Given a Latin square L and a subset F of its autotopism group U(L), we study in this paper some properties and results which partial Latin squares contained in L inherit from U(L), by using F. In particular, we define the concept of F-critical set of L and we ask ourselves about the smallest one contained in L

Falcón Ganfornina, Raúl Manuel

idUS. Depósito de Investigación Universidad de Sevilla

Study of Critical Sets in Latin Squares by usingthe Autotopism GroupR. M. Falco´n 1Departamento de Matema´tica Aplicada I.E. T. S. de Arquitectura. Universidad de Sevilla.Avda. Reina Mercedes 2. 41012 - Sevilla (Spain).AbstractGiven a Latin square L and a subset F of its autotopism group U(L), we study inthis paper some properties and results which partial Latin squares contained in Linherit from U(L), by using F. In particular, we define the concept of F-critical setof L and we ask ourselves about the smallest one contained in L.Keywords: Latin Square, Autotopism Group.1 IntroductionA Latin square L of order n is a n×n array with elements chosen from the setN = {0, 1, ..., n− 1}, such that each symbol occurs precisely once in each rowand each column. The set of Latin squares of order n is denoted by LS(n). IfL = (li,j) ∈ LS(n), the orthogonal array representation of L is the set of n2triples {(i, j, li,j) : i, j ∈ N}. The previous set is identified with L and so, itis written (i, j, li,j) ∈ L, for all i, j ∈ N . It is said that L is an entropic Latinsquare if llij lst = llisljt , for all i, j, s, t ∈ N .1 Email: rafalgan@us.esPreprint of: Electronic Notes in Discrete Mathematics 29 (2007), pp. 503-507. DOI 10.1016/j.endm.2007.07.078 Let Sn be the symmetric group on N . An isotopism of Latin squares oforder n is then a triple Θ = (α, β, γ) ∈ In = Sn × Sn × Sn. If we apply Θto a Latin square L ∈ LS(n), it is verified that α, β and γ are respectively,permutations of rows, columns and symbols of L. The resulting square LΘis also a Latin square and it is said to be isotopic to L. In particular, ifL = (li,j), then LΘ = {(i, j, γ(lα−1(i),β−1(j)): 0 ≤ i, j ≤ n − 1}. An isotopismwhich maps L to itself is an autotopism. The stabilizer subgroup of L in S3nis its autotopism group, U(L) = {Θ ∈ In : LΘ = L}.A partial Latin square, P , of order n, is a n×n array with elements chosenfrom a set of n symbols, such that each symbol occurs at most once in eachrow and in each column. The set of partial Latin squares of order n is denotedby PLS(n). The size of P , |P |, is the number of non-blank cells. If L ∈ LS(n)we will denote by Li,j the partial Latin square contained in L such that theunique filled cell of Li,j is (i, j, li,j). Thus, given P ∈ PLS(n) we can ever finda subset IP ⊆ N ×N such that P =∪(i,j)∈IPLi,j.It is said that P can be completed to a Latin square L ∈ LS(n) if P ⊆ L. IfL is the unique one in such conditions, it is said that P is uniquely completableto L and it is denoted P ∈ UC(L). If besides any proper subset of P can becompleted at least to two distinct Latin squares it is said that P is a criticalset of L and it is denoted P ∈ CS(L). Given L ∈ LS(n), scs(L) denotes thesize of the smallest critical set of L and scs(n) denotes the minimum of scs(L)for all L ∈ LS(n). Analogously, lcs(L) denotes the size of the largest criticalset of L and lcs(n) denotes the maximum of lcs(L) for all L ∈ LS(n).2 Extended autotopisms of partial Latin squaresAn isotopism of partial Latin squares of order n will be a triple Θ = (α, β, γ) ∈In, where γ(∅) = ∅.Lemma 2.1 Let P ∈ PLS(n) be contained in a Latin square L ∈ LS(n) andlet Θ ∈ In. The following asserts are verified:a) PΘ is also in PLS(n) and∣∣PΘ∣∣ = |P |,b) If Q ∈ PLS(n) verifies that P ⊆ Q, then PΘ ⊆ QΘ.c) If P can be completed to L, then PΘ can be completed to LΘ. ✷Lemma 