In 1975 Stein conjectured that in every n × n array filled with the numbers 1,...,n with every number occuring exactly n times, there is a partial transversal of size n − 1. In this note we show that this conjecture is false by constructing such arrays without partial transverals of size n−(1/42)ln(n)

Pokrovskiy, Alexey

Sudakov, Benny

English

Birkbeck Institutional Research Online

BIROn - Birkbeck Institutional Research OnlinePokrovskiy, Alexey and Sudakov, Benny (2019) A counterexample to Stein’sEqui-n-square conjecture. Technical Report. Birkbeck College, University ofLondon, London, UK.Downloaded from: http://eprints.bbk.ac.uk/32252/Usage Guidelines:Please refer to usage guidelines at http://eprints.bbk.ac.uk/policies.html or alternativelycontact lib-eprints@bbk.ac.uk.PROCEEDINGS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 00, Number 0, Pages 000–000S 0002-9939(XX)0000-0A COUNTEREXAMPLE TO STEIN’S EQUI-n-SQUARECONJECTUREALEXEY POKROVSKIY AND BENNY SUDAKOV(Communicated by Patricia Hersh)Abstract. In 1975 Stein conjectured that in every n × n array filled withthe numbers 1, . . . , n with every number occuring exactly n times, there is apartial transversal of size n − 1. In this note we show that this conjecture isfalse by constructing such arrays without partial transverals of size n− 142lnn.1. IntroductionLatin squares are combinatorial objects introduced by Euler in the 18th century.An order n Latin square is an n×n array filled with n symbols such that no symbolappears twice in the same row or column. A partial transversal is a collection ofcells of the Latin square which do not share the same row, column or symbol.Starting with Euler (see [9, 12]), questions about transversals in Latin squares wereextensively studied. The most natural question about them is “how large a partialtransversal one can guarantee to find in every n× n Latin square?” A well knownconjecture of Ryser, Brualdi, and Stein [7, 14, 15] is that the answer should ben− 1.Notice that n × n Latin squares have the property that every symbol occursprecisely n times. Over 40 years ago, Stein conjectured that this condition on itsown is sufficient to guarantee a partial transversal of size n− 1. In [15], an equi-n-square is defined to be an n×n array filled with n symbols such that every symboloccurs precisely n times, and it is conjectured that every equi-n-square has a partialtransversal of size n− 1.Conjecture 1.1 (Stein, [15]). Let S be an n×n array filled with the symbols 1, . . . , nso that each number occurs exactly n times. Then S has a partial transversal ofsize n− 1.Attempts to prove this conjecture have lead to the development of importanttools in extremal combinatorics. Stein’s Conjecture was the setting of the firstapplication of the Lopsided Lovasz Local Lemma [8]—Erdo˝s and Spencer introducedthis variant of the local lemma to show that every n × n array with ≤ (n − 1)/16occurences of every symbol has a size n transversal. Later Alon, Spencer, and Tetalishowed that if there are ≤ n occurences of every symbol and n is a power of 2, thenthe square can be completely decomposed into size n transversals. When working in2010 Mathematics Subject Classification. Primary 05B15.Research supported in part by SNSF grant 200021-175573.Research supported in part by SNSF grant 200021-175573.c©XXXX American Mathematical Society12 ALEXEY POKROVSKIY AND BENNY SUDAKOVequi-n-squares, the best currently known result is due to Aharoni, Berger, Kotlar,and Ziv [3]—they used topological methods to show that such arrays always have apartial transversal of size 2n/3. This improved on an earlier result of Stein [15] whoused the probabilistic method to show that a partial transversal of size (1− e−1)nexists in every equi-n-squareIn this note we produce counterexamples to Conjecture 1.1.Theorem 1.2. For all sufficiently