9. Write a complete working Python program that uses a 2D list (a list of lists)
ID: 643280 • Letter: 9
Question
9. Write a complete working Python program that uses a 2D list (a list of lists) to represent a 3x3 tic-tac-toe. The program shall randomly initialize each of the 9 cells with the numbers 1 (for Xs),-1 (for Os), and 0 (for empty cells). The program shall ask the user to make the next move by entering the row and the column as integer numbers (without seeing the current state of the game, so it?s more like battleship rather than actual tic-tac-toe). If the designated location is available, the program shall mark it with an X, display the list as a matrix (row-by-row), and exit. Otherwise, the program shall display the list as a matrix (row-by-row), print an error message, and exit. There is no need to use GUI or classes, but all your code must be in the function main().Explanation / Answer
def do_tests(case): exec(eval("test_case_" + str(case))) print "Test Completed" def read_int(prompt, min_value, max_value): # Returns a user-provided integer. # Repeats prompt until valid data is provided. # read_int("Enter Selection", 0, 2) displays: # Enter Selection [0-2]: _ if min_value > max_value: raise ValueError, "read_int: min_value cannot be greater than max_value" while True: try: val = input(prompt + " [" + str(min_value) + "-" + str(max_value) + "]: ") if val >= min_value and val True ## all_have_value(['x','x','x']) => True ## all_have_value([3, 9, 2]) => True ## all_have_value(['x','o','-']) => False ## all_have_value(['-','x','-']) => False ## all_have_value([]) => False def all_have_value(lst): if len(lst) == 0: return False has_value = True for x in lst: has_value = has_value and x != '-' return has_value ## Step 1 Test Code ## test_case_1 = """ print "Step 1 Test Cases" print all_have_value(['x','o','x']) print all_have_value(['x','x','x']) print all_have_value([3, 9, 2]) print all_have_value(['x','o','-']) print all_have_value(['-','x','-']) print all_have_value([])""" # Uncomment the following line to run test cases. do_tests(1) ## Step 2 - all_equal ## --------------------------------------------------------- ## Return True if all elements have values (not '-') and are equal. ## Return False if the list is empty, or any element is '-'. ## ## You may assume the list is length 3, but you must also ## handle the case where it is empty (null). ## ## all_equal(['x', 'x', 'x']) => True ## all_equal(['o', 'o', 'o']) => True ## all_equal([5, 5, 5]) => True ## all_equal(['x', 'o', 'o']) => False ## all_equal(['-', '-', '-']) => False ## all_equal([]) => False def all_equal(lst): if len(lst) == 0: return False prev_value = None for x in lst: ##print "%s %s" % (x, prev_value) if x != prev_value and prev_value != None: return False elif x == '-': return False else: prev_value = x return True ## Step 2 Test Code ## test_case_2 = """ print "Step 2 Test Cases" print all_equal(['x', 'x', 'x']) print all_equal(['o', 'o', 'o']) print all_equal([5, 5, 5]) print all_equal(['x', 'o', 'o']) print all_equal(['-', '-', '-']) print all_equal([])""" # Uncomment the following line to run test cases. do_tests(2) ## Provided Procedures - select_col, select_row ## ------------------------------------------------------------------- ## select_row returns the row with index i in a 2D list. ## select_col returns the column with index i in a 2D list. ## ## Notes: ## Indexes start at 0. 0 is the first row, 1 is the second, etc. ## A 3-row by 2-column box has rows 0-2 and columns 0-1. ## The item in the third row (row 2) and second column (column 1) ## is referenced using lst[2][1]. ## Assumes i is a valid index for the given list. ## ## Usage: ## l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] ## select_row(l, 0) => [1, 2, 3] ## select_row(l, 2) => [7, 8, 9] ## select_col(l, 0) => [1, 4, 7] ## select_col(l, 2) => [3, 6, 9] ## select_col([[1,2],[3,4],[5,6],[7,8]], 1) => [2, 4, 6, 8] def select_row(lst, i): return lst[i] def select_col(lst, i): return [lst[idx][i] for idx in range(len(lst))] def num_rows(lst): return len(lst) def num_cols(lst): return len(lst[0]) # If the above is confusing, here are some other examples: # def select_col(lst, i): # new_lst = [] # for row in lst: # new_lst = new_lst + row[i] # return new_lst # # In Scheme: # (define (select_col lst i) # (map (lambda (row) (list-ref i row)) lst)) ## Provided Procedures - select_main_diag, select_counter_diag ## ------------------------------------------------------------ ## See lab writeup for detailed information