2.2 Given L ∈ LS(n), let Θ1,Θ2 ∈ U(L) be two distinct autotopismsof L. Let us consider Li1,j1 , Li2,j2 ∈ PLS(n) with (i1, j1) ̸= (i2, j2). ThenLΘ1i1,j1 ̸= LΘ1i2,j2and LΘ1i1,j1 ̸= LΘ2i1,j1. ✷Now, let us consider L ∈ LS(n) and let F ⊆ U(L). If P ∈ PLS(n) canbe completed to L, we will define P F ∈ PLS(n) as P F =∪Θ∈F PΘ. Then, wewill say that P F is an extended autotopy of P .Lemma 2.3 Let us suppose L ∈ LS(n), P ∈ PLS(n) contained in L andF ⊆ U(L). Then,∣∣P F∣∣ ≤ |P | · |F|. ✷Example 2.4 Let L =(0 11 0)∈ LS(2). So, U(L) = {(Id, Id, Id) , ((Id, (01),(01)), ((01), Id, (01)), ((01), (01), Id)}. Let us take now by example F ={(Id, Id, Id) , (Id, (01), (01))}, P =(0 ∗∗ ∗)∈ PLS(2) and Q =(0 ∗∗ 1)∈ PLS(2).Then, we can prove that P F =(0 1∗ ∗)and P U(L) = L = QF = QU(L). ✁In general, given L ∈ LS(n), there does not exist a subset F of U(L) andP ∈ PLS(n) such that P ⊂ L and P F = L. This is due to that the most ofLatin squares has only the trivial autotopism group [1], U(L) = {(Id, Id, Id)}.We can therefore ask ourselves about conditions under which we can obtain asimilar result to Example 2.4:Theorem 2.5 Every entropic Latin square is an extended autotopy of each ofits proper partial Latin squares. ✷Example 2.6 Let L =(0 1 21 2 02 0 1)∈ LS(3). It is entropic and verifies that|U(L)| = 18 [1]. Let F = {(αs, αt, αlst)}s,t∈N , where α0 = Id, α1 = (012), α2 =(021). So, |F| = 9 and PΘ = L for all P ∈ PLS(3) contained in L. ✁3 Critical sets by considering U(L)Given L ∈ LS(n) and F ⊆ U(L), let < F > be the subgroup of U(L) generatedby composing the elements of F ∪ F−1, where F−1 = {Θ−1 = (α−1, β−1, γ−1) :Θ = (α, β, γ) ∈ F} ⊆ U(L). Now, given P ∈ PLS(n) contained in L, letus denote by F(P ) the partial Latin square P<F>. We will then say thatP is uniquely F-completable to L and it will be denoted by P ∈ UCF(L) ifF(P ) ∈ UC(L). We will say that C is a F-critical set of L if C ∈ UCF(L) andP ̸∈ UCF(L) for all P ⊂ C. So, we are interested in the smallest size scsF(L)of a F-critical set of L.Lemma 3.1 Let L ∈ LS(n). The next assertions are verified:a) Given F1,F2 ⊆ U(L) such that F1 ⊆ F2, then scsF2(L) ≤ scsF1(L).b) If F ⊆ U(L) is such that | < F > | ≥ lcs(L), then scsF(L) = 1. ✷Lemma 3.2 Let L ∈ LS(n), P ∈ PLS(n) contained in L and F ⊆ U(L). LetC ∈ CS(L) be such that |C| = scs(L). If C ⊆ F(P ) then, scsF(L) ≤ |P |. So,scsF(L) ≤ scs(L), for all F ⊆ U(L), such that F ̸= ∅. ✷Proposition 3.3 Let L ∈ LS(n), P ∈ PLS(n) contained in L and F ⊆ U(L).Then, scsF(L) = minP∈PLS(n) {|P | : ∃C ∈ CS(L) such that C ⊆ F(P )}. ✷Example 3.4 Let L =0 1 2 3 41 2 3 4 02 3 4 0 13 4 0 1 24 0 1 2 3 ∈ LS(5), C =0 1 ∗ ∗ ∗1 ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ 2∗ ∗ ∗ 2 3 ∈ CS(L) and P =0 1 ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ 1 2∗ ∗ ∗ ∗ ∗ ∈ PLS(5). Let F = {((04321), Id, (04321))} ⊆ U(L). So, C ⊆P ∪ P F =0 1 ∗ ∗ ∗1 2 ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ 1 2∗ ∗ ∗ 2 3 and therefore, scsF(L) ≤ 4 < 6 = scs(L). ✁3.1 An algorithm to obtain an upper bound of scsF(L)Lemma 3.5 Let L = (lij) ∈ LS(n), F ⊆ U(L), P ∈ PLS(n) be containedin L such that |P | = scsF(L) and C ∈ CS(L) be contained in F(P ). For alli, j ∈ N , there exist (s, t, lst) ∈ C and Θ ∈ < F > such that LΘi,j = Ls,t. ✷Lemma 3.6 Let L ∈ LS(n), F ⊆ U(L), P ∈ PLS(n) be contained in L suchthat |P | = scsF(L) and C ∈ CS(L) be contained in F(P ). Then P ⊆ F(C). ✷In general, given L = (lij) ∈ LS(n), P =∪(i,j)∈IPLi,j ∈ PLS(n) containedin L and F ⊆ U(L), we must be interested in an algorithm which allows us toobtain the number scsF(L). To do it, let F(P ) =∪(i,j)∈IF(P )Li,j ∈ PLS(n),which is contained in L. Given (i, j, lij) ∈ F(P ), let us consider:SPi,j = {(s, t, lst) ∈ P such that Li,j ⊆ F(Ls,t)} ⊆ P.Lemma 3.7 Let L ∈ LS(n), P,Q ∈ PLS(n) be both contained