我试图以递归方式填充树,但我的代码只填写一个深度长度,然后退出。即每个节点只有一个孩子。有什么东西没有考虑到吗?
public static void populate(Node n, int depth, String player){
System.out.println("Depth: " + depth);
if(player.equalsIgnoreCase("X"))
player = "O";
else
player = "X";
int j = 0;
System.out.println("empty spots: " + ((Board)n.getData()).noOfEmpty());
for(int i=0; i<((Board)n.getData()).noOfEmpty(); i++){
if(((Board)n.getData()).getSquare(j).equalsIgnoreCase("X")
|| ((Board)n.getData()).getSquare(j).equalsIgnoreCase("O"))
j++;
else{
Board tmp = new Board(((Board)n.getData()), j, player);
Node newNode = new Node(tmp);
tree.insert(n, newNode);
populate(newNode, depth-1, player);
}
}
}
P.S。我检查noOfEmpty()
返回值,该值应确定节点应具有的子节点数。
编辑:@eznme按要求填写完整的代码:
public class MinMax {
protected static Tree tree;
public static void createTree(Board b){
tree = new Tree();
tree.setRoot(new Node(b));
populate(tree.getRoot(), 5, "X");
//System.out.println("printing tree");
//tree.print(1);
}
public static void populate(Node n, int depth, String player){
System.out.println("Depth: " + depth);
if(player.equalsIgnoreCase("X"))
player = "O";
else
player = "X";
int j = 0;
System.out.println("empty spots: " + ((Board)n.getData()).noOfEmpty());
for(int i=0; i<((Board)n.getData()).noOfEmpty(); i++){
if(((Board)n.getData()).getSquare(j).equalsIgnoreCase("X")
|| ((Board)n.getData()).getSquare(j).equalsIgnoreCase("O"))
j++;
else{
Board tmp = new Board(((Board)n.getData()), j, player);
Node newNode = new Node(tmp);
tree.insert(n, newNode);
populate(newNode, depth-1, player);
}
}
}
}
import java.util.ArrayList;
/**
*
* @author Greg
*/
public class Node {
protected Object data;
protected int score; //fields to be used by the MaxMin class
protected ArrayList<Node> children;
//constructors
public Node(){
children = new ArrayList(0);
data = null;
}
public Node(Object obj){
children = new ArrayList(0);
data = obj;
}
public void setChild(Node n){
//EFFECT: set the ith child to node t
children.add(n);
}
public void setChildren(Node[] t){
//EFFECT: copy the array t, into the array children, effectively
// setting all the chidern of this node simultaneouly
int l = children.size();
for(int i=0; i<t.length; i++){
children.add(l, t[i]);
}
}
public void setData(Object obj){
//EFFECT: set the date of this node to obj, and also set the number of
// children this node has
data = obj;
}
public Node getChild(int i){
//EFFECT: returns the child at index i
return children.get(i);
}
public int noOfChildren(){
//EFFECT: return the length of this node
return children.size();
}
public Object getData(){
//EFFECT: returns the data of this node
return data;
}
@Override
public String toString(){
//EFFECT: returns the string form of this node
return "" + data.toString() + "\nwith " + noOfChildren()+ "\n";
}
public boolean isLeaf(){
if(children.size()==0)
return true;
return false;
}
public void setScore(int scr){
score = scr;
}
public int getScore(){
return score;
}
}
public class Tree {
private Node root;
public Tree(){
setRoot(null);
}
public Tree(Node n){
setRoot(n);
}
public Tree(Object obj){
setRoot(new Node(obj));
}
protected Node getRoot(){
return root;
}
protected void setRoot(Node n){
root = n;
}
public boolean isEmpty(){
return getRoot() == null;
}
public Object getData(){
if(!isEmpty())
return getRoot().getData();
return null;
}
public Object getChild(int i){
return root.getChild(i);
}
public void setData(Object obj){
if(!isEmpty())
getRoot().setData(obj);
}
public void insert(Node p,Node c){
if(p != null)
p.setChild(c);
}
public void print(int mode){
if(mode == 1) pretrav();
else if(mode == 2) postrav();
else
System.out.println("yeah... mode 1 or 2...nothing else, try agn");
}
