Java 在Tictoe minimax算法中实现alpha-beta修剪
在我的方法newminimax49中，我有一个minimax算法，该算法利用了本文中向我建议的其他通用改进。该方法使用一个简单的启发式电路板评估函数。我的问题基本上是关于alpha-beta修剪，即我的minimax方法是否使用alpha-beta修剪。据我所知，我相信这是真的，然而，我用来实现它的东西似乎太简单了，不可能是真的。此外，其他人建议我使用alpha-beta剪枝，正如我所说的，我认为我的minimax方法已经做到了，这让我相信我在这里做的是其他事情。这是我的新minimax49：Java 在Tictoe minimax算法中实现alpha-beta修剪,java,algorithm,artificial-intelligence,tic-tac-toe,minimax,Java,Algorithm,Artificial Intelligence,Tic Tac Toe,Minimax,在我的方法newminimax49中，我有一个minimax算法，该算法利用了本文中向我建议的其他通用改进。该方法使用一个简单的启发式电路板评估函数。我的问题基本上是关于alpha-beta修剪，即我的minimax方法是否使用alpha-beta修剪。据我所知，我相信这是真的，然而，我用来实现它的东西似乎太简单了，不可能是真的。此外，其他人建议我使用alpha-beta剪枝，正如我所说的，我认为我的minimax方法已经做到了，这让我相信我在这里做的是其他事情。这是我的新minimax49：
//This method returns a 2 element int array containing the position of the best possible
//next move and the score it yields. Utilizes memoization and supposedly alpha beta
//pruning to achieve better performance. Alpha beta pruning can be seen in lines such as:
/*if(bestScore==-10)
break;*/
//This basically means that if the best score achieved is the best possible score
//achievable then stop exploring the other available moves. Doing thing I believe
//I'm applying the same principle of alpha beta pruning.
public int[] newminimax49(){
int bestScore = (turn == 'O') ? +9 : -9; //X is minimizer, O is maximizer
int bestPos=-1;
int currentScore;
//boardShow();
String stateString = "";
for (int i=0; i<state.length; i++)
stateString += state[i];
int[] oldAnswer = oldAnswers.get(stateString);
if (oldAnswer != null)
return oldAnswer;
if(isGameOver2()!='N'){
//s.boardShow();
bestScore= score();
}
else{
//s.boardShow();
int i=0;
for(int x:getAvailableMoves()){
if(turn=='X'){ //X is minimizer
setX(x);
//boardShow();
//System.out.println(stateID++);
currentScore = newminimax49()[0];
revert(x);
if(i==0){
bestScore = currentScore;
bestPos=x;
if(bestScore==-10)
break;
}
else if(currentScore<bestScore){
bestScore = currentScore;
bestPos=x;
if(bestScore==-10)
break;
}
}
else { //O is maximizer
setO(x);
//boardShow();
//System.out.println(stateID++);
currentScore = newminimax49()[0];
revert(x);
//boardShow();
if(i==0){
bestScore = currentScore;
bestPos=x;
if(bestScore==10)
break;
}
else if(currentScore>bestScore){
bestScore = currentScore;
bestPos = x;
if(bestScore==10)
break;
}
}
i++;
}
}
int[] answer = {bestScore, bestPos};
oldAnswers.put (stateString, answer);
return answer;
}
//Sets an X at a certain location and updates the turn, countX and lastAdded variables
public void setX(int i){
state[i]='X';
DDState[RowCol.get(i)[0]][RowCol.get(i)[1]]='X';
turn='O';
countX++;
lastAdded=i;
}
//Sets an O at a certain location and updates the turn, countO and lastAdded variables
public void setO(int i){
state[i]='O';
DDState[RowCol.get(i)[0]][RowCol.get(i)[1]]='O';
turn='X';
countO++;
lastAdded=i;
}
public void revert(int i){
