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Reinforcement Learning for Two-Player Games

How does Tic-Tac-Toe differ from the maze problem?

  • Different state and action sets.
  • Two players rather than one.
  • Reinforcement is 0 until end of game, when it is 1 for win, 0 for draw, or -1 for loss.
  • Maximizing sum of reinforcement rather than minimizing.
  • Anything else?

Representing the Q Table

The state is the board configuration. There are $3^9$ of them, though not all are reachable. Is this too big?

It is a bit less than 20,000. Not bad. Is this the full size of the Q table?

No. We must add the action dimension. There are at most 9 actions, one for each cell on the board. So the Q table will contain about $20,000 \cdot 9$ values or about 200,000. No worries.

Instead of thinking about the Q table as a three-dimensional array, as we did last time, let's be more pythonic and use a dictionary. Use the current state as the key, and the value associated with the state is an array of Q values for each action taken in that state.

We still need a way to represent a board.

How about an array of characters? So

 X |   | O
 ---------
   | X | O
 ---------
 X |   |

would be

 board = np.array(['X',' ','O', ' ','X','O', 'X',' ',' '])

The initial board would be

 board = np.array([' ']*9)

We can represent a move as an index, 0 to 8, into this array.

What should the reinforcement values be?

How about 0 every move except when X wins, with a reinforcement of 1, and when O wins, with a reinforcement of -1.

For the above board, let's say we, meaning Player X, prefer move to index 3. In fact, this always results in a win. So the Q value for move to 3 should be 1. What other Q values do you know?

If we don't play a move to win, O could win in one move. So the other moves might have Q values close to -1, depending on the skill of Player O. In the following discussion we will be using a random player for O, so the Q value for a move other than 8 or 3 will be close to but not exactly -1.

Agent-World Interaction Loop

For our agent to interact with its world, we must implement

  1. Initialize Q.
  2. Set initial state, as empty board.
  3. Repeat:
    1. Agent chooses next X move.
    2. If X wins, set Q(board,move) to 1.
    3. Else, if board is full, set Q(board,move) to 0.
    4. Else, let O take move.
    5. If O won, update Q(board,move) by (-1 - Q(board,move))
    6. For all cases, update Q(oldboard,oldmove) by Q(board,move) - Q(oldboard,oldmove)
    7. Shift current board and move to old ones.

Now in Python

In [ ]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
from copy import copy

Let's write a function to print a board in the usual Tic-Tac-Toe style.

In [ ]:
def printBoard(board):
    print('{}|{}|{}\n-----\n{}|{}|{}\n-----\n{}|{}|{}'.format(*board))

board = np.array(['X',' ','O', ' ','X','O', 'X',' ',' '])
printBoard(board)

Let's write a function that returns True if the current board is a winning board for us. We will be Player X. What does the value of combos represent?

In [ ]:
def winner(board):
    combos = np.array((0,1,2, 3,4,5, 6,7,8, 0,3,6, 1,4,7, 2,5,8, 0,4,8, 2,4,6))
    return np.any(np.logical_or(np.all('X' == board[combos].reshape((-1,3)), axis=1),
                                np.all('O' == board[combos].reshape((-1,3)), axis=1)))          
In [ ]:
board = np.array(['X',' ','O', ' ','X','O', 'X',' ',' '])
printBoard(board)
winner(board)
In [ ]:
board[3] = 'X'
printBoard(board)
winner(board)

How can we find all valid moves from a board? Just find all of the spaces in the board representation

In [ ]:
np.where(board == ' ')
In [ ]:
np.where(board == ' ')[0]

And how do we pick one at random and make that move?

In [ ]:
board = np.array(['X',' ','O', ' ','X','O', 'X',' ',' '])
validMoves = np.where(board == ' ')[0]
move = np.random.choice(validMoves)
boardNew = copy(board)
boardNew[move] = 'X'
print('From this board')
printBoard(board)
print('\n  Move',move)
print('\nresults in board')
printBoard(boardNew)

If X just won, we want to set the Q value for the previous state (board) to 1, because X will always win from that state and that action (move).

First we must figure out how to implement the Q table. We want to associate a value with each board and move. We can use a python dictionary for this. We know how to represent a board. A move can be an integer from 0 to 8 to index into the board array for the location to place a marker.