public void pretrav(){
pretrav(getRoot());
}
protected void pretrav(Node t){
if(t == null)
return;
System.out.println(t.getData()+" \n");
for(int i=0; i<t.noOfChildren(); i++)
pretrav(t.getChild(i));
}
public void postrav(){
postrav(getRoot());
}
protected void postrav(Node t){
if(t == null)
return;
System.out.print(t.getData()+" ");
for(int i=0; i<t.noOfChildren(); i++)
pretrav(t.getChild(i));
System.out.print(t.getData()+" ");
}
}
public class Board {
boolean isFull = false; // a check to see if the board is full
String[] grid = new String[9]; //an array represting the 9 square on a board
int hV;
String MIN, MAX;
public Board(){
for(int i=0; i<grid.length;i++)
grid[i] = Integer.toString(i);
hV = heuristicValue(this);
}
public Board(Board b, int x, String player){
this.grid = b.getBoard();
if(!(grid[x].equalsIgnoreCase("X")|| grid[x].equalsIgnoreCase("X")))
grid[x] = player;
}
public boolean setSquare(String player, int position){
/*
EFFECT:set a square on the board to either a X or a O, debending on the player
PRECON: square (x,y) is empty
POATCON: square (x,y) has player 'symbol'
*/
boolean isValidPlay = false;
try{
//as a sanity
Integer.parseInt(grid[position]);
grid[position] = player;
isValidPlay = true;
}catch(NumberFormatException e){
System.out.println("positon " + position + "is already occupied");
}
return isValidPlay;
}
public boolean endGame(){
/*
* EFFECT: check to see if the game have been won or drawn
*/
if(ticTacToe(0,1,2)){
//System.out.println("Player " + grid[0] + " wins");
return true;
}
else if(ticTacToe(3,4,5)){
//System.out.println("Player " + grid[3] + " wins");
return true;
}
else if(ticTacToe(6,7,8)){
//System.out.println("Player " + grid[6] + " wins");
return true;
}
else if(ticTacToe(0,4,8)){
//System.out.println("Player " + grid[0]+ " wins");
return true;
}
else if(ticTacToe(0,3,6)){
//System.out.println("Player " + grid[0]+ " wins");
return true;
}
else if(ticTacToe(1,4,7)){
//System.out.println("Player " + grid[1] + " wins");
return true;
}
else if(ticTacToe(2,5,8)){
//System.out.println("Player " + grid[2] + " wins");
return true;
}else if(ticTacToe(2,4,6)){
//System.out.println("Player " + grid[2] + " wins");
return true;
}
else
return isDrawn();
}
public boolean ticTacToe(int x, int y, int z){
/*
* check is x, y and z has the same value
*/
try{
Integer.parseInt(grid[x]);
return false;
}catch(NumberFormatException e){
if( grid[x].equals(grid[y])
&& grid[x].equals(grid[z]))
return true;
else
return false;
}
}
public String getSquare(int i){
return grid[i];
}
@Override
public String toString(){
String msg = "";
for(int i=0; i<grid.length; i++){
msg = msg + grid[i] + " ";
if(i==2 || i==5)
msg = msg+ "\n";
}
return msg;
}
public boolean isDrawn(){
/*
* check to see if there are any 'free' spaces on the board, if there are any
* return false, else return true
*/
for(int i=0; i<grid.length; i++){
try{
Integer.parseInt(grid[i]);
return false;
}catch(NumberFormatException e){
}
}
System.out.println("Game drawn");
return true;
}
public String[] getBoard(){
return grid;
}
public int noOfEmpty(){
//EFFECT: returns the number of empty squares
int count = 0;
for(int i=0; i<grid.length; i++)
if (!(grid[i].equalsIgnoreCase("X") || grid[i].equalsIgnoreCase("O")))
count++;
return count;
}
public int heuristicValue(Board b){
String MAX = "X", MIN = "O";
/*
* calculate a value that will be used as a heuristic function
* the function works for ever X in a row WITHOUT O: 1 point,
* for two X in a row WITHOUT a O: 5 points
* and 3 X in a row: 100 points
*/
//System.out.println("Computing heuristic");
//System.out.println("Computing horizontals");
int hCount = 0;
//sum up the horizontals
