2012 IEEE 27th Conference on Computational Complexity
On Sunflowers and Matrix Multiplication
Noga Alon Amir Shpilka Christopher Umans Sackler School of Mathematics and Faculty of Computer Science Computing and Mathematical Sciences Blavatnik School of Computer Science Technion-Israel Institute of Technology California Institute of Technology Tel Aviv University Haifa, Israel Pasadena, CA 91125 Tel Aviv 69978, Israel Email: [email protected] Email: [email protected] and Institute for Advanced Study Princeton, New Jersey, 08540 Email: [email protected]
Abstract—We present several variants of the sunflower |F| > (k − 1)s · s! then F contains a k-sunflower. conjecture of Erdos˝ and Rado [ER60] and discuss the relations They conjectured that actually many fewer sets are among them. needed. We then show that two of these conjectures (if true) imply negative answers to questions of Coppersmith and Winograd Conjecture 1 (Classical sunflower conjecture [ER60]): [CW90] and Cohn et al [CKSU05] regarding possible ap- For every k>0 there exists a constant ck such that the proaches for obtaining fast matrix multiplication algorithms. following holds. Let F be an arbitrary family of sets of Specifically, we show that the Erdos-Rado˝ sunflower conjec- size s from some universe U.If|F| ≥ cs then F contains ture (if true) implies a negative answer to the “no three k disjoint equivoluminous subsets” question of Coppersmith a k-sunflower. and Winograd [CW90]; we also formulate a “multicolored” This (and in particular the case k =3) is one of the most n sunflower conjecture in Z3 and show that (if true) it implies a well known open problems in combinatorics and despite negative answer to the “strong USP” conjecture of [CKSU05] a lot of attention it is still open today (see e.g. [Erd71], (although it does not seem to impact a second conjecture [Erd75], [Erd81]). in [CKSU05] or the viability of the general group-theoretic approach). A surprising consequence of our results is that the Coppersmith-Winograd conjecture actually implies the Cohn This conjecture has applications in combinatorial number et al. conjecture. theory and the study of Turan´ type problems in extremal Zn The multicolored sunflower conjecture in 3 is a strength- graph theory ([Fu91]), as well as in other areas in combina- ening of the well-known (ordinary) sunflower conjecture in n torics including the investigation of explicit constructions of Z3 , and we show via our connection that a construction from [CKSU05] yields a lower bound of (2.51 ...)n on the size of Ramsey graphs ([AP01]). A close variant has been applied the largest multicolored 3-sunflower-free set, which beats the in Circuit Complexity ([Ju01], see also [Ra85], [AB87]). current best known lower bound of (2.21 ...)n [Edel04] on the n size of the largest 3-sunflower-free set in Z3 . B. Matrix multiplication and conjectures implying an Keywords-Matrix Multiplication, Sunflower Conjecture O(n2+ ) time algorithm
I. INTRODUCTION A fundamental algorithmic problem in computer science asks to compute the product of two given n × n matrices. A. Sunflowers This problem received a lot of attention since the seminal A k-sunflower (also called a Δ-system of size k)isa work of Strassen [Str69] that showed that one can do better collection of k sets, from some universe U, that have the than the simple row-column multiplication. For many years, same pairwise intersections. This notion was first introduced the best known result was due to Coppersmith and Wino- in [ER60] and proved itself to be a very useful tool in grad, who gave an O(n2.376...) time algorithm for multiply- combinatorics, number theory and computer science ever ing two n × n matrices [CW90]. Very recently, this running since. See, e.g., [Fu91], [Ju01], [AP01]. time has been lowered, by Stothers and Williams [S10], A basic problem concerning sunflowers is how many [W11], and the world record now stands at O(n2.373...) sets do we need in order to guarantee the existence of a [W11]. It is widely believed by experts that one should k-sunflower. Erdos˝ and Rado proved the following bound be able to multiply n × n matrices in time O(n2+ ), for [ER60], [ER69]. every >0 (see e.g., [Gat88], [BCS97]). It is a major open Theorem 1.1 (Erdos˝ and Rado [ER60]): Let F be an ar- problem to achieve such an algorithm (which we term fast bitrary family of sets of size s from some universe U.If matrix multiplication).