large n, there exist n × n arrays filled with thesymbols 1, . . . , n so that each symbol occurs exactly n times, which have no partialtransversals larger than n− 142 lnn.We remark that a corollary of the above theorem is that Erdo˝s and Spencer’sresult cannot be strengthened to hold when the number of occurrences of eachsymbol in the array is “≤ n − 185 lnn” rather than “≤ (n − 1)/16”. To see thisconsider the n×n array S from Theorem 1.2 and append ⌊ 184 lnn⌋ rows and columnsto obtain a new array S′. Fill the newly created entries with (arbitrarily many)previously unused symbols using every symbol ≤ n times. Now S′ is an n′ × n′array for n′ = n +⌊184 lnn⌋with ≤ n ≤ n′ − 185 lnn′ occurrences of each symbol(using that n is sufficiently large). It cannot have a size n′ transversal, since sucha transversal would intersect S in at least n′ − 2(n′ − n) ≥ n− 142 lnn entries, andS was chosen to have no transversal of size n− 142 lnn.The above theorem is proved in the next section. These counterexamples stillleave open the possibility of Stein’s Conjecture holding in some asymptotic sense.In Section 3 we discuss some possible asymptotic versions of it.2. ProofOur proof relies on the fact that the sequence at =1√thas the property thatbt = at∑ti=1 ai converges as t → ∞ while ct =∑ti=1 a2i diverges. The followinglemma proves this in a way which will be convenient to apply.Lemma 2.1. For an integer n ≥ 1060, consider the sequence xt =⌊13√nt⌋. Thefollowing hold for all t.xtt∑i=1xi ≤ n4(2.1)n∑i=1x2i ≥n lnn10(2.2)Proof. We’ll use the fact that for the decreasing function f(x) we have∫ baf(x)dx ≤∑bi=a f(i) ≤∫ ba−1 f(x)dx. This implies13t∑i=1√ni=√n3(1 +t∑i=21√i)≤√n3(1 +∫ t11√xdx)=√n3(1 + 2(√t− 1))≤ 2√nt3(2.3)A COUNTEREXAMPLE TO STEIN’S EQUI-n-SQUARE CONJECTURE 3Figure 1. An illustration of a counterexample to Conjecture 1.1.The colours in the array represent symbols in the array S. Forclarity we use the same colour for symbols in each of N1, . . . , Nn,and A. The picture is not entirely to scale since in the actual arrayin Theorem 1.2, R∗ × C∗ takes up a far larger proportion of thesquare.Now (2.1) comes from using (2.3) and xt ≤ 13√nt to getxtt∑i=1xi ≤ 13√nt(t∑i=113√ni)≤ 13√nt· 2√nt3≤ n4For (2.2) we haven∑i=1x2i =n∑i=1⌊13√ni⌋2≥ n9n∑i=11i− 23n∑i=1√ni≥ n9∫ n11xdx− 4n3=n lnn9− 4n3≥ n lnn10The first inequality comes from⌊13√ni⌋ ≥ 13√ni − 1. The second inequality uses(2.3) and the fact that decreasing functions have∫ baf(x)dx ≤∑bi=a f(i). The lastinequality uses 1 ≤ lnn120 which holds for n ≥ 1060. Now we construct counterexamples to Conjecture 1.1.Proof of Theorem 1.2. Let n ≥ 1060, and consider an n×n array with rows r1, . . . ,rn, and columns c1, . . . , cn. For a set of rows R and a set of columns C, we denotethe rectangle formed by R and C by R × C = {(ri, cj) : ri ∈ R, cj ∈ C}. Letxi be the sequence from Lemma 2.1. Let n0 be the largest number for whichxn0 6= 0, and notice that n0 =⌊n9⌋. From (2.1) and the integrality of xi, wehave∑n0i=1 xi ≤ n4xn0 ≤n4 . Partition {r1, . . . , rn} into sets R1 ∪ · · · ∪ Rn0 ∪ R∗and C into sets C1 ∪ · · · ∪ Cn0 ∪ C∗ with |Ri| = |Ci| = xi =⌊13√ni⌋and |R∗| =|C∗| = n −∑n0i=1 xi ≥ 3n/4. Let Fi = Ri × Ci, Hi = Ri × (C∗ ∪ ⋃j>i Cj), andJi = (R∗ ∪⋃j>iRj)× Ci. Notice that S is the disjoint union of the sets Fi, Hi, Jifor i = 1, . . . , n0, and R∗ × C∗.4 ALEXEY POKROVSKIY AND BENNY SUDAKOVNotice that |Fi| = |Ri||Ci| = x2i . Using (2.1) we have|Hi| = |Ji| = |Ri|(n−∑j≤i|Cj |) = xin− xi∑j≤ixj ≥ n(xi − 14).In particular this means that |Hi ∪ Ji| > n(2xi − 1).We now fill S with the n symbols 1, . . . , n so that each symbol occurs exactly ntimes. First split {1, . . . , n} into sets N1, . . . , Nn0 with |Ni| = 2xi− 1, a set B with|B| = ⌊ 120 lnn⌋ ≥ 121 lnn, and a set A with |A| = n − |B| − |N1| − · · · − |Nn0 | (tosee that such a partition is possible, notice that from∑n0i=1 xi ≤ n4 and n ≥ 1060,we have |B| + |N1| + · · · + |Nn0 | =⌊120 lnn⌋+∑n0i=1(2xi − 1) ≤ 120 lnn + n2 < n).Fill S as follows.