in L andF ⊆ U(L). If Q ⊆ F(P ) and P ⊆ F(Q), then P =∪(i,j)∈IQSPi,j and Q =∪(i,j)∈IPSQi,j. ✷Theorem 3.8 Let L ∈ LS(n) and F ⊆ U(L). Then scsF(L) is equal to:minC∈CS(L)min|P | : ∃IP ⊆ IF(C) being P =∪(i,j)∈IPLi,j , C =∪(i,j)∈IPSCi,j .Proof. Let C ∈ CS(L) and IP ⊆ IF(C) be such that C =∪(i,j)∈IPSCi,j, beingP =∪(i,j)∈IPLi,j ⊆ F(C). So, given (s, t, lst) ∈ C, there exists Θ ∈ < F >and (i, j) ∈ IP such that LΘs,t = Li,j. Then, LΘ−1i,j = Ls,t and so, (s, t, lst) ∈F−1(P ) = F(P ). Therefore, we have that C ⊆ F(P ). So, from Proposition3.3, scsF(L) is smaller than the signaled minimum. Now, let P ∈ PLS(n)contained in L be such that |P | = scsF(L). From Proposition 3.3, there existC ∈ CS(L) contained in F(P ). Besides, from Lemma 3.6, P ⊆ F(C). So,Lemma 3.7 involves C =∪(i,j)∈IPSCi,j, being IP ⊆ IF(C), and therefore, byusing Proposition 3.3 again, scsF(L) is bigger than the signaled minimum. ✷The computation of the minimum of the previous theorem allows us toobtain scsF(L) but it can be an arduous process. In a concrete case, a firstupper bound of scsF(L) can be given by the following way: let C ∈ CS(L)be such that |C| = scs(L). Let us obtain F(C) and next all the sets SCi,j.If the cardinality of all these sets is one, then we cannot improve the upperbound of scs(L) by using this critical set C. In the other case, let n1 > 1 bethe maximum of the mentioned cardinalities and let us take SCi1,j1 a set withcardinality n1. Let us now fixe (s1, t1, ls1t1) ∈ SCi1,j1and let Θ1 ∈ F be suchthat LΘ1s1,t1 = Li1,j1 . Let us then consider C1 = C \(SCi1,j1 \ {(s1, t1, ls1t1)})andP1 = CΘ11 ⊆ CΘ1 . So, |P1| = |C1| < |C| and it can be seen that C ⊆ F(P1) andtherefore, scsF(L) ≤ |C1| < scs(L). Now we can take the same procedure withC1 instead of C. If the cardinality of all the corresponding SC1i,j is one, then wecannot improve the upper bound of scsF(L) by using this method with C. Inthe other case, we take SC1i2,j2 a set with cardinality n2 > 1, the maximum ofthe mentioned cardinalities, and we fixe (s2, t2, ls2t2) ∈ SC1i2,j2. Let us observethat it is necessary to take (s2, t2, ls2t2) = (s1, t1, ls1t1) whenever it is possible.So, we consider C2 = C1 \(SC1i2,j2 \ {(s2, t2, ls2t2)})⊆ C1, being Θ2 ∈ F suchthat LΘ2s2,t2 = Li2,j2 . Let P2 = CΘ22 ⊆ CΘ21 . By construction, C1 ⊆ F(P2) andso, P1 ⊆ F(C1) ⊆ F(F(P2)) = F(P2). Finally, C ⊆ F(P1) ⊆ F(F(P2)) = F(P2).Therefore, scsF(L) ≤ |P2| = |C2| < |C1|. We repeat all this procedure untilwe find that the maximum cardinality of all the corresponding sets Si,j is one.Example 3.9 In Example 3.4, where F(C) =0 1 ∗ 3 41 2 ∗ 4 02 3 ∗ 0 13 4 ∗ 1 24 0 ∗ 2 3, we can see that,for all i ∈ N ,∣∣SCi,j∣∣ ={1, if j = 1 or 3,2, if j = 0 or 4.. Let us then take for exam-ple SC0,0 = {(0, 0, 0), (1, 0, 0)} and let us consider (0, 0, 0) ∈ SC0,0. So, C1 =C \ L1,0 =0 1 ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ 2∗ ∗ ∗ 2 3. Besides, it will be P1 = C1 because LId0,0 = L0,0.So, C ⊆ F(P1) = F(C) and scsF(L) ≤ |C1| = 5. Now, for all i ∈ N ,∣∣SC1i,j ∣∣ ={1, if j = 0, 1 or 3,2, if j = 4.. Let us take SC13,4 = {(3, 4, 2), (4, 4, 3)} andlet us consider (3, 4, 2) ∈ SC0,0. So, C2 = C1 \ L3,4 =0 1 ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ 2∗ ∗ ∗ 2 ∗ = P2, asLId3,4 = L3,4. So, C ⊆ F(P2) = F(C) and scsF(L) ≤ |C1| = 4, as we have seenin Example 3.4. Thus, SC2i,j = 1 for all i, j and so, the algorithm finishes. ✁References[1] B. D. McKay, A. Meynert and W. Myrvold, Small Latin Squares, Quasigroupsand Loops, Journal of Combinatorial Designs 15 (2007) 98-119.

Study of critical sets in latin squares by using the autotopism group

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Study of critical sets in latin squares by using the autotopism group

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