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Latin Squares, Cubes and Hypercubes

Latin Squares, Cubes and Hypercubes. Jerzy Wojdy ł o Southeast Missouri State University March 31, 2007. Definition and Examples. A Latin square is a square array in which each row and each column consists of the same set of entries without repetition. Existence.

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Latin Squares, Cubes and Hypercubes

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  1. Latin Squares, Cubes and Hypercubes Jerzy WojdyłoSoutheast Missouri State University March 31, 2007

  2. Definition and Examples • A Latin square is a square array in which each row and each column consists of the same set of entries without repetition. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  3. Existence • Do Latin squares exist for every nZ+? • Yes. • Consider the addition table (the Cayley table) of the group Zn. • Or, more generally, consider the multiplication table of an n-element quasigroup. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  4. Operations on Latin Squares • Isotopism of a Latin square L is a • permutation of its rows, • permutation of its columns, • permutation of its symbols. (These permutations do not have to be the same.) • L is reduced iff its first row is [1, 2, …, n] and its first column is [1, 2, …, n]T. • L is normal iff its first row is [1, 2, …, n]. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  5. Enumeration of LS • How many Latin squares ( rectangles) are there? • If order  11Brendan D. McKay, Ian M. Wanless, “The number of Latin squares of order eleven” 2004(?) (show the table on page 5)http://en.wikipedia.org/wiki/Latin_square#The_number_of_Latin_squares • Order 12, 13, … open problem. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  6. Enumeration of LS n! (n-1)! times the number of reduced Latin squares Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  7. Orthogonal Latin Squares • Two nn Latin squares L=[lij] and M =[mij] are orthogonal iff the n2 pairs (lij, mij) are all different. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  8. Orthogonal LS - Useful Property • TheoremTwo Latin squares are orthogonal iff their normal forms are orthogonal. (You can permute symbols so both LS have the first row [1, 2, …, n]) • No two 22 Latin squares are orthogonal. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  9. Orthogonal Latin Squares • This 44 Latin square does not have an orthogonal mate. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  10. Orthogonal LS – History 1782 • Leonhard Euler • The problem of 36 officers, 6 ranks, 6 regiments. His conclusion: No two 66 LS are orthogonal. • Additional conjecture: no two nn LS are orthogonal, where n  Z+, n  2 (mod 4). • Case n = 2 is obvious. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  11. Orthogonal LS – History 1782 • Leonhard EulerRecherches sur une nouvelle espèce de quarrés magiques, Originally published in Verhandelingen uitgegeven door het zeeuwsch Genootschap der Wetenschappen te Vlissingen 9, Middelburg 1782, pp. 85-239 Also available in: Commentationes Arithmeticae 2, 1849, pp. 302-361Opera Omnia: Series 1, Volume 7, pp. 291-392 Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  12. Orthogonal LS – History 1900-01 • Gaston Tarry verified case n=6.Le problème de 36 officiers. Compte Rendu de l'Assoc. Français Avanc. Sci. Naturel 1, 122-123, 1900.Le problème de 36 officiers. Compte Rendu de l'Assoc. Français Avanc. Sci. Naturel 2, 170-203, 1901. • Two years of Sundays. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  13. Orthogonal LS – History 1959 • In 1959, Bose and Shrikhande constructed a pair of orthogonal Latin squares of order 22. • Then Parker constructed a pair of orthogonal Latin squares of order 10. • Picture (next slide) orhttp://www.cecm.sfu.ca/organics/papers/lam/paper/html/NYTimes.html Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  14. Orthogonal LS – NYT 4/26/1959 Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  15. Orthogonal LS – History 1960 • 1960 R.C. Bose, S.S. Shrikhande, E.T. Parker, Further Results on the Construction of Mutually Orthogonal Latin Squares and the Falsity of Euler's Conjecture, Canadian Journal of Mathematics, vol. 12 (1960), pp. 189-203. • There exists a pair of orthogonal LS for all nZ+, with exception of n = 2 and n = 6. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  16. Mutually Orthogonal LS (MOLS) • A set of LS that are pairwise orthogonal is called a set of mutually orthogonal Latin squares (MOLS). • TheoremThe largest number of nn MOLS is n1. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  17. Mutually Orthogonal LS (MOLS) • TheoremIf p is prime, then there are p1pp-MOLS. • ProofConstruction of Lk=[akij], k =1, 2, …, p1: akij = ki + j (mod p).  • CorollaryIf n = pt, p prime, then there are n1nn-MOLS. • Open problemIf there are n1nn-MOLS, then n = pt, p prime. