A Lower Bound for the Length of a Partial Transversal in a Latin Square, Revised Version

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1 A Lower Bound for the Length of Prtil Trnsversl in Ltin Squre, Revised Version Pooy Htmi nd Peter W. Shor Deprtment of Mthemtil Sienes, Shrif University of Tehnology, P.O.Bo , Tehrn, Irn Deprtment of Mthemtis, Msshusetts Institute of Tehnology, Cmridge, MA 02139, USA Astrt It is proved tht every n n Ltin squre hs prtil trnsversl of length t lest n O(log 2 n). The previous ppers proving these results [inluding one y the seond uthor] not only ontined n error, ut were sloppily written nd quite diffiult to understnd. We hve orreted the error nd improved the lrity. 1 Introdution A Ltin squre of order n is n n n rry of ells eh ontining one of n distint symols suh tht in eh row nd olumn every symol ppers etly one. We define prtil trnsversl of length j s set of n ells with etly one in eh row nd olumn nd ontining etly j distint symols (this differs from the usul definition in tht n j etr ells re dded). Koksm [5] showed tht Ltin squre of order n hs prtil trnsversl of length t lest (2n + 1)/3. This ws improved y Drke [3] to 3n/4 nd then simultneously y Brouwer et l. [1] nd y Woolright [8] to n n. This ws in turn improved y Shor [7] to n 5.53 log 2 n nd then y Fu et l. [4], who optimized the prmeters in [7] to slightly improve the onstnt. One of us (P.H.) disovered ug in [7] tht lso ffets [4]. This pper fies the ug, whih ws used y the reversl of inequlity (26) in [7]. We still otin n n O(log 2 n) lower ound, leit with worse onstnt thn in [7, 4]. This is well elow Bruldi s onjeture of n 1, nd Ryser s of n for odd n [2, 6]. The proof in this pper is muh the sme s in [7] eept for the lst prt of Setion 4. The erlier prt of the pper hs een revised to improve the lrity of the presenttion. 1

2 d Squre 1 Figure 1: Squre 1: An emple of the opertor #. In this se, we reple the ells (1,1) nd (3,3) in the prtil trnsversl on the digonl with the ells (1,3) nd (3,1) to otin nother prtil trnsversl, lso of length 4. 2 The Opertion # Given prtil trnsversl T of length n k, with k 2, one n find nother prtil trnsversl of equl or greter length in the following mnner: Choose two ells in T, sy ells (i 1, j 1 ) nd (i 2, j 2 ), suh tht T {(i 1, j 1 ), (i 2, j 2 )} ontins n k distint symols. These two ells n either ontin two distint duplited symols, or two ourrenes of the sme symol, provided this symol ppers in the trnsversl t lest three times. Reple these two ells with the ells (i 1, j 2 ) nd (i 2, j 1 ). Sine we hose ells ontining duplited symols, the new prtil trnsversl hs length t lest n k, s eh of the symols in the originl trnsversl is represented in one of the unhnged ells (see Squre 1). We ll this opertion #, nottion hosen for its shpe. We now give motivting emple of the use of the opertion #, y pplying it to show tht every Ltin squre of order 6 hs prtil trnsversl of length t lest 5. Consider ounteremple. Assume for now tht the longest prtil trnsversl hs length 4. The squre must thus hve prtil trnsversl ontining multiset of symols either of the form (,,,,, d) or the form (,,,,, d). Let us nlyze the se where it ontins (,,,,, d) (See Squre 1). We ssume tht this prtil trnsversl is on the digonl, nd ll it T 0. We n pply # to the ells (1, 1) nd (3, 3) in T 0 to get new prtil trnsversl T 1. By our hypotheses, the new ells (1, 3) nd (3, 1) in T 1 must ontin symol hosen from the set {, d}. By symmetry, we only need to nlyze two ses here: either oth symols re the sme or there is one nd one d. We will nlyze the se where they re oth s. We n pply # to the ells (1, 3) nd (5, 5) in T 1 to otin new prtil trnsversl T 2 (s shown 2

