1. HW 6: Problems 2 & 3 Simulating Connect 4

2. HW 6: Overview Connect 4: Variation of Tic-Tac-Toe – Board: Vertical 7x6 – Two players take alternating move Move = Placing checkers on the board Winning = 4 checkers in a row (vert., horz., diag.) – Move constraints: On top of each other, or Start a new column Problems: Board class (p2), Player class (p3)

3. Problem 2: Connect 4 Board Variables Board Representation: – Two-dimensional List – Width & height vary (i.e. not just 7x6) Board Class variables – Variable data storing the 2D list – Variable height storing the number of rows – Variable width storing the number of columns

4. Problem 2: Connect 4 Board Method: Constructor The Constructor __init__ : – Return nothing – Take in the board dimension – Set board height, width, & initialize data elements def __init__( self, width, height ): self.width = width self.height = height self.data = [] for row in range( self.height ): boardRow = [] for col in range( self.width ): boardRow += [' '] self.data += [boardRow]

5. Problem 2: Connect 4 Board Method: Representation Method __repr__ : Print out the board def __repr__(self): s = '' for row in range(self.height): s += '|' for col in range(self.width): s += self.data[row][col]+'|' s += '\n' #print out separator... (your code here) #print out column numbers... (your code here) return s | | | | | | | | --------------- 0 1 2 3 4 5 6

6. Problem 2: Connect 4 Board Method: addMove Method addMove : take in column & symbol >>> b = Board(7,5) >>> b.addMove(0, 'X') >>> b.addMove(1, 'O') >>> b.addMove(1, 'X') >>> b.addMove(3, 'O') | | | | | | | | | |X| | | | | | |X|O| |O| | | | --------------- 0 1 2 3 4 5 6 def addMove(self, col, ox ): if allowMove(col): # find the first row in # col without a checker... (your code here) # put ox in this position... (your code here)

7. Problem 2: Connect 4 Board Methods: clear, delMove, allowMove, isFull Method clear(self) : Clear all data to ' ' Method delMove(self, col) : – Opposite of addMove – Delete the top checker of column col – Do nothing if column is empty Method allowMove(self, col) : – False if col is an invalid column number, or full – True otherwise Method isFull(self) checks if the board is full

8. Problem 2: Connect 4 Board Method: setBoard Method setBoard : take in the string of moves >>> b = Board(7,5) >>> b.setBoard('02353') | | | | | | | | | | | |O| | | | |O| |X|O| |X| | --------------- 0 1 2 3 4 5 6 def setBoard(self,moveStr): ch = 'O' for colStr in moveStr: col = int(colStr) if 0<=col<=self.width: self.addMove(col,ch) if ch == 'X': ch = 'O' else: ch = 'X'

9. Problem 2: Connect 4 Board Methods: winFor Method winFor(self, ox) : – Check if there are 4 ox ’s in a run – Horizontal, vertical, and diagonal – Possible direction: Anchor checkers = checks that may start run Check each of these anchor checkers in 8 directions

10. Problem 2: Connect 4 Board Methods: hostGame Method hostGame(self) : – Runs until the game ends: A player wins The board is full (draw) – Alternatively ask player 'O' and 'X' to move – Print out the board before asking to move – Invalid move = ask the same player to move again – Valid move = place checker & check for win – At game end report the winner (or tie)

11. EC Problem 3: Connect 4 Player Overview Goal: Creating an automated player Player class: – Examine the board – Find appropriate move – May look for several moves ahead The variables: At least the following – Character ox – Tie breaking type tbt (either 'LEFT', 'RIGHT' or 'RANDOM', more detail later) – Number of moves ahead ply

12. EC Problem 3: Connect 4 Player The Essential Methods Construction __init__(self,ox,tbt,ply) Representation __repr__(self) def __init__( self, ox, tbt, ply ): self.ox = ox self.tbt = tbt self.ply = ply def __repr__( self ): s = 'Player for ' + self.ox + '\n' s += 'with tiebreak type: ' + self.tbt + '\n' s += 'and ply == ' + str(self.ply) + '\n\n' return s

13. EC Problem 3: Connect 4 Player The Required Methods Opposite character oppCh(self) : – Returns 'X' if own character is 'O', vice versa Evaluating Board scoreBoard(self, b) : – Given a board b, return 100.0, 50.0 and 0.0 respectively for win, tie, and lose situation Tie-breaking: tiebreakMove(self, scores) : – Score = list of floating point – Returns the column with the highest score – Multiple highest score: select one based on tbt – Example: tiebreakMove([0,50,0,50]) returns 1 if self.tbt == ‘LEFT’

14. EC Problem 3: Connect 4 Player The Brain: scoreFor Returns a list of scores: – The c th score = goodness after a move to column c – Goodness = what happens after self.ply moves Using recursion: – Base case 1: A full column = score of -1.0 – Base case 2: 0-ply = state of the game right now (hence, all non-full columns have the same score) – Base case 3: Game-over = state of the game right now – Recursive case: Make a move into a corresponding column, change the (local) board, evaluate recursively

15. 0-ply scores for O : col 0col 1col 2col 3col 4col 5col 6 1-ply scores for O: col 0col 1col 2col 3col 4col 5col 6 2-ply scores for O : col 0col 1col 2col 3col 4col 5col 6 3-ply scores for O : col 0col 1col 2col 3col 4col 5col 6 50 10050 0 0 100 0 0 0 050 100 00 b ‘X’ ‘O’ EC Problem 3: Connect 4 Player

16. EC Problem 3: Connect 4 Player Put them all together Choose the next move: nextMove(self,b) : – Return the best column to move to – Wrapper method (heavy lifting is already done) Play the game playGame(self,px,po) : – Modify hostGame from the Board class – Instead of getting user input, it use moves getting from px and po alternatively – Additional twist: if one of px and po is a string human, then get the input from the user instead