On completing three cyclic transversals to a latin square
aa r X i v : . [ m a t h . C O ] D ec On completing three cyclic transversals to alatin square
Nicholas J. CavenaghSchool of Mathematics and StatisticsThe University of New South WalesSydney 2052, AustraliaCarlo H¨am¨al¨ainenDepartment of MathematicsThe University of QueenslandQLD 4072, AustraliaAdrian M. NelsonSchool of Mathematics and StatisticsUniversity of SydneyNSW 2006, AustraliaOctober 27, 2018
Abstract
Let P be a partial latin square of prime order p > P be a partiallatin square of the form: P = { ( i, c + i, s + i ) , ( i, c ′ + i, s ′ + i ) , ( i, c ′′ + i, s ′′ + i ) | ≤ i < p } for some distinct c, c ′ , c ′′ and some distinct s, s ′ , s ′′ . In this paper weshow that any such P completes to a latin square which is diagonallycyclic. A latin square of order n is an n × n array of symbols such that each cellcontains one symbol and each symbol occurs once in each row and once ineach column. In this paper, rows, columns and symbols are taken from theset N = { , , . . . , n − } and are always calculated modulo n . We use thenotation i ◦ j to denote the symbol in cell ( i, j ) of a latin square L = L ◦ .We may denote L a set of ordered (row, column, symbol) triples of the form( i, j, i ◦ j ). A partial latin square , as its name suggests, is a partially filled-in n × n array of symbols such that each cell contains at most one symbol andeach symbol occurs at most once in each row and at at most once in eachcolumn.A latin square L ◦ is said to be diagonally cyclic if n is odd and for eachcell ( i, j ), i ◦ j = k implies that ( i + 1) ◦ ( j + 1) = k + 1. Diagonally cyclicsquares of even order do not exist; for a nice proof see [6]. Henceforth weassume that n is odd.Let j be coprime to some odd n and j = 1. If we define ◦ j by 0 ◦ j i = ij (mod p ) for each i , a valid diagonally cyclic latin square is generated, whichwe denote by B n,j . The following diagram shows B , .0 3 1 4 23 1 4 2 01 4 2 0 34 2 0 3 12 0 3 1 4Clearly a diagonally cyclic latin square is defined by the ordering of sym-bols in row 0. However, not all orderings of the first row will complete toa diagonally cyclic latin square, as symbols may not be repeated within acolumn. In fact, row 0 “generates” a diagonally cyclic latin square if andonly if the operation f ( i ) = 0 ◦ i − i is a permutation of the set N .Let L be a diagonally cyclic latin square of order n . For any constantinteger c , we define L ⊕ c to be the diagonally cyclic latin square formed bycycling the rows of L by c (modulo n ). That is, L ⊕ c = { ( i + c, j, k ) | ( i, j, k ) ∈ L } . ◦ ◦ standard order . Similarly we define L ⊕ c and L ⊕ c tobe the diagonally latin squares formed by adding constant c to each columnor symbol, respectively.Let L be a diagonally cyclic latin square in standard order. For any c coprime to n , we define L × c to be the diagonally cyclic latin squaregenerated by operation ◦ c , where i ◦ c j = ( ic − ◦ jc − ) c (mod n ), for each i , j such that 0 ≤ i, j ≤ n −
1. (Informally, the first row of L × c is formed bymultiplying each column and symbol of the first row of L by c .) Again, sucha transformation is invertible. For the curious reader, other equivalences ofdiagonally cyclic latin squares are given in Lemma 2 . L is a diagonally cyclic latin square of prime order p containing the partial latin square P defined as in the abstract. Consider theisomorphic latin square L ′ = ((( L ⊕ ( − s )) ⊕ ( − c )) × (( c ′ − c ) − ) . Observe that L ′ is in standard order, with symbol ( s ′ − s ) / ( c ′ − c ) in row 0and column 1.In Section 4, when we prove the main result, we thus assume, withoutany loss of generality, that 0 ◦ ◦ j .In any diagonally cyclic latin square L , the set of symbols L ( α ) = { ( i, α + i, ◦ α + i ) | ≤ i ≤ n − } contains each row, each column and eachsymbol exactly once and is thus a transversal . We denote L ( α ) as a cyclictransversal of L . In fact, the cyclic transversals L ( α ), 0 ≤ α ≤ n − i ◦ j = j − i .The research in this paper is motivated by the following problem posedby Alspach and Heinrich [1]: For each k , does there exist an N ( k ) such that if k transversals of a partial latin square of order n > N ( k ) are prescribed, thesquare can always be completed? The existence of idempotent latin squaresfor every order n = 2 shows that N (1) = 3. H¨aggkvist and Daykin [3] haveshown that every partial n × n latin square where each row, column andsymbol is used at most 2 − √ n times is completable whenever n is divisibleby 16. However it is as yet unconfirmed that N (2) exists. Gr¨uttm¨uller [6]showed that N ( k ) ≥ k −