3 d d y d d Squre 2 Squre 3 Figure 2: Squre 2: Another emple of the opertion #. In this, we reple the the ells (1,3) nd (5,5) in the prtil trnsversl indited in old with the ells ontining symols nd y. If this Ltin squre hd no prtil trnsversl of length greter thn 4, we must hve tht {, d} nd y {, d}. Squre 3: After two more pplitions of #, we know tht if the Ltin squre hd no prtil trnsversl of length greter thn 4, it would hve to look like this, with the symols hosen from the set {,,, d}. in Squre 2), nd we disover tht the symols in (1, 5) nd (5, 3) must e hosen from the set {,, d}. Now, strting from T 0 gin we n pply # to the ells (1, 1) nd (4, 4) to otin prtil trnsversl T 3, nd we disover tht the ells (1, 4) nd (4, 1) must oth ontin d. (See Squre 3.) We n now pply # to the ells (1, 4) nd (6, 6) in T 3 to otin T 4, nd disover tht the symols in (1, 6) nd (6, 4) must e hosen from the set {,, }. We now know tht our Ltin squre looks like Squre 3, where the s re symols from the set {,,, d}. The first row ontins five distint symols from the set {,,, d}, ontrdition y the pigeonhole priniple. The se where the longest prtil trnsversl hs length less thn 4, s well s the ses where the ell (3, 1) in Squre 1 is d insted of nd the se where the digonl is (,,,,, d), n e hndled using very similr nlysis, whih we will not present here. We define prtil Ltin squre s n n n squre with some of its ells ontining symols (the others we ll empty) suh tht no symol ppers twie in ny row or olumn. A prtil trnsversl of n n n prtil Ltin squre is set of n non-empty ells, one from eh row nd olumn. We sy this prtil trnsversl hs length j if it ontins j distint symols. An m m susqure S of n n n prtil Ltin squre S is the set of m 2 ells in some suset of m rows nd some suset of m olumns of S, where some nonempty ells in S my possily e repled y empty 3

4 ells in S. Consider Ltin squre with prtil trnsversl of mimum length n k, with k 2. By pplying # to this prtil trnsversl, we will get other prtil trnsversls, whose length must lso e n k nd whose set of symols is the sme s the first. Applying # repetedly to these prtil trnsversls, we eventully will otin set of suh prtil trnsversls losed under #. All of these prtil trnsversls ontin the sme set of n k distint symols, so y ignoring ll ells eept those in this set of prtil trnsversls, we otin prtil Ltin squre S ontining n k symols nd set T of prtil trnsversls of S losed under #. We will ll this pir (S, T ) prtil Ltin squre stisfying A k. More formlly, we hve: Definition. An n n prtil Ltin squre stisfying A k is prtil Ltin squre, together with nonempty set T of prtil trnsversls of S of length n k. Eh non-empty ell must pper in t lest one of the prtil trnsversls in T. The set T of prtil trnsversls must form onneted grph under the opertion #, nd must e losed under the opertion #. For prtil Ltin susqure (S, T ) stisfying A k of prtil Ltin squre (S, T ) stisfying A k, in ddition to the properties ontined in the ove definition, we lso require n inhertne property. Nmely, we require T to e suset of the set T restrited to S, i.e., tht T {T S : T T }. Note tht Bruldi s onjeture (tht ll Ltin squres of order n hve prtil trnsversl of length t lest n 1) does not pper to rule out the eistene of prtil Ltin squres stisfying A 2, or A k for k > 2. If S is n n n prtil Ltin squre stisfying A k, we n onstrut n (n + 1) (n + 1) prtil Ltin squre S stisfying A k, y dding n etr row nd olumn tht onsist entirely of empty ells eept for the (n + 1, n + 1) ell, whih ontins new symol. The prtil trnsversls in T re those in T ugmented y the ell (n + 1, n + 1). Together with the se nlysis on Ltin squres of size 6 skethed erlier, this oservtion implies tht ny prtil Ltin squre stisfying A 2 must hve size t lest 7. In terminology we will e defining lter in this pper, this mens tht n 2 7. (1) We use the properties of miniml Ltin squre stisfying A k to otin set of inequlities, nd then use these inequlities to derive our min result. We first prove lemm: Lemm 1 Given prtil Ltin squre (S, T ) stisfying A k suh tht no susqure stisfies A k, then no ell is ontined in ll prtil trnsversls in T. Tht is, given non-empty ell (i, j) nd prtil trnsversl ontining (i, j), y sequene of opertions #, one n otin prtil trnsversl in T not ontining (i, j). 4