In [ ]:
Q = {}  # empty table
Q[(tuple(board),1)] = 0
Q
In [ ]:
Q[(tuple(board),1)]

What if we try to look up a Q value for a state,action we have not encountered yet? It will not be in the dictionary. We can use the get method for the dictionary, that has a second argument as the value returned if the key does not exist.

In [ ]:
board[1] = 'X'
Q[(tuple(board),1)]
In [ ]:
Q.get((tuple(board),1), 42)

Now we can set the Q value for (board,move) to 1.

In [ ]:
Q[(tuple(board),move)] = 1

If the board is full and we have a draw, then the previous state and action should be assigned 0.

In [ ]:
Q[(tuple(board),move)] = 0

If the board is not full, better check to see if O just won. If O did just win, then we should adjust the Q value of the previous state and X action to be closer to -1, because we just received a -1 reinforcement and the game is over.

In [ ]:
rho = 0.1 # learning rate
Q[(tuple(board),move)] += rho * (-1 - Q[(tuple(board),move)])

If nobody won yet, let's calculate the temporal difference error and use it to adjust the Q value of the previous board,move. We do this only if we are not at the first move of a game.

In [ ]:
step = 0
if step > 0:
    Q[(tuple(boardOld),moveOld)] += rho * (Q[(tuple(board),move)] - Q[(tuple(boardOld),moveOld)])

Initially, taking random moves is a good strategy, because we know nothing about how to play Tic-Tac-Toe. But, once we have gained some experience and our Q table has acquired some good predictions of the sum of future reinforcement, we should rely on our Q values to pick good moves. For a given board, which move is predicted to lead to the best possible future using the current Q table?

In [ ]:
validMoves = np.where(board == ' ')[0]
print('Valid moves are',validMoves)
Qs = np.array([Q.get((tuple(board),m), 0) for m in validMoves]) 
print('Q values for validMoves are',Qs)
bestMove = validMoves[np.argmax(Qs)]
print('Best move is',bestMove)

To slowly transition from taking random actions to taking the action currently believed to be best, called the greedy action, we slowly decay a parameter, $\epsilon$, from 1 down towards 0 as the probability of selecting a random action. This is called the $\epsilon$-greedy policy.

In [ ]:
def epsilonGreedy(epsilon, Q, board):
    validMoves = np.where(board == ' ')[0]
    if np.random.uniform() < epsilon:
        # Random Move
        return np.random.choice(validMoves)
    else:
        # Greedy Move
        Qs = np.array([Q.get((tuple(board),m), 0) for m in validMoves]) 
        return validMoves[ np.argmax(Qs) ]
    
epsilonGreedy(0.8, Q, board)

Now write a function to make plots to show results of some games. Say the variable outcomes is a vector of 1's, 0's, and -1's, for games in which X wins, draws, and loses, respectively.

In [ ]:
outcomes = np.random.choice([-1,0,1],replace=True,size=(1000))
outcomes[:10]
In [ ]:
def plotOutcomes(outcomes,epsilons,maxGames,nGames):
    if nGames==0:
        return
    nBins = 100
    nPer = int(maxGames/nBins)
    outcomeRows = outcomes.reshape((-1,nPer))
    outcomeRows = outcomeRows[:int(nGames/float(nPer))+1,:]
    avgs = np.mean(outcomeRows,axis=1)
    plt.subplot(3,1,1)
    xs = np.linspace(nPer,nGames,len(avgs))
    plt.plot(xs, avgs)
    plt.xlabel('Games')
    plt.ylabel('Mean of Outcomes\n(0=draw, 1=X win, -1=O win)')
    plt.title('Bins of {:d} Games'.format(nPer))
    plt.subplot(3,1,2)
    plt.plot(xs,np.sum(outcomeRows==1,axis=1),'g-',label='Wins')
    plt.plot(xs,np.sum(outcomeRows==-1,axis=1),'r-',label='Losses')
    plt.plot(xs,np.sum(outcomeRows==0,axis=1),'b-',label='Draws')
    plt.legend(loc="center")
    plt.ylabel('Number of Games\nin Bins of {:d}'.format(nPer))
    plt.subplot(3,1,3)
    plt.plot(epsilons[:nGames])
    plt.ylabel('$\epsilon$')
In [ ]:
plt.figure(figsize=(8,8))
plotOutcomes(outcomes,np.zeros(1000),1000,1000)

Finally, let's write the whole Tic-Tac-Toe learning loop!

