Ramsey numbers of cubes versus clques Davd Conlon Jacob Fox Choongbum Lee Benny Sudakov Abstract The cube graph Q n s the skeleton of the n-dmensonal cube. It s an n-regular graph on 2 n vertces. The Ramsey number r(q n, K s ) s the mnmum N such that every graph of order N contans the cube graph Q n or an ndependent set of order s. Burr and Erdős n 983 asked whether the smple lower bound r(q n, K s ) (s )(2 n ) + s tght for s fxed and n suffcently large. We make progress on ths problem, obtanng the frst upper bound whch s wthn a constant factor of the lower bound. Introducton For graphs G and H, the Ramsey number r(g, H) s defned to be the smallest natural number N such that every red/blue edge-colorng of the complete graph K N on N vertces contans a red copy of G or a blue copy of H. One obvous constructon, noted by Chvátal and Harary [6], whch gves a lower bound for these numbers s to take χ(h) dsjont red clques of sze G and to connect every par of vertces whch are n dfferent clques by a blue edge. If G s connected, the resultng graph contans nether a red copy of G nor a blue copy of H, so that r(g, H) ( G )(χ(h) ) +. Burr [8] strengthened ths bound by notng that f σ(h) s the smallest color class n any χ(h)-colorng of the vertces of H, we may add a further red clque of sze σ(h), obtanng r(g, H) ( G )(χ(h) ) + σ(h). Followng Burr and Erdős [8, ], we say that a graph G s H-good f the Ramsey number r(g, H) s equal to ths bound. If G s a famly of graphs, we say that G s H-good f all suffcently large graphs n G are H-good. When H = K s, where σ(k s ) =, we smply say that G or G s s-good. The classcal result on Ramsey goodness, whch predates the defnton, s the theorem of Chvátal [5] showng that all trees are s-good for any s. On the other hand, the famly of trees s not H-good for every graph H. For example [3], a constructon of K 2,2 -free graphs due to Brown [7] allows one to show that there s a constant c < 2 such that r(k,t, K 2,2 ) t + t t c Mathematcal Insttute, Oxford OX 3LB, Unted Kngdom. Emal: davd.conlon@maths.ox.ac.uk. Research supported by a Royal Socety Unversty Research Fellowshp. Department of Mathematcs, MIT, Cambrdge, MA 0239-4307. Emal: fox@math.mt.edu. Research supported by a Smons Fellowshp, an MIT NEC Corp. award and NSF grant DMS-06997. Department of Mathematcs, UCLA, Los Angeles, CA, 90095. Emal: choongbum.lee@gmal.com. Research supported n part by a Samsung Scholarshp. Department of Mathematcs, UCLA, Los Angeles, CA 90095. Emal: bsudakov@math.ucla.edu. Research supported n part by NSF grant DMS-085, by AFOSR MURI grant FA9550-0--0569 and by a USA-Israel BSF grant.
for t suffcently large. Ths s clearly larger than ( K,t )(χ(k 2,2 ) ) + σ(k 2,2 ) = t + 2. In an effort to determne what propertes contrbute to beng Ramsey good, Burr and Erdős [9, ] conjectured that f was fxed then the famly of graphs wth bounded maxmum degree should be s-good for any s (and perhaps even H-good for all H). Ths conjecture holds for bpartte graphs H [2] but s false n general, as shown by Brandt [6]. He proved that for 0 almost every - regular graph on a suffcently large number of vertces s not even 3-good. Hs result (and a smlar result n [28]) actually prove somethng stronger, namely that f a graph G has strong expanson propertes then t cannot be 3-good. On the other hand, t has been shown f a famly of graphs exhbts poor expanson propertes then t wll tend to be good [, 28]. To state the relevant results, we defne the bandwdth of a graph G to be the smallest number l for whch there exsts an orderng v,..., v n of the vertces of G such that every edge v v j satsfes j l. Ths parameter s known to be ntmately lnked to the expanson propertes of the graph. In partcular, any bounded-degree graph wth poor expanson propertes wll have sublnear bandwdth [4]. The frst such result, shown by Burr and Erdős [], states that for any fxed l the famly of connected graphs wth bandwdth at most l s s-good for any s. Ths result was recently extended by Allen, Brghtwell and Skokan [], who showed that the set of connected graphs wth bandwdth at most l s H-good for every H. Ther result even allows the bandwdth l to grow at a reasonable rate wth the sze of the graph G. If G s known to have bounded maxmum degree, ther results are partcularly strong, sayng that for any and any fxed graph H there exsts a constant c such that f G s a graph on n vertces wth maxmum degree and bandwdth at most cn then G s H-good. Many of the orgnal problems of Burr and Erdős [] have now been resolved [28], but one that remans open s to determne whether the famly of hypercubes s s-good for every s. The hypercube Q n s the graph on vertex set {0, } n where two vertces are connected by an edge f and only f they dffer n exactly one coordnate. Ths famly of graphs has sublnear bandwdth but does not have bounded degree, so the result of Allen, Brghtwell and Skokan does not apply. To get a frst bound for r(q n, K 3 ), note that a smple greedy embeddng mples that any graph wth maxmum degree d and at least dn + 2 n vertces has a copy of Q n n ts complement. Suppose now that the edges of a complete graph have been 2-colored n red and blue and there s nether a blue trangle nor a red copy of Q n. Then, snce the blue neghborhood of any vertex forms a red clque, the maxmum degree n blue s at most 2 n. Hence, the graph must have at most (2 n )n + 2 n < 2 n (n + ) vertces. We may therefore conclude that r(q n, K 3 ) 2 n (n + ). It s not hard to extend ths argument to show that for any s there exsts a constant c s such that r(q n, K s ) c s 2 n n s 2. Ths s essentally the best known bound. Here we mprove ths bound, obtanng the frst upper bound whch s wthn a constant factor of the lower bound. Theorem.. For any natural number s 3, there exsts a constant c s such that r(q n, K s ) c s 2 n. The orgnal queston of Burr and Erdős [] relates to s-goodness but t s natural to also ask whether the famly of cubes s H-good for any H. For bpartte H, ths follows drectly from a result of Burr, Erdős, Faudree, Rousseau and Schelp [2]. Our result clearly mples that for any H, there s a constant c H such that r(q n, H) c H 2 n. For trangles, the rough dea of the proof s to show that f a red/blue edge-colorng of K N does 2
