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wel
3 years ago
6

Your computer uses 4 bits to represent decimal numbers (0, 1, 2, 3 and so on) in binary. What is the SMALLEST number for which a

n overflow error occur?
Computers and Technology
1 answer:
myrzilka [38]3 years ago
3 0

Answer: C

Explanation:

I believe this is because you cannot represent the number sixteen with only four binary bits. The highest you can go is 15.

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Write code to play a Tic-tac-toe tournament. Tic-tac toe is a game for two players who take turns marking the spaces with Xs and
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Answer:

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The following code is written in Python and is a full Two player tic tac toe game on a 3x3 grid that is represented by numbers per square.

# Making all of the main methods of the game

board = [0,1,2,

        3,4,5,

        6,7,8]

win_con = [[0,1,2],[3,4,5],[6,7,8],

           [0,3,6],[1,4,7],[2,5,8],

           [0,4,8],[2,4,6]] # possible 3-in-a-rows

def show():

   print(board[0],'|',board[1],'|',board[2])

   print('----------')

   print(board[3],'|',board[4],'|',board[5])

   print('----------')

   print(board[6],'|',board[7],'|',board[8])

def x_move(i):

   if board[i] == 'X' or board[i] == 'O':

       return print('Already taken!')

   else:

       del board[i]

       board.insert(i,'X')

def o_move(i):

   if board[i] == 'X' or board[i] == 'O':

       return print('Already taken!')

   else:

       del board[i]

       board.insert(i,'O')  

 

# Creating the main loop of the game

while True:

   turn_num = 1

   board = [0,1,2,3,4,5,6,7,8]

   print('Welcome to Tic-Tac-Toe!')

   print('AI not implemented yet.')

   while True:

       for list in win_con: #check for victor

           xnum = 0

           onum = 0

           for num in list:

               if board[num] == 'X':

                   xnum += 1

               elif board[num] == 'O':

                   onum += 1

               else:

                   pass

           if xnum == 3 or onum == 3:

               break

       if xnum == 3 or onum == 3: # break loops

           break

       if turn_num > 9: # Check if there are any more moves available

           break

       show()

       if turn_num % 2 == 1:

           print('X\'s turn.')

       else:

           print('O\'s turn.')

       move = int(input('Choose a space. '))

       if turn_num % 2 == 1:

           x_move(move)

       else:

           o_move(move)

       turn_num += 1

   if xnum == 3:  #If game ends

       print('X Won!')

   elif onum == 3:

       print('O Won!')

   else:

       print('Draw!')

   play_again = input('Play again? Y or N ')

   if play_again == 'Y' or play_again == 'y':

       continue

   else:

       break

7 0
2 years ago
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