fits the heuristic.
For the rest of this section, we assume that s ≥ 5. Let c = s^15 s^ and suppose that N ≥ cs 2 n^ = s^15 s 2 2 n. We will later use the following estimate.
Lemma 4.1. For every positive integer s,
i=
is 2 i^ ≤^2 s
s.
Proof. Let (x)t = (x − 1) · · · (x − t + 1), Xt =
i=
it 2 i^ , and^ Yt^ =^
i=
(i)t 2 i^ for non-negative integers^ t. The Stirling number S(t, k) of the second kind is the number of ways to partition a set of t objects into k nonempty subsets. These numbers satisfy the following well-known identity xt^ =
∑t k=0 S(t, k)(x)k (see, e.g., [32], Chapter 1.4). This implies the identity
Xt =
∑^ t
k=
S(t, k)Yk.
By taking the derivative k times of both sides of the equality (1 − z)−^1 =
i≥ 0 z
i, note that
k! · (1 − z)−(k+1)^ =
i≥ 1
i(i − 1) · · · (i − k + 1)zi−k.
By multiplying both sides by zk^ and substituting z = 1/2 we have that Yk = 2k!. This, together with the above identity, implies that Xt =
∑t k=0 2 k!S(t, k). Although it will be not be needed, we remark that there is an explicit formula Xt = 2
∑t k=
∑k j=0(−1)
k−j (k j
jt^ which follows from substituting in
the well-known identity S(t, k) = (^) k^1!
∑k j=0(−1) k−j (k j
jt. Let Ts be the number of partitions of a set of s objects into labelled nonempty subsets. By counting over the size k of the partition, we have the identity Ts =
∑s k=0 k!S(s, k) =^ Xs/2. As each partition counted by^ Ts^ is determined by the vector of labels of the sets containing each object, and there are at most s such labels for each partition, we have Ts ≤ ss, and the desired inequality follows.
4.1 Preprocessing the coloring
Let S 0 = {[N ]} and [N ] be the unique set of level zero and codimension zero (denoted as (S) = 0 and d(S) = 0). We construct the levels one at a time. Once we finish constructing S− 1 , for each set S′^ ∈ S− 1 , we use the following procedure to construct sets belonging to the-th level S`:
For each d = 0, 1 , 2 , · · · , log n + 2 log c, if there exists a set S ⊂ S′^ which induces a blue Ks−-free graph of order exactly cs−^ · 2 n−d(S ′)−d , then arbitrarily choose one, add it to the family S, and remove the vertices of S from S′. We define the-codimension d(S) of such a set as d(S) = d, and for i = 1, · · · , ` − 1, we define the i-codimension di(S) as di(S) = di(S′). Let the codimension of S be d(S) =
∑`
i=1 di(S).^ When there are no more such sets, continue to the next value of d. If, after running through all values of d,
S∈S`,S⊂S′^ |S|^ <^
|S′| 2 , then add the set of remaining vertices to^ S, and declare it to be an exceptional set with-codimension zero (note that this set has size at least |S′| 2 =^
cs−(`−1)· 2 n−d(S′) 2 ≥^ c
s−` (^) · 2 n−d(S′)).
Let S =
⋃s− 2 =0 S^ (we suppose that^ S^ is a multi-family and if a set appears multiple times we distinguish them by their levels). The following proposition is similar to Proposition 3.1. We omit its proof.
Proposition 4.1. (i) For 1 ≤ ≤ s − 2 and a set S′^ ∈ S− 1 ,
S∈S`,S⊂S′^ |S| ≥^
|S′|
(ii) For integers 1 ≤ ≤ s − 2 and i ≥ 1 , let S′^ be a set of level − 1 , and let
X =
S∈S,S⊂S′,d(S)≥i
S.
Then each vertex v ∈ S′^ has at most 2 cs−`^ · 2 n−d(S ′)−i blue neighbors in X. (iii) For every set S ∈ Ss− 2 , the subgraph induced by S has maximum blue degree at most 2 n−d(S) n.
4.2 Tiling the cube
In this subsection, we find an (s − 1)-tiling of Qn. Recall that in the previous section, we had to control the blue edge densities between adjacent cubes in the tiling. The parameter that governed the control of these densities was defined in terms of the ρ-codimension, where ρ was the level of adjacency of these cubes. Below we generalize this concept.
Definition 4.1. Let C be an (s − 1)-tiling, and let C, C′^ ∈ C be two adjacent cubes.
(i) The level of adjacency ρ(C, C′) is the minimum such that the cubes of level containing C and C′^ are distinct. We say that C and C′^ are ρ-adjacent if ρ(C, C′) = ρ. (ii) The dominating parameter δ(C, C′) is max{dρ(C), dρ(C′)}, where ρ = ρ(C, C′).