1093-0159/12 $26.00 © 2012 IEEE 214 DOI 10.1109/CCC.2012.26 In their paper, Coppersmith and Winograd proposed an We emphasize that the message of this paper with respect to approach towards achieving fast matrix multiplication. They approaches to matrix multiplication is not entirely negative. showed that the existence of an Abelian group and a In particular, not all of the statements labeled “conjectures” subset of it that satisfy certain conditions, imply that their in this paper have equal weight. After laying out the various techniques can yield an O(n2+ ) time algorithm. connections, we speculate on the relative likelihood of the A new approach for matrix multiplication that is based various conjectures, in Section IV. on group representation theory was suggested by Cohn and Umans [CU03]. In a subsequent work [CKSU05], Cohn D. Organization et al. gave several algorithms based on the framework We organize the paper as follows. In Section II we discuss of [CU03], and were even able to match the result of different sunflower conjectures and give the relations among [CW90]. Similarly to Coppersmith and Winograd, Cohn et them. In Section III we present the questions raised in al. formulated two conjectures regarding the existence of [CW90], [CKSU05] and show their relation to sunflowers. certain combinatorial structures that, if true, would yield In Section IV we discuss these implications. O(n2+ ) time matrix-multiplication algorithms. E. Notation C. Our results Z { − } We relate variants of the sunflower conjecture to each For an integer m we denote by m = 0,...,m 1 other, and to two of the aforementioned matrix multiplication the additive group modulo m. [n] stands for the set [n]= { } conjectures. Our main results are as follows: 1,...,n . All logarithms are taken in base 2. We will often use direct products of families of sets. For families F ⊂ • We prove (Theorem 3.2) that if Conjecture 1 is 1 Zn1 and F ⊂ Zn2 , we define their direct product to be the true then no Abelian group and subset satisfying the m 2 m family F ×F = {u◦v | u ∈F and v ∈F }, where u◦v Coppersmith-Winograd conjecture exists. Since Con- 1 2 1 2 is the concatenation of u and v,overZn1+n2 . When each F jecture 1 is well-known and widely believed, this is a m i is a family of subsets from some universe U , we define the strong indication that the Coppersmith-Winograd con- i direct product analogously, over the disjoint union U U . jecture may be false. 1 2 • We formulate a new variant of the sunflower conjecture II. SUNFLOWER CONJECTURES (Conjecture 6) and prove (Theorem 3.7) that it contra- dicts one of two conjectures in [CKSU05], regarding Definition 2.1: We say that k subsets A1,...,Ak,ofa ∀ ∩ ∩k the existence of “Strong Uniquely Solvable Puzzles.” universe U, form a k-sunflower if i = jAi Aj = =1A . Thus any construction of Strong Uniquely-Solvable Conjecture 1 states that for every k there is an integer s Puzzles that would yield an exponent 2 algorithm for ck such that any ck sets of size s contain a k-sunflower. matrix multiplication must disprove this variant of the The conjecture is open even for k =3, which is the case sunflower conjecture, which helps explain the difficulty that we are most interested in, in this paper. This case is of the problem. We stress though, that this does not rule also the main one studied in most existing papers on the out the possibility of obtaining fast matrix multiplica- conjecture, and it is believed that any proof of the conjecture tion algorithms using the Cohn-Umans framework. In for k =3is likely to provide a proof of the general particular, in [CKSU05] Cohn et al. propose a second case as well. Currently the best known result is given in direction that seems not to contradict the variants of the the following theorem of Kostochka, improving an earlier sunflower conjecture that are considered here. estimate of [Spe77]. • We discover via connections proved in this paper, that Theorem 2.2 (Kostochka [Ko97]): There exists a con- the Coppersmith-Winograd conjecture actually implies stant c such that every family of s-sets of size at least · log log log s s the Strong Uniquely-Solvable Puzzles conjecture of cs! ( log log s ) contains a 3-sunflower. [CKSU05]. Thus the latter conjecture is formally easier The following conjecture of Erdos˝ and Szemeredi´ [ES78] to prove. concerns 