Mutually orthogonal binary frequency squares

Thomas Britz ; Nicholas J. Cavenagh ; Adam Mammoliti ; Ian M. Wanless¹

¹Monash University

The electronic journal of combinatorics, Tome 27 (2020) no. 3

Résumé

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 $k\le(n-1)^2$, 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 $2^{n^2/4-O(n\log n)}$ isomorphism classes of complete sets of MOFS$(n)$. For $2<n\equiv2\pmod4$ 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 $k\in\{1,17\}$ or $5\le k\le 15$. We show that up to isomorphism there is a unique $1$-maxMOFS$(n)$ if $n\equiv2\pmod4$, whereas no $1$-maxMOFS$(n)$ exists for $n\equiv0\pmod4$. We also prove that there exists a set of $5$-maxMOFS$(n)$ for each order $n\equiv 2\pmod{4}$ where $n\geq 6$.

Détail

DOI : 10.37236/9373

Classification : 05B20, 05B15
Mots-clés : Hadamard matrix, isomorphism classes