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  1. PO
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    <p>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, <code>EvaluateStaticPosition</code> should return a value indicating how good the board state is for the first player. Whose turn it is is not relevant.</p> <pre><code>int EvaluateStaticPosition(stateT state) { if(CheckForWin(state, FIRST_PLAYER)) { return WINNING_POSITION; } if(CheckForWin(state, Opponent(FIRST_PLAYER))) { return LOSING_POSITION; } return NEUTRAL_POSITION; } </code></pre> <p>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.</p> <pre><code>moveT MiniMax(stateT state) { moveT bestMove; int i = 0; if (state.whoseTurn == FIRST_PLAYER){ i = MaxMove(state, bestMove); } else{ i = MinMove(state,bestMove); } cout&lt;&lt;"i is "&lt;&lt;i&lt;&lt;endl; return bestMove; } </code></pre> <p>Finally, you have some problems inside of <code>MinMove</code> and <code>MaxMove</code>. when you assign <code>curRating</code> in either one, you shouldn't pass in <code>bestMove</code> as the second argument to <code>MaxMove</code> or <code>MinMove</code>. It will then put the <em>opponent's</em> best move into <code>bestMove</code>, which doesn't make sense. Instead, declare an <code>opponentsBestMove</code> 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 <code>bestMove</code> within <code>MinMove</code>, so you should do so inside the <code>if(curRating &lt; v)</code> block.</p> <pre><code>int MaxMove(stateT state, moveT &amp;bestMove) { if(GameIsOver(state)) { return EvaluateStaticPosition(state); } vector&lt;moveT&gt; moveList; GenerateMoveList(state, moveList); int nMoves = moveList.size(); int v = -1000; for(int i = 0 ;i&lt;nMoves; i++) { moveT move = moveList[i]; MakeMove(state, move); moveT opponentsBestMove; int curRating = MinMove(state, opponentsBestMove); if (curRating &gt; v) { v = curRating; bestMove = move; } RetractMove(state, move); } return v; } int MinMove(stateT state, moveT &amp;bestMove) { if(GameIsOver(state)) { return EvaluateStaticPosition(state); } vector&lt;moveT&gt;moveList; GenerateMoveList(state, moveList); int nMoves = moveList.size(); int v = 1000; for(int i = 0 ; i&lt;nMoves; i++) { moveT move = moveList[i]; MakeMove(state , move); moveT opponentsBestMove; int curRating = MaxMove(state,opponentsBestMove); if(curRating &lt; v) { v = curRating; bestMove = move; } RetractMove(state, move); } return v; } </code></pre> <p>At this point you should have an unbeatable AI!</p> <pre><code>The final position looks like this: O | O | X ---+---+--- X | X | O ---+---+--- O | X | X Cat's game. </code></pre> <hr> <p><strong>An alternative method</strong> 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.</p> <p>Change <code>EvaluateStaticPosition</code> back to its original form, so that it gives a score based on how good the board state is for the current player.</p> <pre><code>int EvaluateStaticPosition(stateT state) { if(CheckForWin(state, state.whoseTurn)) { return WINNING_POSITION; } if(CheckForWin(state, Opponent(state.whoseTurn))) { return LOSING_POSITION; } return NEUTRAL_POSITION; } </code></pre> <p>Delete <code>MinMove</code>, since we only care about maximizing. Rewrite <code>MaxMove</code> 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.</p> <pre><code>int MaxMove(stateT state, moveT &amp;bestMove) { if(GameIsOver(state)) { return EvaluateStaticPosition(state); } vector&lt;moveT&gt; moveList; GenerateMoveList(state, moveList); int nMoves = moveList.size(); int v = -1000; for(int i = 0 ;i&lt;nMoves; i++) { moveT move = moveList[i]; MakeMove(state, move); moveT opponentsBestMove; int curRating = -MaxMove(state, opponentsBestMove); if (curRating &gt; v) { v = curRating; bestMove = move; } RetractMove(state, move); } return v; } </code></pre> <p>Since <code>MaxMove</code> is used for both players, we no longer need to distinguish among players in the <code>MiniMax</code> function.</p> <pre><code>moveT MiniMax(stateT state) { moveT bestMove; int i = 0; i = MaxMove(state, bestMove); cout&lt;&lt;"i is "&lt;&lt;i&lt;&lt;endl; return bestMove; } </code></pre>
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    1. POReturn bestMove in minimax algorithm for tictactoe
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    1. COIf I am not mistaken , did you put bestMove = move out in the MinMove intentionally when curRating < v .
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    2. COYes, I did intentionally put `bestMove = move` in the `MinMove` function, inside the `curRating < v` block. That is the logical place for it; if a move's rating is the lowest found so far, then that move is the best move. At least until another even lower rating is found.
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    3. COMan that works , I wish I could give you a hug ; wherever you are ,thanks . Let me test further , wait .
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