井字游戏中的Minimax算法 [英] The Minimax algorithm in Tic Tac Toe

问题描述

嘿朋友，

我试图在我的Tic Tac Toe程序中实现Minimax算法.

不幸的是，我在德国Wikipedia网站上使用了有关该定律的示例，但它不起作用:(

最终您可以给我小费，因为我的大脑没主意了.这是我的代码中的方法...等等实际上是不言自明的:

私有TicTacToeMove SavedMove = null;

        公共列表< TicTacToeMove> GetPossibleMoves(TicTacToe电路板，TicTacToePlayer电流)
        {
            列表< TicTacToeMove> moves = new List< TicTacToeMove>();

            对于(int x = 0; x< board.Width; x ++)
            {
                为(int y = 0; y< board.Height; y ++)
                {
                    如果(board.Get(x，y)== null)
                    {
                        move.Add(new TicTacToeMove {X = x，Y = y，Sign = current.Sign});
                    }
                }
            }

            退回动作；
        }

        私有TicTacToePlayer GetOpponent(当前TicTacToePlayer)
        {
            如果(当前== _engine.PlayerOne)
            {
                return _engine.PlayerTwo;
            }

            返回_engine.PlayerOne;
        }

        private int评估(TicTacToeBoard板，TicTacToePlayer当前，TicTacToePlayer对手)
        {
            如果(board.HasWinner()== current.Sign)
            {
                返回+15;
            }

            如果(board.HasWinner()==对手.签名)
            {
                返回-20;
            }
            
            返回0;
        }

        

        public int Max(TicTacToeBoard板，TicTacToePlayer当前电流，int深度)
        {
            TicTacToeBoard clonedBoard = board.Clone();
            TicTacToePlayer对手= GetOpponent(当前);

            如果(深度== 0 || clonedBoard.IsFull()||！string.IsNullOrEmpty(clonedBoard.HasWinner()))
            {
                返回Evaluate(clonedBoard，当前，对手)；
            }


            int maxWert = 5;
            列表< TicTacToeMove> moves = GetPossibleMoves(board，current);
            一会儿(moves.Count> 0)
            {
                TicTacToeMove move = moves [0];
                clonedBoard.Set(move.X，move.Y，move.Sign);
                int wert = Min(clonedBoard，对手，深度-1);
                moves.RemoveAt(0);

                如果(wert> maxWert)
                {
                    maxWert =湿润；
                    SavedMove =移动；

                    如果(maxWert == 15)
                        move.Clear();
                }
            }

            返回maxWert;
        }

        

        公共int最小值(TicTacToeBoard板，TicTacToePlayer电流，int深度)
        {
            TicTacToeBoard clonedBoard = board.Clone();
            TicTacToePlayer对手= GetOpponent(当前);

            如果(深度== 0 || clonedBoard.IsFull()||！string.IsNullOrEmpty(clonedBoard.HasWinner()))
            {
                返回Evaluate(clonedBoard，当前，对手)；
            }


            int minWert = int.MaxValue;

            列表< TicTacToeMove> moves = GetPossibleMoves(clonedBoard，当前)；
            一会儿(moves.Count> 0)
            {
                TicTacToeMove move = moves [0];
                clonedBoard.Set(move.X，move.Y，move.Sign);
                int wert = Max(clonedBoard，对手，深度-1);
                moves.RemoveAt(0);

                如果(wert< minWert)
                {
                    minWert =湿润；
                    //SavedMove = move;
                }
            }

            返回minWert;
        }

        公用TicTacToeMove CalculateMoveHard(TicTacToe电路板，TicTacToePlayer当前)
        {
            int分数=最大值(板，当前，5)；

            如果(SavedMove == null)
                返回SavedMove; // (画?！)
            别的
                返回SavedMove;
        } 

如果有人对我有一个主意< 3

解决方案

/SL设计器，Visual Studio指导自动化工具包，开发人员文档和帮助系统以及Visual Studio编辑器. >

此致

private TicTacToeMove SavedMove = null; public List<TicTacToeMove> GetPossibleMoves(TicTacToeBoard board, TicTacToePlayer current) { List<TicTacToeMove> moves = new List<TicTacToeMove>(); for (int x = 0; x < board.Width; x++) { for (int y = 0; y < board.Height; y++) { if (board.Get(x, y) == null) { moves.Add(new TicTacToeMove { X = x, Y = y, Sign = current.Sign }); } } } return moves; } private TicTacToePlayer GetOpponent(TicTacToePlayer current) { if (current == _engine.PlayerOne) { return _engine.PlayerTwo; } return _engine.PlayerOne; } private int Evaluate(TicTacToeBoard board, TicTacToePlayer current, TicTacToePlayer opponent) { if (board.HasWinner() == current.Sign) { return +15; } if (board.HasWinner() == opponent.Sign) { return -20; } return 0; } public int Max(TicTacToeBoard board, TicTacToePlayer current, int depth) { TicTacToeBoard clonedBoard = board.Clone(); TicTacToePlayer opponent = GetOpponent(current); if (depth == 0 || clonedBoard.IsFull() || !string.IsNullOrEmpty(clonedBoard.HasWinner())) { return Evaluate(clonedBoard, current, opponent); } int maxWert = 5; List<TicTacToeMove> moves = GetPossibleMoves(board, current); while (moves.Count > 0) { TicTacToeMove move = moves[0]; clonedBoard.Set(move.X, move.Y, move.Sign); int wert = Min(clonedBoard, opponent, depth - 1); moves.RemoveAt(0); if (wert > maxWert) { maxWert = wert; SavedMove = move; if (maxWert == 15) moves.Clear(); } } return maxWert; } public int Min(TicTacToeBoard board, TicTacToePlayer current, int depth) { TicTacToeBoard clonedBoard = board.Clone(); TicTacToePlayer opponent = GetOpponent(current); if (depth == 0 || clonedBoard.IsFull() || !string.IsNullOrEmpty(clonedBoard.HasWinner())) { return Evaluate(clonedBoard, current, opponent); } int minWert = int.MaxValue; List<TicTacToeMove> moves = GetPossibleMoves(clonedBoard, current); while (moves.Count > 0) { TicTacToeMove move = moves[0]; clonedBoard.Set(move.X, move.Y, move.Sign); int wert = Max(clonedBoard, opponent, depth - 1); moves.RemoveAt(0); if (wert < minWert) { minWert = wert; //SavedMove = move; } } return minWert; } public TicTacToeMove CalculateMoveHard(TicTacToeBoard board, TicTacToePlayer current) { int score = Max(board, current, 5); if (SavedMove == null) return SavedMove; // (Draw?!) else return SavedMove; }

It would be wonderful if anyone have an idea for me <3

解决方案

Hi LucasProgrammer,

This forum is discussing Visual Studio WPF/SL Designer, Visual Studio Guidance Automation Toolkit, Developer Documentation and Help System, and Visual Studio Editor.

Your issue is related to C# develop, I will move this thread to corresponding forum for a professional answer.

Sincerely,

Oscar

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