1. Computational results on latin squares of smallorder support the conjecture that N ( k ) = 4 k − C ( k ) such that if k cyclically generated diagonals of a partiallatin square of odd order n ≥ C ( k ) are prescribed, the square can always becompleted? Gr¨uttm¨uller showed that C (2) = 3 ([6]) and that C ( k ) ≥ k − k ≥ C (3) may be equal to 9.Specifically, we show (in Theorem 7) that if P is a partial latin square ofprime order p > P = { ( i, c + i, s + i ) , ( i, c ′ + i, s ′ + i ) , ( i, c ′′ + i, s ′′ + i ) | ≤ i < p } for some distinct c , c ′ , c ′′ and some distinct s , s ′ , s ′′ , then P has a completionto a (diagonally cyclic) latin square.In our proof, in Section 2 we first identify a method to reorder cer-tain cyclically generated transversals within B p,j to form a new diagonallycyclic latin square. This reordering or trade is algebraically defined and thetransversals are based on some linear transformation of the quadratic residuesmod p . We are thus able to redefine the problem above in terms of simul-taneous equations, where instead of a precise solution to each equation, weinstead require some information about which coset the solution belongs to.Here cosets are taken from the multiplicative group of the field of prime order p . When p is large enough (specifically, when p ≥ Diagonally cyclic latin squares have a number of intriguing equivalences.Firstly, diagonally cyclic latin squares of order n are equivalent to transver-sals in the latin square B n , which is precisely the operation table for theintegers modulo n . To see this, let L be a diagonally cyclic latin squarewith operation ◦ . Let P ⊆ B n be the partial latin square defined by P = { (0 ◦ i − i, i, ◦ i ) | ≤ i ≤ n − } . Then P is a transversal. Conversely,4et P be a transversal of B n . Then, for each triple ( r, c, r + c (mod n )) ∈ P ,define 0 ◦ c = r + c . Then ◦ generates a valid diagonally cyclic latin square.Another equivalence involves placing n semi-queens on a toroidal n × n chessboard such that no two semi-queens may attack each other. A semi-queen attacks any piece in its row and column, but only on the ascending diagonals; i.e. those which begin in the lower left and finish in the upperright. On a toroidal chessboard these diagonals “wrap around” in the obviousfashion. Thus if we replace the chessboard with the operation table for theintegers modulo n , we see that each queen must lie on a different symbol,and no queens may share a common row or a common column. So a set of n semi-queens which may not attack each other on the next turn is equivalentto a transversal within B n . For more detail on the semi-queen problem, werefer the reader to [9].Diagonally cyclic latin squares in standard order are also equivalent to complete mappings of the cyclic group. We refer the reader to [7] for moredetail on complete mappings. B n,j via number theory Consider the diagonally cyclic latin square B , below. If we replace eachsymbol in bold with its subscript, we obtain another diagonally cyclic latinsquare. In fact, the bold symbols in the first row are precisely the quadraticresidues modulo 11, and the subscripts in the first row are obtained by mul-tiplication by 4. This example is part of a general construction given in thefollowing theorem. 5 xample 1.
10 5 0
34 10 5 Theorem 2.
Let ( a, b, c ) be a solution to the equation: (1 − j ) a m + jb m ≡ c m (mod p ) (1) where p is a prime, j = 1 , j is coprime to p , m ≥ , m divides p − and a m , b m and c m are non-zero and pairwise distinct (mod p ). Let α = 0 and γ beconstants modulo p . Next, replace each symbol in the first row of B p,j of theform αβ m + γ (where β = 0 ) with the symbol α ( b/c ) m β m + γ. Then this new first row generates a valid diagonally cyclic latin square.Proof.