not contan a blue trangle then t may be tled wth a collecton of red clques whch have low blue densty between them. A red copy of Q n may then be found by nsertng subcubes nto each of the red clques and patchng them together usng the fact that the blue densty between these dfferent clques s low. Throughout the argument, t s very mportant to keep close control on the sze and number of the red clques as the defnton of low blue densty n a partcular bpartte subgraph wll depend on the sze of the two clques formng ts endponts. For K 4, the method s an extenson of ths dea, except the tlng wll now consst of clques whch are free of blue trangles, agan wth low blue densty between dfferent clques. In order to make ths useful for embeddng cubes, we then have to perform a second level of tlng, splttng each such clque nto red subclques wth low densty between them, as was already done for trangles. In general, when we consder K s, there wll be s 2 levels of tlng to keep track of. Ths makes the bookkeepng somewhat complex. Accordngly, we have chosen to present the proof n stages, frst consderng trangles, then K 4 and only then the general case. Although ths leads to some redundances, t allows us to ntroduce the addtonal concepts needed for each step at a reasonable pace. More generally, one may ask whch famles of graphs are s-good for all s. A powerful result proved by Nkforov and Rousseau [28] shows that graphs wth small separators are s-good. They used ths result to resolve a number of the orgnal questons of Burr and Erdős [] regardng Ramsey goodness. Let the degeneracy d(g) of a graph G be the smallest natural number d such that every nduced subgraph of G has a vertex of degree at most d. Furthermore, we say that a graph G has a (t, η)-separator f there exsts a vertex subset T V (G) such that T t and every connected component of V (G)\T has sze at most η V (G). The result of Nkforov and Rousseau then says that for any s 3, d and 0 < γ <, there exsts η > 0 such that the class G of d-degenerate graphs G wth a ( V (G) γ, η)-separator s s-good. We wll apply ths theorem together wth the Alon-Seymour-Thomas separator theorem for graphs wth a forbdden mnor [2] to show that for any s 3 any famly of graphs wth a forbdden mnor s s-good. A graph H s sad to be a mnor of G f H can be obtaned from a subgraph of G by contractng edges. By an H-mnor of G, we mean a mnor of G whch s somorphc to H. For a graph H, let G H be the famly of graphs whch do not contan a H-mnor. Theorem.2. For every fxed graph H, the class G H of graphs G whch do not contan an H-mnor s s-good for all s 3. In partcular, snce the famly of planar graphs conssts exactly of those graphs whch do not contan K 5 or K 3,3 as a mnor, we have the followng corollary. Corollary.. The famly of planar graphs s s-good for all s 3. A mnor-closed famly G s a collecton of graphs whch s closed under takng mnors. The graph mnor theorem of Robertson and Seymour [29] states that any mnor-closed famly of graphs may be characterzed by a fnte collecton of forbdden mnors. We say that a mnor-closed famly s nontrval f t s not the class consstng of all graphs. Note that the followng corollary s an mmedate consequence of Theorem.2 as any nontrval mnor-closed famly G s a subfamly of G H, where H s a graph not n G. Corollary.2. Any nontrval mnor-closed famly of graphs G s s-good for all s 3. 3
We wll begn, n Secton 2, by studyng the Ramsey number of cubes versus trangles. We wll then show, n Secton 3, how our arguments extend to K 4 before treatng the general case n Secton 4. In Secton 5, we wll prove Theorem.2. We wll conclude wth some further remarks. All logarthms are base 2 unless otherwse ndcated. For the sake of clarty of presentaton, we systematcally omt floor and celng sgns whenever they are not crucal. We also do not make any serous attempt to optmze absolute constants n our statements and proofs. 2 Trangle versus cube The argument works for n 6. Consder a colorng of the edges of the complete graph K N on the vertex set [N] = {, 2,, N} for N 7000 2 n wth two colors, red and blue, and assume that there are no blue trangles. We wll prove that ths colorng contans a red Q n. 2. Preprocessng the colorng For each d = 0,, 2,, log n + 3, we use the followng procedure to construct a famly S of subsets of [N] (note that log n + 3 n for n 6): If there exsts a set S whch nduces a red clque of order exactly 4 2 n d, then arbtrarly choose one, add t to the famly S, and remove the vertces of the clque from [N]. We defne the codmenson d(s) of such a set as d(s) = d. When there are no more such red clques, contnue to the next value of d. In the end, f we have S S S N 2, then we let S be our famly. Otherwse, f S S S < N 2, we add the set of remanng vertces to S, and declare t to have codmenson zero (note that ths set has sze at least N 2 4 2n ). In ether of the cases, we have S S S N 2. Note that we also have the followng propertes. Proposton 2.. () For an nteger, let X = S S,d(S) S. Then each vertex v [N] has at most 2 n +3 blue neghbors n X. () For every set S S, the subgraph nduced by S has maxmum blue degree at most 2n d(s) 2n. Proof. () Gven and a set X defned as above, note that no subset of X nduces a red clque of sze at least 4 2 n + = 2 n +3, snce such a set would have been added to S n the prevous round. On the other hand, snce there are no blue trangles, the blue neghborhood of every vertex nduces a red clque. Therefore, for every v [N], v has at most 2 n +3 blue neghbors n X. () The clam s trvally true for d > 0, snce every set of codmenson at least s a red clque. For d = 0, f S s not a red clque, then t s the set of remanng vertces n our procedure. Thus, there are no subsets of S whch nduce a red clque of sze at least 4 2 n (log n+3), snce such a set would have been added earler to the famly S. On the other hand, snce there are no blue trangles, the neghborhood of every vertex nduces a red clque. Therefore, every vertex of S has at most 4 2 n (log n+3) 2n 2n blue neghbors n S. 2.2 Tlng the cube Our strategy s to frst decompose the cube Q n nto smaller cubes, and then to embed t pece by pece nto K N, placng one of the subcubes n each of the subsets from S. We represent the subcubes 4