3-sunflowers inside [n]. • We show via connections established in this paper that Conjecture 2 (Sunflower conjecture in {0, 1}n [ES78]): a construction from [CKSU05] yields a lower bound There exists >0 such that any family F of subsets of [n] of (2.51 ...)n on the size of the largest multicolored 3- (n ≥ 2) of size |F| ≥ 2(1− )·n contains a 3-sunflower. sunflower-free set, which beats the current best known Note the difference between Conjectures 1 and 2. While lower bound of (2.21 ...)n [Edel04] on the size of the Conjecture 1 concerns s-sets from an unbounded universe, Zn largest 3-sunflower-free set in 3 . Conjecture 2 does not restrict the sets to be of the same size, • We show that the Erdos-Rado˝ sunflower conjecture, and and instead demands that they all come from an n element another well-known sunflower conjecture of Erdos˝ and set. Szemeredi´ [ES78] can be viewed in a uniform way as An interesting fact is that Conjecture 1 implies Conjec- Zn conjectures about sunflowers in D. ture 2. This was proved in [ES78] (see also [DEGKM97] for
215 a somewhat sharper estimate). For completeness, we include By taking the direct product of F with itself several times a short proof. (more accurately, (c − 1) · m times, for some integer m) Theorem 2.3 ([ES78]): Assume that for k =3there we may assume that r =(c − c)/(c − 1)n is an integer. exists a constant c such that every family of s-sets of size The probability a fixed n-subset of [cn] contains a random at least cs contains a 3-sunflower. Set =1/4c. Then any r-subset of [cn] is F |F| ≥ (1− )n family of subsets of [n] of size 2 contains a n cn−r r n−r 3-sunflower. 1 ( −δ)n 1 cn = cn 2 n (1+o(1))H( 2 −δ)·n Proof: Recall that i=0 i =2 , r n − − − − where H(x)= x log(x) (1 x)log(1 x) is the entropy Thus, the expected number of n-subsets in F containing a 1 − − 2 function. As √H( 2 δ) < 1 δ , for δ small enough, we get random r-subset of [cn] is at least F 1 (1− )·n that for δ = 2 , must contain at least 2 sets 4δn 1− cn−r cn−r 1− of size s for some s ∈ [( 1 − δ) · n, ( 1 + δ) · n]. From now cn − − cn − r 2 2 · n r ≥ n r ≥ , on we shall only consider those sets of size exactly s in F. cn cn − n n 2 n r Let α>0 be some small number to be determined later, for =Θ( ) (the constant depends only on c, c ). Select such that α · n is an integer. As each s-set contains exactly s n an -set achieving at least this expectation and remove it subsets of size s − αn, and there are such r s−αn s−αn from the subsets containing it, and the universe, and set sets, one (s − αn)-set belongs to at least n = n − r. The resulting family F consists of n -subsets s n+αn − − 1 − · s−αn − · ( αn ) of cn r = c (n r)=c n , and if it contains a 3-sunflower, (1 ) n · 1 (1 ) n · 2 n = 2 n+αn F ⇒ 4δn ( s ) then so does . Thus H(c ) H(c) for all c >c. 4δn s−αn n+αn Next, we show how to increase the density of the sets in 1 (1− )·n · ( αn ) > 4δn 2 2n+αn (1) a family, by finding many that are contained in subset of α · − − − =2H( 1+α ) (1+α)n αn n o(n) (2) the universe, and restricting the universe to that subset. As 1+α − − before, let F be a family of n-subsets of [cn] of cardinality n(α log( α ) α ) > 2 (3) cn 1− n , and fix a rational number 1
216 Note that this definition is equivalent to that of a k- v ,u ,w form a 3-sunflower in {0, 1}dn then v, u, w form Zn sunflower of sets, if we assume that the universe is parti- a 3-sunflower in D (the converse is not necessarily true). tioned into n pairwise disjoint blocks, each of size D, and From our choice of parameters it follows that |F | = |F| ≥ every set contains exactly one element in each block. D(1− )n ≥ 2(1− 0)dn. Hence, F contains a 3-sunflower and Zn F Conjecture 3 (Sunflower conjecture in D): For every k therefore so does . ⇒ there is an absolute constant bk so that for every D and every Conjecture 4 Conjecture 2: Assume that Conjecture 2 n Zn n any set of at least bk vectors in D contains a k-sunflower. is false. By Theorem 2.4, for infinitely many n, for each ≤ ≤ F cn 1− 0/2 While this conjecture seems different from the classical 2 c D0 there is a family c of at least n sunflower conjecture, it turns out that they are equivalent. n-subsets of [cn]. We now describe a family of vectors in ZD0n Theorem 2.6: If Conjecture 1 holds for ck then Conjec- D0 . We use the phrases “0-set,” “1-set,” etc... to refer to ture 3 is true