Java i'；我坚持使用alpha-beta剪枝算法实现
我正在尝试为9男子莫里斯游戏实施游戏AI。 到目前为止，我的董事会如下所示：Java i'；我坚持使用alpha-beta剪枝算法实现,java,algorithm,alpha-beta-pruning,Java,Algorithm,Alpha Beta Pruning,我正在尝试为9男子莫里斯游戏实施游戏AI。 到目前为止，我的董事会如下所示： public class board { public node []gNode = null; ... // so the table has 24 nodes, for 9 men morris game: gNode = new node[24]; ... int evalua
public class board
{
public node []gNode = null;
... // so the table has 24 nodes, for 9 men morris game:
gNode = new node[24];
...
int evaluateBoard(); // evaluates the current board (tokens)
}
public class node
{
node() // constructor
{ ... }
// setting current node's neighbours (maximum 4 neighbours)
void setNeighbours(int left, int right, int top, int bottom)
{ ... }
short gOccupiedByTeam = renderer.TEAM_NOTEAM; // info if this node is occupied by a token (and a wich team this token belongs to)
short []gNeighbourId = null; // info about this node neighbours (can be max. 4 in a 9Men's morris game)
short gInternalID = -1; // board's IDs (from 0..23)
short gTokenID = -1; // this node can be occupied by a token. (from 0 .. 8) -see below the token class.
short gNodeScore = -1; // a dummy node score.
vector3 gLocation = null; // 3d coordinates for this node.
}
public class token
{
token(vector3 startpos, short nodeId) // Constructor.
{ ... }
public physx gPhysX = null; // 3d coordinates , velocity , accel. for this Token.
public boolean bIsAlive = false; // is this token alive ? (or eliminated?)
public boolean bFormsMill = false; // does it form a Mill?
public short gNodeID = -1; // "link" this token with a gNodeID (when placing a token on current board). See above the node class. This represents a link ID to that node.
public short gTokenMill1 = -1; // used when this token forms a mill (with gTokenMill1 token!)
public short gTokenMill2 = -1; // same.
}
public int getBestMove(board board, int depth, int alpha, int beta, boolean bIsPlayer)
{
// if depth reached, return current's board's Evaluation (a score).
if (depth == 0) return board.evaluateBoard(bIsPlayer);
// is it Player's turn ? (max?)
if (bIsPlayer)
{
// QUESTIONS:
// retrevie all possible "boards" below ! (all new possible token moves)
// 1. here i should generate a new board with 1st possible move (for player token1) ?? ... then a second new board with 2nd possible move still for token1 ? .. and so on until no possible moves for token1?
// (remembering that a token can move in 4 available spots - wich are a neighbour?)
//
// 2. the problem is that if i generate 4 new boards as per token 1 above let's say, then it will "eat" lot of memory for all 18 tokens and a function recursion depth of 5 for example, right ?
// 3. how do i fix point 2?
ArrayList<board> possible_boards = board.getAllPossibleBoards();
// 4. ok, some possible boards were generated, loop thru them starting with the first one and calling recursively this function, is it right ?
for(board iterator: possible_boards)
{
alpha = Math.max(alpha, getBestMove(iterator, depth - 1, alpha, beta, !bIsPlayer));
if (beta < alpha)
{
break;
}
}
// 5. how do i return best move to main calling function ? (wich token is it best move from all of these board's moves ?
return alpha;
}
else
{
ArrayList<board> possible_boards = board.getAllPossibleBoards();
for(board iterator: possible_boards)
{
beta = Math.min(beta, getBestMove(iterator, depth - 1, alpha, beta, !bIsPlayer));
if (beta < alpha)
{
break;
}
}
return beta;
}
}
再次非常感谢saeedn 一般来说，您的代码是可以的，但是您应该记住一些要点 首先，您应该检查节点（这里是一个棋盘）是否为最终节点（有人赢得了游戏），然后检查深度是否等于零。如果有人在该状态下获胜，您可能希望分别为exmaple MAXINT和MININT返回一个大值（用于赢得max player）和一个小值（用于赢得min player） 为避免高内存消耗，请勿生成所有可能的电路板。生成一块电路板，对其进行递归调用，然后生成另一块电路板并搜索该电路板，依此类推。这样，您只需在每个堆栈帧中为一个状态使用内存。这在具有高分支因子的搜索中至关重要 最后，您应该记录max player的棋盘更新分数（更新alpha） 有关更多说明，请参见我的伪代码：
if ( board.checkWinner(bIsPlayer) ) return board.evaluateBoard(bIsPlayer);
// if depth reached, return current's board's Evaluation (a score).
if (depth == 0) return board.evaluateBoard(bIsPlayer);
board chosenBoard;
if (bIsPlayer)
{
// You should implement this method, or write your board generation code here
// returns false if no more boards could be generated
board nextBoard;
while( board.generateNextPossibleBoard(nextBoard) )
{
int v = getBestMove(iterator, depth - 1, alpha, beta, !bIsPlayer));
if ( v > alpha )
{
alpha = v;
chosenBoard = nextBoard; // return this chosenBoard by reference ;)
}
if (beta < alpha)
{
break;
}
}
return alpha;
}
else
{
// The same for beta except you don't need to update chosenBoard :)
}
if（board.checkWinner（bIsPlayer））返回board.evaluateBoard（bIsPlayer）；
//如果达到深度，返回当前板的评估（分数）。
如果（深度==0）返回板。评估板（bIsPlayer）；
董事会选择董事会；
如果（双星玩家）
{
//您应该实现此方法，或者在此处编写电路板生成代码
//如果无法生成更多板，则返回false
下一块板；
while（板生成下一个板）
{
intv=getBestMove（迭代器，深度-1，alpha，beta，！bIsPlayer））；
如果（v>alpha）
{
α=v；
chosenBoard=nextBoard；//通过引用返回此chosenBoard；）
}
如果（β<α）
{
打破
}
}
返回α；
}
其他的
{
//测试版也一样，只是不需要更新chosenBoard:）
}
evaluateBoardboard.evaluateBoard（bIsPlayer）evaluateBoardif ( board.checkWinner(bIsPlayer) ) return board.evaluateBoard(bIsPlayer);
// if depth reached, return current's board's Evaluation (a score).
if (depth == 0) return board.evaluateBoard(bIsPlayer);
board chosenBoard;
if (bIsPlayer)
{
// You should implement this method, or write your board generation code here
// returns false if no more boards could be generated
board nextBoard;
while( board.generateNextPossibleBoard(nextBoard) )
{
int v = getBestMove(iterator, depth - 1, alpha, beta, !bIsPlayer));
if ( v > alpha )
{
alpha = v;
chosenBoard = nextBoard; // return this chosenBoard by reference ;)
}
if (beta < alpha)
{
break;
}
}
return alpha;
}
else
{
// The same for beta except you don't need to update chosenBoard :)
}