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实际问题
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好吧,真正的问题不是Alpha-Beta修剪与最小算法。问题是,在树上中只能提供最佳解决方案时,minimax算法将提供正确的价值,但是多个孩子具有最佳价值,其中一些孩子不应该具有此价值。
我猜最终的问题是,词根节点的最有效方法是什么最有效的方法(在领带的情况下可能是多重的(。
该算法会产生正确的值,但是即使某些动作显然是错误的,多个节点都与该值联系在一起。
示例:ticktacktoe
-|-|O
-|X|-
-|X|-
将产生值为:(0,1(和(1,0(,我的启发式
值为-0.06(0,1(是正确的值,因为它会阻止我的x,但(0,1(是错误的,因为下一步我可以将x放在(0,1(和获胜。
中。当我运行相同的算法时,没有
if(beta<=alpha)
break;
它仅返回带有值-0.06
的(0,1(----------------------
最初发布的问题,现在只是糖
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我已经花了几天的时间试图弄清楚为什么我的最小算法有效,但是当我添加alpha beta pruning时,它行不通。我了解他们应该给出相同的结果,甚至对此进行了快速测试。我的问题是,为什么我的实施不产生相同的结果?
这是在Android中实现的TIC Tak Toe。有时我可以在
时击败算法if(beta<=alpha) break;
没有评论,但是当评论出来时,它是不败的。
private static double minimax(Node<Integer,Integer> parent, int player, final int[][] board, double alpha, double beta, int depth) {
List<Pair<Integer, Integer>> moves = getAvailableMoves(board);
int bs = getBoardScore(board);
if (moves.isEmpty() || Math.abs(bs) == board.length)//leaf node
return bs+(player==X?-1:1)*depth/10.;
double bestVal = player == X ? -Integer.MAX_VALUE : Integer.MAX_VALUE;
for(Pair<Integer, Integer> s : moves){
int[][] b = clone(board);
b[s.getFirst()][s.getSecond()]=player;
Node<Integer, Integer> n = new Node<>(bs,b.hashCode());
parent.getChildren().add(n);
n.setParent(parent);
double score = minimax(n,player==O?X:O,b,alpha,beta, depth+1);
n.getValues().put("score",score);
n.getValues().put("pair",s);
if(player == X) {
bestVal = Math.max(bestVal, score);
alpha = Math.max(alpha,bestVal);
} else {
bestVal = Math.min(bestVal, score);
beta = Math.min(beta,bestVal);
}
/*
If i comment these two lines out it works as expected
if(beta<= alpha)
break;
*/
}
return bestVal;
}
现在,由于小搜索树而言,这对于Tick Tok Toe来说不是问题,但是我随后为Checkers开发了它,并注意到了同样的现象。
private double alphaBeta(BitCheckers checkers, int depth, int absDepth, double alpha, double beta){
if(checkers.movesWithoutAnything >= 40)
return 0;//tie game//needs testing
if(depth == 0 || checkers.getVictoryState() != INVALID)
return checkers.getVictoryState()==INVALID?checkers.getBoardScore()-checkers.getPlayer()*moves/100.:
checkers.getPlayer() == checkers.getVictoryState() ? Double.MAX_VALUE*checkers.getPlayer():
-Double.MAX_VALUE*checkers.getPlayer();
List<Pair<Pair<Integer, Integer>, Pair<Integer, Integer>>> moves;
if(absDepth == maxDepth)
moves = (List<Pair<Pair<Integer, Integer>, Pair<Integer, Integer>>>) node.getValues().get("moves");
else
moves = checkers.getAllPlayerMoves();
if(moves.isEmpty()) //no moves left? then this player loses
return checkers.getPlayer() * -Double.MAX_VALUE;
double v = checkers.getPlayer() == WHITE ? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
for(Pair<Pair<Integer, Integer>, Pair<Integer, Integer>> i : moves){
BitCheckers c = checkers.clone();
c.movePiece(i.getFirst().getFirst(),i.getFirst().getSecond(),i.getSecond().getFirst(),i.getSecond().getSecond());
int newDepth = c.getPlayer() == checkers.getPlayer() ? depth : depth - 1;
if(checkers.getPlayer() == WHITE) {
v = Math.max(v, alphaBeta(c, newDepth, absDepth - 1, alpha, beta));
alpha = Math.max(alpha,v);
}else {
v = Math.min(v, alphaBeta(c, newDepth, absDepth - 1, alpha, beta));
beta = Math.min(beta,v);
}
if(absDepth == maxDepth) {
double finalScore = v;
for(Node n : node.getChildren())
if(n.getData().equals(i)){
n.setValue(finalScore);
break;
}
}
/*
If i comment these two lines out it works as expected
if(beta<= alpha)
break;
*/
}
return v;
}
我用PVS对其进行了测试,并给出与α-beta修剪相同的结果,即不像minimax一样好。
public double pvs(BitCheckers checkers, int depth, int absDepth, double alpha, double beta){
if(checkers.movesWithoutAnything >= 40)
return 0;//tie game//needs testing
if(depth == 0 || checkers.getVictoryState() != INVALID)
return checkers.getVictoryState()==INVALID?checkers.getBoardScore()-checkers.getPlayer()*moves/100.:
checkers.getPlayer() == checkers.getVictoryState() ? Double.MAX_VALUE*checkers.getPlayer():
-Double.MAX_VALUE*checkers.getPlayer();
List<Pair<Pair<Integer, Integer>, Pair<Integer, Integer>>> moves;
if(absDepth == maxDepth)
moves = (List<Pair<Pair<Integer, Integer>, Pair<Integer, Integer>>>) node.getValues().get("moves");
else
moves = checkers.getAllPlayerMoves();
if(moves.isEmpty()) //no moves left? then this player loses
return checkers.getPlayer() * -Double.MAX_VALUE;
int j = 0;
double score;
for(Pair<Pair<Integer, Integer>, Pair<Integer, Integer>> i : moves){
BitCheckers c = checkers.clone();
c.movePiece(i.getFirst().getFirst(),i.getFirst().getSecond(),i.getSecond().getFirst(),i.getSecond().getSecond());
int newDepth = c.getPlayer() == checkers.getPlayer() ? depth : depth - 1;
double sign = c.getPlayer() == checkers.getPlayer()? -1 : 1;
if(j++==0)
score = -pvs(c,newDepth,absDepth-1,sign*-beta,sign*-alpha);
else {
score = -pvs(c,newDepth, absDepth-1,sign*-(alpha+1),sign*-alpha);
if(alpha<score || score<beta)
score = -pvs(c,newDepth,absDepth-1,sign*-beta,sign*-score);
}
if(absDepth == maxDepth) {
double finalScore = score;
for(Node n : node.getChildren())
if(n.getData().equals(i)){
n.setValue(finalScore);
break;
}
}
alpha = Math.max(alpha,score);
if(alpha>=beta)
break;
}
return alpha;
}
没有αβ修剪的检查器很好,但不是很好。我知道使用Alpha-Beta的工作版本可能真的很棒。请帮助修复我的alpha-beta修剪。
我知道它应该给出相同的结果,我的问题是为什么我的实施不给出相同的结果?
