The problem of Euler/Tarry revisited
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The problem of Euler/Tarry revisited Dieter Betten1 Received: 4 June 2020 / Accepted: 1 November 2020 © The Author(s) 2020
Abstract The problem of Euler/Tarry concerning 36 officers can be formulated in mathematical terms: Can a latin square of order 6 have an orthogonal square, or equivalently, are there 6 pairwise disjoint transversals? This was first answered (in the negative) by Tarry (1900/01). We prove the following Theorem: If a latin square of order 6 admits a reflection, i. e. an automorphism of order two which fixes the main diagonal elementwise, then it has no orthogonal square. We list the 12 isomorphism types of latin squares of order 6 and see: they all admit such a reflection. So we get a solution of the Euler problem without the tedious task of tracing the transversals. Keywords Orthogonal arrays · Latin squares · Room squares · Orthogonal designs · configurations
1 Recalling some notions We recall Euler’s 36 officers Problem: Can 36 officers, drawn from 6 different ranks and also from 6 different regiments, be ranged in a square so that in each line (both horizontal and vertical) there are 6 officers of different rank and different regiments? This was a famous problem at Euler’s time, and Euler wrote a paper concerning it, Euler (1782). The first proof of the non-existence was by Tarry (1901), so we may call this problem now the Euler/Tarry problem. The problem can be formulated in mathematical terms by the existence of a MOL(6) (a pair of orthogonal latin squares LQ(6) of order 6), Denes and Keedwell (1974). Further proofs for the non-existence were given, also very short ones, (Betten 1983; Beth et al. 1985; Stinson 1984). We give an example (Fig. 1): In each cell of the (6 × 6)-array, there is a pair of numbers. We call the first number the digit (corresponding to the rank) and the second number the transversal (corresponding to the regiment).
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Dieter Betten [email protected] Christian-Albrechts-University Kiel Department of Mathematics, Ludewig-Meyn-Str. 4, 24118 Kiel, Germany
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Beitr Algebra Geom Fig. 1 Example of a MOL(6)
Fig. 2 Incidence structure of the MOL(6)
We can describe this as an incidence structure on 24 variables with 4 subsets of order 6 (the 6-blocks) and 36 subsets of order 4 (the 4-blocks), Fig. 2. The four 6blocks are R = {r1 , r2 , . . . , r6 } (the six rows), C = {c1 , c2 , . . . , c6 } the six columns, D = {d1 , d2 , . . . , d6 } (the six digits) and T = {t1 , t2 , . . . , t6 } (the six transversals). Each 4-block has the form (r , c, d, t), r ∈ R, c ∈ C, d ∈ D, t ∈ T . The main condition is: each pair of distinct variables in on exactly one block. In the last line of the first figure we had to put two question marks since a MOL(6) does not exist. We also marked a special 4-block (r , c, d, t) = (2, 4, 3, 5). This can be seen in Fig. 2 as the 4-block with O instead of X. In the standard notation (Fig. 1) the sets of rows R and columns C are distinguished: they define the (6 × 6)-array, where the 36 pairs (d, t) ∈ D × T are inserted. But in the notation o
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