Coloring a Generalized Latin Square

Question

Suppose we have an $n \times n$ array, and there is a decomposition $\mathcal{A}$ of its coordinates $a_{i,j}$ into sets $A_m$ as follows:

1. If $a_{i,j} \in A_m$, then $a_{j,i} \in A_m$. So they're symmetrical pieces of the array (across the diagonal).

2. If $a_{i,j}, a_{i,k} \in A_m$ and $j \neq k$ then $a_{j,k} \in A_m$. Similarly, if $a_{j,i}, a_{k,i} \in A_m$ and $j \neq k$ then $a_{j,k} \in A_m$. So the relations are transitive, in a sense.

3. If a diagonal coordinate $a_{i,i}$ is contained in $A_m$, then $A_m = \lbrace a_{i,i} \rbrace$ is a singleton.

The question is this: Thinking of the $A_m$'s as chunks of the $n \times n$ array, can we apply a color to each $A_m$ so that the array satisfies a 'Latin Square' condition using at most $n$ colors, i.e. within any row or column no color appears more than once unless they belong to the same $A_m$? So, every entry in any fixed $A_M$ will all have the same color, and we color the square 'up to $\mathcal{A}$.'

This is a restatement of a known hypergraph problem (after using some observations I made) that doesn't have much associated theory to attack it with.

Here are a couple of examples for $n=8$:

In the first example we have

$A_1 = \lbrace a_{1,2}, a_{2,1}, a_{1,3}, a_{3,1}, a_{2,3}, a_{3,2} \rbrace$,

$A_2 = \lbrace a_{1,4}, a_{4,1}, a_{1,5}, a_{5,1}, a_{4,5}, a_{5,4} \rbrace$,

$A_3 = \lbrace a_{6,7}, a_{7,6}, a_{6,8}, a_{8,6}, a_{7,8}, a_{8,7} \rbrace$ and the rest are pairs $\lbrace a_{i,j}, a_{j,i} \rbrace$ or diagonal singletons. An $8$-coloring is shown.

In the second example, we have

$A_1 = \lbrace \text{non-diagonal entries with indices from 1-5} \rbrace$

$A_2 = \lbrace a_{1,6}, a_{6,1}, a_{1,7}, a_{7,1}, a_{1,8}, a_{8,1}, a_{6,7}, a_{7,6}, a_{6,8}, a_{8,6}, a_{7,8}, a_{8,7} \rbrace$ and the rest are pairs $\lbrace a_{i,j}, a_{j,i} \rbrace$ or diagonal singletons. A $7$-coloring is shown.

It starts getting much more complex once you get up into double-digit values of $n$, the way the 'chunks' can be enmeshed with each other.

Also, please double-check that the colorings I gave are actually right. I screwed it up a jillion times haha

Answer 1

Executive Summary

The problem here is one of graph coloring, specifically that of coloring the "chunks" (vertices) of a decomposition as described in the Question. An $n\times n$ array $A$ gives a set of $n^2$ unit squares, collected into disjoint subsets (chunks) $A_m$ that cover the array and satisfy three rules. Two chunks are adjacent vertices of a simple graph $\mathcal G$ when they both contain entries of a common row (equiv. a common column, by symmetry).

The least number of colors necessary to assign adjacent vertices different colors is called the chromatic number $\chi(\mathcal G)$ of graph $\mathcal G$. The problem asks if $\chi(\mathcal G) \le n$ for all graphs obtained by partitioning those $n\times n$ array entries in accordance with the three rules of decomposition.

But this is true only when $n\le 4$. For each $n\gt 4$ we will construct a decomposition whose graph $\mathcal G$ contains an $(n+1)$-clique, so that its chromatic number $\chi(\mathcal G) \gt n$. Thus for $n = 5$ and greater, there exist counterexamples to the claim.

In our figures we will use numbers to label colors and letters to label chunks with more than two squares. That is, unlabelled squares are considered to be in singleton chunks if on the diagonal or in symmetric pairs otherwise. Furthermore, instead of referring to a square as an entry of a matrix $a_{i,j}$, we instead identify such a square simply as $(i,j)$.