for bk = ck. Similarly, if Conjecture 3 holds the coordinates of a given vector that have 0s, 1s, etc... Our F for bk, then Conjecture 1 is true with ck = e · bk. vectors have as their 0-sets the subsets in family D0 .For Proof: Conjecture 1 ⇒ Conjecture 3: Let U = Z and each 0-set, we have vectors whose 1-sets (which are subsets − denote by p1,...,pn the first n prime numbers. Given v ∈ of the remaining (D0 1)n coordinates) are given by family Zn { 1+v1 1+vn } F − . Then for each pattern of s and s, we have vectors D define Sv = p1 ,...,pn . Clearly Sv is an n-set. D0 1 0 1 n It is not hard to see that a collection F⊆Z contains a whose 2-sets (which are subsets of the remaining (D0 −2)n D F k-sunflower if and only if the corresponding family SF = coordinates) are given by family D0−2, and so on, until we − F {Sv | v ∈F}contains a k-sunflower. define the (D0 2)-sets using family 2. The remaining n − Conjecture 3 ⇒ Conjecture 1: Given a family F of cs coordinates of each vector are then set to D0 1. It is clear subsets of U we shall define a corresponding family inside that if three vectors u, v, w in this family form a 3-sunflower, Zs then their 0-sets coincide (since F is 3-sunflower-free), D, for some large D. Indeed, we can assume w.l.o.g. that D0 s F |U|≤s · c . For convenience assume that U =[D] for D ≤ and then their 1-sets coincide (since D0−1 is 3-sunflower- s · cs. Pick a map from [D] to [s] uniformly at random from free), and so on, until we conclude that u = v = w. Thus ZD0n all such maps. For a given s-set A ⊂ [D], the probability our family in D0 is 3-sunflower free, and it has cardinality that A was mapped injectively to [s] is exactly s!/ss >e−s. at least Thus, there exists a map f :[D] → [s], that is 1 − 1 on at 1− 0/2 D0n (D0 − 1)n (D0 − 2)n 2n s F F˜ ··· least (c/e) of the sets in . Denote those sets by and n n n n consider any such set A. Define the vector vA =(v1,...,vs) − − 1 0/2 1 ∩ D0n where vi = f (i) A. Namely, the ith coordinate of v is = the unique element of A that was mapped to i by f. Denote n, n, n, ··· ,n − − − the new family by vF . It is not hard to see that F˜ contains ≥(∗) D0n(1 o(1))(1 0/2) ≥ (1 0)D0n D0 D0 , a k-sunflower if and only if vF does. As |vF | = |F|˜ ≥ ∗ (c/e)s = bs, where b = c/e, the result follows. where ( ) follows easily from Stirling’s theorem. The following is a weaker version of Conjecture 3, for The assertion of Conjecture 4 may well hold for small the special case k =3. values of D as well. In particular, the case D =3 Conjecture 4 (Weak sunflower conjecture in Zn ): There attracted a considerable amount of attention as in this case D Zn a 3-sunflower in the group 3 is equivalent to a 3-term is an >0 so that for D>D0 and n>n0, any set of at least D(1− )n vectors in Zn contains a 3-sunflower. arithmetic progression in this group. D Conjecture 5 (Weak sunflower conjecture in Zn): There The main difference from Conjecture 3 is that we allow 3 is an >0 so that for n>n, any set of at least 3(1− )n the number of sets to scale with D. It is clear that Conjec- 0 vectors in Zn contains a 3-sunflower. ture 3 immediately implies Conjecture 4. Next we show that 3 An obvious modification of the previous proof (replacing Conjecture 4 is in fact equivalent to Conjecture 2. any symbol of [D] by its base-3 representation) shows that Theorem 2.7: If Conjecture 2 holds for then Conjec- 12 0 2 Conjecture 5 implies Conjecture 4. The best known upper ≥ 0 ture 4 holds for = 0/2, D0 2 and n>n0.If bound for the cardinality of a subset of Zn that contains ≥ 3 Conjecture 4 holds for 0 and D0 3 and n>n0 then no -sunflower is the recent n 1+ bound, for some 3 O(3 /n ) Conjecture 2 holds for some 0 > 0. small >0, of [BK11] that improves upon the previous Proof: Conjecture 2 ⇒ Conjecture 4: Let d be such 2 · 3n/n bound of [Me95]. The best known lower bound is d ≥ n that d/2 D (by Stirling’s approximation, d =logD + (2.217 ...) [Edel04]. 1 2 log log D +3 suffices). Pick an arbitrary 1-1 map f from It is natural to guess that Conjecture 5 is true when it is [D] to subsets of size d/2 in {0, 1}d. Given a family F⊂ seen as a variant of Conjecture 4. On the other hand, one Zn ∈F ∈{ }dn D map every vector v to v 0, 1 in the natural might guess that Conjecture 5 is false when it is viewed as a ◦ ◦···◦ ◦ (1− )n Zn way. I.e. v = f(v1) f(v2) f(vn), where stands for variant of the assertion that sets of size D in D have concatenation. Call the resulting family F . It is clear that if a 3-term arithmetic progression, because the latter statement