In [ ]:
from IPython.display import display, clear_output
In [ ]:
maxGames = 50000
rho = 0.2
epsilonDecayRate = 0.9999
epsilon = 1.0
graphics = True
showMoves = not graphics

outcomes = np.zeros(maxGames)
epsilons = np.zeros(maxGames)
Q = {}

if graphics:
    fig = plt.figure(figsize=(10,10))

for nGames in range(maxGames):
    epsilon *= epsilonDecayRate
    epsilons[nGames] = epsilon
    step = 0
    board = np.array([' '] * 9)  # empty board
    done = False
    
    while not done:        
        step += 1
        
        # X's turn
        move = epsilonGreedy(epsilon, Q, board)
        boardNew = copy(board)
        boardNew[move] = 'X'
        if (tuple(board),move) not in Q:
            Q[(tuple(board),move)] = 0  # initial Q value for new board,move
        if showMoves:
            printBoard(boardNew)
            
        if winner(boardNew):
            # X won!
            if showMoves:
                print('        X Won!')
            Q[(tuple(board),move)] = 1
            done = True
            outcomes[nGames] = 1
            
        elif not np.any(boardNew == ' '):
            # Game over. No winner.
            if showMoves:
                print('        draw.')
            Q[(tuple(board),move)] = 0
            done = True
            outcomes[nGames] = 0
            
        else:
            # O's turn.  O is a random player!
            moveO = np.random.choice(np.where(boardNew==' ')[0])
            boardNew[moveO] = 'O'
            if showMoves:
                printBoard(boardNew)
            if winner(boardNew):
                # O won!
                if showMoves:
                    print('        O Won!')
                Q[(tuple(board),move)] += rho * (-1 - Q[(tuple(board),move)])
                done = True
                outcomes[nGames] = -1
        
        if step > 1:
            Q[(tuple(boardOld),moveOld)] += rho * (Q[(tuple(board),move)] - Q[(tuple(boardOld),moveOld)])
            
        boardOld, moveOld = board, move # remember board and move to Q(board,move) can be updated after next steps
        board = boardNew
        
        if graphics and (nGames % (maxGames/10) == 0 or nGames == maxGames-1):
            fig.clf() 
            plotOutcomes(outcomes,epsilons,maxGames,nGames-1)
            clear_output(wait=True)
            display(fig);

if graphics:
    clear_output(wait=True)
print('Outcomes: {:d} X wins {:d} O wins {:d} draws'.format(np.sum(outcomes==1), np.sum(outcomes==-1), np.sum(outcomes==0)))

How can we examine the Q function that predicts the future for every board and move?

In [ ]:
Q[(tuple([' ']*9),0)]
In [ ]:
Q[(tuple([' ']*9),1)]
In [ ]:
Q.get((tuple([' ']*9),0), 0)
In [ ]:
[Q.get((tuple([' ']*9),m), 0) for m in range(9)]
In [ ]:
board = np.array([' ']*9)
Qs = [Q.get((tuple(board),m), 0) for m in range(9)]
printBoard(board)
print()
print('''{:5.2f} | {:5.2f} | {:5.2f}
---------------------
{:5.2f} | {:5.2f} | {:5.2f}
---------------------
{:5.2f} | {:5.2f} | {:5.2f}'''.format(*Qs))
In [ ]:
def printBoardQs(board,Q):
    printBoard(board)
    Qs = [Q.get((tuple(board),m), 0) for m in range(9)]
    print()
    print('''{:5.2f} | {:5.2f} | {:5.2f}
---------------------
{:5.2f} | {:5.2f} | {:5.2f}
---------------------
{:5.2f} | {:5.2f} | {:5.2f}'''.format(*Qs))
In [ ]:
board[0] = 'X'
board[1] = 'O'
printBoardQs(board,Q)
In [ ]:
board[4] = 'X'
board[3] = 'O'
printBoardQs(board,Q)
In [ ]:
board[0] = 'X'
board[4] = 'O'
printBoardQs(board,Q)
In [ ]:
board[2] = 'X'
board[1] = 'O'
printBoardQs(board,Q)
In [ ]:
board[7] = 'X'
board[3] = 'O'
printBoardQs(board,Q)
In [ ]:
board[5] = 'X'
board[6] = 'O'
printBoardQs(board,Q)