2021
DOI: 10.1016/j.jcta.2021.105403
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Latin squares with maximal partial transversals of many lengths

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Cited by 3 publications
(3 citation statements)
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“…Then Evans [14] constructed an infinite family of Latin squares which simultaneously have maximal partial transversals of each of the permissible lengths. Subsequently, Evans et al [15] showed that there exists a Latin square of order n which has maximal partial transversals of each permissible length if and only if n {3, 4} ∉ and n 2 (mod 4) ≢ . In Latin squares, it is easy to find a partial transversal of length n 2 ⌈ ∕ ⌉ using a greedy algorithm.…”
Section: Every M Nmentioning
confidence: 99%
“…Then Evans [14] constructed an infinite family of Latin squares which simultaneously have maximal partial transversals of each of the permissible lengths. Subsequently, Evans et al [15] showed that there exists a Latin square of order n which has maximal partial transversals of each permissible length if and only if n {3, 4} ∉ and n 2 (mod 4) ≢ . In Latin squares, it is easy to find a partial transversal of length n 2 ⌈ ∕ ⌉ using a greedy algorithm.…”
Section: Every M Nmentioning
confidence: 99%
“…. , n k ] that minimises n k 2 [11] 3 [11,17] 4 [11,17,28] 5 [11,17,31,41] 6 [11,17,28,46,58] 7 [11,17,28,42,64,78] 8 [11,17,28,42,63,90, 107] 9 [11,17,28,46,58,91,122,140] 10 [11,17,28,42,64,78,122,157,177] 11 [11,17,28,42,63,90,107,165,204,226] 12 [11,17,…”
Section: Long Partial Transversalsmentioning
confidence: 99%
“…Then Evans [16] constructed an infinite family of Latin squares which simultaneously have maximal partial transversals of each of the permissible lengths. Subsequently, Evans et al [17] showed that there exists a Latin square of order n which has maximal partial transversals of each permissible length if and only if n / ∈ {3, 4} and n ≡ 2 (mod 4). In Latin squares, it is easy to find a partial transversal of length ⌈n/2⌉ using a greedy algorithm.…”
Section: Introductionmentioning
confidence: 99%