Rosetta Code

Magic squares of odd order

Build a 3x3 magic square with the classic Siamese method.

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Sourcerosettacode/popular/magic_squares_of_odd_order.vibe
def blank_square(size: int) -> array<array<int>>
  rows: array<array<int>> = []
  row = 0
  while row < size
    current: array<int> = []
    column = 0
    while column < size
      current.push(0)
      column = column + 1
    end
    rows.push(current)
    row = row + 1
  end
  rows
end

def set_cell(square: array<array<int>>, row: int, column: int, value: int) -> array<array<int>>
  current = square.fetch(row)
  current[column] = value
  square[row] = current
  square
end

def magic_square(size: int) -> array<array<int>>
  square = blank_square(size)
  row = 0
  column = size // 2
  value = 1

  while value <= size * size
    square = set_cell(square, row, column, value)
    next_row = row - 1
    next_column = column + 1

    if next_row < 0
      next_row = size - 1
    end
    if next_column >= size
      next_column = 0
    end

    if square.fetch(next_row)[next_column] != 0
      row = row + 1
      if row >= size
        row = 0
      end
    else
      row = next_row
      column = next_column
    end

    value = value + 1
  end

  square
end

def run -> array<array<int>>
  magic_square(3)
end

Output

Click Run example to see the result.

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