With those conventions it is only necessary to label chunks with more than two squares, as this counterexample for $n=5$ illustrates:

$$ \begin{array}{|c|c|c|c|c|} \hline \phantom{B} & & & & \\ \hline & & A & & B \\ \hline & A & & B & \\ \hline & & B & & A \\ \hline & B & & A & \\ \hline \end{array} $$

Since the unlabelled off-diagonal chunks of the first row and column form an $(n-1)$-clique, we need only provide two additional chunks, here labelled $A$ and $B$, and check that are adjacent to each other and to all the former (first row/column) chunks.

All counterexamples except $n=6$ use exactly two such off-diagonal chunks of more than two squares to embed an $(n+1)$-clique into the graph.

The case $n=6$ embeds a $7$-clique using seven chunks of size four squares:

$$ \begin{array}{|c|c|c|c|c|c|} \hline & A & B & C & D & \\ \hline A & & E & F & G & D \\ \hline B & E & & G & F & C \\ \hline C & F & G & & E & B \\ \hline D & G & F & E & & A \\ \hline & D & C & B & A & \\ \hline \end{array} $$

Proof that $\mathbf n$ colors suffice for $\mathbf{n\le 4}$

We begin with this observation:

Lemma 1 A decomposition of an $n\times n$ array into chunks (as allowed by the rules) gives a graph that can be $n$-colored if and only if its off-diagonal chunks can be $n$-colored.

Proof of Lemma 1

One direction, that if the whole graph can be $n$-colored, then the off-diagonal chunks can be $n$-colored, is easy.

For the other direction note that a diagonal chunk (singleton) is adjacent (by symmetry) to at most $n-1$ off-diagonal chunks. So if the off-diagonal chunks can be $n$-colored, then the remaining diagonal parts can also be $n$-colored (because each of them has at most $n-1$ neighbors). QED

When $n=1,2,3$, there are at most $n$ off-diagonal chunks, so $n$ colors suffice for all the off-diagonal chunks (and for the graph as a whole).

When $n=4$ there are six upper diagonal squares, so at most six off-diagonal chunks. Consider the figure below (the addition table of abelian group $\mathbb Z_2 + \mathbb Z_2$) as a coloring of the decomposition into chunks of minimal size, diagonal singletons and off-diagonal symmetric pairs, i.e. the symmetric latin square of order $4$:

$$ \begin{array}{|c|c|c|c|} \hline 0 & 1 & 2 & 3 \\ \hline 1 & 0 & 3 & 2 \\ \hline 2 & 3 & 0 & 1 \\ \hline 3 & 2 & 1 & 0 \\ \hline \end{array} $$

As those six off-diagonal chunks can be $3$-colored, the array decomposition as a whole (including the diagonal chunks) can be $4$-colored.

What about decompositions that include chunks with more than two squares? We would need at least five off-diagonal chunks in order for them to require more than four colors.

A decomposition with five off-diagonal chunks would mean exactly one chunk has more than two squares, and that chunk will have two squares above the diagonal and two squares below.

If we check which pairs of upper-diagonal squares can be combined in a chunk without forcing a third upper-diagonal square by transitivity, we find each those pairs are ones that already have the same color in the table above. The same $3$-coloring works as before!

Therefore any $4\times 4$ decomposition requires no more than four colors for its graph.

Counterexamples for all $\mathbf{n \gt 4}$

We already exhibited counterexamples for $n=5,6$. Decompositions will be constructed similar to the one for $n=5$ for all larger values of $n$. Start by putting off-diagonal squares of the first row and column into $n-1$ symmetric pair chunks, each consisting of just the two squares. That gives us an $(n-1)$-clique.

We then define two additional chunks using only the off-diagonal squares of the remaining rows and columns. We arrange that these two chunks are adjacent to each other and to each of the chunks in the $(n-1)$-clique. This gives an $(n+1)$-clique, proving that the chromatic number of the graph is more than $n$.

