我正在尝试使用简化的minimax算法制作无与伦比的Tic Tac Toe游戏。代码如下：
private static int findBestMove(String[][] board, boolean comp) {
// comp returns true if the computer is the one looking for the best move
// findBestMove is always called by the program as findBestMove(board, true)
// since the computer is the only one that uses it
// If the board in its current state is a win for the
// player, return -1 to indicate a loss
if (playerWon(board)) return -1;
// If the board in its current state is a win for the
// computer, return 1 to indicate a win
if (compWon(board)) return 1;
// If the board in its current state is a tie
// return 0 to indicate a tie
if (tie(board)) return 0;
// Set the default possible outcome as the opposite of what
// the respective player wants
int bestPossibleOutcome = comp ? -1 : 1;
// Loop through the board looking for empty spaces
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++)
// Once an empty space is found, create a copy of the board
// with that space occupied by the respective player
if (board[i][j].equals(" ")) {
String[][] newBoard = new String[3][3];
for (int a = 0; a < 3; a++) {
System.arraycopy(board[a], 0, newBoard[a], 0, 3);
}
newBoard[i][j] = comp ? "O" : "X";
// Recursively call findBestMove() on this copy
// and see what the outcome is
int outCome = findBestMove(newBoard, !comp);
// If this is the computer's turn, and the outcome
// is higher than the value currently stored as the
// best, replace it
if (comp && outCome > bestPossibleOutcome) {
bestPossibleOutcome = outCome;
// r and c are instance variables that store the row
// and column of what the computer's next move should be
r = i;
c = j;
// If this is the player's turn, and the outcome
// is lower than the value currently stored as the
// best, replace it
} else if (!comp && outCome < bestPossibleOutcome) {
bestPossibleOutcome = outCome;
}
}
}
}
// Return the ultimate value deemed to be the best
return bestPossibleOutcome;
}
这个想法是，在我运行此程序之后，实例变量r和c应该分别包含计算机最佳运行的行和列。但是，该程序仅成功防止了大约一半时间的损失，我无法确定另一半是否很幸运，或者该程序是否确实在工作。
我知道计算机将对每种情况做出的响应与每种游戏完全相同。很好。
如果有人想运行该程序，我将在下面提供完整的课程：
import java.util.Scanner;
public class TicTacToe {
private static int r;
private static int c;
private static void printBoard(String[][] board) {
System.out.println(" 0 1 2");
System.out.println("0 " + board[0][0] + " | " + board[0][1] + " | " + board[0][2] + " ");
System.out.println(" ---+---+---");
System.out.println("1 " + board[1][0] + " | " + board[1][1] + " | " + board[1][2] + " ");
System.out.println(" ---+---+---");
System.out.println("2 " + board[2][0] + " | " + board[2][1] + " | " + board[2][2] + " ");
}
private static boolean playerWon(String[][] board) {
return playerHasThreeInCol(board) || playerHasThreeInDiag(board) || playerHasThreeInRow(board);
}
private static boolean playerHasThreeInRow(String[][] board) {
for (int i = 0; i < 3; i++) {
if (board[i][0].equals(board[i][1]) && board[i][0].equals(board[i][2]) && board[i][0].equals("X")) return true;
}
return false;
}
private static boolean playerHasThreeInCol(String[][] board) {
for (int i = 0; i < 3; i++) {
if (board[0][i].equals(board[1][i]) && board[0][i].equals(board[2][i]) && board[0][i].equals("X")) return true;
}
return false;
}
private static boolean playerHasThreeInDiag(String[][] board) {
if (board[0][0].equals(board[1][1]) && board[0][0].equals(board[2][2]) && board[0][0].equals("X")) return true;
return board[0][2].equals(board[1][1]) && board[0][2].equals(board[2][0]) && board[0][2].equals("X");
}
private static boolean compWon(String[][] board) {
return compHasThreeInCol(board) || compHasThreeInDiag(board) || compHasThreeInRow(board);
}
private static boolean compHasThreeInRow(String[][] board) {
for (int i = 0; i < 3; i++) {
