Trades and defining sets in Latin squares and related combinatorial - - PDF document

Trades and defining sets in Latin squares and related combinatorial arrays

Nicholas J. Cavenagh University of Waikato July 9, 2019

Outline

Synopsis Latin trades Defining sets History of lower bounds for sds(n) Equivalent ideas from computational learning theory Critical sets Arrays with frequency constraints Full Latin arrays (0,1)-arrays with constant row and column sum

Trades and defining sets

Synopsis

For a Latin square or other discrete mathematical structure, we can ask the following inter-related questions:

• What is the minimum amount of information needed

to define a structure uniquely?

• What is the minimum amount of information possible

to change one structure to another of the same order?

• When can a partial structure be completed to one

with specified parameters? In turn, these questions inform fundamental questions on enumeration and randomization of combinatorial structures.

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Trades and defining sets

MANY THINGS ARE CONNECTED BUT NOT EVERYTHING

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Trades and defining sets

Latin trades

Let L1 and L2 be two Latin squares of the same order. 1 2 3 4 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 L1 = L(Z5) 3 2 1 4 1 2 3 4 2 4 3 1 3 1 4 2 4 1 2 3 L2 Consider a Latin square to be a set of ordered (row, column, symbol) triples; for example L1 = {(r, c, r + c) | r, c ∈ Z5}. Then the “difference” between L1 and L2 can be described by (L1 \ L2, L2 \ L1). The distance between L1 and L2 is given by |L1 \ L2| = 8.

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Trades and defining sets

We say that L1 \ L2 is a Latin trade with disjoint mate L2 \ L1. 1 3 3 4 4 1 L1 \ L2 3 1 4 3 1 4 L2 \ L1 Alternatively, a Latin trade T is a partially filled Latin square with disjoint mate T ′ such that: (i) T and T ′ occupy the same filled cells; (ii) T ∩ T ′ = ∅; (iii) Each row/column of T contains the same set of cells as the corresponding row/column of T ′.

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Trades and defining sets

These two “definitions” of a Latin trade are not exactly the same!! 1 2 2 1 T 2 1 1 2 T ′ Neither T nor T ′ embed in a Latin square of order 3.

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Trades and defining sets

Defining sets

A defining set for a Latin square L of order n is a subset D

• f L with unique completion to a Latin square of order n.

1 1 2 2 3 D It is conjectured that the smallest size for a defining set for a Latin square of order n, denoted by sds(n), is ⌊n2/4⌋. This is true for n ≤ 8 (Bean, 2005).

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Trades and defining sets

Lemma A subset D of a Latin square L is a defining set for L if and only if L \ D contains no Latin trade. 1 2 3 4 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 L1 3 2 1 4 1 2 3 4 2 4 3 1 3 1 4 2 4 1 2 3 L2 For example, the elements in blue are not a defining set.

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Trades and defining sets

Any two rows (or columns or symbols) of a Latin square form a trade. 1 2 3 4 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 L1 1 2 3 4 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 L2 Corollary The smallest defining set size sds(n) ≥ n − 1.

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Trades and defining sets

History of lower bounds for sds(n)

• sds(n) ≥ n + 1, (Cooper, et al., 1994; Fu, et al.,

1995)

• sds(n) ≥ ⌊(7n − 3)/6⌋, (Fu, Fu, Rodger, 1997)
• sds(n) ≥ ⌊(4n − 8)/3⌋, (Horak, Aldred, Fleischner,

2002)

• sds(n) ≥ Ω(n!(n − 3)!) ∼ n log log n (Hermiston,

2018)

• sds(n) ≥ n⌊(log n)1/3/2⌋ (C., 2007)
• sds(n) ≥ Ω(n3/2) (C., Ramadurai, 2017)
• sds(n) ≥ n2/104 (Hatami and Qian, 2018)

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Trades and defining sets

The prime Omega function Ω(n) is the sum of the exponents in the prime factorization of the integer n. The following is a great table to show students.... n 2 3 4 5 6 7 8 9 10 ⌊n2/4⌋ 1 2 4 6 9 12 16 20 25 Ω(n!(n − 3)!) 1 2 4 6 9 12 16 20 23

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Trades and defining sets

sds(n) ≥ Ω(n3/2) (C., Ramadurai, 2017) This result is related to the fact that every Latin square has a trade of size O(√n). If p is prime, L(Zp) has no trade smaller than e log p + 2 (Drápal). It is conjectured this is best possible.