5 Proof. Suppose there is ell (i, j) ontined in ll prtil trnsversls. We will ll this fied ell. Let e the symol in this ell. If ppers nywhere else in the prtil Ltin squre, there is trnsversl ontining oth s (the seond ppers in some prtil trnsversl sine every non-empty ell does, nd this prtil trnsversl must ontin the first sine ll prtil trnsversls do). We n then pply # to this prtil trnsversl to otin prtil trnsversl without the fied ell, ontrdition. We re left with the se where does not pper nywhere else in the prtil Ltin squre. Now, y deleting the row nd olumn ontining the, one finds susqure stisfying A k, ontrdition of the hypothesis. We hve just proved tht no ell in miniml squre stisfying A k is fied, so given non-empty ell in suh squre, there is prtil trnsversl in T ontining oth tht ell nd nother ell with the sme symol (otherwise, the grph of prtil trnsversls would not e onneted y #). We n hoose ny filled ell, sy (1, 1), nd hoose prtil trnsversl T 0 through it tht duplites the symol in it, sy. Now, let T T e the set of prtil trnsversls ontining t lest two s, inluding the one in ell (1, 1). Consider the onneted omponent of T whih is generted y sequene of opertions # strting with T 0. This omponent orresponds preisely to the set of prtil trnsversls generted y # strting from T 0 (1, 1) in the susqure formed y deleting the first row nd olumn. Tking this set of prtil trnsversls gives n (n 1) (n 1) prtil Ltin squre stisfying A k 1. Note tht this susqure my hve some empty ells whih were filled in the originl n n squre. Lemm 2 In n (n 1) (n 1) prtil Ltin squre stisfying A k 1 indued s desried ove from n n n prtil Ltin squre stisfying A k, the prtil trnsversls generted y # must hve fied ell, i.e., there is some ell tht ppers in ll of these prtil trnsversls. Proof. We ssume without loss of generlity tht the prtil trnsversl T 0 ontining two s disussed ove is the digonl. Suppose tht there re no fied ells in the (n 1) (n 1) prtil Ltin squre. Then there must e prtil trnsversl T 1 in this smller squre whih ontins the ell (i, i) s well s nother ell with the sme symol. The prtil trnsversl T 1 := T 1 (1, 1) must lso pper in the n n squre. Now, in T 1, either (1, 1) nd (i, i) ontin two different duplited symols, or there re t lest three s in T 1 nd (1, 1) nd (i, i) oth ontin. In either se, we n pply # to T 1, deleting the ells (1, 1) nd (i, i) nd otining the ells (1, i) nd (i, 1) (See Squre 4). Sine i ws ritrry, this gives us n filled ells in the first row nd olumn of the squre, ontrdition sine there re only n k distint symols. We now etend the nlysis of Lemm 2 to prove the following. Theorem 1 In prtil Ltin squre stisfying A k suh tht no susqure stisfies A k, there re t lest n k 1 + k filled ells in eh row nd olumn, where n k 1 is the size of the smllest susqure stisfying A k 1. 5

6 1 1 i 1 1 i j i i d d j Squre 4 Squre 5 Figure 3: Squre 4: Illustrtion for the proof of Lemm 2. By pplying # s shown, we find the first row ontins only non-empty ells, ontrdition. Squre 5: Illustrtion for the proof of Theorem 1. If n element lies in the first row ove the smll squre nd lso on the digonl elow nd to the right of the smll squre, we find non-empty ell in the first olumn, elow the smll squre, s shown. Proof. Consider prtil trnsversl T 0, whih we will ssume is long the digonl, nd ell within it, sy (1, 1), ontining duplited symol. Now, hold this ell fied, nd onsider the (n 1) (n 1) prtil Ltin squre stisfying A k 1 generted y the opertion # s ove. Let us ssume tht m ells of T 0 re not fied, nd re in rows nd olumns 2 through m + 1. Note tht this m m susqure stisfies A k 1. By the sme resoning s in the ove lemm, there is trnsversl with duplited symol in ell (i, i), for ll i, 2 i m + 1. Applying #, we find tht there is symol in ells (1, i) nd (i, 1), for 2 i m + 1. There re only m (k 1) symols in the m m susqure stisfying A k 1 ontining rows nd olumns 2 through m + 1, leving (k 1) symols in (1, i), 2 i m + 1, whih re not in the susqure stisfying A k 1 (See Squre 5). Note tht some of these symols my pper in the m m susqure in the originl prtil Ltin squre, ut they do not pper in the set T of prtil trnsversls ssoited with this susqure. Suppose one of these k 1 symols, sy, is in the (1, i) ell. There is in the originl prtil trnsversl T 0, nd sine it is not in the susqure, it must e in ell (j, j), for some j > m + 1. Moreover, there is prtil trnsversl of the smll squre with duplite letter in ell (i, i), sy. (Note this letter ould e, the sme s in ell (1, 1), in whih se there re three s in the orresponding prtil trnsversl of the n n squre). We n now pply # to remove the (1, 1) nd the (i, i) ells, nd we find tht the (1, i) nd (i, 1) ells re filled. Now, the in the (j, j) ell nd the symol in the (i, 1) re oth duplites, so y pplying # gin 6

