Return bestMove in minimax algorithm for tictactoe
I have tried to code the minimax algorithm for tic-tac-toe given in Russel Norvig's book on Artificial Intelligence. It had everything except that the way to return the bestMove to the user. I am trying hard to return the bestMove, but cannot decide when to choose the bestMove. Help, anyone?
moveT MiniMax(stateT state)
{
moveT bestMove;
max_move(state,bestMove);
return bestMove;
}
int max_move(stateT state,int & bestMove)
{
int v = -10000;
if(GameIsOver(state))
{
return EvaluateStaticPosition(state);
}
vector<moveT> moveList;
GenerateMoveList(state, moveList);
int nMoves = moveList.size();
for(int i = 0 ; i < nMoves ; i++)
{
moveT move = moveList[i];
MakeMove(state, move);
int curValue = min_move(state,bestMove);
if(curValue > v)
{
v = curValue;
bestMove = move;
}
RetractMove(state, move);
}
return v;
}
int min_move(stateT state, int &bestMove)
{
int v = 10000;
if(GameIsOver(state))
{
return EvaluateStaticPosition(state);
}
vector<moveT> moveList;
GenerateMoveList(state, moveList);
int nMoves = moveList.size();
for(int i = 0 ; i < nMoves; i++)
{
moveT move = moveList[i];
MakeMove(state, move);
int curValue = max_move(state,depth+1,bestMove);
if(curValue < v)
{
curValue = v;
}
RetractMove(state, move);
}
return v;
}
P.S.: There are other pseudocode for finding the minmax value. However, they are focused on tic-tac-toe only, I am trying to extend it to other games. Thanks.
Update : The whole code can be found here : http://ideone.com/XPswCl
Answer
In the simplest version of minimax, the first player wishes to maximize his score, and the second player wishes to minimize the first player's score.
Since both first and second player only care about the first player's score, EvaluateStaticPosition should return a value indicating how good the board state is for the first player. Whose turn it is is not relevant.
int EvaluateStaticPosition(stateT state)
{
if(CheckForWin(state, FIRST_PLAYER))
{
return WINNING_POSITION;
}
if(CheckForWin(state, Opponent(FIRST_PLAYER)))
{
return LOSING_POSITION;
}
return NEUTRAL_POSITION;
}
Now, when you want the move that's best for the first player, call MaxMove. When you want the move that's best for the second player, call MinMove.
moveT MiniMax(stateT state)
{
moveT bestMove;
int i = 0;
if (state.whoseTurn == FIRST_PLAYER){
i = MaxMove(state, bestMove);
}
else{
i = MinMove(state,bestMove);
}
cout<<"i is "<<i<<endl;
return bestMove;
}
Finally, you have some problems inside of MinMove and MaxMove. when you assign curRating in either one, you shouldn't pass in bestMove as the second argument to MaxMove or MinMove. It will then put the opponent's best move into bestMove, which doesn't make sense. Instead, declare an opponentsBestMove object and pass that as the second argument. (You won't actually be using the object or even looking at its value afterwards, but that's ok). With that change, you never assign anything to bestMove within MinMove, so you should do so inside the if(curRating < v) block.
int MaxMove(stateT state, moveT &bestMove)
{
if(GameIsOver(state))
{
return EvaluateStaticPosition(state);
}
vector<moveT> moveList;
GenerateMoveList(state, moveList);
int nMoves = moveList.size();
int v = -1000;
for(int i = 0 ;i<nMoves; i++)
{
moveT move = moveList[i];
MakeMove(state, move);
moveT opponentsBestMove;
int curRating = MinMove(state, opponentsBestMove);
if (curRating > v)
{
v = curRating;
bestMove = move;
}
RetractMove(state, move);
}
return v;
}
int MinMove(stateT state, moveT &bestMove)
{
if(GameIsOver(state))
{
return EvaluateStaticPosition(state);
}
vector<moveT>moveList;
GenerateMoveList(state, moveList);
int nMoves = moveList.size();
int v = 1000;
for(int i = 0 ; i<nMoves; i++)
{
moveT move = moveList[i];
MakeMove(state , move);
moveT opponentsBestMove;
int curRating = MaxMove(state,opponentsBestMove);
if(curRating < v)
{
v = curRating;
bestMove = move;
}
RetractMove(state, move);
}
return v;
}
At this point you should have an unbeatable AI!
The final position looks like this:
O | O | X
---+---+---
X | X | O
---+---+---
O | X | X
Cat's game.
An alternative method takes advantage of the fact that tic-tac-toe is a zero-sum game. In other words, at the end of the game, the sum of the scores of the players will equal zero. For a two player game, this means that one player's score will always be the negative of the other player's. This is convenient for us, since minimizing the other player's score is then identical to maximizing one's own score. So instead of one player maximizing his score and one player minimizing the other player's score, we can just have both players attempt to maximize their own score.
Change EvaluateStaticPosition back to its original form, so that it gives a score based on how good the board state is for the current player.
int EvaluateStaticPosition(stateT state)
{
if(CheckForWin(state, state.whoseTurn))
{
return WINNING_POSITION;
}
if(CheckForWin(state, Opponent(state.whoseTurn)))
{
return LOSING_POSITION;
}
return NEUTRAL_POSITION;
}
Delete MinMove, since we only care about maximizing.
Rewrite MaxMove so that it chooses the move that gives the opponent the worst possible score. The score for the best move is the negative of the other player's worst score.
int MaxMove(stateT state, moveT &bestMove)
{
if(GameIsOver(state))
{
return EvaluateStaticPosition(state);
}
vector<moveT> moveList;
GenerateMoveList(state, moveList);
int nMoves = moveList.size();
int v = -1000;
for(int i = 0 ;i<nMoves; i++)
{
moveT move = moveList[i];
MakeMove(state, move);
moveT opponentsBestMove;
int curRating = -MaxMove(state, opponentsBestMove);
if (curRating > v)
{
v = curRating;
bestMove = move;
}
RetractMove(state, move);
}
return v;
}
Since MaxMove is used for both players, we no longer need to distinguish among players in the MiniMax function.
moveT MiniMax(stateT state)
{
moveT bestMove;
int i = 0;
i = MaxMove(state, bestMove);
cout<<"i is "<<i<<endl;
return bestMove;
}