• For each symbol in Ni, place it n times into Hi ∪ Ji (|Hi ∪ Ji| > n|Ni|ensures that this is possible).• For each symbol in B, place it n times into F1 ∪ · · · ∪ Fn0 . This is possiblesince (2.2) implies∑n0i=1 x2i =∑ni=1 x2i ≥ 110n lnn ≥ n|B|.• Place the symbols fromA arbitrarily into the remaining entries of S (makingsure that there are exactly n occurances of each symbol).Suppose, for the sake of contradiction, that we have a partial transversal T ofsize > n− 142 lnn.Claim 2.2. If T contains s entries in Fi, then T contains at most 2xi− 2s entriesin Hi ∪ JiProof. Suppose that (ra1 , cb1), . . . , (ras , cbs) ∈ T ∩Fi. Then since T is a transversal,T cannot have any other entries in rows raj or columns cbj for j = 1, . . . , s. Recallthat Hi and Fi are both contained in the xi rows Ri, which implies ra1 , . . . , ras ∈ Ri.This means that Hi ∩ T must be contained in the xi − s rows Ri \ {ra1 , . . . , ras}.Since T has at most one entry in each row we have |T ∩Hi| ≤ xi − s. By the sameargument we have |T ∩ Ji| ≤ xi − s. Since |B| ≥ 121 lnn, T must contain at least 142 lnn of the symbols of B. Lettingzi = |T ∩ Fi|, we have∑n0i=1 zi ≥ 142 lnn. By the claim, for all i, T has at most2xi − 2zi entries in Hi ∪ Ji, and so uses at most 2xi − 2zi symbols in Ni (since thesymbols in Ni only occur in Hi∪Ji). Since |Ni| = 2xi−1, this means that T doesn’thave any entries of at least |Ni| − (2xi − 2zi) = 2zi − 1 symbols of Ni. Summingup, we have that T doesn’t use at least∑n0i=1 min(2zi − 1, 0) ≥∑n0i=1 zi ≥ 142 lnnsymbols. This contradicts |T | > n− 142 lnn. 3. Concluding remarksHere we make some remarks about directions for further research following thedisproof of Conjecture 1.1.Asymptotic versions of Stein’s Conjecture. The counterexamples constructedin this note still leave the possibility of Stein’s Conjecture holding in some asymp-totic sense. There are two natural asymptotic weakenings of the conjecture whichmay still be true. The first is to ask whether, in the setting of Stein’s Conjecture,there always a size n− o(n) partial transversal. This would strengthen the resultsof Stein [15] and of Aharoni, Berger, Kotlar, and Ziv [3].It is possible to show that this asymptotic version of Stein’s Conjecture holdswith a mild additional condition on the square — that no symbol appears too oftenA COUNTEREXAMPLE TO STEIN’S EQUI-n-SQUARE CONJECTURE 5in a row or column. To prove this we will use the following version of Ro¨dl’s Nibble(see eg. [5]).Theorem 3.1 (Ro¨dl). Fix  > 0, r ∈ N, the following holds for sufficiently largen and d. Let H be an r-uniform, d-regular, n-vertex hypergraph with every pair ofvertices u, v having d(u, v) ≤ o(n). Then H has a matching with (1− )n/r edges.Using the above result we can prove an asymptotic version of Stein’s Conjecturewhen no symbol appears too often in a row or column.Corollary 3.2. Fix  > 0. Let S be an n× n array filled with the symbols 1, . . . , nso that each number occurs exactly n times in the square, and at most o(n) timesin every row and column. Then S has a partial transversal of size (1− )n.Proof. We define a 3-uniform, 3-partite hypergraph H as follows. The vertex setof H is {r1, . . . , rn, c1, . . . , cn, s1, . . . , sn}. The edges of H are exactly triples of theform {ri, cj , sk} with the (i, j)th entry of S being k. We claim that H satisfiesthe assumptions of Theorem 3.1. Notice that H is a 3-uniform, n-regular, and 3n-vertex hypergraph. For a pair of vertices u, v we have d(u, v) ≤ 1 unless one of u, vis in {r1, . . . , rn, c1, . . . , cn} and the other in {s1, . . . , sn}. Notice that d(ri, sj) andd(ci, sj) are equal to the number of occurances of symbol j in row i and column irespectively. By assumption, both of these quantities are at