Rectangles in latin squares

TL;DR

This paper explores the minimal modifications needed to transform one Latin square into another, focusing on changing at least four entries and providing methods to identify and implement these changes.

Contribution

It introduces a method for finding and changing four entries in a Latin square to obtain another, advancing understanding of Latin square transformations.

Findings

Minimum of four entries must be changed to obtain a different Latin square.

Provides a systematic way to identify these entries.

Demonstrates how to perform the changes effectively.

Abstract

To get another from a given latin square, we have to change at least 4 entries. We show how to find these entries and how to change them.

Equations

$$\varphi=\left(\begin{array}[]{ccccccccc}1&2&3&4&5&6&7&8\\ 3&1&2&5&4&6&8&7\end{array}\right)=(132.45.6.78.)$$

$$\begin{array}[]{c|cccccccc}\cdot&1&2&3&4&5&6&7\\ \hline\cr 1&1&2&3&4&5&6&7\\ 2&2&1&\framebox{4}&3&7&5&\framebox{6}\\ 3&3&6&5&1&4&7&2\\ 4&4&5&2&7&6&1&3\\ 5&5&4&7&6&2&3&1\\ 6&6&7&1&2&3&4&5\\ 7&7&3&\framebox{6}&5&1&2&\framebox{4}\end{array}$$

$$\begin{array}[]{cccc}x\circ_{1}y=z\longleftrightarrow x\cdot z=y\\ x\circ_{2}y=z\longleftrightarrow z\cdot y=x\\ x\circ_{3}y=z\longleftrightarrow z\cdot x=y\\ x\circ_{4}y=z\longleftrightarrow y\cdot z=x\\ x\circ_{5}y=z\longleftrightarrow y\cdot x=z\\ \end{array}$$

$$L_{x}=\left(\begin{array}[]{ccccccccc}\ldots&y&\ldots&u&\ldots\\ \ldots&a&\ldots&b&\ldots\end{array}\right),\ \ \ \ \ L_{z}=\left(\begin{array}[]{ccccccccc}\ldots&y&\ldots&u&\ldots\\ \ldots&b&\ldots&a&\ldots\end{array}\right).$$

$$L_{x}L_{z}^{-1}=\left(\begin{array}[]{ccccccccc}\ldots&b&\ldots&a&\ldots\\ \ldots&a&\ldots&b&\ldots\end{array}\right).$$

$$\begin{array}[]{llllll}L_{2}=(12.34.576.),&&L_{3}=(1354.267.),&&L_{4}=(1473256.),\\[3.0pt] L_{5}=(1524637.),&&L_{6}=(1642753.),&&L_{7}=(1745.236.).\end{array}$$

$$\begin{array}[]{llllll}L_{1}L_{2}^{-1}=(12.34.576.),&&L_{2}L_{4}^{-1}=(15.24.367.),&&L_{3}L_{4}^{-1}=(17.25643.),\\[3.0pt] L_{4}L_{5}^{-1}=(13.267.54.),&&L_{2}L_{7}^{-1}=(17253.46.).&&\end{array}$$

$$\begin{array}[]{c|cccccccc}\cdot&1&2&3&4&5&6\\ \hline\cr 1&1&2&3&4&5&6\\ 2&2&1&4&5&6&3\\ 3&3&4&2&6&1&5\\ 4&4&5&6&2&3&1\\ 5&5&6&1&3&2&4\\ 6&6&3&5&1&4&2\end{array}$$

$$x\circ y=\left\{\begin{array}[]{ccl}a\cdot d&&{\rm if}\ x=a,\ y=b,\\ a\cdot b&&{\rm if}\ x=c,\ y=b,\\ c\cdot b&&{\rm if}\ x=c,\ y=d,\\ c\cdot d&&{\rm if}\ x=a,\ y=d,\\ x\cdot y&&{\rm in\;other\;cases}.\end{array}\right.$$

$$\begin{array}[]{c|cccccccc}\circ&1&2&3&4&5&6&7\\ \hline\cr 1&1&2&3&4&5&6&7\\ 2&2&1&\framebox{6}&3&7&5&\framebox{4}\\ 3&3&6&5&1&4&7&2\\ 4&4&5&2&7&6&1&3\\ 5&5&4&7&6&2&3&1\\ 6&6&7&1&2&3&4&5\\ 7&7&3&\framebox{4}&5&1&2&{\framebox{6}}\end{array}$$

Full text

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has the rectangle $\langle 2,3,7,7\rangle$. Other rectangles of this quasigroup are calculated in Example 2.