7 P S j S k Squre 6 Figure 4: The region P in the proof of Theorem 2 we find tht the (j, 1) ell nd the (i, j) ell re filled. (Agin, if oth ells (j, j) nd (i, 1) ontin, there re three s in the prtil trnsversl.) Thus, we know tht the (j, 1) ell is filled. Sine there re t lest k 1 symols in the (1, i) ells, 2 i m + 1, whih re not in the susqure stisfying A k 1, we n pply the sme proess to otin k 1 filled ells in the first olumn in or elow the (m+2)nd row. This gives t lest m + k filled ells in the first olumn, sine the first m + 1 ells re lso filled. Now, m n k 1, euse m is the size of susqure stisfying A k 1, nd n k 1 ws the size of the miniml suh susqure. Sine (1, 1) ws n ritrry ell in our originl prtil trnsversl, this rgument shows tht t lest n k 1 + k ells re filled in eh row nd olumn. If we let n k = n e the size of the originl prtil Ltin squre stisfying A k, then this Theorem shows tht n k n k 1 + 2k, (2) sine the lrger squre hs n k k distint symols, of whih t lest n k 1 +k pper in eh row nd olumn. 3 An Inequlity Let S k e squre stisfying A k suh tht no susqure stisfies A k. It ws shown in Setion 2 tht there must e susqure stisfying A k 1, Choose S k 1 to e the smllest susqure of S k stisfying A k 1 nd, reursively, S m to e the smllest susqure of S m+1 stisfying A m, until the sequene ends t S 2. Let n j e the size of S j. Theorem 2 In S k, s defined ove, for ll j < k, (n k n j )(2n j + n k 1 2n k + 2k j) n j (n j n j 1 2j). (3) Proof. Consider Squre 6. We will ount the numer of filled ells in the retngle P in two different wys. First, there re n k n j olumns in P, nd sine 7

8 eh olumn of S k hs t lest n k 1 +k filled ells, we hve t lest n k 1 +k (n k n j ) filled ells in eh olumn of P, nd t lest (n k n j )(n k 1 + n j n k + k) filled ells in P. We will ll the symols in S j old symols nd those in S k nd not in S j new symols. There re n j j old symols nd n k k n j + j new symols. There re n j rows in P. In eh row of S j there re t lest n j 1 + j old symols. Sine there re only n j j distint old symols, this leves t most n j j (n j 1 + j) old symols in eh row of P, for totl of t most n j (n j n j 1 2j) filled ells ontining old symols in P. There re n k k n j +j new symols, nd n k n j olumns in P. Thus, there re t most (n k n j )(n k k n j + j) filled ells ontining new symols in P. Adding the numer of ells with old nd with new symols in P, we get upper ound for the numer of filled ells in P. Setting this upper ound greter thn or equl to the lower ound, nd simplifying, we otin the inequlity (3) ove. 4 The Min Result Suppose we hve Ltin squre with no prtil trnsversl of length more thn n l. By the previous setions, we hve sequene n 2 < n 3 <... < n l stisfying the Inequlities (1), (2), (3), from the previous setion. Reiterting these inequlities, we hve tht if 2 i l nd 1 j < k l, then n 2 7, (1) n i n i 1 + 2i, (2) (n k n j )(2n j + n k 1 2n k + 2k j) n j (n j n j 1 2j). (3) We will now derive the inequlity k log 2 n k from the Inequlity (3). We first prove the following lemm. Lemm 3 Either n j 4 5 n k or n j n j (n k n j 1 ). Proof. Letting d k := n k n k 1, d j := n j n j 1, (4) we hve, from (3) d j 2j n k n j n j (2n j d k n k + 2k j). (5) 8

9 The diretion of the inequlity lets us remove the lower order terms, giving Now, we ssume tht n j 4 5 n k. This gives d j n k n j n j (2n j d k n k ). (6) d k = n k n k 1 n k n j 1 5 n k, (7) n k + d k 6 5 n k, (8) n k + d k 3 2 n j. (9) Comining (6) nd (9) gives d j 1 2 (n k n j ), (10) By the definition of d j, we hve n j n j n k 1 2 n j, (11) so nd 3 2 n j 3 2 n j (n k n j 1 ). (12) n j n j (n k n j 1 ), (13) ompleting the proof. giving Now, suppose tht n k < 5 4 n j, so 1 3 (n k n j 1 ) n j n j 1, (14) n k n j 2 3 (n k n j 1 ). (15) Sine (15) holds for ll j where j < k, nd n k < 5 4 n j, y indution we get, 1 n k n k 1 ( ) k j 2 (n k n j 1), (16) 3 or k j log 3/2 (n k n j 1 ). (17) Now, suppose in ddition tht k j 1 > log 3/2 n j 4, (18) 9