To justify constructions of the additional chunks we prove two small results:

Prop. 1 Let $P$ be a partition of the index set: $$ I_n = \{1,\ldots,n\},\; n\gt 1$$ As is well-known, $P$ induces an equivalence relation $\mathcal R_P$ on indexes $I_n$ via $i\sim j$ iff $i,j$ both belong to the same part of $P$. If any part of $P$ is larger than a singleton, then the greatest irreflexive subset of $\mathcal R_P$:

$$ \mathcal C = \mathcal R_P \setminus \{(i,i) \mid 1\le i \le n \} $$

is an off-diagonal chunk permitted by the three rules. Conversely any off-diagonal chunk $\mathcal C$ so permitted corresponds uniquely to the greatest irreflexive subset of an equivalence relation obtained via:

$$ \mathcal R_P = \mathcal C \cup \{(i,i) \mid 1\le i \le n \} $$

Proof of Prop. 1

We check that once the diagonal points are excluded from the equivalence relation $\mathcal R_P$, we have a nonempty set satisfying the three rules. First rule is symmetry:

$$ (i,j) \in \mathcal C \iff (j,i) \in \mathcal C $$

which follows from the symmetric property of an equivalence relation. Second rule is transitivity for off-diagonal squares:

$$ (i,j),(i,k) \in \mathcal C \implies (j,k) \in \mathcal C $$ whenever $j\neq k$. This follows from the transitivity of an equivalence relation. Finally the third rule excludes diagonal points from $\mathcal C$ because some part of $P$ contains two distinct indexes $i,j$ and therefore the off-diagonal point $(i,j)$ belongs to $\mathcal C$. We obviously guarantee the exclusion of diagonal points from $\mathcal C$ in our construction, and note $(i,j) \in \mathcal C$ proves it to be nonempty.

The recovery of $\mathcal R_P$ by adjoining the diagonal points to irreflexive relation $\mathcal C$ is evident, as are the conditions (symmetry, reflexivity, transitivity) that conversely make $\mathcal R_P$ an equivalence relation based on the three rules that a chunk must satisfy. QED

Next we extend the above correspondence of index partition $P$ with off-diagonal chunk $C$ to a correspondence between a suitable family of partitions and a complete decomposition into chunks.

Prop. 2 Let $\{ P_m\}_{m=1}^M$ be a finite family of partitions of $I_n, n\gt 1$, each with a part of size greater than one. Let $\{\mathcal C_m\}_{m=1}^M$ be the corresponding off-diagonal chunks of the preceding proposition. Then the following are equivalent: (i) Every $2$-subset $\{i,j\}\subset I_n$ is contained in exactly one part of some partition $P_m$ and not in any part of any other partition.
(ii) The chunks $\mathcal C_m$ together with the diagonal chunks $\{(i,i)\}, 1\le i \le n$ form a decomposition of the $n\times n$ array.
Conversely any decomposition of the $n\times n$ array into a family of chunks $\{\mathcal C_m\}_{m=1}^M$ is produced by some such a family of partitions.

Sketch of proof of Prop. 2

Since we already know the off-diagonal chunks $\mathcal C_m$ and the diagonal chunks $\{(i,i)\}, 1\le i \le n$ satisfy the three rules, the crux of the result is to show these partition the squares of the $n\times n$ array. As the diagonal chunks are disjoint from the off-diagonal chunks, attention should be directed at why the family of off-diagonal chunks $\{\mathcal C_m\}_{m=1}^M$ partitions the off-diagonal squares of the array. This is the implication of condition (i) on the partitions $P_m$, that each off-diagonal square $(i,j), i\neq j$ belongs to the one and only one $\mathcal C_m$ for which the $2$-subset $\{i,j\}$ is contained in a part of $P_m$.

Conversely the disjointness of off-diagonal chunks and their coverage of the off-diagonal squares implies property (i) of the family of partitions.

Practice with these re-presentations of structures will be developed in the next section, as cases $n=7,8,9,10$ serve as a basis for induction to all higher orders. We state without proof a further observation about the adjacency of chunks from the viewpoint of their corresponding partitions:

Prop. 3 Two disjoint off-diagonal chunks $\mathcal C$ and $\mathcal C'$ of a decomposition are adjacent if and only if their corresponding partitions $P,P'$ have respective nonsingleton parts $s\in P$ and $s'\in P'$ that have nonempty intersection.

By (i) of the previous proposition this nonempty intersection can only be a singleton. However more than one part of $P$ may so intersect with one or more parts of $P'$, signifying that the chunks are adjacent in more than one row and column. We will see examples.

Counterexample decomposition for $\mathbf{n = 7}$

Recall that we start with the $6$-clique of off-diagonal chunks taken from the first row and column of the array. We additionally require two chunks formed from the remaining rows and columns with mutual adjacency and adjacency to those initial six chunks.

