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      Mutually orthogonal binary frequency squares

      Britz, Thomas; Cavenagh, Nicholas J.; Mammoliti, Adam; Wanless, Ian M.
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      9373-PDF file-32639-3-10-20200710.pdf
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      DOI
       10.37236/9373
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      Britz, T., Cavenagh, N. J., Mammoliti, A., & Wanless, I. M. (2020). Mutually orthogonal binary frequency squares. The Electronic Journal of Combinatorics, 27(3). https://doi.org/10.37236/9373
      Permanent Research Commons link: https://hdl.handle.net/10289/13697
      Abstract
      A frequency square is a matrix in which each row and column is a permutation of the same multiset of symbols. We consider only binary frequency squares of order n with n/2 zeros and n/2 ones in each row and column. Two such frequency squares are orthogonal if, when superimposed, each of the 4 possible ordered pairs of entries occurs equally often. In this context we say that a set of k-MOFS (n) is a set of k binary frequency squares of order n in which each pair of squares is orthogonal.

      A set of k-MOFS (n) must satisfy k≤(n−1)², and any set of MOFS achieving this bound is said to be complete. For any n for which there exists a Hadamard matrix of order n we show that there exists at least 2 ⁿ²/⁴−ᴼ ⁽ⁿˡᵒᵍⁿ ⁾ isomorphism classes of complete sets of MOFS (n). For 2<n≡2(mod 4) we show that there exists a set of 17-MOFS(n) but no complete set of MOFS(n).

      A set of k –maxMOFS (n) is a set of k-MOFS(n) that is not contained in any set of (k+1)-MOFS(n). By computer enumeration, we establish that there exists a set of k-maxMOFS(6) if and only if k∈{1,17} or 5≤k≤15. We show that up to isomorphism there is a unique 1-maxMOFS(n) if n≡2(mod4), whereas no 1-maxMOFS(n) exists for n≡0(mod4). We also prove that there exists a set of 5-maxMOFS(n) for each order n≡2(mod4) where n≥6.
      Date
      2020
      Type
      Journal Article
      Publisher
      The Electronic Journal of Combinatorics
      Rights
      © The authors. Released under the CC BY-ND license (International 4.0).
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      • Computing and Mathematical Sciences Papers [1454]
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