if (board[i][0].equals(board[i][1]) && board[i][0].equals(board[i][2]) && board[i][0].equals("O")) return true;
}
return false;
}
private static boolean compHasThreeInCol(String[][] board) {
for (int i = 0; i < 3; i++) {
if (board[0][i].equals(board[1][i]) && board[0][i].equals(board[2][i]) && board[0][i].equals("O")) return true;
}
return false;
}
private static boolean compHasThreeInDiag(String[][] board) {
if (board[0][0].equals(board[1][1]) && board[0][0].equals(board[2][2]) && board[0][0].equals("O")) return true;
return board[0][2].equals(board[1][1]) && board[0][2].equals(board[2][0]) && board[0][2].equals("O");
}
private static boolean tie(String[][] board) {
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (board[i][j].equals(" ")) return false;
}
}
return true;
}
private static int findBestMove(String[][] board, boolean comp) {
if (playerWon(board)) return -1;
if (compWon(board)) return 1;
if (tie(board)) return 0;
int bestPossibleOutcome = comp ? -1 : 1;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (board[i][j].equals(" ")) {
String[][] newBoard = new String[3][3];
for (int a = 0; a < 3; a++) {
System.arraycopy(board[a], 0, newBoard[a], 0, 3);
}
newBoard[i][j] = comp ? "O" : "X";
int outCome = findBestMove(newBoard, !comp);
if (comp && outCome > bestPossibleOutcome) {
bestPossibleOutcome = outCome;
r = i;
c = j;
} else if (!comp && outCome < bestPossibleOutcome) {
bestPossibleOutcome = outCome;
}
}
}
}
return bestPossibleOutcome;
}
private static void go() {
Scanner input = new Scanner(System.in);
String[][] board = new String[3][3];
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
board[i][j] = " ";
}
}
printBoard(board);
for (int i = 0;; i++) {
if (i % 2 == 0) {
while (true) {
System.out.println("Enter position: ");
String position = input.nextLine();
int row, column;
try {
row = Integer.parseInt(position.substring(0, 1));
column = Integer.parseInt(position.substring(1, 2));
} catch (Exception e) {
System.out.println("Invalid entry. ");
continue;
}
if (row < 0 || row > 2 || column < 0 || column > 2) {
System.out.println("That position is not on the board. ");
continue;
}
if (!board[row][column].equals(" ")) {
System.out.println("That space is already taken. ");
continue;
}
board[row][column] = "X";
break;
}
} else {
System.out.println("\nMy move: ");
findBestMove(board, true);
board[r][c] = "O";
}
printBoard(board);
if (playerWon(board)) {
System.out.println("You win!");
break;
} else if (compWon(board)) {
System.out.println("I win!");
break;
} else if (tie(board)) {
System.out.println("Tie game");
break;
}
}
}
public static void main(String[] args) {
go();
}
}
我并没有要求任何人为我重写整个内容，但是如果您可以指出任何明显的错误或将我指向正确的方向，那将不胜感激。我也欢迎您提出任何建议或意见。
答案 0 :(得分：0)
我还没有对其进行广泛的测试，但是我相信我已经解决了这个问题。新代码如下所示：
private static void findBestMove(String[][] board) {
double bestMove = Double.NEGATIVE_INFINITY;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (board[i][j].equals(" ")) {
board[i][j] = "O";
double score = minimax(board, false);
board[i][j] = " ";
if (score > bestMove) {
bestMove = score;
r = i;
c = j;
}
}
}
}
}
private static double minimax(String[][] board, boolean comp) {
if (playerWon(board)) {
return -1;
}
if (compWon(board)) {
return 1;
}
if (tie(board)) return 0;
double bestScore;
if (comp) {
bestScore = Double.NEGATIVE_INFINITY;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (board[i][j].equals(" ")) {
board[i][j] = "O";
double score = minimax(board, false);
board[i][j] = " ";
bestScore = Math.max(score, bestScore);
}
}
}
} else {
bestScore = Double.POSITIVE_INFINITY;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
if (board[i][j].equals(" ")) {
board[i][j] = "X";
double score = minimax(board, true);
board[i][j] = " ";
bestScore = Math.min(score, bestScore);
}
}
}
}
return bestScore;
}
我从下一个移动坐标设置器中抽象出了minimax算法，我认为这可能有所作为。否则，它非常相似。