10 then ontrdition. n j log 3/2 4 < k j 1 log 3/2(n k n j ), (19) n j 4 < n k n j (20) 5 4 n j < n k, (21) If k j log 3/2 n j, then sine log 3/2 n j > log 3/2 n j log 3/ , we hve tht (18) holds, implying tht n k 5 4 n j. We now let k 4 = 2, nd k i = k i 1 + log 3/2 (n ki 1 ). (22) By indution, we otin tht for l k i, ( ) i+1 5 n l, (23) 4 where the se se follows from n 2 7 > (5/4) 5. We know tht n ki 1 < n k for k i 1 < k. So from (22), if k i 1 < k, we hve k i k i 1 + log 3/2 n k. (24) We n now prove the following Lemm. We will speify the et vlue of lter. Lemm 4 For 1/2, either or 1 log 3/2 n k k 1 2, (25) log 5/4 n k > k 1 2. (26) Proof. If log 3/2 n k k 1 2, we hve (25). Otherwise suppose tht log 3/2 n k < k 1 2. (27) Then from (24), for ll k i 1 < k we hve k i < k i 1 + k 1 2. (28) Let j e the minimum integer suh tht k j k. Summing (28) over i gives k k j < k 4 + (j 4)k 1 2, (29) 10

11 whih rerrnges to j > k k (30) k1/2 This shows tht for minimum j suh tht k j k we hve Using (23) with i = j 1 nd (31) we otin j > k 2 k 1/2 + 4 > 1 k 1 2. (31) n kj 1 ( ) j 5 > 4 ( ) 1 5 k1/2. (32) 4 We know tht n k > n kj 1 (euse k > k j 1 ). Then giving (26). log 5/4 n k > 1 k 1 2, (33) We n now mke the left-hnd side of the two equtions in Lemm 4 equl y setting = log 5/ log 3. This gives 4 2 from whih follows: 1 log n k > k 1/2, (34) log 5 log Theorem 3 Every Ltin squre hs prtil trnsversl of length t lest n log 2 n. (35) Here (log 5 4 log 3 2 ) 1. No serious ttempt hs een mde to optimize this onstnt. As ws remrked in [7] the Inequlity (3) nnot imply nything etter thn n log 2 n, sine the sequene n k = 2 k stisfies (3). Let us tke the opportunity to remrk tht Inequlities (1 3) nnot in ft hieve ound etter thn n O(log 2 n). If we let κ = k 1 2 nd γ = k κ 2, then the sequene n k = β κ α 3κ γ for suffiiently lrge 1 nd β α stisfies these inequlities nd hs growth rte of n k = e O(k1/2). 11

12 5 Aknowledgments The uthors would like to thnk the referees for reful reding of this pper nd eeptionlly helpful omments. The seond uthor is indeted to Her Ryser, who introdued him to this prolem over 27 yers go. Referenes [1] A. E. Brouwer, A. J. de Vries, nd R. M. A. Wiering. A lower ound for the length of prtil trnsversls in Ltin squre. Nieuw Arh. Wisk., 24(3): , [2] R. A. Bruldi nd H. J. Ryser. Comintoril Mtri Theory. Cmridge Univ. Press, [3] D. A. Drke. Miml sets of Ltin squres nd prtil trnsverls. J. Sttist. Plnn. Inferene, 1: , [4] H. L. Fu, S. C. Lin, nd C. M. Fu. The length of prtil trnsversl in Ltin squre. J. Com. Mth. Comin. Comput., 43:57 64, [5] K. K. Koksm. A lower ound for the order of prtil trnsversl in Ltin squre. J. Comintoril Theory Ser. A, 7:94 95, [6] H. J. Ryser. Neuere Proleme der Komintorik. In Vorträge üer Komintorik, pges 69 91, Oerwolfh, Germny, July Mtemtishes Forshungsinstitute Oerwolfh. [7] P. W. Shor. A lower ound for the length of prtil trnsversl in Ltin squre. J. Comintoril Theory Ser. A, 33(1):1 8, [8] D. E. Woolright. An n n Ltin squre hs trnsversl with t lest n n distint elements. J. Comintoril Theory Ser. A, 24: ,