of Q n by usng vectors n {0,, } n. For example, the vector (0,,,, ) wll represent the subcube {(0,, x,, x n 2 ) : (x,, x n 2 ) {0, } n 2 }. We seek a tlng of Q n, whch we defne as a collecton of vertex-dsjont cubes that covers all the vertces of Q n and uses only cubes that have all ther fxed coordnates at the start of the vector. That s, they are of the form C = (a, a 2,, a d,,,, ) for some d 0 and a, a 2,, a d {0, }. We wll call such a cube specal and let d be the codmenson d(c) of C. We say that two dsjont cubes C and C are adjacent, f there are vertces x C and x C whch are adjacent n Q n. Note that the followng lst of propertes holds for such cubes and a tlng composed from them. Proposton 2.2. () If two specal cubes C and C ntersect, then we have C C or C C. () Two dsjont cubes C = (a,, a d,,, ) and C = (b,, b d,,, ) for d d are adjacent f and only f (a,, a d ) and (b,, b d ) dffer n exactly one coordnate. () For a tlng C and a specal cube C C of codmenson d, there are at most d other specal cubes C C of codmenson d(c ) d whch are adjacent to C. Proof. () Let d = d(c), d = d(c ) and, wthout loss of generalty, suppose that d d. Suppose that (a,, a n ) C C. Then, snce we are only consderng specal cubes, we must have C = (a,, a d,,, ) and C = (a,, a d,,, ). However, we then have C C. () If (a,, a d ) = (b,, b d ), then we have C C, and thus we may assume that ths s not the case. Cubes C and C are adjacent f and only f there are two vectors v = (a,, a d, x d+,, x n ) and w = (b,, b d, y d +,, y n ) n {0, } n whch dffer n exactly one coordnate. However, snce the two vectors restrcted to the frst d coordnates are already dfferent, ths can happen only f (a,, a d ) and (b,, b d ) dffer n exactly one coordnate. Moreover, one can see that f ths s the case, then there s an assgnment of values to x and y j so that v and w ndeed dffer n exactly one coordnate. () Snce C has codmenson d, there are only d coordnates that one can flp from C to obtan a cube adjacent to C. As we have already mentoned, our tlng C of the cube Q n wll be constructed n correspondence wth the famly S constructed n Secton 2.. We wll construct the tlng C by fndng cubes of the tlng one by one. We slghtly abuse notaton and use C also to denote the partal tlng, where only part of the cube Q n s covered. At each step, we wll fnd a subcube C whch covers some non-covered part of Q n, and assgn t to some set S C S. We say that such an assgnment s proper f the followng propertes hold. Proper assgnment.. If C has codmenson d, then S C has codmenson d. 2. Suppose that C s adjacent to some cube C already n the tlng, and that C s assgned to S C. Then the bpartte graph nduced by S C and S C contans at most S C S C 6δ 2 blue edges, where δ = max{d(s), d(s )}. We use S C to denote the set n S to whch C s assgned. Our algorthm for fndng the tlng C and the correspondng sets n S s as follows. Tlng Algorthm. At each step, consder all possble specal cubes C whch 5
(a) are dsjont from the cubes C C, (b) have d(c) d(c ) for all C C, and (c) for whch there exsts a set S S whch has not yet been assgned and such that assgnng C to S s a proper assgnment. Take a cube C 0 of mnmum codmenson satsfyng (a),(b),(c), and add t to the tlng. Assgn C 0 to the set S C0 S gven by (c). The followng proposton shows that the algorthm wll termnate successfully. Proposton 2.3. If the tlng s not complete, then the algorthm always chooses a cube from a non-empty collecton. Proof. If S contans a set of codmenson zero, then the algorthm wll choose the whole cube Q n n the frst step and assgn t to a set of codmenson zero. In ths case we have a trval tlng, and there s nothng to prove. Thus, we may assume that S contans no set of codmenson zero. If ths s the case, note that for all C C, we have S C = 4 C. Let S be the subfamly of S consstng of sets whch are already assgned to some cube n C, and let S = S \ S. Note that S S S = S S S S S S N 2 C C S C 3500 2 n 4 2 n 3496 2 n. For each, let S = {S S : d(s) = }. If S 323 for all, then we have S = S 32 3 4 2 n = 28 2 n 3 2 = 3328 2n, whch s a contradcton. Therefore, there exsts an ndex for whch S > 323. We have > 0 snce S contans no set of codmenson zero. Snce the tlng s not complete, there exsts a vertex (a,, a n ) Q n whch s not yet covered by any of the cubes already n C. Consder the cube C = (a,, a,,, ). By Proposton 2.2, there are at most cubes n our partal tlng C of codmenson at most whch are adjacent to ths cube. Let C,, C j for j be these cubes, and fx an ndex a j. We say that a set S S s bad for S Ca f there are at least S 6 2 Ca S blue edges between S Ca and S. We clam that there are at most 32 2 sets n S whch are bad for S C a. Let X = S S S and note that, by Proposton 2. (), there are at most S C a 2 n +3 blue edges between the sets S Ca and X. Each set S S whch s bad for S C a accounts for at least S 6 2 Ca S = 2n +3 S 32 2 Ca such blue edges (note that S = 4 2 n ). Therefore, n total, there are at most 32 2 sets n S whch are n bad relaton wth S C a, as clamed above. Snce S > 323, there exsts a set S S whch s not bad for any of the sets S C a for a j. Suppose that n the prevous step, we embedded some cube of codmenson d. In order to show that C satsfes (a) and (b), t suffces to verfy that d, snce d mples the fact that C s dsjont from all the other cubes n C (f C ntersects some other cube, then that cube must contan C and therefore also contans (a,, a n ) by Proposton 2.2). Furthermore, f ths s the case, assgnng 6
C to S s a proper assgnment (thus we have (c)), snce C,, C j are the only cubes adjacent to C already n the tlng. Suppose, for the sake of contradcton, that < d and consder the tme t mmedately after we last embedded a cube of codmenson at most. At tme t, snce (a,, a n ) was not covered, the cubes n C are dsjont from C. Moreover, the set of cubes of codmenson at most whch are adjacent to C s the same as at the current tme. Therefore, C could have been added to the tlng at tme t as well, and ths contradcts the fact that we always choose a cube of mnmum codmenson. Thus we have d as clamed. Note that as an outcome of our algorthm, we obtan a tlng C for whch every par of adjacent cubes C, C C have at most 6δ 2 S C S C blue edges between S C and S C, where δ = max{d(c), d(c )}. 