用python编写递归算法，定义一个函数，只提供num_of_row一个参数，要求输出num_of_row行的杨辉三角，输出到Triangle.txt中，不需要返回值。并且Triangle.txt保存在我的桌面上

可以使用递归函数和回溯算法来解决九宫格数独问题。以下是一个简单的 Python 代码示例： ```python def solve(board): """ Solve the Sudoku puzzle using recursive backtracking. :param board: 2D list of integers representing the Sudoku puzzle :return: boolean indicating if the puzzle is solved """ empty_cell = find_empty(board) if not empty_cell: return True row, col = empty_cell for num in range(1, 10): if is_valid(board, row, col, num): board[row][col] = num if solve(board): return True board[row][col] = 0 return False def find_empty(board): """ Find an empty cell in the Sudoku puzzle. :param board: 2D list of integers representing the Sudoku puzzle :return: tuple containing the row and column of an empty cell, or None if there are no empty cells """ for row in range(9): for col in range(9): if board[row][col] == 0: return (row, col) return None def is_valid(board, row, col, num): """ Check if a number can be placed in a given cell in the Sudoku puzzle. :param board: 2D list of integers representing the Sudoku puzzle :param row: integer representing the row of the cell to check :param col: integer representing the column of the cell to check :param num: integer representing the number to check :return: boolean indicating if the number can be placed in the cell """ # Check row for i in range(9): if board[row][i] == num: return False # Check column for i in range(9): if board[i][col] == num: return False # Check square square_row = (row // 3) * 3 square_col = (col // 3) * 3 for i in range(square_row, square_row + 3): for j in range(square_col, square_col + 3): if board[i][j] == num: return False return True ``` 使用该函数，您可以通过以下方式调用它： ```python board = [ [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0] ] solve(board) for row in board: print(row) ``` 该代码将打印已解决数独的完整板。

当然可以，这里我会为你展示如何用 Python 编写广度优先搜索（BFS）和深度优先搜索（DFS）算法来解决经典的八数码问题（又称数独）。我们首先假设已经有一个函数 `check_sudoku` 来检查给定的数组是否满足数独规则。 ```python import collections # 定义数独的空格 def empty_board(): return [[0] * 9 for _ in range(9)] # 检查给定的数独布局是否有效 def check_sudoku(board): # 逻辑省略，你可以自行实现九宫格划分、行、列和宫的遍历以及数字重复性的检查 pass # 广度优先搜索（BFS） def bfs_solve(board): visited = set() queue = collections.deque([[i // 9, i % 9, board.copy()] for i in range(81) if board[i] == 0]) # 初始化队列 while queue: row, col, current_board = queue.popleft() # 弹出队首元素 if (row, col) not in visited: visited.add((row, col)) for num in range(1, 10): # 尝试填入数字 if is_valid_move(row, col, num, current_board): current_board[row][col] = num if check_sudoku(current_board): return current_board # 找到解决方案，返回 else: queue.append([row, col, current_board]) # 无效则回溯 return None # 没有找到解决方案 # 深度优先搜索（DFS） def dfs_solve(board): def backtrack(row, col): nonlocal found_solution if row == 9: found_solution = True return for num in range(1, 10): if is_valid_move(row, col, num, board): board[row][col] = num if check_sudoku(board): solution.append(board.copy()) found_solution = True else: backtrack(row + 1, col) # 递归尝试下一个位置 board[row][col] = 0 # 回溯，撤销当前选择 found_solution = False solution = [] backtrack(0, 0) return solution if found_solution else None # 返回所有解决方案列表 # 判断当前位置填入num是否合法 def is_valid_move(row, col, num, board): # 省略对行、列和宫的检查逻辑 pass # 示例用法 board = empty_board() solution_bfs = bfs_solve(board) solution_dfs = dfs_solve(board) if solution_bfs: print("BFS Solution:") print(solution_bfs) else: print("BFS couldn't find a solution.") if solution_dfs: print("\nDFS Solutions:") for sol in solution_dfs: print(sol) else: print("DFS couldn't find a solution.") ```