Note that the level of adjacencies of two cubes C and C′^ is at most min{(C),(C′)}. The following proposition is similar to Proposition 3.2 and we omit its proof.
Proposition 4.2. Let C be an (s − 1)-tiling, and let C ∈ C be a level special cube of codimension d. For each ρ = 1, 2 , · · · , and ′^ = 1, 2 , · · · , s − 2 , the number of special cubes of level′^ and codimension at most d which are ρ-adjacent to C is at most dρ(C).
Our (s−1)-tiling C of the cube Qn will be constructed in correspondence with the family S constructed in Section 4.1. We will construct C by finding cubes of the tiling one by one. We slightly abuse notation and use C also to denote the ‘partial’ (s − 1)-tiling, where only part of the cube Qn is covered. At each step, we find a subcube C which covers some non-covered part of Qn, and assign it to some set SC ∈ S. We say that such an assignment is proper if the following properties hold.
Proper assignment.
(C) =(SC ) and, for =(C), we have d(C) = d(SC ).- If C ⊂ C′^ for C′^ ∈ C, then SC ⊂ SC′.
- Suppose that C is adjacent to some cube C′^ already in the tiling, where C′^ is assigned to SC′^ , and that C and C′^ have level ρ adjacency. Then the number of blue edges in the bipartite graph induced by SC and SC′^ is at most |SC ||SC′^ | · (4s^2 δ(C, C′))
(C)+(C ′)− 2 s .
from which we have
S∈FA |S| ≤^
1 4 s^2 δA |SC`−^1 | ≤^
1 4 s^2 dρ(C− 1 ) |SC−^1 |. Let F′′^ be the subfamily of F of sets which are already assigned to some cube in C. There can be no sets of relative codimension zero in F′′^ since this implies that (a 1 , · · · , an) is covered by a cube in C. It thus follows that for every C ∈ Csuch that C ⊂ C− 1 , we have |SC | = cs−`|C|, and ∣∣ ∣
S∈F′′
S
∣ =^
S∈F′′
|S| ≤ cs−^ · |C− 1 |.
Let S′^ = F \ (F′′^ ∪
ρ
A∈A(ρ)^ FA) be the subfamily of^ F^ of sets which are not assigned to any cubes yet and are good for all the cubes in A(ρ)^ for all ρ. Since |SC− 1 | ≥ cs−+1|C`− 1 |, we have
∣∣ ∣
S∈S′
S
S∈F
S
S∈F′′
S
∣ −^
ρ
A∈A(ρ)
S∈FA
S
|SC`− 1 |
− cs−^ · |C− 1 | −
ρ
sdρ(C`− 1 ) ·
|SC`− 1 |
4 s^2 dρ(C`− 1 )
cs−`+ 5
· |C`− 1 |.
For each i ≥ 0, let S i′ = {S ∈ S′^ : d(S) = i} = {S ∈ S′^ : d(S) = d(C− 1 ) + i}. Suppose that |S′ i| ≤ (4s^2 i)^2 s+1^ for all i. Then, since we have |S| = cs−2 n−d(C−^1 )−i^ = cs−^ · 2 −i|C− 1 | for all S ∈ S′ i, ∣∣ ∣∣ ∣
S∈S′
S
i
S∈S i′
|S| ≤
i
(4s^2 i)^2 s+1^ · cs−^ · 2 −i|C− 1 |.
By Lemma 4.1, we have ∑
i
(4s^2 i)^2 s+1 2 −i^ = (2s)^4 s+^
i
i^2 s+1/ 2 i^ ≤ (2s)^4 s+2^ · 2(2s + 1)^2 s+1^ < s^15 s/5 = c/ 5.
Therefore,
∣⋃ S∈S′ S
∣ (^) < cs− 5 +1· |C− 1 |, which is a contradiction. Thus, there exists an index i for
which |S i′| > (4s^2 i)^2 s+1.
Let C = (a 1 , · · · , ad(C− 1 )+i, ∗, · · · , ∗) and consider it as a level cube. By Proposition 4.2, there are at most s·i cubes in C with codimension at most d(C) which have level adjacency with C (remember that these may have level higher than). Let A()^ be the family consisting of these cubes, and note that the dominating parameter of C and A for A ∈ A()^ is always i. For a cube A ∈ A(), we say that a set S ∈ S i′ is bad for A if there are at least (4s^2 i)(SA)+(S)−^2 s|SA||S| = (4s^2 i)(SA)+`−^2 s|SA||S| blue edges between SA and S. Otherwise, we say that S is good for A. We claim that there are at most (4s^2 i)^2 s^ sets in S i′ which are bad for each fixed A.