2.3 Imposng a maxmum degree condton Havng constructed the tlng C and made the assgnment of the cubes to sets n S, we now wsh to mpose certan maxmum degree condtons between the sets S C for C C. For a set C C of codmenson d = d(c), let C,, C j be the cubes of codmenson at most d whch are adjacent to C. Note that we have j d by Proposton 2.2. Moreover, there are at most S 6d 2 C S Ca blue edges between S C and S Ca for all a j. Now, for each a, remove all the vertces n S C whch have at least 8d S C a blue neghbors n S Ca, and let T C be the subset of S C left after these removals. Snce there are at most S 6d 2 C S Ca blue edges between S C and S Ca, for each ndex a, we remove at most S C 2d vertces from S C, and thus the resultng set T C s of sze at least S C 2 2 2 n d. Note that all the vertces n T C have blue degree at most 8d S C a 4d T C a n the set T Ca. That s, we have the followng property. Maxmum degree condton. For each par of adjacent cubes C, C C wth d(c) d(c ), every vertex n T C has at most 4d(C) T C blue neghbors n the set T C. 2.4 Embeddng the cube We now show how to embed Q n. Recall that Q n was tled by cubes n C, each correspondng to a subset from the famly S. We wll greedly embed these cubes one by one nto ther assgned sets from the famly S, n decreasng order of ther codmensons. If there are several cubes of the same codmenson, then we arbtrary choose the order between them. For each C C, we wll greedly embed the vertces of C nto the set T C S C. Let d = d(c). Suppose that we are about to embed x C and let f : Q n [N] denote the partal embeddng of the cube Q n obtaned so far. Note that x has at most d neghbors x, x 2,, x j (for j d) whch are already embedded and belong to a cube other than C. Snce we have only embedded cubes of codmenson at least d, the maxmum degree condton mposed n Secton 2.3 mples that all the vertces f(x ) have blue degree at most 4d T C n the set T C. Together, these neghbors forbd at most 4 T C vertces of T C from beng the mage of x. In addton, x has at most n d neghbors y,, y k (for k n d) whch are already embedded and belong to C. By Proposton 2., each vertex f(y ) has blue degree at most 2n d 2n n the set T C. Together, these neghbors forbd at most 2n d 2 vertces of T C from beng the mage of x. Fnally, 7
there are at most 2 n d vertces n T C whch are mages of some other vertex of C that s already embedded. Therefore, the number of vertces n T C nto whch we cannot embed x s at most 4 T C + 2 2n d + (2 n d ), whch s less than T C snce T C 2 2 n d. Hence, there exsts a vertex n T C whch we can choose as an mage of x to extend the current partal embeddng of the cube. Repeatng ths procedure untl we fnsh embeddng the whole cube Q n completes the proof. 3 Clques of order 4 The argument for general clques s smlar to that for trangles gven n the prevous secton. However, there are several new concepts nvolved and, to slowly develop the necessary concepts, we frst provde a proof of the next case, whch s K 4 versus a cube. Recall that n the trangle case, we started by teratvely fndng sets whch formed a red clque (see Secton 2.). Red clques were a natural choce, snce the blue neghborhood of every vertex formed a red clque. Ether we were able to fnd large red clques or we were able to restrct the maxmum blue degree of vertces n some way (see Proposton 2.). If one attempts to employ the same strategy for the K 4 case, then the natural choce of sets that we should take nstead of red clques are blue trangle-free sets, snce the blue neghborhood of every vertex now forms a blue trangle-free set. Suppose that we found a famly S of blue trangle-free sets. Snce blue K 3 -free sets are not as powerful as red clques for embeddng subgraphs, we repeat the whole argument wthn each blue trangle-free set S S, to obtan red clques whch are subsets of S. By so dong, we obtan a second famly of sets S 2, consstng of red clques. We refer to sets n S l as level l sets, and for a set S S l, we defne ts level as l(s) = l. In order to fnd an embeddng of the cube Q n usng a strategy smlar to that used n the trangle case, we wsh to fnd a tlng of Q n, and for each cube C n the tlng, a red clque S C S 2 so that for two adjacent cubes C and C, the sets S C and S C stand n good relaton. However, drectly fndng such a tlng and an assgnment s somewhat dffcult snce we do not have good control on the blue edges between the sets n S 2. To be more precse, suppose that we are gven S, S S and subsets S 2 S, S 2 S n S 2. Then the control on the blue edges between S 2 and S 2 s nherted from the control on the blue edges between S and S, and thus depends on the relatve szes of S and S, not on the relatve szes of S 2 and S 2 as n the trangle case (unless S = S ). To crcumvent ths dffculty, we wll need to mantan tght control on the edge densty between dfferent sets n S as well as those n S 2. We seek a double tlng, whch s defned to be a par C = C C 2 of tlngs satsfyng the property that for every C 2 C 2 there exsts a cube C C such that C 2 C (n other words, C 2 s a refned tlng of C ). We refer to cubes n C l as level l cubes, and for a cube C C l, we defne ts level as l(c) = l. Our goal s to fnd, for each l =, 2, an assgnment of cubes n C l to sets n S l. The followng are the key new concepts nvolved n the K 4 case. Defnton 3.. () For a cube C C, we defne ts -codmenson as d (C) = d(c). For a cube C C 2 contaned n a cube C C, we defne ts -codmenson as d (C) = d(c ) and ts 2-codmenson as d 2 (C) = d(c) d(c ). 8
() Suppose C, C C are adjacent. We say that they have level adjacency f the level cubes contanng C and C are dfferent. Otherwse, they have level 2 adjacency (note that C and C are both level 2 cubes n the second case). After fndng the double tlng, n order to fnd an embeddng of the cube, we only consder the level 2 tlng (as explaned above, we need to go through a level tlng n order to obtan a nce level 2 tlng). For two adjacent cubes C, C C 2, the defnton of the correspondng sets S C, S C beng n good relaton (see the defnton of proper assgnment n Secton 2.2) now depends on the type of adjacency they have. If they have level adjacency, then the good relaton wll be defned n terms of ther -codmensons, and f they have level 2 adjacency, then t wll be n terms of ther 2-codmensons. By dong so, we can overcome the above mentoned dffculty of not havng enough control on the blue edges between S C and S C. Afterwards, we proceed as n the prevous secton, by mposng a maxmum degree condton between cubes n C 2 and then embeddng the cube. We now provde the detals of the proof. Our argument works for n 32. Consder a colorng of the edges of the complete graph K N on the vertex set [N] for N 2 46 2 n wth two colors, red and blue, and assume that there s no blue K 4. We prove that ths colorng contans a red Q n. 3. Preprocessng the colorng Note that we have n 2 log n + 2 for n 32. We use the followng procedure to construct our frst level S of subsets of [N]: For each