Here are the six partitions of $I_7$ for those initial six chunks:

$$ P_1 = \{\{1,2\},\{3\},\{4\},\{5\},\{6\},\{7\}\} \\ P_2 = \{\{1,3\},\{2\},\{4\},\{5\},\{6\},\{7\}\} \\ P_3 = \{\{1,4\},\{2\},\{3\},\{5\},\{6\},\{7\}\} \\ P_4 = \{\{1,5\},\{2\},\{3\},\{4\},\{6\},\{7\}\} \\ P_5 = \{\{1,6\},\{2\},\{3\},\{4\},\{5\},\{7\}\} \\ P_6 = \{\{1,7\},\{2\},\{3\},\{4\},\{5\},\{6\}\} $$

Because of property (i) of partitions, the next two must contain singleton part $\{1\}$, to avoid having any $2$-subset appear in a part of more than one partition. So the next two partitions will be defined by the parts that are contained in $\{2,3,4,5,6,7\}$. In order to have the adjacencies we want between them and the initial six partitions, indexes $2$ through $7$ must appear in nonsingleton parts (by Prop. 3) of both of them.

Here's one way to make that work:

$$ P_7 = \{\{1\},\{2,3\},\{4,5\},\{6,7\}\} \\ P_8 = \{\{1\},\{2,5\},\{3,6\},\{4,7\}\} \\ $$

Check that chunks $\mathcal C_7$ and $\mathcal C_8$ are adjacent: yes, since doubleton part $\{2,3\}$ of $P_7$ has nonempty intersection with doubleton part $\{2,5\}$ of $P_8$. In other words the last two chunks are adjacent in row (and in column) $2$. As well this shows both chunks are adjacent to the first of our initial chunks, the one that holds square $(1,2)$ (or resp. $(2,1)$.

These kinds of checks are easily programmed (using sorted lists without repetition to model sets), but for humans it is easier to check visually. Here are the chunks $\mathcal C_7$ and $\mathcal C_8$ labelled as $G$ and $H$ respectively, and the first six chunks labelled $A,B,C,D,E,F$ for clarity (this one time):

$$\begin{array}{|c|c|c|c|c|c|c|} \hline & A & B & C& D & E & F \\ \hline A & & G & & H & & \\ \hline B & G & & & & H & \\ \hline C & & & & G & & H \\ \hline D & H & & G & & & \\ \hline E & & H & & & & G \\ \hline F & & & H & & G & \\ \hline \end{array}$$

The figure makes it obvious that each row after the first contains a square of the $G$ chunk and one of the $H$ chunk (and of course the same is true of columns). So we do have an $8$-clique embedded into a decomposition of the $7\times 7$ array. We don't need to worry about filling up the rest of the array decomposition; they can be just the diagonal singleton chunks and the symmetric pair doubletons, and no matter how they might be colored, at least eight colors will always be required because of the embedded $8$-clique.

Counterexample decompositions for $\mathbf{n = 8,9,10,\ldots}$

While we included the singleton parts (equivalence classes) in the partitions for $n=7$, this is unnecessary for our purposes. It only served to be able to refer to all of them as partitions of $I_n$. But the only parts that contribute to off-diagonal chunks are those of size two or more. We dispense with the singletons going forward working only with reduced partitions $$P^* = P \setminus \{\{i\} \mid 1\le i\le n\}$$ that exclude the singletons.

Case $\mathbf{n=8}$

Start the construction for $n=8$ as before, with seven first row and column off-diagonal chunks, giving these seven reduced partitions:

$$ P_1^* = \{\{1,2\}\} \\ P_2^* = \{\{1,3\}\} \\ P_3^* = \{\{1,4\}\} \\ P_4^* = \{\{1,5\}\} \\ P_5^* = \{\{1,6\}\} \\ P_6^* = \{\{1,7\}\} \\ P_7^* = \{\{1,8\}\} $$

To get a $9$-clique we need two more off-diagonal chunks that are adjacent to each other and to those just listed. This is a little tricky, amounting to two partitions of the indexes greater than $1$:

$$I_8 \setminus \{1\} = \{2,3,4,5,6,7,8\}$$

where all parts are of size greater than one and the parts of one partition intersect the parts of the other in not more than one index.