217 Conj. 3 ks T hm. 2.6 +3Conj. 1
T hm. 2.3 Conj. 6 +3Conj. 5 +3Conj. 4 ks T hm. 2.7 +3Conj. 2
Figure 1. Sunflower conjectures
{ } Zn is false for large D. We include a simple construction, due x, y, z is a sunflower in 3 , and we say that two ordered to Salem and Spencer [SS42]: sunflowers (x, y, z) and (a, b, c) are disjoint if x = a, y = b, Theorem 2.8: For all >0,ifD>22/ and n is and z = c. We say that a collection of ordered triples (1− )n Zn sufficiently large, there is a set of D vectors in D contains a multicolored sunflower if it contains three triples, that contains no 3-term arithmetic progressions. (x(1),y(1),z(1)), (x(2),y(2),z(2)), (x(3),y(3),z(3)), not all Proof: Set d = (D − 1)/2, and for simplicity assume equal, for which {x(1),y(2),z(3)} form a sunflower. | F Zn Zn (d +1)n. Let be the set of all vectors in D with equal Conjecture 6 (multicolored sunflower conjecture in 3 ): numbers of the elements {0, 1, 2,...,d}. Then There exists >0 so that for n>n0, every collection F⊆Zn × Zn × Zn of at least (1− )n ordered sunflowers n 3 3 3 3 |F| = contains a multicolored sunflower. n/(d +1),n/(d +1),...,n/(d +1) Note that if F⊆Zn is 3-sunflower-free, then {(x, x, x): (1−o(1))n 3 =(d +1) x ∈F}is a collection of ordered sunflowers containing no − − ≥ D(1 /2 o(1))n multicolored sunflower. Thus Conjecture 6 implies Conjec- − >D(1 )n ture 5. Figure 1 describes the connections between the different (if (d+1) n, we take vectors with a near-equal distribution, conjectures. and the calculation is essentially unchanged). Suppose we have u, v, w ∈Ffor which u + w =2v III. MATRIX MULTIPLICATION (i.e., u, v, w form a 3-term arithmetic progression). Since In [CW90] Coppersmith and Winograd gave what was all entries in u, v, w are at most d this equation holds in the until very recently the asymptotically fastest algorithm for integers as well. Then, consider the set I of coordinates i multiplying two n×n matrices. Their argument fell short of 2+ for which vi = d. Since all entries in u, w are at most d, providing an O(n ) time algorithm; however they proved it must be that ui = wi = d for all i ∈ I. But now we that if a certain structure (that we define next) exists then have accounted for all of the entries equal to d in all three their techniques can yield such an algorithm. vectors. The same argument then gives that all three vectors Definition 3.1 ([CW90]): An Abelian group G (with at are d − 1 in exactly the same set of coordinates, etc... We least two elements) and a subset S of G satisfy the no three conclude u = v = w. disjoint equivoluminous subsets property if: whenever T1, All of the conjectures discussed so far, which posit the T2 and T3 are three disjoint subsets of S, not all empty, existence of a k-sunflower in sufficiently large families of they cannot all have the same sum in G: sets or vectors, have multicolored variants, in which we g = g or g = g. imagine that members of the family are colored with k col- ∈ ∈ ∈ ∈ ors, and we are interested only in multicolored k-sunflowers g T1 g T2 g T2 g T3 – ones formed with exactly one set from each color class. In Coppersmith and Winograd showed that if “... we can find this setting one can trivially avoid k-sunflowers by having a sequence of pairs G, S with the no three disjoint equivolu- all but one color class include a distinguished element of minous subset property, such that log(|G|)/|S| approaches the universe, and the remaining color class exclude that 0” then for every >0 there is an O(n2+ ) time fast matrix element. Instead, the correct (it seems to us) and non- multiplication algorithm. However, the next claim, whose trivial formulation demands that the family is itself the proof is very simple, shows that if Conjecture 2 is true then disjoint union of multicolored k-sunflowers, and then we there is no such sequence. are interested only in multicolored k-sunflowers whose sets Theorem 3.2: If Conjecture 2 holds with 0 then if G, S come from more than one of these sunflowers. have the no three disjoint equivoluminous subset property The only multicolored variant we will need below is the then |S|≤log(|G|)/ 0. | | multicolored version of Conjecture 5. To formulate it, we Proof: Given S, consider all its 2S subsets. For ⊆ will be discussing collections of ordered triples of vectors each subset T S, compute σ(T )= g∈T g. Clearly, Zn Zn ∈ |S| | | in 3 (instead of collections of vectors in 3 ). We say there is some g G such that at least 2 / G subsets that such a triple (x, y, z) is an ordered sunflower if the set T satisfy σ(T )=g.Now,iflog(|G|)/|S| < 0 then