d = 0,, 2,, log n+8, f there exsts a set S whch nduces a blue trangle-free graph of order exactly 2 8 2 n d, then arbtrarly choose one, add t to the famly S, and remove the vertces of S from [N]. We defne the -codmenson d (S) of such a set as d (S) = d. When there are no more such sets, contnue to the next value of d. If, after runnng through all values of d, S S S < N 2, then add the set of remanng vertces to S, and declare t to be an exceptonal set wth -codmenson zero (note that ths set has sze at least N 2 28 2 n ). Now we perform a smlar decomposton for each set n S to construct our second level S 2. For each set n S S, consder the followng procedure (suppose that S has codmenson d = d(s )):. If S s not exceptonal: for each d = 0,, 2,, log n + 3, f there exsts a subset S S whch nduces a red clque of order exactly 8 2 n d d, then arbtrarly choose one, add t to the famly S 2, and remove the vertces of the clque from S. We defne the 2-codmenson of such a set as d 2 (S) = d and the -codmenson as d (S) = d. When there are no more such red clques, contnue to the next value of d. If, after runnng through all values of d, S S 2,S S S < S 2, then add the set of remanng vertces to S 2, and declare t to be an exceptonal set wth 2-codmenson zero and -codmenson d (note that ths set has sze at least S 2 8 2 n d ). 2. If S s exceptonal: add S to the famly S 2. Defne the 2-codmenson of S S 2 as zero and ts -codmenson as zero. Let S = S S 2 (we suppose that S s a mult-famly and, thus, f a set s n both S and S 2, then we have two copes of ths set n S, however they can be dstngushed by ther levels). For a set S S, defne ts codmenson as d(s) = l(s) d (S). 9
Proposton 3.. () S S S N 2, and for every S S, we have S S 2,S S S S 2. () For an nteger, let X = S S,d (S) S. Then each vertex v [N] has at most 29 2 n blue neghbors n X. () For a set S S and an nteger, let X = S. S S 2,S S,d 2 (S) Then each vertex v S has at most 6 2 n d (S ) blue neghbors n X. (v) For every set S S 2, the subgraph nduced by S has maxmum blue degree at most 2n d(s) n. Proof. Part () s clear and we omt the proof of (), snce ts proof s smlar to that of part () of Proposton 2.. () Suppose that S s an exceptonal set. Then S s the unque set S satsfyng S S 2 and S S. Snce d 2 (S ) = 0, X s an empty set for every. Thus the concluson follows. Now suppose that S s not an exceptonal set. Snce S nduces a blue trangle-free set, a blue neghborhood n S of a vertex v S forms a red clque. Thus f a vertex v S has more than 6 2 n d (S ) blue neghbors n X, then nsde the neghborhood of v, we can fnd a set nducng a red clque of sze at least 8 2 n d (S ) ( ). However, ths set must have been added n the prevous step. Therefore, there are no such vertces. (v) Suppose that S S wth S S. If both S and S are not exceptonal sets, then S nduces a red clque, and thus the concluson follows. Suppose that S s exceptonal, and S s not. Then as n (), we can see that there s no vertex n S whch has at least 8 2 n d (S) (log n+3) = 2n d(s) n blue neghbors n S. One can smlarly handle the case when S s exceptonal, snce we have S = S n ths case. 3.2 Tlng the cube The followng proposton s smlar to Proposton 2.2 (we omt ts proof). Proposton 3.2. Let C = C C 2 be a double tlng, and let C C be a specal cube of codmenson d. () If C s a level cube, then for each l =, 2, there are at most d (C) specal cubes of level l and codmenson at most d whch are adjacent to C (and C has level adjacency wth all these cubes). () If C s a level 2 cube, then there are at most d 2 (C) specal cubes of codmenson at most d whch have level 2 adjacency wth C (they necessarly are of level 2), and for l =, 2, at most d (C) cubes of level l and codmenson at most d whch have level adjacency wth C. Our double tlng C of the cube Q n wll be constructed n correspondence wth the famly S constructed n Secton 3.. We wll construct C by fndng cubes of the tlng one by one. We slghtly abuse notaton and use C also to denote the partal double tlng, where only part of the cube Q n s covered. Ideally, we would lke to construct C by constructng the level tlng C frst, and then the level 2 tlng C 2. However, as we wll soon see, t turns out that constructng the tlng n ncreasng order of codmenson s more effectve than n ncreasng order of level. At each step, we fnd a 0
subcube C whch covers some non-covered part of Q n, and assgn t to some set S C S. We say that such an assgnment s proper f the followng propertes hold. Proper assgnment.. l(c) = l(s C ) and, for l = l(c), we have d l (C) = d l (S C ). 2. If C 2 C wth C C and C 2 C 2, then S C2 S C. 3. Suppose that C s adjacent to some cube C whch s already n the tlng, where C s assgned to S C and C and C have level ρ adjacency. Then the number of blue edges n the bpartte graph nduced by S C and S C s at most S C S C (8δ) l(c)+l(c ) 6, where δ = max{d ρ (C), d ρ (C )}. We use S C to denote the set n S to whch C s assgned. Our algorthm for fndng the tlng C and the correspondng sets n S s as follows. Tlng Algorthm. At each step, consder all possble specal cubes C whch (a) can be added to C to extend the partal tlng, (b) have d(c) d(c ) for all C C, and (c) for whch there exsts a set S S whch has not yet been assgned and such that assgnng C to S gves a proper assgnment. Take a cube C 0 of mnmum codmenson satsfyng (a),(b),(c), and add t to the tlng. Assgn C 0 to the set S C0 S gven by (c). Condton (a) s equvalent to sayng that ether C s dsjont to all cubes n C, or s contaned n some cube n C and s dsjont to all cubes n C 2. The followng proposton shows that the algorthm wll termnate successfully. Proposton 3.3. If the tlng s not complete, then the algorthm always chooses a cube from a non-empty collecton. Proof. Suppose that n the prevous step we embedded some cube of codmenson d (let d = 0 for the frst teraton of the algorthm). Snce the tlng s not complete, there exsts a vertex (a,, a n ) Q n whch s not covered twce. Case : (a,, a n ) s not covered by any of the cubes n C. Let S be the subfamly of S consstng of sets whch are already assgned to some cube n C, and let S = S \ S. If S contans a set of codmenson zero, then the correspondng cube s the whole cube Q n, and t contradcts the fact that (a,, a n ) s not covered. Thus we may assume that there s no set of codmenson zero n S, from whch t follows that S C = 2 8 C for all C C. Note that S S S = S S S S S S N 2 C C S C 2 45 2 n 2 8 2 n > 2 44 2 n.