Let Xi =
S∈S i′ S^ and note that, by Proposition 4.1, there are at most^ |SA| ·^2 c s−` (^) · 2 n−d(C) (^) blue
edges between the sets SA and Xi. Each set S ∈ S i′ which is bad for A accounts for at least
(4s^2 i)(SA)+−^2 s|SA||S| ≥ c s−`· 2 n−d(C) (4s^2 i)^2 s−^1 |SA|^ such blue edges (note that^ |S|^ =^ c
s−(^) · 2 n−d(C) (^) and ≥ 1).
Therefore, in total, there are at most (4s^2 i)^2 s^ sets in S i′ which are bad for A, as claimed above. Since there are at most si cubes in A()^ and |S i′| > (4s^2 i)^2 s+1, there exists a set S ∈ S i′ which is good for all the cubes A ∈ A().
In order to show that C satisfies (a) and (b), it suffices to verify that d(C) ≥ d, since this implies the fact that C is disjoint from all the other cubes of level at least (note that C ⊂ C− 1 , and if C intersects some other cube of level at least `, then that cube must contain C and therefore also
contains (a 1 , · · · , an) by Proposition 2.2). Furthermore, if this is the case, assigning C to S is a proper assignment (thus we have (c)).
Now suppose, for the sake of contradiction, that d(C) < d and consider the time t immediately after we last embedded a cube of codimension at most d(C). At time t, since d(C) ≥ d(C− 1 ), the cube C− 1 was already embedded and, since there are no cubes of level covering (a 1 , · · · , an), the cubes in C are disjoint from C. For ρ ≤ − 1, a cube in the embedding at time t, which is of codimension at most d(C), is ρ-adjacent to C if and only if it is ρ-adjacent to C. Hence, the family of cubes which are adjacent to C at time t is a subfamily of
⋃`
ρ=1 A
(ρ). Therefore, C could have been added
to the tiling at time t as well, and this contradicts the fact that we always choose a cube of minimum codimension. Thus we have d(C) ≥ d as claimed.
Note that as an outcome of our algorithm, we obtain a tiling C such that for every pair of adjacent cubes C, C′^ ∈ C, we have control on the number of blue edges between SC and SC′ (as given in the definition of proper assignment).
4.3 Imposing a maximum degree condition
As in the previous sections, we now impose certain maximum degree conditions between the sets SC for C ∈ Cs− 2. For a set C ∈ Cs− 2 of codimension d = d(C) and relative codimensions d= d(C), 1 ≤ ≤ s − 2, recall that we have a set SC ∈ S such that |SC | ≥ c^2 · 2 n−d. Let A(ρ)^ be the family of cubes in Cs− 2 with codimension at most d which have level ρ adjacency with C. By Proposition 4.2, we have |A(ρ)| ≤ dρ(C) for each ρ = 1, · · · , s − 2. For each A ∈ A(ρ), let δA = δ(C, A) and note that δA = max{dρ(A), dρ(C)} ≥ dρ(C). Also, since(A) + `(C) − 2 s = −4, there are at most 1 (4s^2 δA)^4 |SC^ ||SA|^ blue edges between^ SC^ and^ SA.
Now for ρ = 1, · · · , s − 2, and each A ∈ A(ρ), remove all the vertices in SC which have at least 1 4 s^2 δA |SA|^ blue neighbors in^ SA, and let^ TC^ be the subset of^ SC^ left after these removals. Since there are at most (^) (4s (^21) δA) 4 |SC ||SA| blue edges between SC and SA, we remove at most (^) (4|sS 2 CδA^ |) 3 vertices from
SC , for each set A ∈ A(ρ). Thus the resulting set TC is of size at least
|TC | ≥ |SC | −
∑^ s−^2
ρ=
dρ ·
|SC |
(4s^2 δρ)^3
|SC |
≥ c · 2 n−d.
For each A ∈ A(ρ), all the vertices in TC have blue degree at most (^4) s (^21) δA |SA| ≤ (^2) s (^21) δA |TA| in the set TA. Thus we obtain the following property.
Maximum degree condition. Let C, C′^ ∈ Cs− 2 be a pair of cubes having level ρ adjacency, and d(C) ≥ d(C′). Then every vertex in TC has at most (^2) s (^2) δ(^1 C,C′) |TC′^ | blue neighbors in the set TC′^.
4.4 Embedding the cube
We now show how to embed Qn. Recall that we found an (s − 1)-tiling C of Qn. We will greedily embed the cubes in the level s − 2 tiling Cs− 2 one by one into their assigned sets from the family Ss− 2 , in decreasing order of their codimensions. If there are several cubes of the same codimension, then we arbitrary choose the order between them.
