javascript minimax算法tic tac toe,并不总是给出最佳的移动
我一直在尝试在我的井字游戏中使用minimax算法,以使我的AI无敌。然而,它并不总是返回最佳移动javascript minimax算法tic tac toe,并不总是给出最佳的移动,javascript,algorithm,tic-tac-toe,minimax,Javascript,Algorithm,Tic Tac Toe,Minimax,我一直在尝试在我的井字游戏中使用minimax算法,以使我的AI无敌。然而,它并不总是返回最佳移动 AI = X; human = "O" var board = ["x", "x", "-", "-", "o", "o", "-", "-", "-"] 它给出了指数[2]以上的板作为最佳移动,这是正确的 然而,在下面的棋盘上,它给出的答案是索引[3],这将允许人类玩家在其回合中获胜 var boardB = ["x","x","
AI = X; human = "O"
var board = ["x", "x", "-",
"-", "o", "o",
"-", "-", "-"]
它给出了指数[2]以上的板作为最佳移动,这是正确的
然而,在下面的棋盘上,它给出的答案是索引[3],这将允许人类玩家在其回合中获胜
var boardB = ["x","x","o",
"-","o","-",
"-","_","-"];
var player = 'x';
var opponent = 'o';
function isMovesLeft(board){
for (var i = 0; i<board.length; i++){
if (board[i] =='-'){
return 'true';
}
else{
return 'false';
}
}
}
function evaluate(){
for (var i = 0; i < board.length; i += 3) {
if (board[i] === board[i + 1] && board[i + 1] === board[i + 2]) {
if (board[i] == player){
return +10;
}
else if(board[i]== opponent){
return -10;
}
}
}
for (var j = 0; j < board.length; j++) {
if (board[j] === board[j + 3] && board[j + 3] === board[j + 6]) {
if (board[j] == player){
return +10;
}
else if(board[j] == opponent){
return -10;
}
}
}
if ((board[4]==board[0] && board[4]==board[8]) || (board[4]==board[2] && board[4]==board[6])) {
if (board[4]==player){
return +10;
}
else if (board[4]==opponent){
return -10;
}
}
return 0;
}
function minimax(board, depth, isMax){
var score = evaluate(board);
if (score == 10){
return score;
}
if (score == -10){
return score;
}
if (isMovesLeft(board) =="false"){
return 0;
}
if (isMax == "true"){
var best = -1000;
for (var i = 0; i< board.length; i++){
if (board[i]=='-'){
board.splice(i, 1, player);
var value = minimax(board, depth+1, "false");
best = Math.max(best, value);
board.splice(i, 1, "-");
}
}
return best;
}
else if (isMax == 'false'){
var best = 1000;
for (var i = 0; i<board.length; i++){
if (board[i]=='-'){
board.splice(i, 1, opponent);
var value = minimax(board, depth+1, "true");
best = Math.min(best, value);
board.splice(i, 1, "-");
}
}
return best;
}
}
function findBestMove(board){
var bestVal = -1000;
var bestMove= -1;
for (var i = 0; i<board.length; i++){
if (board[i]=='-'){
board.splice(i, 1, player);
var moveVal = minimax(board, 0, "false");
board.splice(i, 1, "-");
if (moveVal > bestVal)
{
bestMove = i;
bestVal = moveVal;
}
}
}
alert("bestVal is : " + bestVal + "<br> best Move is : " + bestMove;)
}
var boardB=[“x”、“x”、“o”,
“-”、“o”、“-”,
"-","_","-"];
var player='x';
var对手='o';
功能isMovesLeft(板){
对于(var i=0;i如果您看到每个i
的moveVal
,您将看到它是零。
这是因为在minimax函数中调用isMovesLeft
,当false时返回0。
程序的问题是isMovesLeft
总是返回false
您应该将其更改为:
function isMovesLeft(board){
for (var i = 0; i<board.length; i++){
if (board[i] =='-'){
return 'true';
}
}
return false;
}
功能isMovesLeft(板){
对于(var i=0;iThanks@Mahesh!它对上面的棋盘有效。但是,如果我再次将棋盘更改为不同的顺序,AI将阻止人类玩家的移动以阻止人类玩家获胜。下面的棋盘答案是索引[3]。var board=[“x”、“o”、“x”,“-”,“o”,“-”,“-”,“”,“-”];
我认为minimax函数只给出了数字最少的索引,在这种情况下[3],当它遇到一个0的最佳分数时。@Marci Banez我试过那块板,它可能对你不起作用,因为在索引7中,你用了“u”(下划线)而不是“-”(连字符)。如果你将isMovesLeft
函数更改为我在回答中提到的,那么你应该得到bestMove
as 7,它有效地阻止了0
获胜。你是对的!!非常感谢!我在这段代码中错过了很多东西。现在一切都很好。但是,你之前提到过,我可以返回深度代码>和分数一起。我想知道我如何实现这一点,我也不太确定它将如何提高AI的效率。