Mutually orthogonal binary frequency squares

TL;DR

This paper investigates the existence, enumeration, and properties of mutually orthogonal binary frequency squares, establishing bounds, existence results, and classifications for various orders and configurations.

Contribution

It introduces new bounds and existence results for complete and maximal sets of mutually orthogonal binary frequency squares, including constructions linked to Hadamard matrices.

Findings

Existence of at least 2^{n^2/4 - O(n log n)} isomorphism classes of complete MOFS(n) when a Hadamard matrix exists.

Existence of a 17-MOFS(n) for certain n, but no complete MOFS(n) for n ≡ 2 mod 4.

Classification of 1-maxMOFS(n) and 5-maxMOFS(n) for specific orders.

Abstract

A \emph{frequency square} is a matrix in which each row and column is a permutation of the same multiset of symbols. We consider only {\em binary} frequency squares of order $n$ with $n/2$ zeroes and $n/2$ ones in each row and column. Two such frequency squares are \emph{orthogonal} if, when superimposed, each of the 4 possible ordered pairs of entries occurs equally often. In this context we say that a $k$-MOFS$(n)$ is a set of $k$ binary frequency squares of order $n$ in which each pair of squares is orthogonal. A $k$-MOFS$(n)$ must satisfy $k\le(n-1)^2$, and any MOFS achieving this bound are said to be \emph{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 MOFS$(n)$. For $2<n\equiv2\pmod4$ we show that there exists a $17$-MOFS$(n)$ but no complete MOFS$(n)$. A $k$-maxMOFS$(n)$ is a $k$-MOFS$(n)$ that is not contained in any $(k+1)$-MOFS$(n)$. By computer enumeration, we establish that there exists a $k$-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 $5$-maxMOFS$(n)$ for each order $n\equiv 2\pmod{4}$ where $n\geq 6$.