Let L be the square array (at this stage we haven’t proved that itis latin) generated by the new first row. Since b m = c m , ( b/c ) m acts as aderangement on the set of m th powers, so the symbols in the first row of L (and thus in all rows) are distinct.It suffices, then, to check that all symbols in the first column are distinct.Within B p,j , symbol αβ m + γ occurs in column j − ( αβ m + γ ) in row 0. Sothis cyclic transversal contains the symbol ( αβ m + γ )(1 − j − ) in column 0.Within L this is replaced by( α ( b/c ) m β m + γ ) − j − ( αβ m + γ ) = αβ m (( b/c ) m − j − ) + γ (1 − j − )= ( αβ m ( a/c ) m + γ )(1 − j − ) . a/c ) m permutes the m th powers, the symbols in the first column(and thus in all columns) are distinct.Observe that 7 + 1 = 2 × . Thus, for m = 2, the triple ( a, b, c ) =(7 , ,
5) is a valid solution of Equation (1) for j = ( p + 1) / p >
7. The earlier Example 1 demonstrates this for p = 11, α = 1 and γ = 0. We now focus on the problem of completing three arbitrary cyclic transversalsto a (diagonally cyclic) latin square. In this section, we restrict ourselves tothe case where p is a prime. We denote the finite field of size p by GF( p ).We may assume, without loss of generality, that we have transversalsgenerated by 0 and j in cells (0 ,
0) and (0 , j = 1 for this to be a valid partial latin square. Our third transversal isarbitrary; assume that it is generated by symbol e in column c . So that nosymbols repeat in a row or in a column, we have as necessary conditions c
6∈ { , } , e
6∈ { , j, c } and e − c = j −
1. If e = jc , then these threetransversals are contained within B p,j and thus complete to a latin square.So we henceforth assume also that e = jc .Our aim is to apply Theorem 2 to B p,j to obtain a latin square which con-tains the above three cyclic transversals. To transform the third transversal,it is sufficient to find m , α , β and γ , with m | p − α = 0, β = 0, such that jc = αβ m + γ (2) e = αxβ m + γ (3)where x = ( b/c ) m for some appropriate solution ( a, b, c ) to Equation 1. Equiv-alently (considering the conditions of Theorem 2), we assume that x is an m th power modulo p , x
6∈ { , } and that if we define F ( x ) = 1 − jx − j , then F ( x ) is also an m th power and F ( x ) / ∈ { , } .7quations (2) and (3) imply that a given x determines γ and the prod-uct αβ m : γ = xjc − ex − , (4) αβ m = e − jcx − . (5)Since x = 1 these equations are well-defined. Given a valid x and m ,it is always possible to choose such α , β and γ . However, in this process oftransformation we do not wish to alter the first two transversals in columns 0and 1.Equivalently, − γ/α and ( j − γ ) /α must not be m th powers modulo p (although they may be equal to 0). But Equation (5) implies that α is anon-zero m th power if and only if ( e − jc ) / ( x −
1) is a non-zero m th power.Thus, the following expressions must not be non-zero m th powers modulo p : G ( x ) = − γ ( x − e − jc = e − xjce − jcH ( x ) = ( j − γ )( x − e − jc = xj (1 − c ) + e − je − jc Note that if H ( x ) = 1 then x = 1. In turn, if x = 1 then F ( x ) = 1. So ifwe assume that H ( x ) = 1 it is unnecessary to specify x = 1 and F ( x ) = 1 asconditions. In summary, we have the following. Lemma 3.
Let p be prime. Let c , e and j be residues mod p such that c
6∈ { , } , j
6∈ { , } and e
6∈ { , j, c, c + j − , jc } . Let x = 0 be an m thpower mod p such that F ( x ) is a non-zero m th power mod p , and G ( x ) and H ( x ) are not non-zero m th powers mod p . Then, the three cyclic transversalsgenerated by ◦ , ◦ j and ◦ c = e can be completed to a diagonallycyclic latin square. If we fix a value of x there will be some inappropriate choices of c , e and j .However, we next show that for sufficiently large primes, it is possible to findsuch a x for any choice of c , e and j that satisfies the necessary conditionsoutlined above.The techniques that follow are similar to those applied to constructingcyclic triplewhist tournaments in [2]. We need the following version of Weil’sTheorem (see [8], Theorem 5.38). Recall that a multiplicative character χ p ) is a homomorphism from the multiplicative group of GF( p ) − { } into the multiplicative group of unitary complex numbers. When Weil’stheorem is applied it is also understood that χ (0) = 0 for any multiplicativecharacter χ . Theorem 4. (Weil’s Theorem) Let χ be a multiplicative character of thefinite field GF( p ) with order m > and let f be a polynomial of GF( p )[ x ] of degree d which is not of the form kg m for some constant k ∈ GF( p ) andsome g ∈ GF( p )[ x ] . Then: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X c ∈ GF( p ) χ ( f ( c )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( d − p / . For m = 2, we will use the quadratic character η defined as follows onGF( p ) − { } : η ( x ) = ( x is a quadratic residue; − m = 2 we can obtain a more explicit version of Weil’s Theorem ([8],Theorem 5.48): Theorem 5.
