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

• nce in each row and each column.

e.g. L =     1 3 4 2 4 2 1 3 2 4 3 1 3 1 2 4     and M =     4 2 3 1 2 4 1 3 3 1 4 2 1 3 2 4     are two latin squares of order 4. N =       1 2 3 4 5 2 1 4 5 3 3 4 5 2 1 4 5 1 3 2 5 3 2 1 4       O =       2 1 4 5 3 1 3 5 4 2 5 2 1 3 4 3 4 2 1 5 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

A latin hypercube of order n and dimension d is an

d factors

• n × n × · · · × n

array in which each line parallel to any axis is a permutation of the same set of n symbols.

Latin hypercubes

A latin hypercube of order n and dimension d is an

d factors

• 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

A latin hypercube of order n and dimension d is an

d factors

• 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

• nce 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. Theorem: [Kochol’95] For 1

2n < k n − 2 there is a

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. Theorem: [Kochol’95] For 1

2n < k n − 2 there is a

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. Theorem: [Kochol’95] For 1

2n < k n − 2 there is a

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. Theorem: For all m 4, there exists a non-completable 2m × 2m × m latin cuboid.

Our results

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

Example of the construction for m = 4

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

Example of the construction for m = 4

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

Extending the non-extendible

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

Extending the non-extendible

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

Our story ends...