Magic square for odd numbers with checking in JavaScript

The magic square, is a matrix with a constant sum in every row, column, diagonal, which is known as magic sum. Each element in the magic square cannot be the same.

Here is a JavaScript solution when size of the sides of square is an odd number.

Rules:

1. Put 1 in the middle of the first row of the matrix.
2. Place the numbers in the matrix in ascending order along the upper left corner.
3. If it is out of bounds, assume that there is a matrix around, put the number in that position.
4. If position is already occupied, skip the position and put it below, then re-place it as usual.

function magicSquare(n) {
  //create an empty array for writing numbers into it and store it in a variable `matrix`
  let matrix = [];
  //create a variable for the horizontal position - `x`
  let x = 0;
  //create a variable for the vertical position- `y`
  let y = (n - 1) / 2;
  //loop to iterate over strings (to create empty arrays equal to the size of the square)
  for (let i = 0; i < n; i++) {
    matrix[i] = [];
    //loop to iterate over columns (to create cells with null values in a previously created arrays)
    for (let j = 0; j < n; j++) {
      matrix[i][j] = 0;
    }
  }
  //filling arrays with numbers - the number of iterations is equal to the square of the matrix size
  for (let i = 0; i < n * n; i++) {
    pos(i + 1);
  }
  //function with filling rules which has as argument `value` equal to the square of the matrix size
  function pos(value) {
    //filling in the cell corresponding to the calculated values of `x` and` y`
    matrix[x][y] = value;
    //create variables to store the current values of `x` and` y` or the current positions of the number
    let tx = x;
    let ty = y;
    //decrease the value of `x` by 1
    x--;
    //if the value of `x` is less than zero, then add to the `x` variable a number equal to the size of the matrix
    if (x < 0) {
      x += n;
    }
    //the same for `y` variable
    y--;
    if (y < 0) {
      y += n;
    }

    //if the number in the cell is not 0
    if (matrix[x][y] !== 0) {
      //then 1 is added to `tx` variable and stored in the `x` variable
      x = tx + 1;
//if `x` variable is equal to the size of the matrix
      if (x === n) {
        //then x equals 0
        x = 0;
      }
      //`y` equals the value` ty`, i.e. last number "address"
      y = ty;
    }
  }
  //after loop of filling the numbers, the filled "matrix" is returned
  return matrix;
}

Check given matrix is magic square or not:

function isMagicSquare(mat) {
  let N = mat.length;
  // `sumd1` and` sumd2` - the sum of the number along the diagonals
  let sumd1 = 0,
    sumd2 = 0;
  for (let i = 0; i < N; i++) {
    // (i, i) - diagonal from top left corner to bottom right
    // (i, N - i - 1) - diagonal from top right to bottom left
    sumd1 = sumd1 + mat[i][i];
    sumd2 = sumd2 + mat[i][N - 1 - i];
  }
  // if the two diagonal sums are not equal, then this is not a magic square
  if (sumd1 != sumd2) return false;

  //similar check for rows and columns
  for (let i = 0; i < N; i++) {
    let colSum = 0;
    let rowSum = 0;
    for (let j = 0; j < N; j++) {
      rowSum += mat[i][j];
      colSum += mat[j][i];
    }
    if (rowSum != colSum || colSum != sumd1) return false;
  }
  return true;
}

For size of the sides of square equals 3:

let oddNumber = 3;

console.log(magicSquare(oddNumber));
console.log(isMagicSquare(magicSquare(oddNumber)));

Output:

[ [ 6, 1, 8 ], 
  [ 7, 5, 3 ],
  [ 2, 9, 4 ] ]
true