For each, let S = {S S : d(s) = }. Suppose that S 24 5 for all. Then, snce each set n S has sze 28 2 n, we have S = S 2 4 5 2 8 2 n = 2 32 2 n 5 2 < 244 2 n, whch s a contradcton. Therefore, there exsts an ndex for whch S > 24 5. We have > 0 snce otherwse the set of codmenson zero would have been added to the tlng n the frst step. Let C = (a,, a,,, ). For each l =, 2, let A l be the famly of cubes of level l n our partal double tlng whch are adjacent to C and have codmenson at most. By Proposton 3.2, we have A l for each l. For a cube A A l, we say that a set S S s bad for A f there are at least S (8) 5 l A S blue edges between S A and S. Otherwse, we say that S s good for A. We clam that there are at most 2 3 4 sets n S whch are bad for each fxed A. Let X = S S S, and note that by Proposton 3., there are at most S A 2 9 2 n blue edges between the sets S A and X. Each set S S whch s bad for A accounts for at least (8) 5 l S A S 28 2 n (8) 4 S A = 26 2 n 4 S A such blue edges (note that S = 2 8 2 n ). Therefore, n total, there are at most 2 3 4 sets n S whch are n bad relaton wth A, as clamed above. Snce A A 2 2 and S > 24 5, there exsts S S whch s good for all cubes n A A 2. In order to show that C satsfes (a) and (b), t suffces to verfy that d, snce d mples the fact that C s dsjont from all the other cubes n C (f C ntersects some other cube, then that cube must contan C and therefore also contans (a,, a n ) by Proposton 2.2). Furthermore, f ths s the case, assgnng C to S s a proper assgnment (thus we have (c)). Now suppose, for the sake of contradcton, that < d and consder the tme t mmedately after we last embedded a cube of codmenson at most. At tme t, snce (a,, a n ) was not covered, the cubes n C are dsjont from C. Moreover, the set of cubes of codmenson at most whch are adjacent to C s the same as at current tme. Therefore, C could have been added to the tlng at tme t as well, and ths contradcts the fact that we always choose a cube of mnmum codmenson. Thus we have d as clamed. Case 2: (a,, a n ) s covered by a cube C C but not by a cube n C 2. In ths case, n order to assgn some cube C C contanng (a,, a n ), to a subset of S C n S 2, there are two types of adjacency that one needs to consder. Let C 2 = (a,, a d,,, ) and temporarly consder t as a level 2 cube. We frst consder the cubes whch are -adjacent to C 2 and remove all the subsets of S C whch are bad for these cubes. The reason we consder cubes whch are adjacent to C 2 nstead of C s because we do not know what C wll be at ths pont (note that, dependng on the choce of C, some of these cubes may not be -adjacent to C). However, snce we choose C C 2 (as we wll show later n the proof), the set of cubes whch are -adjacent to C wll be a subset of the set of cubes whch are -adjacent to C 2. Havng removed all bad subsets, we have enough nformaton to determne C. Then, by consderng the relaton of the remanng subsets of S C to cubes whch are 2-adjacent to C, we can fnd a set S C that can be assgned to C. Let d = d(c ) and let A () be the subfamly of our partal embeddng C consstng of cubes whch are -adjacent to C 2. Note that, by Proposton 3.2(), there are at most d cubes whch are -adjacent 2
to C 2 n each level, and thus A () 2d. Let F = {S S 2 : S S C }. By Proposton 3., we have S F S S C 2. We say that a set S F s bad for a cube A A () f there are at least (8δ A ) l(s)+l(sa) 6 S S A = S S A blue edges between S and S (8δ A ) 4 l(a) A (where δ A = max{d, d (A)}). Otherwse, we say that S s good for A. For each fxed A, let F A be the subfamly of F consstng of sets whch are bad for F. By the properness of the assgnment up to ths pont, we know that there S are at most C S A blue edges between S (8δ A ) 5 l(a) C and S A for every A A (). Therefore, by countng the number of blue edges between S C and S A n two ways, we see that S S A (8δ S F A ) 4 l(a) S C S A (8δ A ) 5 l(a), A from whch we have S F A S 8δ A S C 8d S C. Let F be the subfamly of F of sets whch are already assgned to some cube n C 2. There are no sets of relatve codmenson zero n F snce ths mples that (a,, a n ) s covered by a cube n C 2. It thus follows that for every C C 2 such that C C, we have S C = 8 C, and S F S = S F S 8 C C C 8 C = 8 2 n d. Let S = F \ (F A A () F A) be the subfamly of F = {S S 2 : S S C } of sets whch are not assgned to any cubes yet and are good for all the cubes n A (). Snce S C 2 8 C = 2 8 2 n d, we have S S S S F S C 2 S S F S A A () S F A S 8 2 n d 2d S C 8d > 2 5 2 n d. For each 0, let S = {S S : d 2 (S) = } = {S S : d(s) = d + }. Suppose that S 283 for all. Then, snce we have S = 8 2 n d for all S S, S = S 28 3 8 2 n d < 2 5 2 n d, whch s a contradcton. Therefore, there exsts an ndex for whch S > 28 3. Let C = (a,, a d +,,, ) and consder t as a level 2 cube. By Proposton 3.2(), there are at most cubes n C whch have level 2 adjacency wth C and have codmenson at most d +. Let A (2) be the famly consstng of these cubes. For a cube A A (2), we say that a set S S s bad for A f there are at least (8) l(sa)+l(s) 6 S A S = S (8) 2 A S blue edges between S A and S. Otherwse, we say that S s good for A. We clam that there are at most 28 2 sets n S whch are bad for each fxed A. Let X = S S S, and note that by Proposton 3., there are at most S A 6 2 n blue edges between the sets S A and X. Each set S S whch s bad for A accounts for at least (8) 2 S A S = 8 2n 64 2 S A = 2n 8 2 S A 3
such blue edges (note that S = 8 2 n ). Therefore, n total, there are at most 28 2 sets n S whch are bad for A, as clamed above. Snce there are at most sets n A (2) and S > 283, there exsts a set S S whch s good for all the cubes A A(2). In order to show that C satsfes (a) and (b), t suffces to verfy that d + d, snce ths mples the fact that C s dsjont from all the other cubes n C 2 (note that C C, and f C ntersects some other cube of level 2, then that cube must contan C and therefore also contans (a,, a n ) by Proposton 2.2). Furthermore, f ths s the case, assgnng C to S s a proper assgnment (thus we have (c)). Now suppose, for the sake of contradcton, that d + < d and consder the tme t mmedately after we last embedded a cube of codmenson at most d +. At tme t, snce d + d = d(c ), the cube C was already embedded and, snce there are no cubes of level 2 coverng (a,, a n ), the cubes n C 2 are dsjont from C. A cube n the partal embeddng at tme t, whch s of codmenson at most d +, s -adjacent to C f and only f t s -adjacent to C 2. Hence, the famly of cubes whch are adjacent to C at tme t s a subfamly of A () A (2). Therefore, C could have been added to the tlng at tme t as well, and ths contradcts the fact that we always choose a cube of mnmum codmenson. Thus we have d + d as clamed. Note that as an outcome of our algorthm, we obtan a tlng C such that for every par of adjacent cubes C, C C, we have control on the number of blue edges between S C and S C (as gven n the defnton of proper assgnment). 