Conversely, let $<x,y,z,u>$ be a rectangle in a group $G$. Then $xyu^{-1}=xuy^{-1}$, and consequently $(yu^{-1})^{2}=e$. So $a=yu^{-1}\neq e$ has order two. ∎

4. Each quasigroup $Q=(Q,\cdot)$ determines five new quasigroups $Q_{i}=(Q,\circ_{i})$ with the operations $\circ_{i}$ defined as follows:

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Such defined (not necessarily distinct) quasigroups are called parastrophes or conjugates of $Q$. Traditionally they are denoted as

$Q_{1}=Q^{-1}=(Q,\backslash),\ \ Q_{2}={}^{-1}\!Q=(Q,/),\ \ Q_{3}={}^{-1}\!(Q^{-1})=(Q_{1})_{2}$,

$Q_{4}=({}^{-1}\!Q)^{-1}=(Q_{2})_{1}$  and  $Q_{5}=({}^{-1}\!(Q^{-1}))^{-1}=((Q_{1})_{2})_{1}=((Q_{2})_{1})_{2}$.

From the above results and results obtained in [7] it follows that a fixed $IP$-quasigroup and all its parastrophes have the same number of rectangles. Similarly, a quasigroup isotopic to a group and its parastrophes.

5. Note that if a commutative quasigroup has a rectangle $\langle x,y,z,u\rangle$ with $x=y$, then also $z=u$. Similarly, $z=u$ implies $x=y$. So, if in a commutative quasigroup $Q$ one of vertices of the rectangle $\langle x,y,z,u\rangle$ lies on the diagonal of the multiplication table of $Q$, then this rectangle is a square and its diagonal coincides with the diagonal of the multiplication table.

In Boolean groups each two elements $x,y$ determine the rectangle $\langle x,y,xy,e\rangle$; each three elements $x,y,z$ determine the rectangle $\langle x,y,z,xyz\rangle$.

An interesting question is how to find rectangles in a given quasigroup. Direct calculation of $\langle x,\,x\!\setminus\!a,\,a/(x\!\setminus\!b),\,x\!\setminus\!b\rangle$ is rather trouble. Below we present simplest method based on left translations.

Let $\langle x,y,z,u\rangle$ be a rectangle in a quasigroup $(Q,\cdot)$. Then, according to the definition, $xy=zu=a$ and $xu=zy=b$. Thus the left translations $L_{x}$ and $L_{z}$ have the form

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Hence

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This means that the permutation $L_{x}L_{z}^{-1}$ used in the construction of indicators (for details see [3]) has the cycle $(a,b)$. Thus vertices $a,b$ of this rectangle are located in the $x$-th row, vertices $b,a$ in the $a/(x\!\setminus\!b)$-th row. Since $L_{z}L_{x}^{-1}=(L_{x}L_{z}^{-1})^{-1}$, the permutation $L_{x}L_{z}^{-1}$ and $L_{z}L_{x}^{-1}$ have the same cycles of the length $2$. So, it is sufficient to calculate $L_{x}L_{z}^{-1}$ for $x<z$ only.

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Consequently,

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In other $L_{x}L_{z}^{-1}$ with $x<z$, there are no cycles of the length $2$.

Below we present cycles $(a,b)$ and rectangles generated by these cycles.

\begin{array}[]{lllllllll}(1,2):\langle 1,1,2,2\rangle,&&&(3,4):\langle 1,3,2,4\rangle,&&&(1,5):\langle 2,2,4,6\rangle,\\[3.0pt] (2,4):\langle 2,1,4,3\rangle,&&&(1,7):\langle 3,4,4,6\rangle,&&&(1,3):\langle 4,6,5,7\rangle,\\[3.0pt] (5,4):\langle 4,1,5,2\rangle,&&&(4,6):\langle 2,3,7,7\rangle.\end{array}

Hence this quasigroup has eight rectangles.

Note that one pair $(a,b)$ can determine several rectangles.

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the pair $(1,2)$ determines three rectangles: $\langle 1,1,2,2\rangle$, $\langle 3,5,5,3\rangle$, $\langle 4,6,6,4\rangle$. Other rectangles are: $\langle 3,3,4,4\rangle$, $\langle 3,3,6,6\rangle$, $\langle 4,4,5,5\rangle$ and $\langle 5,5,6,6\rangle$.

6. A quasigroup $(Q,\circ)$ is a rectangle transformation of a quasigroup $(Q,\cdot)$ if in $(Q,\cdot)$ there exists a rectangle $\langle a,b,c,d\rangle$ such that

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Two rectangles $\langle x,y,z,u\rangle$ and $\langle x^{\prime},y^{\prime},z^{\prime},u^{\prime}\rangle$ of a quasigroup $(Q,\cdot)$ are equivalent if there exists an autotopism $(\alpha,\beta,\gamma)$ of $(Q,\cdot)$ such that $\alpha(x)=x^{\prime}$, $\beta(y)=y^{\prime}$, $\alpha(z)=z^{\prime}$ and $\beta(u)=u^{\prime}$.

Observe that by the converse rectrangle tansformations we obtain the same quasigroup. So, by the rectangle transformation from two non-isotopic quasigroups we can obtain isotopic quasigroups. Hence, the converse of Theorem 2 is not true.