state[i]='-';
DDState[RowCol.get(i)[0]][RowCol.get(i)[1]]='-';
if(turn=='X'){
turn = 'O';
countO--;
}
else {
turn = 'X';
countX--;
}
}
所以，这看起来像是阿尔法-贝塔删减吗？如果不是的话，我怎么能做到呢？你已经在使用某种“简化”的阿尔法-贝塔：目前，每当一个玩家找到一个获胜的位置时，你都在删减 一个合适的AB会给自己传递一个Alpha值和一个Beta值，以分别确定玩家将达到的最小值和最大值。在那里，每当分数低于或等于对方玩家当前的“最坏情况”时，你就会删减
在您的情况下，您将不仅能够删减获胜分数（如您目前所做的），还可以删减某些分数为0的分数 是的，这似乎是互联网上的通用词。问题是，对于使用简单求值函数的minimax方法，没有太多关于如何实现它的信息。你有没有想过我将如何做一些类似于你在这里为我的newminimax49方法所建议的事情？@Omar只需以更复杂的计算方式实现它：向方法中添加两个整数参数（alpha/beta），以正确的alpha-beta方式更新这些值，你会没事的。要了解更多信息，我仍然建议使用chessprogramming wiki。。。
public char isGameOver2(){
char turnOpp;
int count;
if(turn=='X'){
count=countO;
turnOpp='O';
}
else {
count=countX;
turnOpp='X';
}
if(count>=n){
//^No win available if each player has less than n seeds on the board
//Checking begins
//DDState[RowCol.get(lastAdded)[0]][RowCol.get(lastAdded)[1]]=turn;
//Check column for win
for(int i=0; i<n; i++){
if(DDState[i][RowCol.get(lastAdded)[1]]!=turnOpp)
break;
if(i==(n-1)){
//DDState[RowCol.get(x)[0]][RowCol.get(x)[1]]='-';
return turnOpp;
}
}
//Check row for win
for(int i=0; i<n; i++){
if(DDState[RowCol.get(lastAdded)[0]][i]!=turnOpp)
break;
if(i==(n-1)){
//DDState[RowCol.get(x)[0]][RowCol.get(x)[1]]='-';
return turnOpp;
}
}
//Check diagonal for win
if(RowCol.get(lastAdded)[0] == RowCol.get(lastAdded)[1]){
//we're on a diagonal
for(int i = 0; i < n; i++){
if(DDState[i][i] != turnOpp)
break;
if(i == n-1){
//DDState[RowCol.get(x)[0]][RowCol.get(x)[1]]='-';
return turnOpp;
}
}
}
//check anti diagonal
for(int i = 0; i<n; i++){
if(DDState[i][(n-1)-i] != turnOpp)
break;
if(i == n-1){
//DDState[RowCol.get(x)[0]][RowCol.get(x)[1]]='-';
return turnOpp;
}
}
//check for draw
if((countX+countO)==(n*n))
return 'D';
}
return 'N';
}
public void boardShow(){
if(n==3){
System.out.println(stateID);
for(int i=0; i<=6;i+=3)
System.out.println("["+state[i]+"]"+" ["+state[i+1]+"]"+" ["+state[i+2]+"]");
System.out.println("***********");
}
else {
System.out.println(stateID);
for(int i=0; i<=12;i+=4)
System.out.println("["+state[i]+"]"+" ["+state[i+1]+"]"+" ["+state[i+2]+"]"+" ["+state[i+3]+"]");
System.out.println("***********");
}
}
public int score(){
if(isGameOver2()=='X')
return -10;
else if(isGameOver2()=='O')
return +10;
else
return 0;
}
//Sets an X at a certain location and updates the turn, countX and lastAdded variables
public void setX(int i){
state[i]='X';
DDState[RowCol.get(i)[0]][RowCol.get(i)[1]]='X';
turn='O';
countX++;
lastAdded=i;
}
//Sets an O at a certain location and updates the turn, countO and lastAdded variables
public void setO(int i){
state[i]='O';
DDState[RowCol.get(i)[0]][RowCol.get(i)[1]]='O';
turn='X';
countO++;
lastAdded=i;
}
public void revert(int i){
state[i]='-';
DDState[RowCol.get(i)[0]][RowCol.get(i)[1]]='-';
if(turn=='X'){
turn = 'O';
countO--;
}
else {
turn = 'X';
countX--;
}
}