3.3 Imposng a maxmum degree condton As n the trangle case, we now mpose certan maxmum degree condtons between the sets S C for C C. It suffces to mpose maxmum degree condtons between sets of level 2. For a set C C 2 of codmenson d = d(c), and relatve codmensons d = d (C), d 2 = d 2 (C), recall that we have a set S C S such that S C 8 2 n d. Let A ρ be the famly of cubes n C 2 wth codmenson at most d whch have level ρ adjacency wth C. For each A A ρ, let δ A,ρ = max{d ρ, d ρ (A)}. By Proposton 3.2(), we have A ρ d ρ for each ρ =, 2. For each A A ρ, there are at most S 64δA,ρ 2 C S A blue edges between S C and S A. Now for ρ =, 2, and each A A ρ, remove all the vertces n S C whch have at least 8δ A,ρ S A blue neghbors n S A, and let T C be the subset of S C left after these removals. Snce there are at most S C S A blue edges between S C and S A, we remove at most S C 8δ A,ρ vertces from S C, for each set A A ρ. Thus the resultng set T C s of sze at least 64δ 2 A,ρ T C S C d S C d 2 S C S C 4 2 n d. 8δ A, 8δ A,2 2 For each A A ρ, all the vertces n T C have blue degree at most 8δ A,ρ S A 4δ A,ρ T A n the set T A. Thus we obtan the followng property. Maxmum degree condton. Let C, C C 2 be a par of cubes havng level ρ adjacency wth d(c) d(c ). Then every vertex n T C has at most 4δ ρ T C blue neghbors n the set T C (where δ ρ = max{d ρ (C), d ρ (C )}). 4
3.4 Embeddng the cube We now show how to embed Q n. Recall that we found a double tlng C of Q n. We wll greedly embed the cubes n the level 2 tlng C 2 one by one nto ther assgned sets from the famly S 2, n decreasng order of ther codmensons. If there are several cubes of the same codmenson, then we arbtrary choose the order between them. Suppose that we are about to embed the cube C C 2. Let d = d(c), d = d (C) and d 2 = d 2 (C). We wll greedly embed the vertces of C nto the set T C S C. Suppose that we are about to embed x C and let f : Q n [N] denote the partal embeddng of the cube Q n obtaned so far. For ρ =, 2, let A ρ be the set of neghbors of x whch are already embedded and belong to a cube other than C that has level ρ adjacency wth C. Note that we have A ρ d ρ for ρ = and 2. Snce we have so far only embedded cubes of codmenson at least d, we have, for each ρ =, 2, that the vertces f(v) for v A ρ have blue degree at most 4d ρ T C n the set T C, by the maxmum degree condton mposed n Secton 3.3. Together, these neghbors forbd at most d vertces of T C from beng the mage of x. T C + d 2 T C 4d 4d 2 2 T C. In addton, x has at most n d neghbors whch are already embedded and belong to C. By Proposton 3., for each such vertex v, f(v) has blue degree at most 2n d n n the set T C. Together, these neghbors forbd at most 2 n d vertces of T C from beng the mage of x. Fnally, there are at most 2 n d vertces n T C whch are mages of some other vertex of C that s already embedded. Therefore, the number of vertces n T C nto whch we cannot embed x s at most 2 T C + 2 n d + (2 n d ) < T C, where the nequalty follows snce T C 4 2 n d. Hence, there exsts a vertex n T C whch we can choose as an mage of x to extend the current partal embeddng of the cube. Repeatng ths procedure untl we fnsh embeddng the whole cube Q n completes the proof. 4 General case In ths secton, we further extend the arguments presented so far to prove the man theorem. The framework s very smlar to that of the prevous sectons, where we fnd multple levels of tlng of Q n. We begn by preprocessng the colorng to fnd famles of sets S = S 0 S S s 2 such that, for l, sets n S l are subsets of sets n S l whch do not contan blue K s l (except for some specal cases). We refer to the sets n S l as level l sets and, for a set S S l, we let ts level be l(s) = l. Note that, unlke n the cases where s = 3, 4, we have an addtonal level, level zero. Level zero wll consst of a sngle set [N] and t s there merely for techncal reasons. We then seek a correspondng (multple level) tlng C of the cube Q n. Defne an (s )-fold tlng (or (s )-tlng, n short) of Q n as a collecton of (s )-tlngs C 0 C C s 2, where C 0 s the trval tlng consstng of the unque cube Q n, and, for all l, C l s a refned tlng of C l (.e. for every C C l, there exsts C C l such that C C ). We refer to cubes n C l as level l cubes and, for a cube C C l, defne ts level as l(c) = l. We wll construct the tlng by fndng cubes C C 5
and assgnng each of them to some set S C S. Informally, ths means that the subcube C of Q n wll be found n the S C part of our graph. Note that the trval level zero cube Q n gets assgned to the trval level zero set [N] and ths fts the heurstc. For the rest of ths secton, we assume that s 5. Let c = s 5s and suppose that N c s 2 n = s 5s2 2 n. We wll later use the followng estmate. Lemma 4.. For every postve nteger s, = s 2 2s s. Proof. Let (x) t = (x ) (x t + ), X t = = t, and Y 2 t = () t = for non-negatve ntegers t. 2 The Strlng number S(t, k) of the second knd s the number of ways to partton a set of t objects nto k nonempty subsets. These numbers satsfy the followng well-known dentty x t = t k=0 S(t, k)(x) k (see, e.g., [32], Chapter.4). Ths mples the dentty X t = t S(t, k)y k. k=0 By takng the dervatve k tmes of both sdes of the equalty ( z) = 0 z, note that k! ( z) (k+) = ( ) ( k + )z k. By multplyng both sdes by z k and substtutng z = /2 we have that Y k = 2k!. Ths, together wth the above dentty, mples that X t = t k=0 2k!S(t, k). Although t wll be not be needed, we remark that there s an explct formula X t = 2 t k k=0 j=0 ( )k j( k j) j t whch follows from substtutng n the well-known dentty S(t, k) = k k! j=0 ( )k j( k j) j t. Let T s be the number of parttons of a set of s objects nto labelled nonempty subsets. By countng over the sze k of the partton, we have the dentty T s = s k=0 k!s(s, k) = X s/2. As each partton counted by T s s determned by the vector of labels of the sets contanng each object, and there are at most s such labels for each partton, we have T s s s, and the desred nequalty follows. 