The content of this note mostly belongs to the first section of TAoCP, Vol4A.

# Definition

Given a square matrix of size $M$ and each element is $\{0, 1, \cdots, M-1 \}$, we construct a matrix such that for the element $i$ in the set $\{0, 1, \cdots, M-1\}$ only appears exactly 1 time on every row and column.

### Examples

We have 16 cards which comprises a combination of 4 ranks, i.e, J, K, Q, A and 4 suits: $\heartsuit, \diamondsuit, \clubsuit, \spadesuit$, we arrange the cards such that for each column or row, we have 4 ranks and 4 suitst (Jacques Ozaman, mathematiques et physiques, Paris 1725).

# Applications

While studying mathematics, sometimes we could not find an obvious application in order to motivate ourself to study the concept. However, since am a computer science engineer who is pragmatic than _mathematical inclination, I am specially interested in concepts which have applications in real life. Regards the latin square, we get several ones:

- Sudoku: the variant of latin square is one of the most popular puzzles in the world.
- Design of experiments.
- Error corecting codes.
- Crytography.

# War Story

In 1779, a puzzle similar to aforemention example attracted the famous mathematician: Leonhard Euler:

Thirty-six officers of six different ranks, taken from six different regiments, want to march in a $6 \times 6$ formation so that each row and each column will contain one officer of each rank and one of each regiment. How can they do it?

Even though being amost blind, Euler was determined to solve the puzzle and tried to generalize the problem for other cases $M = 1, 2, \cdots$. However, when $M\mod 4 = 2$, he could not solve it. Later, he stated a conjenture that there is no solution for $M \mod 4 = 2$.

However, the puzzle Euler tackled is quite different compared to the definition given in the previous section. More precisely, let $\heartsuit = 1, \diamondsuit = 2, \clubsuit = 3, \spadesuit = 4$, and we divide ranks and suits into 2 seperated matrices, we got two Latin squares:

$$ \begin{bmatrix} J& A& K& Q\newline Q& K& A& J\newline A& J& Q& K\newline K& Q& J& A \end{bmatrix} $$

and

$$ \begin{bmatrix} 2& 1& 4& 3\newline 4& 3& 2& 1\newline 3& 4& 1& 2\newline 1& 2& 3& 4 \end{bmatrix} $$

Fig. 1 becomes:

$$ \begin{bmatrix} J2& A1& K4& Q3\newline Q4& K3& A2& J1\newline A3& J4& Q1& K2\newline K1& Q2& J3& A4 \end{bmatrix} $$

Two Latin sqaures is orthogonal iff when we combine two elements with the same position together, each element in the new matrix is unique. The puzzle of Euler is: given a Latin sqaure, how do we find another Latin matrix which is orthogonal to the given one.

The combined matrix is called as a Graeco-Latinx matrix.

In order to prove Euler’s conjenture, mathematics spent almost 180 years with the support from computers. In 1957, Paige and Tompkins programmed on SWAC to find a counterexample with $M=10$. After 10 hours running without any results, they terminated the program.

In 1960, Parker et al. proved that it is possible to find an orthogonal Latin square when $M \geq 6$ and disprove the conjenture. In addition, they also wrote a program on UNIVAC 1206 which is able to generate the matrix. The program only took nearly 1 hour to find a solution of input in 1957:

```
0 1 2 3 4 5 6 7 8 9
1 8 3 2 5 4 7 6 9 0
2 9 5 6 3 0 8 4 7 1
3 7 0 9 8 6 1 5 2 4
4 6 7 5 2 9 0 8 1 3
5 0 9 4 7 8 3 1 6 2
6 5 4 7 1 3 2 9 0 8
7 4 1 8 0 2 9 3 5 6
8 3 6 0 9 1 5 2 4 7
9 2 8 1 6 7 4 0 3 5
```

One instant of the solution (TAoCP):

```
0 2 8 5 9 4 7 3 6 1
1 7 4 9 3 6 5 0 2 8
2 5 6 4 8 7 0 1 9 3
3 6 9 0 4 5 8 2 1 7
4 8 1 7 5 3 6 9 0 2
5 1 7 8 0 2 9 4 3 6
6 9 0 2 7 1 3 8 4 5
7 3 5 1 2 0 4 6 8 9
8 0 2 3 6 9 1 7 5 4
9 4 3 6 1 8 2 5 7 0
```

# Transvervals

Parker utilized a concepted called transversal which was previously proposed by Euler. A transersal is a way to choose a sequence of $M$ elements such that there is only 1 element in each row, each column and is unique in the sequence.

For example: 0859734216, it means that we choose 0 in column 1, 8 in column 2, 5 in column 3, etc. Supposed that $k$ is the first item of a transversal, based on following elements, we replace them by $k$. Given $M$ transversals of $k = 0, 1, \cdots, M-1$, we are able to construct an orthogonal matrix.

The algorithm below counts the number of transversals of a Latin square.

```
typedef vector<vector<int>> mat;
class transversal {
private:
int n;
vector<bool> row;
vector<bool> nums;
mat input;
int cnt;
public:
transversal(const mat& in_): n(in_.size()), input(in_),
row(n, false), nums(n, false), cnt(0) {
assert(in_.size() == in_[0].size());
}
void print() { // debugging
for (auto& rows: input) {
for (auto& e: rows) cout << e << " ";
cout << endl;
}
}
int compute(int idx) {
cnt = 0;
fill(row.begin(), row.end(), false);
fill(nums.begin(), nums.end(), false);
nums[input[idx][0]] = true;
row[idx] = true;
comp(1);
return cnt;
}
private:
void comp(int idx) { //idx = index of column
if (idx == n) {
cnt++;
return;
}
for (int i = 0; i < n; ++i) {
if (!row[i] && !nums[input[i][idx]] ) {
row[i] = true;
nums[input[i][idx]] = true;
comp(idx+1);
row[i] = false;
nums[input[i][idx]] = false;
}
}
}
};
```

From the above input, we get the output:

```
79 96 76 87 70 84 83 75 95 63
```

It is exactly the same as in the book. We later can modify the code to find all transversals given an input. After that, for every transversal $k$, we choose a set such that there is no duplicate elements which becomes the output matrix. The details of the algorithm will be discussed in later posts about Dancing Links.

# References

- Knuth, Donald E. The Art of Computer Programming, Volume 4A: Combinatorial Algorithms, Part 1. Pearson Education India, 2011.
- https://en.wikipedia.org/wiki/Latin_square
- http://elscorcho.50webs.com/latinsquares.pdf