Graeco-Latin Squares and a Mistaken Conjecture of Euler

Klyve, Dominic; Stemkoski, Lee

doi:10.2307/27646265

Cited by 9 publications

(5 citation statements)

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“…4.3], McKay and Wanless [96] and McKay, Meynert and Myrvold [94]. It is possible that Clausen found R 6 in 1842 (see [80] for a discussion). The value of R 12 is currently unknown, but the estimate R 12 ≈ 1.62 • 10 44 was one of the estimates given by McKay and Rogoyski [95].…”

Section: Introductionmentioning

confidence: 99%

The Many Formulae for the Number of Latin Rectangles

Stones¹

2010

Electron. J. Combin.

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A $k \times n$ Latin rectangle $L$ is a $k \times n$ array, with symbols from a set of cardinality $n$, such that each row and each column contains only distinct symbols. If $k=n$ then $L$ is a Latin square. Let $L_{k,n}$ be the number of $k \times n$ Latin rectangles. We survey (a) the many combinatorial objects equivalent to Latin squares, (b) the known bounds on $L_{k,n}$ and approximations for $L_n$, (c) congruences satisfied by $L_{k,n}$ and (d) the many published formulae for $L_{k,n}$ and related numbers. We also describe in detail the method of Sade in finding $L_{7,7}$, an important milestone in the enumeration of Latin squares, but which was privately published in French. Doyle's formula for $L_{k,n}$ is given in a closed form and is used to compute previously unpublished values of $L_{4,n}$, $L_{5,n}$ and $L_{6,n}$. We reproduce the three formulae for $L_{k,n}$ by Fu that were published in Chinese. We give a formula for $L_{k,n}$ that contains, as special cases, formulae of (a) Fu, (b) Shao and Wei and (c) McKay and Wanless. We also introduce a new equation for $L_{k,n}$ whose complexity lies in computing subgraphs of the rook's graph.

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Section: Introductionmentioning

confidence: 99%

The Many Formulae for the Number of Latin Rectangles

Stones¹

2010

Electron. J. Combin.

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“…Two permutation matrices that bring U Γ to P 1 U Γ P 2 = U Γ block , shown in Fig. D2, can be written as two vectors consisting of 36 integers, [6,2,36,24,13,29,22,10,32,27,1,17,31,26,3,19,23,9,18,5,12,33,28,16,11,34,25,15,20,7,14,30,8,4,35,21], 3,4,9,10,7,8,1,2,27,28,33,34,16,15,…”

Section: Discussionmentioning

confidence: 99%

“…On the other hand, he observed that Graeco-Latin squares of order 2 do not exist, and was unable to construct an arrangement of order 6. This resulted in one of his famous conjectures -see [34].…”

Section: -Leonhard Eulermentioning

confidence: 99%

9 × 4 = 6 × 6: Understanding the Quantum Solution to Euler’s Problem of 36 Officers

Życzkowski

Bruzda

Rajchel-Mieldzioć

et al. 2023

J. Phys.: Conf. Ser.

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The famous combinatorial problem of Euler concerns an arrangement of 36 officers from six different regiments in a 6×6 square array. Each regiment consists of six officers each belonging to one of six ranks. The problem, originating from Saint Petersburg, requires that each row and each column of the array contains only one officer of a given rank and given regiment. Euler observed that such a configuration does not exist. In recent work, we constructed a solution to a quantum version of this problem assuming that the officers correspond to superpositions of quantum states. In this paper, we explain the solution which is based on a partition of 36 officers into nine groups, each with four elements. The corresponding quantum states are locally equivalent to maximally entangled two-qubit states, hence each quantum officer is represented by a superposition of at most 4 classical states. The entire quantum combinatorial design involves 9 Bell bases in nine complementary 4–dimensional subspaces.

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Section: -Leonard Eulermentioning

confidence: 99%

Section: Appendix D Block Diagonal Form Of Ame and Bell Basesmentioning

confidence: 99%

9 $\times$ 4 = 6 $\times$ 6: Understanding the quantum solution to the Euler's problem of 36 officers

Życzkowski¹,

Bruzda²,

Rajchel-Mieldzioć³

et al. 2022

Preprint

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The famous combinatorial problem of Euler concerns an arrangement of 36 officers from six different regiments in a 6 × 6 square array. Each regiment consists of six officers each belonging to one of six ranks. The problem, originating from Saint Petersburg, requires that each row and each column of the array contains only one officer of a given rank and given regiment. Euler observed that such a configuration does not exist. In recent work, we constructed a solution to a quantum version of this problem assuming that the officers correspond to quantum states and can be entangled. In this paper, we explain the solution which is based on a partition of 36 officers into nine groups, each with four elements. The corresponding quantum states are locally equivalent to maximally entangled two-qubit states, hence each officer is entangled with at most three out of his 35 colleagues. The entire quantum combinatorial design involves 9 Bell bases in nine complementary 4-dimensional subspaces.

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Graeco-Latin Squares and a Mistaken Conjecture of Euler

Cited by 9 publications

References 0 publications

The Many Formulae for the Number of Latin Rectangles

The Many Formulae for the Number of Latin Rectangles

9 × 4 = 6 × 6: Understanding the Quantum Solution to Euler’s Problem of 36 Officers

9 $\times$ 4 = 6 $\times$ 6: Understanding the quantum solution to the Euler's problem of 36 officers

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