Almost every Latin square has a decomposition into transversals

TL;DR

The paper proves that almost every Latin square of order $n\equiv 2\pmod 4$ admits a decomposition into $n$ disjoint transversals, overturning Euler's conjecture in a typical-case sense for large $n$. It reduces transversals to rainbow perfect matchings in an optimally coloured $K_{n,n}$ and develops a three-phase framework: (A) a sparse absorption schematic that enables local corrections, (B) a realisation phase that constructs near-matching absorption structures via the semi-random method and $L$-links, and (C) a final covering/balancing step that completes the decomposition into perfect rainbow matchings. The analysis leverages advanced probabilistic tools (Rödl nibble, deletion method, concentration inequalities) and intricate hypergraph constructions to control degrees and codegrees, ensuring high-probability success. The work connects Latin-square transversals to resolvable designs and hypergraph decompositions, providing a robust probabilistic path to near-resolvable structures in large random instances with potential applicability to broader design theory problems.

Abstract

In 1782, Euler conjectured that no Latin square of order $n\equiv 2\; \textrm{mod}\; 4$ has a decomposition into transversals. While confirmed for $n=6$ by Tarry in 1900, Bose, Parker, and Shrikhande constructed counterexamples in 1960 for each $n\equiv 2\; \textrm{mod}\; 4$ with $n\geq 10$. We show that, in fact, counterexamples are extremely common, by showing that if a Latin square of order $n$ is chosen uniformly at random then with high probability it has a decomposition into transversals.

Figures (9)

• Figure 1: Two Latin squares decomposed into transversals. On the left, the addition group of integers $\mod 9$ is given, which is then decomposed into transversals indicated by integers in the top right, starting with the transversal along the leading diagonal (marked by 0) which is then moved to the right by $1 \mod 9$ 8 times to create 8 new transversals. On the right, Bose, Parker, and Shrikhande's example of a Latin square of order $10$ with a transversal decomposition BS.
• Figure 2: A simple $\{(i,x),(j,y)\}$-switcher.
• Figure 3: For each $i\in [n]$, we have a partition of $A\cup B$ into $S_i\cup X_i\cup Y_i\cup Z_i$ (which is the same for individuals $i$ in the same tribe), a partition $S_i=U_i\cup V_i\cup W_i$ (which is the same for individuals $i$ in the same family) and disjoint sets $R_i,T_i\subset U_i$ (which are distinct for each individual $i \in [n]$).
• Figure 4: a) Each cycle $C_j$ we consider must be rainbow, for otherwise we would replace, for example, the orange edges $\vec{u_1u_2}$ and $\vec{u_3u_4}$ with orange edges $\vec{u_1u_4}$ and $\vec{u_3u_2}$. . b) For each rainbow cycle $C_j$, we take a set $E_j=\{v_4v_1,v_1v_3,v_3v_2\}$ of edges whose addition allows an (undirected) decomposition into triangles, and put a directed 2-cycle with colour $i_e$ on each $e\in E_j$.
• Figure 5: Each directed triangle $x_jy_jz_j$ is replaced by a collection of 2-cycles, where each vertex has balanced in- and out-degree in each colour except for $x_j,y_j,z_j$ which maintain the same in- and out-degree in each colour.

Theorems & Definitions (121)

• Theorem 1.1
• Theorem 2.1
• Theorem 2.2
• Theorem 2.3
• Theorem 2.4
• Corollary 2.5
• proof
• Lemma 2.6
• Lemma 2.7
• Definition 3.1