Non-extendible Latin cuboids Ian Wanless Monash University, - PowerPoint PPT Presentation

Non-extendible Latin cuboids Ian Wanless Monash University, Australia D. Bryant, N. J. Cavenagh, B. Maenhaut, K. Pula & IMW, Non-extendible latin cuboids, SIAM J. Discrete Math. 26 , (2012) 239–249.

Latin squares

Latin squares A Latin square of order n has n symbols, each occurring exactly once in each row and each column. e.g.     1 3 4 2 4 2 3 1 4 2 1 3 2 4 1 3     L = and M =     2 4 3 1 3 1 4 2     3 1 2 4 1 3 2 4 are two latin squares of order 4.     1 2 3 4 5 2 1 4 5 3 2 1 4 5 3 1 3 5 4 2         N = 3 4 5 2 1 O = 5 2 1 3 4         4 5 1 3 2 3 4 2 1 5     5 3 2 1 4 4 5 3 2 1 are two latin squares of order 5.

Latin cubes If we stack up n latin squares which pairwise never “agree” in any position...

Latin cubes If we stack up n latin squares which pairwise never “agree” in any position...

Latin cubes If we stack up n latin squares which pairwise never “agree” in any position... ...then we get a latin cube .

Latin hypercubes d factors � �� � A latin hypercube of order n and dimension d is an n × n × · · · × n array in which each line parallel to any axis is a permutation of the same set of n symbols.

Latin hypercubes d factors � �� � A latin hypercube of order n and dimension d is an n × n × · · · × n array in which each line parallel to any axis is a permutation of the same set of n symbols. d = 1: permutations d = 2: latin squares d = 3: latin cubes . . .

Latin hypercubes d factors � �� � A latin hypercube of order n and dimension d is an n × n × · · · × n array in which each line parallel to any axis is a permutation of the same set of n symbols. d = 1: permutations d = 2: latin squares d = 3: latin cubes . . . A n × n × k latin cuboid has each of n symbols occurring exactly once in each row and column, and at most once in each stack.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square. We say that a n × n × k latin cuboid is extendible if it is contained in a n × n × ( k + 1) latin cuboid.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square. We say that a n × n × k latin cuboid is extendible if it is contained in a n × n × ( k + 1) latin cuboid. ...and completable if it is contained in a n × n × n latin cube.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square. We say that a n × n × k latin cuboid is extendible if it is contained in a n × n × ( k + 1) latin cuboid. ...and completable if it is contained in a n × n × n latin cube. Every n × n × 1 or n × n × ( n − 1) latin cuboid is completable.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square. We say that a n × n × k latin cuboid is extendible if it is contained in a n × n × ( k + 1) latin cuboid. ...and completable if it is contained in a n × n × n latin cube. Every n × n × 1 or n × n × ( n − 1) latin cuboid is completable. [Kochol’95] For 1 2 n < k � n − 2 there is a Theorem: non-completable n × n × k latin cuboid.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square. We say that a n × n × k latin cuboid is extendible if it is contained in a n × n × ( k + 1) latin cuboid. ...and completable if it is contained in a n × n × n latin cube. Every n × n × 1 or n × n × ( n − 1) latin cuboid is completable. [Kochol’95] For 1 2 n < k � n − 2 there is a Theorem: non-completable n × n × k latin cuboid. He conjectured that all non-completable latin cuboids are more than “half-full”.

Extendibility Theorem: [Hall’45] Every k × n latin rectangle extends to an n × n latin square. We say that a n × n × k latin cuboid is extendible if it is contained in a n × n × ( k + 1) latin cuboid. ...and completable if it is contained in a n × n × n latin cube. Every n × n × 1 or n × n × ( n − 1) latin cuboid is completable. [Kochol’95] For 1 2 n < k � n − 2 there is a Theorem: non-completable n × n × k latin cuboid. He conjectured that all non-completable latin cuboids are more than “half-full”. This conjecture is false: e.g. The two 5 × 5 latin squares N and O stack to form a non-extendible 5 × 5 × 2 latin cuboid.

Our results Every 6 × 6 × 2 latin cuboid is extendible, but they are not all completable.

Our results Every 6 × 6 × 2 latin cuboid is extendible, but they are not all completable. For all m � 4, there exists a non-completable Theorem: 2 m × 2 m × m latin cuboid.

Our results Every 6 × 6 × 2 latin cuboid is extendible, but they are not all completable. For all m � 4, there exists a non-completable Theorem: 2 m × 2 m × m latin cuboid. For all even m > 2, there exists a non-extendible latin Theorem: cuboid of dimensions (2 m − 1) × (2 m − 1) × ( m − 1).