on these procedures. def select_main_diag(array) : return [array[i][i] for i in range(len(array))] def select_counter_diag(array) : return [array[len(array)-i-1][i] for i in range(len(array))] ## Step 3 - Find The Winner ## ----------------------------------------------------------------------------- ## Check each column, row, and diagonal to see if there has been a winner. ## Return 'x' or 'o' if there is a winner. ## Return "Draw" if there is no winner and the board is full. ## Return False if there is no winner and it is still possible to make a move. ## ## You may assume the board is 3x3. ## Use the procedures above to simplify this problem. def winner(board): for row_n in range(0, num_rows(board)): row = select_row(board, row_n) if( all_equal( row ) ): return row[0] for col_n in range(0, num_cols(board)): col = select_col(board, col_n) if all_equal(col): return col[0] main_diag = select_main_diag(board) if all_equal(main_diag): return main_diag[0] counter_diag = select_counter_diag(board) if all_equal(counter_diag): return counter_diag[0] all_full = True for row_n in range(0, num_rows(board)): for col_n in range(0, num_cols(board)): if board[row_n][col_n] == '-': ## is there a way to break 2 levels? all_full = False if all_full: return "Draw" return False ## Step 3 Test Code ## test_case_3 = """ print "Step 5 Test Cases" print winner([['-','-','-'], ['-','-','-'], ['-','-','-']]) print winner([['x','x','x'], ['-','-','-'], ['-','-','-']]) print winner([['o', '-','-'], ['-', 'o','-'], ['-','-','o']]) print winner([['-','-','x'], ['-','-','x'], ['-','-', 'x']]) print winner([['x','o','x'], ['x','o','o'], ['o','x','x']]) print winner([['x','o','x'], ['x','o','o'], ['o','-','x']])""" # Uncomment the following line to run test cases. do_tests(3) ## Provided Procedure - display_board ## -------------------------------------------------------------------------- ## Displays the tic-tac-toe board. For example: ## ## x | o | - ## -----+-----+----- ## o | x | - ## -----+-----+----- ## - | o | x ## def display_board(board): def display_row(row) : print " " + row[0] + " | " + row[1] + " | " + row[2] display_row(board[0]) for i in range(1,3) : print "---+---+---" display_row(board[i]) ## Step 4 - play_game ## --------------------------------------------------------------- ## This procedure ties all of the above procedures together to ## let a user play the game. Most of the structure is provided, ## as is the loop to get the user input. ## ## You need to write the code to check if the selected position ## is available, make the changes to the board, and switch ## between Player 1 and Player 2. def play_game(): # Create Board new_board = [['-', '-', '-'], ['-', '-', '-'], ['-', '-', '-']] # Display Header print "-~ Tic-Tac-Toe ~-" # Initialize Variables current_player = 1 # Start with Player 1 (x) current_row = -1 # Temporary Values for Row/Column current_col = -1 # Start Main Loop while winner(new_board) == False: display_board(new_board) # Get Input From User print "Player " + str(current_player) + "'s Turn: " current_row = read_int("Select Row", 0, 2) current_col = read_int("Select Col", 0, 2) # Check if Position is Empty (Empty Positions Contain '-') # If it is, mark it for current_player and switch player. # If already taken, display warning and let current player try again. # Notes: # current_player is either 1 or 2 (not 'x' or 'o'). # If current_player == 1, then you will place a 'x' on the board. # If current_player == 2, then you will place a 'o' on the board. # Then switch the value of current_player. # # If the position selected was already taken, just print a message # notifying the player of an Invalid Selection. # No other action is required because the while loop will start # again with the same current_player value and the same board. # This allows the player to try again. def current_piece(): if current_player == 1: return 'x' else: return 'o' def next_player(): if current_player == 1: return 2 else: return 1 if new_board[current_row][current_col] == '-': new_board[current_row][current_col] = current_piece() current_player = next_player() else: print "Position is already taken, try again." # Display Final Result of Game display_board(new_board) if winner(new_board) == 'x': print "Player 1 has won!" if winner(new_board) == 'o': print "Player 2 has won!" if winner(new_board) == "Draw": print "Game has resulted in a draw." # To test your code, uncomment the following line. play_game()Related Questions
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