218 |S| (1− 0)|S| − n Zn 2 /|G| > 2 . Therefore, in this case, Conjecture 2 local strong USPs of size (C o(1)) in 3 for infinitely many values of n. implies the existence of a 3-sunflower T1, T2 and T3 such ∩ It is not hard to see that a local strong USP is a strong USP that σ(T1)=σ(T2)=σ(T3). Let T = T1 T2. Set \ (Lemma 6.1 in [CKSU05]). More interestingly, we have the Ti = Ti T . By our construction we have that the Ti’s are disjoint and σ(Ti)=g − σ(T ). Hence, they violate the the strong USP capacity is achieved by local strong USPs no three disjoint equivoluminous subsets property. Hence, (Proposition 6.3 in [CKSU05]). In particular, if the strong we must have |S|≤log(|G|)/ 0 for the property to hold. USP capacity is at least c, then for infinitely many n, there F⊆Zn − n is a local strong USP 3 of size (c o(1)) . Cohn et We now turn to discussing a question that was formulated al. conjectured that the strong USP capacity (and thus also by Cohn et al. [CKSU05]. Denote by Sym(U) the group of the local strong USP capacity) is equal to the USP capacity. permutations of a set U. As shown in [CKSU05], this would imply an O(n2+ ) time Definition 3.3 (Uniquely solvable puzzle [CKSU05]): algorithm for matrix multiplication. A uniquely solvable puzzle (USP) of width n is a Conjecture 7 (Conjecture 3.4 in [CKSU05]): The strong F⊆Zn USP capacity (and the local strong USP capacity) equals subset 3 satisfying the following property: For all 2/3 permutations π0,π1,π2 ∈ Sym(F), either π0 = π1 = π2 or 3/2 . else there exist u ∈F and i ∈ [n] such that at least two of Next, we show that Conjecture 6 implies that Conjecture 7 (π0(u))i =0, (π1(u))i =1, and (π2(u))i =2hold. is false. In the proof, it will be convenient if our local strong The USP capacity is the largest constant C such that there USPs have equal numbers of 0’s, 1’s, and 2’s in each vector, exist USPs of size (C − o(1))n and width n for infinitely and the next lemma shows this can be assumed without loss many values of n. of generality. Quoting [CKSU05]: “The motivation for the name Lemma 3.6: If the strong USP capacity is at least c, then F⊆Z3n uniquely solvable puzzle is that a USP can be thought of as for infinitely many n, there is a local strong USP 3 ∈F a jigsaw puzzle. The puzzle pieces are the sets {i : ui =0}, with each v having equal numbers of 0’s, 1’s, and 2’s, − 3n {i : ui =1}, and {i : ui =2} with u ∈F, and the of cardinality at least (c o(1)) . ⊆ Zm puzzle can be solved by permuting these types of pieces Proof: We begin with local strong USP U 3 of − m according to π0, π1, and π2, respectively, and reassembling cardinality at least N =(c o(1)) , which exists because them without overlap into triples consisting of one piece the capacity is at least c. Now, form the strong USP V ⊆ ZmN of each of the three types. The definition requires that the 3 by taking all vectors that are the concatenation (in puzzle must have a unique solution.” some order) of the N vectors of U. Each vector v ∈ V Cohn et al. observed that the USP capacity is at most now has the same distribution of 0’s, 1’s and 2’s. It is easy 3/22/3 and noticed that a construction of Coppersmith to see that the local strong USP property is preserved when 2/3 Z and Winograd implies that the capacity is at least 3/2 cyclically permuting 3. The direct product of the three local [CW90]. They thus concluded that the USP capacity equals strong USPs obtained from V by these three transformations 2/3 F⊆Z3mN ∈F 3/2 . For the purpose of obtaining fast matrix multiplica- is a local strong USP 3 with each v having tion algorithms, [CKSU05] require a structure that is more an equal number of 0’s, 1’s and 2’s. Its cardinality is restrictive than USP, which they call a strong USP. 3 3 3 N(1−o(1)) ≥ − mN(1−o(1)) Definition 3.4 (Strong uniquely solvable puzzle): A (N!) = N (c o(1)) F⊆Zn strong USP is a USP 3 in which the defining =(c − o(1))3mN property is strengthened as follows: For all permutations