most o(n).By Theorem 3.1, H has a matching M with (1 − )n edges. Let T be the setof entries in S corresponding to the edges of M . Notice that T doesn’t have morethan one entry with any row, column or symbol because M doesn’t have more thanone edge through any vertex. Thus T is the required partial transversal. The above corollary shows that an asymptotic version of Stein’s Conjecture holdswhen no symbol is repeated more than o(n) times in any row or column. It is easy tocheck that in the arrays constructed in Theorem 1.2, each symbol is repeated O(√n)in every row or column. Thus Corollary 3.2 applies to the squares constructed inTheorem 1.2 to show that they have partial transversals of size n− o(n). It wouldbe interesting to know it this holds without any restriction on symbol repetitionsi.e. if Corollary 3.2 holds without the “each number occurs at most o(n) times inevery row and column” condition.A second asymptotic version of Steins Conjecture one can look for is to find thelargest α so that every n × n square with ≤ αn occurances of every symbol has asize n transversal. Erdo˝s and Spencer [8] proved that α = 1/16 suffices, but it islikely that α can be as large as 1 − o(1). Again, one can ask an easier questionby adding an extra condition forbidding symbol repetitions in rows and columns.Such a result is true and will be proved in [13]:Theorem 3.3 (Montgomery, Pokrovskiy, Sudakov, [13]). Let S be an n× n arrayfilled with the symbols so that each symbol occurs ≤ (1−o(1))n times in the square,and no symbol is repeated in a row or column. Then S has a transversal.In [13] something stronger is actually shown — that under the assumptions ofthe above Theorem the square has (1 − o(1))n disjoint transversals. Theorem 3.3shows that if we completely forbid symbol repetitions in rows and columns, thenthe asymptotic version of Stein’s Conjecture holds. Again, it would be interestingto know it Theorem 3.3 holds without any restrictions on repetitions in rows andcolumns.6 ALEXEY POKROVSKIY AND BENNY SUDAKOVVariations of Stein’s Conjecture. After Stein made his conjecture, many au-thors have proposed strengthenings and variations of Conjecture 1.1. The construc-tion in this note can be used to disprove most of these other conjectures as well.Sometimes this is immediate e.g. the conjecture in [10] and Conjecture 1.9 in [1] aredirect strengthenings Conjecture 1.1 and so are false by Theorem 1.2. Sometimesone needs to modify our construction slightly to disprove related conjectures. Forexample, Hahn suggested the following conjecture.Conjecture 3.4 (Hahn, [11]). In every edge-colouring of Kn with ≤ n/2− 1 edgesof each colour, there is a rainbow path of length n− 1.Here “rainbow path” means a path in the graph all of whose edges have thesame colour. The relationship between this and Stein’s Conjecture is that to everysymmetric n × n array S, one can assign an edge-coloured complete graph Kn bycolouring edge ij with the symbol in the (i, j)th entry of S (since S is symmetric,this gives a well-defined colouring of Kn). It is easy to see that under this corre-spondence, partial transversals in Kn correspond to rainbow maximum degree 2subgraphs in Kn. Thus Conjecture 3.4 would imply that every symmetric n × narray S with ≤ n − 1 occurrences of each symbol has a partial transversal of sizen−1. The proof of Theorem 1.2 can easily be run so it gives a symmetric array i.e.we get graphs satisfying the assumptions of Conjecture 3.4 without rainbow pathslonger than n− 142 lnn. Here is another conjecture about rainbow subgraphs.Conjecture 3.5 (Aharoni, Barat, Wanless, [2]). Let G be a bipartite graph with> ∆(G) + 1 edges of each colour. Then G has a rainbow matching using everycolour.To see the relationship between this and Stein’s Conjecture: from an n×n arrayS, build a coloured Kn,n with vertices {x1, . . . , xn, y1, . . . , yn} by colouring the edgexiyj ∈ Kn,n by the symbol in the (i, j)th entry of S. It is easy to see that underthis correspondance a transversal in S corresponds to a rainbow matching in Kn,n.Consider the n×n array S from Theorem 1.2 and the corresponding coloured Kn,n.Since S has n copies of each symbol, the corresponding Kn,n has n edges of eachcolour. Since ∆(Kn,n) = n, this is just short of the assumption of Conjecture 3.5(and so of disproving the conjecture).However, it is easy to modify the construction slightly to actually get a coun-terexample e.g. by deleting the edges of the form xiyi and xiyi+1 (mod n) in Kn,n.We use the proof of Theorem 1.2 to get a colouring of this (n − 2)-regular graphwith n − 2 colours such that each colour has n edges, but there is no rainbowmatching of size n − 142 lnn. This corresponds to deleting two diagonals in thesquare S in Theorem 1.2, and checking that the proof still works if we omit theseentries: The only parts that need to be checked are that the sets Fi and Hi ∪ Jihave enough room to fit the colours they must contain. Specifically we need tocheck that |Hi∪Ji| ≥ n(2xi−1) and |F1∪ · · ·∪Fn0 | ≥ n|B| still hold after deletingthe two diagonals from these sets. This is indeed true because there is room in theinequalities we used for |Hi ∪ Ji| and |F1 ∪ · · · ∪ Fn0 |.The fact that our constructions can disprove Conjecture 3.5 and Conjecture 1.9in [1] was first noticed by Alon. He also found some interesting further modificationsof our construction, see [4] for details.A COUNTEREXAMPLE TO STEIN’S EQUI-n-SQUARE CONJECTURE 7Acknowledgemnt. The authors would like to thank Noga Alon for pointing us tothe conjectures in [1, 2], and for other remarks greatly improving the presentationof this paper.References1. R. Aharoni, N. Alon, E. Berger, M. Chudnovsky, D. Kotlar, M. Loebl, and R. Ziv. Fairrepresentation by independent sets In A Journey Through Discrete Mathematics, 31–58 ,Springer, Cham., 20172. R. Aharoni, J. Barat and I. Wanless. Multipartite hypergraphs achieving equality in Rysersconjecture. Graphs and Combinatorics., 32:1–15, 2016.3. R. Aharoni, E. Berger, D. Kotlar, and R. Ziv. On a conjecture of Stein. Abh. Math. Semin.Univ. Hambg., 87:203–211, 2017.4. N. Alon. Personal communication. 2018.5. N. Alon and J. Spencer. The Probabilistic Method. John Wiley & Sons , 1990.6. N. Alon, J. Spencer, and P. Tetali. Covering with latin transversals. Disc. Appl. Math., 57:1–10, 1995.7. R. A. Brualdi and H. J. Ryser. Combinatorial matrix theory. Cambridge University Press,1991.8. P Erdo˝s and J Spencer. Lopsided Lova´sz local lemma and Latin transversals. Disc. Appl.Math., 30:151–154, 1991.9. L. Euler. Recherches sur une nouvelle espe´ce de quarre´s magiques. Verh. Zeeuwsch. Gennot.Weten. Vliss., 9:85–239, 1782.10. J. De´nes. Research problems. Periodica Mathematica Hungarica,, 17:245–246, 1986.11. G. Hahn and C. Thomassen. Path and cycle sub-Ramsey numbers and an edge-colouringconjecture. Discrete Math., 62:29–33, 1986.12. A.D. Keedwell and J. De´nes. Latin Squares and their Applications. Elsevier Science, 2015.13. R. Montgomery, A. Pokrovskiy, and B. Sudakov. Decompositions into spanning rainbow struc-tures. preprint, 2018.14. H. Ryser. Neuere probleme der kombinatorik. Vortra¨ge u¨ber Kombinatorik, Oberwolfach, 69–91, 1967.15. S. K. Stein. Transversals of Latin squares and their generalizations. Pacific J. Math., 59:567–575, 1975.Department of Mathematics, ETH, 8092 Zurich, Switzerland.E-mail address: dr.alexey.pokrovskiy@gmail.comDepartment of Mathematics, ETH, 8092 Zurich, Switzerland.E-mail address: benjamin.sudakov@math.ethz.ch

A counterexample to Stein's Equi-n-square conjecture

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A counterexample to Stein's Equi-n-square conjecture

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