Do K33-Free Latin Squares Exist?
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arXiv
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| Natura: | Preprint |
| Pubblicazione: |
2023
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| _version_ | 1866912844509872128 |
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| author | Krotov, Aleksandr D. Krotov, Denis S. |
| author_facet | Krotov, Aleksandr D. Krotov, Denis S. |
| contents | We discuss the problem of existence of latin squares without a substructure consisting of six elements $(r_1,c_2,l_3)$, $(r_2,c_3,l_1)$, $(r_3,c_1,l_2)$, $(r_2,c_1,l_3)$, $(r_3,c_2,l_1)$, $(r_1,c_3,l_2)$. Equivalently, the corresponding latin square graph does not have an induced subgraph isomorphic to $K_{3,3}$. The exhaustive search [Brouwer, Wanless. Universally noncommutative loops. 2011] says that there are no such latin squares of order $3$--$7$, $9$--$11$ and there are only two $K_{3,3}$-free latin squares of order $8$, up to equivalence. We repeat the search, establishing also the number of latin $m$-by-$n$ rectangles for each $m$ and $n$ less than or equal to $11$. As a switched combination of two orthogonal latin squares of order $8$, we construct a $K_{3,3}$-free (universally noncommutative) latin square of order $16$.
We also consider a similar problem for orthogonal latin squares, proving that there are both $K_{4,4}$-free and non-$K_{4,4}$-free linear pairs of orthogonal latin squares for each odd prime-power order larger than~$5$.
Keywords: latin square; transversal; trade; pattern avoiding; eigenfunction; universally noncommutative loop. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2304_07157 |
| institution | arXiv |
| publishDate | 2023 |
| record_format | arxiv |
| spellingShingle | Do K33-Free Latin Squares Exist? Krotov, Aleksandr D. Krotov, Denis S. Combinatorics 05B15, 05E30 We discuss the problem of existence of latin squares without a substructure consisting of six elements $(r_1,c_2,l_3)$, $(r_2,c_3,l_1)$, $(r_3,c_1,l_2)$, $(r_2,c_1,l_3)$, $(r_3,c_2,l_1)$, $(r_1,c_3,l_2)$. Equivalently, the corresponding latin square graph does not have an induced subgraph isomorphic to $K_{3,3}$. The exhaustive search [Brouwer, Wanless. Universally noncommutative loops. 2011] says that there are no such latin squares of order $3$--$7$, $9$--$11$ and there are only two $K_{3,3}$-free latin squares of order $8$, up to equivalence. We repeat the search, establishing also the number of latin $m$-by-$n$ rectangles for each $m$ and $n$ less than or equal to $11$. As a switched combination of two orthogonal latin squares of order $8$, we construct a $K_{3,3}$-free (universally noncommutative) latin square of order $16$. We also consider a similar problem for orthogonal latin squares, proving that there are both $K_{4,4}$-free and non-$K_{4,4}$-free linear pairs of orthogonal latin squares for each odd prime-power order larger than~$5$. Keywords: latin square; transversal; trade; pattern avoiding; eigenfunction; universally noncommutative loop. |
| title | Do K33-Free Latin Squares Exist? |
| topic | Combinatorics 05B15, 05E30 |
| url | https://arxiv.org/abs/2304.07157 |