π0,π1,π2 ∈ Sym(F), either π0 = π1 = π2 or else for infinitely many values of m. there exist u ∈Fand i ∈ [n] such that exactly two of Theorem 3.7: If the strong USP capacity is at least c, (π0(u))i =0, (π1(u))i =1, and (π2(u))i =2hold. then for infinitely many N, there exists a collection of at 2/3 − N ZN × ZN × The strong USP capacity is the largest constant C such least (2 c o(1)) ordered sunflowers in 3 3 n ZN that there exist strong USPs of size (C − o(1)) and width 3 that contains no multicolored sunflower. In particular, if n for infinitely many values of n. Conjecture 6 is true for 0, then the strong USP capacity is − A variant of strong USPs that seems easier to reason at most (3/22/3)1 0 . ⊆ Zm about, but potentially harder to construct, is what [CKSU05] Proof: We begin with local strong USP U 3 of call a local strong USP. Expressing their definition slightly cardinality at least N =(c − o(1))m, which exists because differently, we see the connection to sunflowers: the capacity is at least c. Now, form the strong USP V ⊆ ZmN Definition 3.5 (local strong USP [CKSU05]): A collec- 3 by taking all vectors that are the concatenation (in F Zn some order) of the N vectors of U. Each vector v ∈ V tion of vectors in 3 is a local strong USP if for every u, v, w ∈F, not all equal, the sets {i : ui =0}, {i : vi =1}, now has the same distribution of 0’s, 1’s and 2’s. It is easy and {i : wi =2} do not form a 3-sunflower. The local strong to see that the local strong USP property is preserved when Z USP capacity is the largest constant C such that there exist cyclically permuting 3. The direct product of the three local
219 strong USPs obtained from V by these three transformations i = zI , j = xJ , and k = yK ,wehave is a local strong USP F⊆Z3mN with each v ∈F having 3 a(v)(x)+b(v)(y)+c(v)(z) an equal number of 0’s, 1’s and 2’s. Its cardinality is 2 2 2 =(j − s) +2(j − s)k + k +2ki + i +2i(j − s) − 3 − 3 (N!)3 = N N(1 o(1)) ≥ (c − o(1))mN(1 o(1)) =(i + j + k − s)2. =(c − o(1))3mN Clearly this function is non-negative, and equals 0 exactly when i + j + k = s. Since each of x, y, and Z determines for infinitely many values of m. two of i, j, k and any two of i, j, k determine the third under F⊆Z3n Proof: Let 3 be a local strong USP for which this constraint (and thus determine the entire triple), the every v ∈Fhas equal numbers of 0’s, 1’s and 2’s. Our goal sunflowers in Δ(v) are pairwise disjoint. The cardinality of F ⊆ × × ⊆ Zn 3 (v) will be to produce a collection S0 S1 S2 ( 3 ) Δ is an easy calculation. such that (1) F is a collection of pairwise disjoint ordered (v) (v) Now, define T0 = v∈F T0 , T1 = v∈F T1 , and sunflowers, and (2) every ordered sunflower in S0 ×S1 ×S2 (v) T2 = ∈F T . We note that any ordered sunflower is in F . Such a collection then contains no multicolored v 2 (x, y, z) ∈ T ×T ×T must have (x, y, z) ∈ T (v) ×T (v) × sunflowers. 0 1 2 0 1 T (v) for some v ∈F, because otherwise by the strong USP For x ∈ Z3n and I ⊆ [n], we denote by x the projection 2 3 I property, there is some coordinate i in which exactly two of of x to the coordinates I. Fix a vector v ∈F and let I = {x ,y ,z } equal 0, and so {x, y, z} cannot be a sunflower. {i : v =0}, J = {j : v =1} and K = {k : v =2}.We i i i i j k We will use this observation, together with Lemma 3.8 to define produce our final construction. (v) (v) (v) T (v) = {x : x =0n,x ∈{1, 2}n,x ∈{1, 2}n} Notice that the a ,b ,c define in a consistent way 0 I J K Z (v) { ∈{ }n n ∈{ }n} functions a, b, c from T0,T1,T2, respectively, to , because T1 = y : yI 1, 2 ,yJ =0 ,yK 1, 2 (v) (v) the T0 are disjoint sets (and the same for the T1 and (v) n n n (v) T = {z : zI ∈{1, 2} ,zJ ∈{1, 2} ,zK =0 }. 2 the T2 ). Now, consider the -fold direct product of the three sets, T , T , T , and the functions A : T → Z, B : We identify {1, 2}n with the integers {0, 1, 2,...,2n − 1} 0 1 2 0 T → Z, and C : T → Z defined by A(X)= a(X ), arbitrarily and denote this lift of w ∈{1, 2}n to the integers 1 2 i i B(Y )= b(Y ), and A(Z)= c(Z ). We claim that by w. i i i i (v) (v) → Z (v) (v) → Z { ∈ × × { } form Define functions a : T0 , b : T1 , Δ= (X, Y, Z) T0 T1 T2 : X, Y, Z (v) (v) → Z } and c : T2 as follows, where s is the integer a sunflower and A(X)+B(Y )+C(Z)=0 3 · 2n/2 +1: is a collection of pairwise disjoint ordered sunflowers with (v) 2 a (x)=((xJ ) − s) +2((xJ ) − s)(−xK ) cardinality at least b(v)(y)=(y )2 +2(y )(−y ) K K I (|F| · 3 · 22n/4) . (v) 2 c (z)=(zI ) +2(zI )((−zJ ) − s). (vi) × (vi) × This is because each (Xi,Yi,Zi) must lie in T0 T1 These functions and the following lemma are used in (vi) ∈F T2 for some vi (as we observed above), and then by [BCS97] (in a presentation of material originally appearing the first part of Lemma 3.8, in [Str87]), to show, in their language, that the matrix ⇔ multiplication tensor has a large diagonal which is a com- A(X)+B(Y )+C(Y )=0 a(Xi)+b(Yi)+c(Zi) binatorial degeneration. We use it for a slightly different =0for all i. purpose here. Thus Δ= Δ(v) . Lemma 3.8 ([BCS97] Lemma 15.31): If v∈F We are not done, however, because T × T × T may (x, y, z) ∈ T (v) × T (v) × T (v) form a sunflower, 0 1 2 0 1 2 contain an ordered sunflower (X, Y, Z) not already in Δ, a(v)(x)+b(v)(y)+c(v)(z) ≥ 0. Moreover, the set and this happens exactly when A(X)+B(Y )+C(Z) > 0. (v) (v) (v) One final trick will fix this problem, and we describe it next. Δ(v) = {(x, y, z) ∈ T × T × T |{x, y, z} form 0 1 2 If a, b, c take values in [−M,M], then A, B, C take values (v) (v) (v) } a sunflower and a (x)+b (y)+c (z)=0 in [−M, M]. Thus there exist α, β, γ such that at least a 1/(2M)3 fraction of the triples (X, Y, Z) in Δ satisfy is a collection of pairwise disjoint ordered sunflowers with A(X)=α, B(Y )=β,C(Z)=γ. So, define S = {X ∈ cardinality at least 3 · 22n/4. 0 T : A(X)=α}, S = {Y ∈ T : B(Y )=β}, S = {Z ∈ We sketch the proof here for completeness. 0 1 1 2 T : C(Z)=γ}, and Proof: When (x, y, z) form an ordered sunflower, we 2 have yI = −zI , xJ = −zJ and xK = −yK , and then setting F =Δ∩ (S0 × S1 × S2).
220 − Then F is a collection of pairwise disjoint ordered sun- nm 1 0 flowers (because Δ is) for which every ordered sunflower > nm/3 (X, Y, Z) ∈ S0×S1×S2 is in F , as desired. The cardinality of F is at least we get, by the assumption that the strong USP capacity is at most (3/22/3)1− 0 , that there are three vectors u ,v ,w (not (|F| · 3 · 22n/4) all equal) in F such that the sets {i : u =0}, {j : v =1} . i j 3 { } (2M) and l : wl =2 form a 3-sunflower. It is not hard to verify that the ‘original’ vectors v, u and w of F form a If the strong USP capacity is at least c, then by Lemma 3-sunflower. Indeed, suppose that in some coordinate i we 3.6, for infinitely many n, there exist balanced, local strong have that the set {u ,v ,w } contains exactly two distinct USPs F⊆Z3n with cardinality at least (c−o(1))3n. Taking i i i 3 values, say u = α, v = α and w = β. Let A and A sufficiently large in the above expression, the theorem i i i α β be the corresponding members of G. By the construction follows. of G, the sets {j | (Aα)j =3} and {j | (Aβ)j =3} are Cohn et al. proved that the strong USP capacity is at least different. Hence, {j | (Aβ)j =3} intersects at least one 22/3 (Proposition 3.8 in [CKSU05]). Applying Theorem 3.7, of the sets {j | (Aα)j =1} or {j | (Aα)j =2}. However, we obtain a lower bound of (24/3 − o(1))n > 2.51n on the since {j | (Aα)j =1} and {j | (Aα)j =2} are disjoint, this maximum cardinality of a collection of ordered sunflowers contradicts the assumption that u ,v ,w form a 3-sunflower. in Zn × Zn × Zn containing no multicolored sunflower. 3 3 3 To complete the proof we argue that we can assume Notice that this is larger than the best known lower bound w.l.o.g. that D = 1 · n , for some integer n divisible on the maximum cardinality of 3-sunflower-free subsets of 3 n/3 by 3. Indeed, we can always find an integer n divisible by Zn − − 3 [Edel04]. 3, such that 1 · n 3
221 ks T hm. 2.6 +3 Conj. 7 falseM Conj. 3 Conj. 1 q 4< MMM qqq MMM T hm.3.7qqq MTM hm.M 3.9 qqq MMM T hm. 2.3 qqq M "* Conj. 6 +3Conj. 5 +3Conj. 4 ks T hm. 2.7 +3Conj. 2 T hm.3.2 +3Conj.[CW90] false
Figure 2. Sunflower conjectures and matrix multiplication conjectures. See the discussion in Section IV. beating the best known construction in the standard setting [DEGKM97] W. A. Deuber, P. Erdos,˝ D. S. Gunderson, A. V. gives a concrete example. Understanding the extra flexibility Kostochka, and A. G. Meyer. Intersection statements afforded constructions in the multicolored setting thus seems for systems of sets. J. Combin. Theory Ser. A 79 (1997), no. 1, 118–132. like a worthwhile direction for future work.
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