For a recent experiment I used the method presented here to generate a balanced latin square for selecting participant condition orders. Rather than do this manually, I wrote a couple of functions to do this automatically for squares of any size. Here’s an implementation in both C# and R!

C#

public static int[,] GetLatinSquare(int n)
{
    // 1. Create initial square.
    int[,] latinSquare = new int[n, n];

    // 2. Initialise first row.
    latinSquare[0, 0] = 1;
    latinSquare[0, 1] = 2;

    for (int i = 2, j = 3, k = 0; i < n; i++)
    {
        if (i % 2 == 1)
            latinSquare[0, i] = j++;
        else
            latinSquare[0, i] = n - (k++);
    }

    // 3. Initialise first column.
    for (int i = 1; i <= n; i++)
    {
        latinSquare[i - 1, 0] = i;
    }

    // 4. Fill in the rest of the square.
    for (int row = 1; row < n; row++)
    {
        for (int col = 1; col < n; col++)
        {
            latinSquare[row, col] = (latinSquare[row - 1, col] + 1) % n;

            if (latinSquare[row, col] == 0)
                latinSquare[row, col] = n;
        }
    }

    return latinSquare;
}

R

LatinSquare <- function(n) {
  # 1. Create initial square.
  sq <- matrix(0, n, n)

  # 2. Initialise first row.
  sq[1, 1] <- 1
  sq[1, 2] <- 2

  j <- 3
  k <- 0

  for (i in 3:n) {
    if (i %% 2 == 0) {
      sq[1, i] <- j
      j <- j + 1
    } else {
      sq[1, i] <- n - k
      k <- k + 1
    }
  }

  # 3. Initialise first column.
  for (i in 2:(n+1)) {
    sq[i - 1, 1] <- i - 1
  }

  # 4. Fill in the rest of the square.
  for (row in 2:n) {
    for (col in 2:n) {
      sq[row, col] <- (sq[row - 1, col] + 1) %% n

      if (sq[row, col] == 0) {
        sq[row, col] = n
      }
    }
  }

  return (sq)
}