for(int i=0; i<9; i=i+3){
int tmpMAX = playerCount(b, MAX,i,i+1,i+2);
int tmpMIN = playerCount(b, MIN,i,i+1,i+2);
//System.out.println(tmpMAX);
//System.out.println(tmpMIN);
if(tmpMIN > 0){
//System.out.println("Min was zero");
}
else if(tmpMAX==1){
//System.out.println("has one");
hCount = hCount + 1;
}
else if(tmpMAX==2){
//System.out.println("was tw0");
hCount = hCount + 5;
}
else if(tmpMAX==3){
//System.out.println("full 100");
hCount = hCount + 100;
}
}
//System.out.println("Computing verticals");
//sum up the verticals
for(int i=0; i<3; i++){
int tmpMAX = playerCount(b, MAX,i,i+3,i+6);
int tmpMIN = playerCount(b, MIN,i,i+3,i+6);
if(tmpMIN > 0){}
else if(tmpMAX==1){
hCount = hCount +1;
}
else if(tmpMAX==2){
hCount = hCount + 5;
}
else if(tmpMAX==3){
hCount = hCount + 100;
}
}
//System.out.println("Computing diagonals");
//sum up diagonals
if(playerCount(b, MIN,0,4,8)==0){
if(playerCount(b, MAX,0,4,8)==1){
hCount = hCount + 1;
}
if(playerCount(b, MAX,0,4,8)==2)
hCount = hCount + 5;
if(playerCount(b, MAX,0,4,8)==3)
hCount = hCount + 100;
}
if(playerCount(b, MIN,2,4,6)==0){
if(playerCount(b, MAX,2,4,6)==1){
hCount = hCount + 1;
}
if(playerCount(b, MAX,2,4,6)==2)
hCount = hCount + 5;
if(playerCount(b, MAX,2,4,6)==3)
hCount = hCount + 100;
}
//System.out.println("Computing completed");
int hV = hCount;
return hV;
}
int playerCount(Board b, String player, int x, int y, int z){
int count = 0;
if(b.getSquare(x).equals(player))
count = count + 1;
if(b.getSquare(y).equals(player))
count = count + 1;
if(b.getSquare(z).equals(player))
count = count + 1;
//System.out.println("playerCount; " + count);
return count;
}
}
import java.io.*;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException{
BufferedReader reader = new BufferedReader(new
InputStreamReader(System.in));
Board thisGame = new Board();
System.out.println("Start \n" + thisGame.toString());
MinMax.createTree(thisGame);
System.exit(0);
}
}
答案 0 :(得分:3)
为了递归地构建一个n-ary树,我会这样做:
public static void populate(Node n, int height){
if(height = 0){
n = new Node();
}else{
n = new Node();
for(int i = 0; i < n.nbChilds(); i++){
populate(n.getChildAt(i), height - 1);
}
}
}
我希望它有所帮助。
使用此算法创建节点的顺序(在二叉树上):
1
2 9
3 6 10 13
4 5 7 8 11 12 14 15
答案 1 :(得分:2)
所以这就是我在你的情况下会做的事情(minimax tic-tac-toe):
术语:
你必须继续尝试所有情况,直到:电路板已满或一名玩家获胜。因此,树的高度为numberOfCells + 1
。
如果我们简化问题并且不担心对称重复:
每个节点都有numberOfcells - nodeDepth
个孩子。
public static void main(String[] args){
Tree t = new Tree();
int nbCells = 9;
t.setRoot(buildTree(new Board(nbCells), 0, -1));
}
public static Node buildTree(Board b, int player, int positionToPlay){
if(player != 0){
b.setCellAt(positionToPlay, player);
}
Node n = new Node(b, b.nbEmptyCells());
int j = 0;
for(int i = 0; i < b.nbCells(); i++){
if(b.getCellAt(i) == 0)
n.setChildAt(j++, buildTree(new Board(b), changePlayer(player), i));
}
return n;
}
public static int changePlayer(int p){
switch(p){
case 0:
return 1;
case 1:
return 2;
case 2:
return 1;
default:
return 0;
}
}
节点类:
public class Node {
private Board board;
private Node[] childs;
public Node(Board b, int nbChilds){
this.board = new Board(b);
this.childs = new Node[nbChilds];
}
public Node getChildAt(int i){
return childs[i];
}
public int nbChilds(){
return childs.length;
}
public void setChildAt(int i, Node n){
this.childs[i] = n;
}
public Board getBoard(){
return this.board;
}
答案 2 :(得分:1)
我认为你的做法是错误的。
首先,你正在做一个循环和递归,除了使用一个没有意义的depth
变量,因为你从来没有检查它的值来结束递归,或者知道一些关于你想做什么。
在循环中使用动态函数本身并不是很好,因为迭代应该从循环的开始很好地定义。
i
在您的背景下无用。
因此,如果我理解你的代码,那么有问题的情况是有3个空方块和4个非空方块,因为你会将i
从0迭代到3,除了递增{{1}之外什么都不做从0到3然后退出,因为j
将达到3。
当然我可能在某些方面有误,因为我不知道i
是什么,它来自哪里?它与n有关吗?什么是董事会。
我希望我的贡献可以帮到你,我鼓励你发布更多细节来澄清漏洞并让我能够帮助你。