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oksian1 [2.3K]
3 years ago
14

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

Os in a 3x3 grid. The purpose of the game is to place three of your marks in a horizontal, vertical or diagonal.
Computers and Technology
1 answer:
-Dominant- [34]3 years ago
7 0

Answer:

Explanation:

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

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Use JavaWrite a program that will simulate a change machine found at cash registers. Input the amount due and amount paid from t
Ksju [112]

Answer:

Here is the JAVA program:

import java.util.Scanner;  //to accept input from user

public class Main { // class name

 public static void main(String [] args) {  // start of main method

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   change = change % 25;  // compute the quarters remaining

   int dimes = change / 10;  //  computes dimes

   change = change % 10;  //  computes dimes remaining

   int nickels = change / 5;  // computes nickels

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   System.out.println("Dimes: " + dimes);  // displays value of dimes

   System.out.println("Nickels: " + nickels);  // displays value of nickels

   System.out.println("Pennies: " + pennies);   }} //displays value of pennies

Explanation:

I will explain the program with an examples.

Suppose the user enters 4.57 as cost of the item and 5.00 as amount paid. Then the program works as follows:

Change is computed as

change = (int)(amount * 100 - itemCost * 100);

This becomes;

change = (int)(5.00 * 100 - 4.57 * 100)

            = 500 - 457

change = 43

Now the change owed is computed as:

change / 100.0 = 43/100.0 = 0.43

Hence change owed = 0.43

Next quarters are computed as:

quarters = change / 25;

This becomes:

quarters = 43/25

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Now the remaining is computed as:

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This becomes:

   change = 43 % 25;

  change = 18

Next the dimes are computed from remaining value of change as:

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dimes = 18 / 10

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Now the remaining is computed as:

   change = change % 10;

This becomes:

   change = 18 % 10

  change = 8

Next the nickels are computed from remaining value of change as:

nickels = change / 5;

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   change = change % 5;

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pennies = change;

pennies = 3

So the output of the entire program is:

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The screenshot of the output is attached.

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