This forces us to use partitions of three+two+two, e.g.

$$ P_8^* = \{\{2,3,4\},\{5,6\},\{7,8\}\} \\ P_9^* = \{\{2,5,7\},\{3,6\},\{4,8\}\} $$

We label only the two chunks formed by the latter reduced partitions as $A$ and $B$:

$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline \phantom{B} & & & & & & & \\ \hline & & A & A & B & & B & \\ \hline & A & & A & & B & & \\ \hline & A & A & & & & & B \\ \hline & B & & & & A & B & \\ \hline & & B & & A & & & \\ \hline & B & & & B & & & A \\ \hline & & & B & & & A & \\ \hline \end{array}$$

It is easy now to check visually that the nine chunks form a clique (so that nine colors at least are needed) for this decomposition of the $8\times 8$ array.

Case $\mathbf{n=9}$

Skipping over the routine contents of the eight chunks in the first row and column (and their reduced partitions), we give reduced partitions for two additional chunks, these two ways to pair up the indexes greater than $1$ :

$$ P_9^* = \{\{2,3\},\{4,5\},\{6,7\},\{8,9\}\} \\ P_{10}^* = \{\{2,5\},\{3,4\},\{6,8\},\{7,9\}\} $$

These are mutually adjacent chunks, and each is adjacent to the first eight chunks. So the decomposition contains a $10$-clique.

This is similar to doubling up the arrangement of chunks we used to decompose $n=5$, as the following figure with respective chunks $A$ and $B$ illustrates:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \phantom{B} & & & & & & & & \\ \hline & & A & & B & & & & \\ \hline & A & & B & & & & & \\ \hline & & B & & A & & & & \\ \hline & B & & A & & & & & \\ \hline & & & & & & A & & B \\ \hline & & & & & A & & B & \\ \hline & & & & & & B & & A \\ \hline & & & & & B & & A & \\ \hline \end{array}$$

Soon we will have occasion to use such an "add-on" pattern to extend our counterexamples to all $n$.

Case $\mathbf{n=10}$

Again we just write two additional chunks, now in the form of reduced partitions of $\{2,3,4,5,6,7,8,9,10\}$:

$P_{10}^* = \{\{2,3,4\},\{5,6,7\},\{8,9,10\}\}$
$P_{11}^* = \{\{2,5,8\},\{3,6,9\},\{4,7,10\}\}$

Each part of one intersects each part of the other in exactly one index, resembling the row and column indexes of latin squares. In any case the figure they form is a bit striking:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \phantom{B} & & & & & & & & & \\ \hline & & A & A & B & & & B & & \\ \hline & A & & A & & B & & & B & \\ \hline & A & A & & & & B & & & B \\ \hline & B & & & & A & A & B & & \\ \hline & & B & & A & & A & & B & \\ \hline & & & B & A & A & & & & B \\ \hline & B & & & B & & & & A & A \\ \hline & & B & & & B & & A & & A \\ \hline & & & B & & & B & A & A & \\ \hline \end{array}$$

Cases $\mathbf{n>10}$

Finally we use induction to extend our counterexamples to all $n\gt 10$. We take $n=7,8,9,10$ decompositions as above as the base cases.

Suppose for $n\ge 7$ a decomposition like those we demonstrated for $n=7,8,9,10$ exists, with $n-1$ chunks formed out of the first row and column, and two suitable additional off-diagonal chunks $A,B$ from the remaining rows and columns forming an $(n+1)$-clique.

We now extend this to a decomposition of the $(n+4)\times (n+4)$ array. Add four more symmetric pair chunks from the first row and column. Enlarge the $A$ chunk by adjoining two more doubleton parts to the corresponding reduced partition:

$$ P_n^* \cup \{\{n+1,n+2\},\{n+3,n+4\}\} $$

Enlarge the $B$ chunk by adjoining two different doubleton parts to its reduced partition:

$$ P_{n+1}^* \cup \{\{n+1,n+3\},\{n+2,n+4\}\} $$

This gives us an $(n+5)$-clique of chunks in a decomposition of the $(n+4)\times (n+4)$ array.

Since our base cases were $n=7,8,9,10$, marching forward in steps $n+4$ as outlined will reach all $n\gt 10$. Note that this trick applied to our $n=5$ counterexample would essentially give our $n=9$ counterexample. The peculiarity of case $n=6$ made our exposition slightly longer!