2006
DOI: 10.1007/s10623-006-8168-9
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The Existence of Latin Squares without Orthogonal Mates

Abstract: A latin square is a bachelor square if it does not possess an orthogonal mate; equivalently, it does not have a decomposition into disjoint transversals. We define a latin square to be a confirmed bachelor square if it contains an entry through which there is no transversal. We prove the existence of confirmed bachelor squares for all orders greater than three. This resolves the existence question for bachelor squares.

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Cited by 42 publications
(50 citation statements)
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“…The general case for n = 4t +3 remained unsettled until two independently obtained results were simultaneously published in 2006. Those results, by Evans [11], and by Wanless and Webb [18], each conclude the existence of a bachelor latin square of order n for all n / ∈ {1, 3}. The actual result due to Wanless and Webb [18] is stated below in Theorem 1.6.…”
Section: Corollary 15mentioning
confidence: 87%
See 2 more Smart Citations
“…The general case for n = 4t +3 remained unsettled until two independently obtained results were simultaneously published in 2006. Those results, by Evans [11], and by Wanless and Webb [18], each conclude the existence of a bachelor latin square of order n for all n / ∈ {1, 3}. The actual result due to Wanless and Webb [18] is stated below in Theorem 1.6.…”
Section: Corollary 15mentioning
confidence: 87%
“…It is a slightly stronger statement. A confirmed bachelor square is a latin square with at least one entry not in any transversal [18]. Theorem 1.6.…”
Section: Corollary 15mentioning
confidence: 99%
See 1 more Smart Citation
“…In the next section we will use a very different argument to construct nonextendible latin cuboids, an argument which is reminiscent of the Δ-lemma arguments in [1,3,4,11].…”
Section: −1 Umentioning
confidence: 99%
“…Another result about transervals showed that for n > 3, there exists a latin square of order n with a cell that is contained in no transversal of the latin square [87]. This implies that there is always some latin square of order n > 3 without an orthogonal mate.…”
Section: Questions 131 Transversals In Latin Squaresmentioning
confidence: 99%