4. Preprocessng the colorng Let S 0 = {[N]} and [N] be the unque set of level zero and codmenson zero (denoted as l(s) = 0 and d(s) = 0). We construct the levels one at a tme. Once we fnsh constructng S l, for each set S S l, we use the followng procedure to construct sets belongng to the l-th level S l : For each d = 0,, 2,, log n + 2 log c, f there exsts a set S S whch nduces a blue K s l -free graph of order exactly c s l 2 n d(s ) d, then arbtrarly choose one, add t to the famly S l, and remove the vertces of S from S. We defne the l-codmenson d l (S) of such a set as d l (S) = d, and for =,, l, we defne the -codmenson d (S) as d (S) = d (S ). Let the codmenson of S be d(s) = l = d (S). When there are no more such sets, contnue to the next value of d. If, after runnng through all values of d, S S l,s S S < S 2, then add the set of remanng vertces to S l, and declare t to be an exceptonal set wth l-codmenson zero (note that ths set has sze at least S 2 = cs (l ) 2n d(s ) 2 c s l 2 n d(s ) ). 6
Let S = s 2 l=0 S l (we suppose that S s a mult-famly and f a set appears multple tmes we dstngush them by ther levels). The followng proposton s smlar to Proposton 3.. We omt ts proof. Proposton 4.. () For l s 2 and a set S S l, S S l,s S S S 2. () For ntegers l s 2 and, let S be a set of level l, and let X = S S l,s S,d l (S) Then each vertex v S has at most 2c s l 2 n d(s ) blue neghbors n X. () For every set S S s 2, the subgraph nduced by S has maxmum blue degree at most 2n d(s) n. S. 4.2 Tlng the cube In ths subsecton, we fnd an (s )-tlng of Q n. Recall that n the prevous secton, we had to control the blue edge denstes between adjacent cubes n the tlng. The parameter that governed the control of these denstes was defned n terms of the ρ-codmenson, where ρ was the level of adjacency of these cubes. Below we generalze ths concept. Defnton 4.. Let C be an (s )-tlng, and let C, C C be two adjacent cubes. () The level of adjacency ρ(c, C ) s the mnmum l such that the cubes of level l contanng C and C are dstnct. We say that C and C are ρ-adjacent f ρ(c, C ) = ρ. () The domnatng parameter δ(c, C ) s max{d ρ (C), d ρ (C )}, where ρ = ρ(c, C ). Note that the level of adjacences of two cubes C and C s at most mn{l(c), l(c )}. The followng proposton s smlar to Proposton 3.2 and we omt ts proof. Proposton 4.2. Let C be an (s )-tlng, and let C C be a level l specal cube of codmenson d. For each ρ =, 2,, l and l =, 2,, s 2, the number of specal cubes of level l and codmenson at most d whch are ρ-adjacent to C s at most d ρ (C). Our (s )-tlng C of the cube Q n wll be constructed n correspondence wth the famly S constructed n Secton 4.. We wll construct C by fndng cubes of the tlng one by one. We slghtly abuse notaton and use C also to denote the partal (s )-tlng, where only part of the cube Q n s covered. At each step, we fnd a subcube C whch covers some non-covered part of Q n, and assgn t to some set S C S. We say that such an assgnment s proper f the followng propertes hold. Proper assgnment.. l(c) = l(s C ) and, for l = l(c), we have d l (C) = d l (S C ). 2. If C C for C C, then S C S C. 3. Suppose that C s adjacent to some cube C already n the tlng, where C s assgned to S C, and that C and C have level ρ adjacency. Then the number of blue edges n the bpartte graph nduced by S C and S C s at most S C S C (4s 2 δ(c, C )) l(c)+l(c ) 2s. 7
We use S C to denote the set n S to whch C s assgned. Our algorthm for fndng the tlng C and the correspondng sets n S s as follows. Tlng Algorthm. At each step, consder all possble cubes C whch (a) can be added to C to extend the partal tlng, (b) have d(c) d(c ) for all C C, and (c) for whch there exsts a set S S whch has not yet been assgned and such that assgnng C to S gves a proper assgnment. Take a cube C 0 of mnmum codmenson satsfyng (a),(b),(c), and add t to the tlng. Assgn C 0 to the set S C0 S gven by (c). The followng proposton shows that the algorthm wll termnate successfully. Proposton 4.3. If the tlng s not complete, then the algorthm always chooses a cube from a non-empty collecton. Proof. In the begnnng, the algorthm wll take Q n as the level 0 tlng, and wll assgn t to [N], whch s the unque set of level 0. Now suppose that n the prevous step we embedded some cube of codmenson d. Snce the tlng s not complete, there exsts a vertex (a,, a n ) Q n whch s not covered (s )-tmes. Suppose that ths vertex s covered l tmes for l (s 2), and let C 0,, C l be the cubes of each level that cover t (note that C 0 C C l ). Let C l = (a,, a d,,, ). We temporarly consder C l as a level l cube. In ths case, n order to assgn some cube C C l contanng (a,, a n ) to a subset of S Cl n S l, we frst consder the cubes whch are ρ-adjacent to C l for ρ l and remove all the subsets of S Cl whch are bad for these cubes. The reason we consder cubes whch are adjacent to C l nstead of C s because we do not know what C wll be at ths pont. Note that, dependng on the choce of C, some of these cubes may not be ρ-adjacent to C. However, snce we choose C C l (as we wll show later n the proof), the set of cubes whch are ρ-adjacent to C wll be a subset of the set of cubes whch are ρ-adjacent to C l. Havng removed these bad subsets, we have enough nformaton to determne C. Then, by consderng the relaton of the remanng subsets of S Cl to cubes whch are l-adjacent to C, we can fnd a set S C that can be assgned to C. For each ρ and, let A (ρ) be the famly of level cubes from our partal embeddng C consstng of cubes whch are ρ-adjacent to C l. Let A (ρ) = A(ρ). By Proposton 4.2, we have A (ρ) d ρ (C l ) for each, and thus A (ρ) s d ρ (C l ). Let F = {S S l : S S Cl }. By Proposton 4., we have S F S S Cl 2. We say that a set S F s bad for a cube A A (ρ) f there are at least (4s 2 δ A ) l(s)+l(sa) 2s S S A = (4s 2 δ A ) l+ 2s S S A blue edges between S and S A (where δ A = δ(c l, A) s the domnatng parameter of C l and A). Otherwse, we say that S s good for A. For each fxed A, let F A be the subfamly of F consstng of sets whch are bad for A. By the properness of the assgnment up to ths pont, we know that there are at most (4s 2 δ A ) l+ 2s S Cl S A blue edges between S Cl between S Cl and S A for every A A (ρ) and S A n two ways, we see that. Therefore, by countng the number of blue edges S F A (4s 2 δ A ) l+ 2s S S A (4s 2 δ A ) l+ 2s S Cl S A, 8