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  18. Mutually Orthogonal LS (MOLS) Let N(n) be the number of MOLS that exist of size n. • Open problem: Find N(n) for n ≠ pt, p prime.For example N(10)  2. Is N(10) = 3? • C. Colbourn and J. Dinitz, Mutually orthogonal latin squares: a brief survey of constructions, Journal of Statistical Planning and Inference 95 (2001) 9-48 • The CRC Handbook of Combinatorial Designs, edited by Charles Colbourn and Jeff Dinitz, CRC Press 1996. (Table of lower bounds for N(n)) Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  19. MOLS Lower Bounds for N(n) Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  20. Completion Problems • When can a pqLatin rectangle with entries in {1, 2, …, n} be completed to a nnLatin square? Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  21. Completion Theorems • TheoremLet p < n. Any pnLatin rectangle with entries in {1, 2, …, n} can be completed to a nnLatin square. • The proof uses Hall’s marriage theorem or transversals to complete the bottom n p rows. The construction fills one row at a time. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  22. Completion Problems • The good: Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  23. Completion Theorems • TheoremLet p, q < n. A pqLatin rectangle R with entries in {1, 2, …, n} can be completed to a nnLatin square iff R(t), the number of occurrences of t in R, satisfiesR(t)  p + q  nfor each t with1  t  n. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  24. Completion Problems • The bad: Where to put “2” in the last column? Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  25. Completion Problems • The ugly (?)a. k. a. sudoku Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  26. Completion Problems • The ugly (?)a. k. a. sudoku Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  27. Latin Hypercube • A Latin k-hypercube of order nis a k-dimensional array Sk=[si1, i2,…, ik] in which each row is a permutation of symbols 1,2,…,n. A row of Sk is an n-tuple of elements si1, i2,…, ik which have identical coordinates at k-1 places. • We cat treat Sk as a function Sk(i1, i2,…, ik). Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  28. Latin Cube of Order n • Latin 2-hypercube = Latin square. • Latin 3-hypercube = Latin cubehas n layers, each a Latin square.Latin cube of order 3 Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  29. Latin Cube of Order n • Latin cube of order 4 Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  30. Orthogonal Hypercubes • A set of k Latin k-hypercubes (of order n)is orthogonal iff, when superimposed, each ordered k-tuple of symbols 1, 2,…, n appears once. • Example (next slide) Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  31. Orthogonal Hypercubes 3 orthogonal Latin cubes of order 4 Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  32. Orthogonal Hypercubes n>2, n≠6 • J. Arkin, E. G. Strauss, Latin k-Cubes, The Fibonacci Quarterly, Vol. 12 (3) (1974): 288-292. • J. Arkin, E. G. Strauss, Orthogonal Latin Systems, The Fibonacci Quarterly, Vol. 19 (3) (1981): 281-293. • M. Trenkler, On Orthogonal Latin p-Dimensional Cubes, Czechoslovak Mathematical Journal, 55 (130) (2005), 725-728. • All produced essentially the same theorem: Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  33. Orthogonal Hypercubes n>2, n≠6 • TheoremThere exists a set of k orthogonal Latin k-hypercubes of order n, n > 2 and n ≠ 6. • Proof (inductive construction)Let S and T be orthogonal Latin squares,let V1k, V2k,…, Vkk, be a set of k orthogonal Latin k-hyprecubes. Then we construct a set (k+1) orthogonal Latin (k+1)-hyprecubes of order n. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  34. Orthogonal Hypercubes n>2, n≠6 • W1k+1 = S(V1k(x1, x2,…, xk), xk+1), W2k+1 = S(V2k(x1, x2,…, xk), xk+1),W3k+1 = S(V3k(x1, x2,…, xk), xk+1),…………………………………Wkk+1 = S(Vkk(x1, x2,…, xk), xk+1),Wk+1k+1 = T(Vkk(x1, x2,…, xk), xk+1) • But S and T are orthogonal LS, so n ≠ 6. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  35. Orthogonal Hypercubes n = 6 • What about n = 6? • J. Kerr, The Existence of k Orthogonal Latin k-Cubes of order 6, The Fibonacci Quarterly Vol. 20. No. 4 (1982): 360-362. • Similar theorem. • Examples of three orthogonal Latin cubes and four orthogonal Latin 4-hypercubes of order 6: Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  36. Orthogonal Hypercubes n = 6 Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  37. Orthogonal Hypercubes n = 6 • TheoremLet A, B and C be orthogonal Latin cubes, let V1k, V2k,…, Vkk, be a set of k orthogonal Latin k-hypercubes of order 6, then … Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  38. Orthogonal Hypercubes n = 6 W1k+2 = A(V1k(x1, x2,…, xk), xk+1, xk+2), W2k+2 = A(V2k(x1, x2,…, xk), xk+1, xk+2), …………………………………… Wkk+2 = A(Vkk(x1, x2,…, xk), xk+1, xk+2), Wk+1k+2 = B(Vkk(x1, x2,…, xk), xk+1, xk+2), Wk+2k+2 = C(Vkk(x1, x2,…, xk), xk+1, xk+2) are a set of (k+2) orthogonal Latin (k+2)-hypercubes. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  39. Orthogonal Hypercubes • CorollaryFor every k > 2 and every n > 2, there exist a set of k orthogonal Latin k-hypercubes of order n. Jerzy Wojdylo, Latin Squares, Cubes and Hypercubes

  40. The End In case you blinked and missed something: http://www2.semo.edu/jwojdylo/research.htm

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