Example of the construction for m = 4 Start with 2 MOLS of order m .     1 3 4 2 4 2 3 1 4 2 1 3 2 4 1 3     L = and M =      . 2 4 3 1 3 1 4 2    3 1 2 4 1 3 2 4

Example of the construction for m = 4 Start with 2 MOLS of order m .     1 3 4 2 4 2 3 1 4 2 1 3 2 4 1 3     L = and M =      . 2 4 3 1 3 1 4 2    3 1 2 4 1 3 2 4 Then form these layers of a non-extendible 7 × 7 × 3 latin cuboid: 1 ∗ 2 ∗ 3 ∗ 4 1 2 3 2 ∗ 3 ∗ 1 ∗ 1 4 3 2 3 ∗ 1 ∗ 2 ∗ 3 2 1 4 2 ∗ 3 ∗ 1 ∗ 1 2 3 4 3 ∗ 1 ∗ 2 ∗ 2 1 4 3 1 ∗ 2 ∗ 3 ∗ 4 3 2 1 3 ∗ 1 ∗ 2 ∗ 2 3 4 1 1 ∗ 2 ∗ 3 ∗ 3 2 1 4 2 ∗ 3 ∗ 1 ∗ 1 4 3 2 1 ∗ 2 ∗ 3 ∗ 2 3 ∗ 3 1 ∗ 2 ∗ 2 ∗ 3 ∗ 4 1 ∗ 3 4 1 4 1 2 1 2 3 2 ∗ 3 ∗ 1 1 ∗ 4 1 ∗ 2 ∗ 3 ∗ 1 ∗ 2 ∗ 3 ∗ 3 2 3 4 1 2 3 4 1 2 3 ∗ 4 1 ∗ 2 ∗ 1 ∗ 2 ∗ 3 ∗ 1 2 1 ∗ 2 ∗ 3 ∗ 1 2 3 2 3 4 3 4 1 3 1 ∗ 2 ∗ 3 ∗ 2 ∗ 3 ∗ 2 1 ∗ 3 ∗ 1 1 ∗ 2 ∗ 4 1 2 3 4 1 2 3 4

Extending the non-extendible Actually we don’t need MOLS. For m = 6 these are close enough:     2 5 3 4 6 6 1 2 3 4 5 ∗ 3 6 4 1 5 5 6 4 2 1 3  ∗        6 5 1 2 4 2 4 6 5 3 1     ∗ L = M =     1 6 3 5 2 4 3 1 6 5 2     ∗     4 1 2 6 3 3 2 5 1 6 4     ∗ 5 4 1 2 3 1 5 3 4 2 6 ∗

Extending the non-extendible Actually we don’t need MOLS. For m = 6 these are close enough:     2 5 3 4 6 6 1 2 3 4 5 ∗ 3 6 4 1 5 5 6 4 2 1 3  ∗        6 5 1 2 4 2 4 6 5 3 1     ∗ L = M =     1 6 3 5 2 4 3 1 6 5 2     ∗     4 1 2 6 3 3 2 5 1 6 4     ∗ 5 4 1 2 3 1 5 3 4 2 6 ∗ And we can do a similar thing for odd m :     2 3 4 5 6 7 7 1 2 3 4 5 6 ∗ 1 5 6 7 4 3 4 7 5 1 2 6 3  ∗        4 6 7 1 5 2 1 2 7 4 6 3 5     ∗     L = 7 5 1 3 2 6 M = 5 6 3 7 1 2 4     ∗     6 1 7 2 3 4 3 5 1 6 7 4 2     ∗     3 7 2 1 4 5 6 3 4 2 5 7 1     ∗ 5 4 6 3 2 1 2 4 6 5 3 1 7 ∗

Open questions ◮ Do the “near MOLS” exist for all large odd orders?

Open questions ◮ Do the “near MOLS” exist for all large odd orders? ◮ Algorithms for completing cuboids

Open questions ◮ Do the “near MOLS” exist for all large odd orders? ◮ Algorithms for completing cuboids ◮ What is the smallest n for which every n × n × 2 latin cuboid is extendible?

Open questions ◮ Do the “near MOLS” exist for all large odd orders? ◮ Algorithms for completing cuboids ◮ What is the smallest n for which every n × n × 2 latin cuboid is extendible? ...completable?

Open questions ◮ Do the “near MOLS” exist for all large odd orders? ◮ Algorithms for completing cuboids ◮ What is the smallest n for which every n × n × 2 latin cuboid is extendible? ...completable? ◮ Asymptotically, what are the “thinnest” non-extendible cuboids?

Open questions ◮ Do the “near MOLS” exist for all large odd orders? ◮ Algorithms for completing cuboids ◮ What is the smallest n for which every n × n × 2 latin cuboid is extendible? ...completable? ◮ Asymptotically, what are the “thinnest” non-extendible cuboids? ...non-completable...

Open questions ◮ Do the “near MOLS” exist for all large odd orders? ◮ Algorithms for completing cuboids ◮ What is the smallest n for which every n × n × 2 latin cuboid is extendible? ...completable? ◮ Asymptotically, what are the “thinnest” non-extendible cuboids? ...non-completable... ◮ Does a d -dimensional latin hypercube of order n have a transversal unless d and n are even?