By J. Denes and A.D. Keedwell (Eds.)

ISBN-10: 0444888993

ISBN-13: 9780444888990

In 1974 the editors of the current quantity released a well-received ebook entitled ``Latin Squares and their Applications''. It incorporated an inventory of seventy three unsolved difficulties of which approximately 20 were thoroughly solved within the intervening interval and approximately 10 extra were in part solved. the current paintings includes six contributed chapters and in addition six extra chapters written via the editors themselves. in addition to discussing the advances that have been made within the material of lots of the chapters of the sooner e-book, this new publication comprises one bankruptcy which offers with an issue (r-orthogonal latin squares) which didn't exist whilst the sooner publication was once written. The luck of the previous publication is proven by means of the 2 or 300 released papers which care for questions raised through it.

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**Extra resources for Latin Squares: New Developments in the Theory and Applications**

**Example text**

It follows that LO a set of at most 2n+2 mutually orthogonal latin squares. 0 As Drake has remarked, no corresponding theorem for the case of a latin square of order 4n+4 with a latin subsquare of the maximum order 2n+2 exists, since, in the particular case 4n+4 = 2". a complete set of mutually orthogonal latin squares exists all of which have latin subsquares of order Zm-l. This set of squares arises from the desarguesian projective plane of order 2m. In his paper (1977). Drake used the above theorem to help him in his attempt to determine the spectrum of pairs of integers (n,k) for which there exists a maximal set of k mutually orthogonal latin squares: that is a set of k mutually orthogonal latin squares which cannot be embedded in a set of k t l such squares.

Of m-1 mutually orthogonal latin squares of order m and, since '/m = 9 n jzh (where m = p t ) is a product of prime powers, Chapter 2 26 2:20 Q* of at least min (pj -1) 3 m-1 mutually orthogonal j #h latin squares of order q = S/m by MacNeish’s theorem. ) From these we can select a subset Q = {Bo,B1,. . Bm-2) squares and thence a set C = latin squares of order mq s as in Drake’s theorem above. ,] = Ca:,,Bt1 of m-1 of m-1 mutually orthogonal Since m does not divide q , the set C has no common transversal by Corollary (1).

J=O i=O d=O i+j=d i=O j=O (mod m) So, However, the left-hand side of this equality can be computed in another way by first grouping together those terms which involve Cf(i,j)Ii+j=d to . -d i j (mob m) 28 2:22 Chapter 2 i t j < m and (itj-d)i+ka = m if i t j z m , m is a factor of the double sum. It follows that m divides $m(m-l)q and so i(m-1)q is an integer. But this is impossible if m is even and q is odd. transversal in this case follows. (5) The non-existence of a fl & of mutu_allv orthogonal latin-3guae-s- which are not eAgndible.

### Latin Squares: New Developments in the Theory and Applications by J. Denes and A.D. Keedwell (Eds.)

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