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Trades and defining sets

Consider the operation table for the integers as an infinite Latin square L(Z) := {(a, b, a + b) | a, b, ∈ Z}. Lemma The Latin square L(Z) has no finite Latin trades. Suppose for the sake of contradiction, there is a finite Latin trade T in L(Z). Let k be the largest integer that appears in T as a symbol. Let i be the smallest integer such that (i, i − k, k) ∈ T. Then (j, i − k, k) ∈ T ′ for some j = i. In fact j < i; otherwise (j, i−k, j+i−k > k) ∈ T. Therefore (j, j − k, k) ∈ T, contradicting the minimality of i.

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Trades and defining sets

Corollary The following partial Latin square has no Latin trades. 1 2 1 2 3 1 2 3 4 1 2 3 4 2 3 4 Corollary The Latin square L(Zn) has a defining set of size ⌊n2/4⌋. Corollary sds(n) ≤ ⌊n2/4⌋.

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Trades and defining sets

The story behind sds(n) ≥ n2/104 (Hatami and Qian, 2018) Theorem A Latin square of order n is equivalent to a partition of the edges of the complete tripartite graph K(n, n, n) into triangles. Label one partite set with the rows, one partite set with the columns and the remaining partite set with the symbols. Each (row, column, symbol) triple in the Latin square corresponds to a triangle in the graph.

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Trades and defining sets

The story behind sds(n) ≥ n2/104 (Hatami and Qian, 2018) Theorem (Barber, Kühn, Lo, Osthus and Taylor 2016; Dukes 2015) Let γ > 0 and n > n0(γ). Every balanced and locally balanced 3-partite graph

• n 3n vertices with minimum degree at least

(101/52+γ)n, admits a triangle decomposition. Corollary (BKLOT, 2016) Let P be a partial Latin square

• f order n > n0 such that every row, column,

and symbol is used at most 0.0288n times. Then P can be completed to a Latin square.

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Trades and defining sets

Proof that sds(n) ≥ n2/104 (Hatami and Qian, 2018): Let P be a partial Latin square of size at most n2/104. It suffices to show that if P completes to a Latin square L, P completes to a Latin square L′ = L (and thus is not a defining set. Let R, C and S be the rows, columns and symbols which are “crowded” (have more than 0.012n entries). Fill all the empty cells in R, C and S by elements of L; obtain a partial Latin square P ′ such that m rows, columns and symbols are all filled for some m ≤ 0.0084n. Add an entry which is not in L. Create corresponding tripartite graph for the complement of P ′, deleting vertices of degree 0. It can be shown that the minimum degree of such a graph is less than 101/52(n − m).

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Trades and defining sets

Conjecture (Daykin and Häggkvist, 1984) Every balanced and locally balanced 3-partite graph on 3n ver- tices with minimum degree at least 3n/2, ad- mits a triangle decomposition. Wanless (2002): The above is the best possible bound. For example: 1 1 1 1 2 3 3 2 2 The above conjecture appears not to be enough to prove that sds(n) ≥ ⌊n2/4⌋.

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Trades and defining sets

Conjecture (Nash-Williams, 1970) Every sufficiently large even graph G with minimum degree 3n/4 (and number of edges divisible by 3) has a triangle decomposition. Best-known upper bound for the above: Minimum degree 0.956n (BKLO, 2015).

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Trades and defining sets

Equivalent ideas from computational learning theory

(Hatami and Qian, 2018) expressed their results also using terminology from computational learning theory. teaching set (witness set, discriminant, specifying set)= defining set

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Trades and defining sets

Critical sets

A critical set for a Latin square L is a minimal defining set for L. The size of the smallest (respectively, largest) critical set in L is denoted by sds(L) (respectively, lcs(L)). Let Ln be the set of Latin squares of order n. Then: sds(n) = min{sds(L) | L ∈ Ln} = sds(n), lcs(n) = max{lcs(L) | L ∈ Ln}. We also define a type of infimum and supremum, respec- tively: inf(n) = max{sds(L) | L ∈ Ln} ≤ lcs(n), sup(n) = min{lcs(L) | L ∈ Ln} ≥ sds(n).

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Trades and defining sets

Theorem (Ghandehari, Hatami, Mahmoodian, 2005) n2 − (e + o(1))n5/3 ≤inf(n) ≤ n2 −

√π 2 n9/6.

Proof of upper bound: Consider a Latin square L of order

• n. Assign each (i, j, k) ∈ L a birth-time xe (independently

uniformly chosen from between 0 and 1). Include each entry in a defining set iff it is not “forced” by elements that were born earlier. For each k′ = k, the probability that at least one of (i, ⋆, k′) or (⋆, j, k′) is born before (i, j, k) is 1 − (1 − xe)2. Thus the probability that (i, j, k) is not included in the defining set is (1 − (1 − xe)2)n−1. So the expected size of the defining set is E(|C|) = n2(1 − 1 (1 − (1 − xe)2)n−1dx) ≤ n2 − √π 2 n9/6. Finally, we can’t all be above average!