要确认它应该给出相同的结果,我进行了快速测试类实现。
public class MinimaxAlphaBetaTest {
public static void main(String[] args) {
Node<Double,Double> parent = new Node<>(0.,0.);
int depth = 10;
createTree(parent,depth);
Timer t = new Timer().start();
double ab = alphabeta(parent,depth+1,Double.NEGATIVE_INFINITY,Double.POSITIVE_INFINITY,true);
t.stop();
System.out.println("Alpha Beta: "+ab+", time: "+t.getTime());
t = new Timer().start();
double mm = minimax(parent,depth+1,true);
t.stop();
System.out.println("Minimax: "+mm+", time: "+t.getTime());
t = new Timer().start();
double pv = pvs(parent,depth+1,Double.NEGATIVE_INFINITY,Double.POSITIVE_INFINITY,1);
t.stop();
System.out.println("PVS: "+pv+", time: "+t.getTime());
if(ab != mm)
System.out.println(ab+"!="+mm);
}
public static void createTree(Node n, int depth){
if(depth == 0) {
n.getChildren().add(new Node<>(0.,(double) randBetween(1, 100)));
return;
}
for (int i = 0; i < randBetween(2,10); i++) {
Node nn = new Node<>(0.,0.);
n.getChildren().add(nn);
createTree(nn,depth-1);
}
}
public static Random r = new Random();
public static int randBetween(int min, int max){
return r.nextInt(max-min+1)+min;
}
public static double pvs(Node<Double,Double> node, int depth, double alpha, double beta, int color){
if(depth == 0 || node.getChildren().isEmpty())
return color*node.getValue();
int i = 0;
double score;
for(Node<Double,Double> child : node.getChildren()){
if(i++==0)
score = -pvs(child,depth-1,-beta,-alpha,-color);
else {
score = -pvs(child,depth-1,-alpha-1,-alpha,-color);
if(alpha<score || score<beta)
score = -pvs(child,depth-1,-beta,-score,-color);
}
alpha = Math.max(alpha,score);
if(alpha>=beta)
break;
}
return alpha;
}
public static double alphabeta(Node<Double,Double> node, int depth, double alpha, double beta, boolean maximizingPlayer){
if(depth == 0 || node.getChildren().isEmpty())
return node.getValue();
double v = maximizingPlayer ? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
for(Node<Double,Double> child : node.getChildren()){
if(maximizingPlayer) {
v = Math.max(v, alphabeta(child, depth - 1, alpha, beta, false));
alpha = Math.max(alpha, v);
}else {
v = Math.min(v,alphabeta(child,depth-1,alpha,beta,true));
beta = Math.min(beta,v);
}
if(beta <= alpha)
break;
}
return v;
}
public static double minimax(Node<Double,Double> node, int depth, boolean maximizingPlayer){
if(depth == 0 || node.getChildren().isEmpty())
return node.getValue();
double v = maximizingPlayer ? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
for(Node<Double,Double> child : node.getChildren()){
if(maximizingPlayer)
v = Math.max(v,minimax(child,depth-1,false));
else
v = Math.min(v,minimax(child,depth-1,true));
}
return v;
}
}
实际上,这确实给出了我期望的α-beta和PV的速度大致相同(PVS较慢,因为儿童是随机顺序的(,并且产生与minimax相同的结果。这证明了算法是正确的,但是无论出于何种原因,我对它们的实施都是错误的。
Alpha Beta: 28.0, time: 25.863126 milli seconds
Minimax: 28.0, time: 512.6119160000001 milli seconds
PVS: 28.0, time: 93.357653 milli seconds
Checkers实施的源代码
pvs
的伪代码alpha beta的伪代码我正在关注
tick钉脚趾实现的完整Souce代码
我认为您可能会误解AB修剪。
ab修剪应该给您与Minmax相同的结果,这只是一种不跌落某些分支机构的方式,因为您知道这一举动会比您检查过的另一个举动要糟,这在您有巨大的树木时会有所帮助。p>另外,Minmax不使用启发式启发式并切断您的搜索始终是不败的,因为您已经计算了到达每个终止状态的所有可能路径。因此,我希望AB修剪和Minmax都无与伦比,所以我认为您的AB修剪有问题。如果您的Minmax不败,那么您的方法也应使用AB修剪。