Let η be the quadratic character of GF( p ) and let f = a x + a x + a ∈ GF( p )[ x ] with p odd and a = 0 . If a − a a = 0 then: X c ∈ GF( p ) η ( f ( c )) = − η ( a ) . We can now show the main theorem of this section:
Theorem 6. If P is a partial latin square of prime order p ≥ comprisingof three arbitrary transversals, then P has a completion to a (diagonallycyclic) latin square.Proof. Setting m = 2, it suffices to show that the conditions of Lemma 3hold for any valid choices of j , c , and e . We wish to show that there existssome x ∈ GF( p ) such that η ( x ) = 1, η ( F ( x )) = 1, ( η ( G ( x )) = − η ( H ( x )) = − x for which η ( x ) = 1, η ( F ( x )) = 1, η ( G ( x )) = − η ( H ( x )) = − p ≥ A is non-empty: A = { x | η ( x ) = η ( F ( x )) = 1 and η ( G ( x )) = η ( H ( x )) = − } . Define J ( x ) = [1 + η ( x )][1 + η ( F ( x ))][1 − η ( G ( x ))][1 − η ( H ( x ))]and S = X x ∈ GF( p ) J ( x ) . If any of x , F ( x ), G ( x ) or H ( x ) equals 0, then J ( x ) ≤
8. If x ∈ A then J ( x ) = 16. For other values of x ∈ GF( p ), J ( x ) = 0. Thus S ≤ | A | + 32.So it suffices to show that S > S and using the fact that η is a homomorphism, we can ex-press S as p plus a series of terms of the form P c ∈ GF( p ) η ( K ( x )), where K ( x )is a product of non-repeated factors from the set { x, F ( x ) , − G ( x ) , − H ( x ) } .For any (non-constant) linear function g ( x ), it is well-known that X x ∈ GF( p ) g ( x ) = 0 . From the necessary conditions on j , c and e , each of the quadratics xF ( x ), xG ( x ), xH ( x ) F ( x ) G ( x ), F ( x ) H ( x ) and G ( x ) H ( x ) have no repeated linearfactors and thus each has a non-zero discriminant.It follows that the conditions of Weil’s theorem hold for each of the termsof the form K ( x ) in the expansion of S . There are (cid:0) (cid:1) quadratic terms,so from Theorem 5, these contribute at least − S . Next, there are (cid:0) (cid:1) cubic terms, so from Theorem 4 with d = 3, these contribute at least − √ p to S . Finally there is one quartic term, so from Theorem 4 with d = 4, this contributes at least − √ p to S . Thus, S ≥ p − √ p −
6. But p − √ p − >
32 for any prime p ≥ By using computational methods for primes less than 191, Theorem 6 fromthe previous section can be improved to the following:10 heorem 7. If P is a partial latin square of prime order p > of the form P = { ( i, c + i, s + i ) , ( i, c ′ + i, s ′ + i ) , ( i, c ′′ + i, s ′′ + i ) | ≤ i < p } for some distinct c, c ′ , c ′′ and some distinct s, s ′ , s ′′ , then P has a completionto a (diagonally cyclic) latin square. Note that in [5] it is shown that three cyclic transversals do not alwayscomplete to diagonally cyclic latin squares of order 7.Two approaches were used to do the small cases computationally. Firstly,we checked all the instances where the conditions of Lemma 3 were satisfiedfor an appropriate m . This approach provided complete solutions for allprimes between 61 and 181 (inclusive), and most solutions for primes between11 and 59, leaving a total of 2076 exceptions (choices of j , c , and e ) for smallerprimes: prime 11 13 17 19 23 29 31 41 47 59 ≤ p ≤
59 were solved using a randomised depth firstsearch algorithm [4]. 11
Case p = 11 Here we present the computational data for p = 11. Each of the followingfour columns contain the exceptions (triples c , e , j ) and the first row of thediagonally cyclic latin square which was found by backtrack search for thegiven triple. The symbol ‘a’ denotes 10. The authors wish to thank Dr. Julian Abel for directing them to [2].12 eferences [1] Brian Alspach and Katherine Heinrich. Matching designs.
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