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Trades and defining sets

Outline of proof for the lower bound: Let k be the largest possible size for a critical set in any Latin square of order n. Then, every Latin square of order n has a defining set of size k. Let Ln be the number of Latin squares of order n. Thus, |Ln| is less than the number of partial Latin squares of size k. Then, using permanents and what-nots (eg: the Van der Waerden conjecture), |Ln| ≥ (n!)2n/nn2. The number of partial Latin squares of size at most k can also be bounded above using permanents.

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Trades and defining sets

Arrays with frequency constraints

Latin squares are on of a general class of arrays where for each row column (and cell), constraints are given on the frequency of symbols. These include:

• (0, 1)-matrices (with prescribed row and column

sums);

• Frequency squares;
• semi-Latin squares;
• multi-Latin squares.

We consider these structures under the umbrella term Latin array. What properties connect Latin arrays of any type?

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Trades and defining sets

Each of these structures is equivalent to the decomposition

• f some tripartite graph into triangles.

For example, let A be a m × n (0, 1)-matrix with row-sum vector (r1, r2, . . . , rm) and column-sum vector (c1, c2, . . . , cn). This corresponds to a decomposition of a tripartite graph G into triangles, where G has partite sets R = {1, 2, . . . , m}, C = {1, 2, . . . , n}, S = {0, 1}, every possible edge exists between R and C, there are:

• ri edges between each i ∈ R and 1 ∈ S,
• n − ri edges between each i ∈ R and 0 ∈ S,
• cj edges between each j ∈ C and 1 ∈ S,
• m − cj edges between each j ∈ C and 0 ∈ S.

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Trades and defining sets

For any Latin array, the size of the smallest trade is 4; such a trade is called an intercalate. 1 2 2 1 T 2 1 1 2 T ′

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Trades and defining sets

At the other extreme, take partite sets of size m, n and t and consider the set of all possible (mnt) triangles in this

• graph. We call the associated array the full (m, n, t)-Latin
• rectangle. For example, if m = 2, n = 3 and t = 4 we have

the full (2, 3, 4)-Latin rectangle: 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 This may look like a boring array, however.....

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Trades and defining sets

Full Latin arrays

Let a Latin array be simple if there are no repeated triangles in the corresponding tripartite graph (no repeats within cells). Then the full (m, n, t)-Latin rectangle includes every possible trade in any simple m×n Latin array with t symbols. Thus, we can intersect any defining set for a full Latin array with another Latin array L (of the same dimensions) to

• btain a defining set for L.

For example, 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 ∩ 1 2 3 4 4 1 2 3 3 4 1 2 2 3 4 1 = 1 2 3 4 1 2 3 4 1

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Trades and defining sets

Unfortunately, this doesn’t improve any of the bounds sds(n), lcs(n), inf(n) or sup(n). This is because defining sets for full Latin squares are always asymptotically “full”: Theorem Any defining set for the full Latin square (n, n, n) has size n3(1 − o(1)).

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Trades and defining sets

(0,1)-arrays with constant row and column sum

For the rest of the talk, we restrict ourselves to A2m, the set

• f 2m × 2m (0, 1)-matrices with constant row and column

sum m. We thus redefine the following in terms of such structures: sds(2m), lcs(2m), inf(2m) or sup(2m). The structure of trades in (0, 1)-matrices is straightforward: these are always disjoint unions of cycles. It can then be shown that any South-East walk in a (0,1)-matrix gives a defining set. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Trades and defining sets

Correspondingly, we have much more precise information about the above:

• sds(2m) = m2 (C., 2013)
• 3m2 −2m ≥ lcs(2m) ≥ 3m2 −4m+2 (C. and Liam

Wright, 2019)

• inf(2m) = 2m2(1−o(1)) (C. and Ramadurai, 2018),

(Bodkin, Liebenau and Wanless, 2019)

• 3m(m − 1)/2 ≥ sup(2m) ≥ 2m2 − m (C. and Liam

Wright, 2019)

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Trades and defining sets

The story behind inf(2m) = m2(1 − o(1)). Let A be an element of A2m with the property that in any subarray, the number of 1’s and 0’s are aysmptotically “similar”. C. and Ramdurai showed that the generating matrix for the Reed-Muller code has this property. Bodkin, Liebenau and Wanless showed that almost all elements of A2m have this property! Then, no matter which SE-walk is used, the defining set created is never too large. It follows that any critical set in A uses close to half the cells!!!

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