如何检查数独板上的所有子网格，而不是单独检查数独板上的子网格

Question

我有一个算法，我正在尝试检查数独板上的所有子网格。但是，我的代码似乎只迭代一次。它只抓取第一个子网格。我不想做第三个嵌套for循环。我还能怎么改进这个

class Sudoku_Checker:
  def __init__(self,board):
    self.board = board

  def board_validater(self,board):
    self.checkSquares(board)

    return self.checkRows(board) == True and self.checkCols(board) == True and self.checkSquares(board) == True

  def checkSquares(self,board):
      compare = [1,2,3,4,5,6,7,8,9]
      hold = []
      row = 0
      column = 0
      for i in range(row, row + 3):
          for j in range(column, column + 3):
              hold.append(board[i][j])
          if len(hold)==9:
              if sorted(hold) == compare:
                  hold =[]
                  continue
              else:
                return False

      return True

board = [  [5, 3, 4, 6, 7, 8, 9, 1, 2],
          [6, 7, 2, 1, 9, 5, 3, 4, 8],
          [1, 9, 8, 3, 4, 2, 5, 6, 0],
          [8, 5, 9, 7, 6, 1, 4, 2, 3],
          [4, 2, 6, 8, 5, 3, 7, 9, 1],
          [7, 1, 3, 9, 2, 4, 8, 5, 6],
          [9, 6, 1, 5, 3, 7, 2, 8, 4],
          [2, 8, 7, 4, 1, 9, 6, 3, 5],
          [3, 4, 5, 2, 8, 6, 1, 7, 9]]
s = Sudoku_Checker(board)
s.board_validater(board)

谢谢你

Answer 1

好吧，使用第三个for循环将是一个简单的修复，但是通过使用numpy，您可以摆脱内部的两个循环

board = np.array(board)
compare = np.arange(9)

for i in range(9):
    subgrid = board[i//3:i//3+3,i%3:i%3].flatten()
    if not np.array_equal(subgrid,compare):
        return False