java - 我试图解决 '15 puzzle' ，但我得到 'OutOfMemoryError'

Question

很难说出这里问的是什么。这个问题是含糊的、模糊的、不完整的、过于宽泛的或修辞性的，无法以目前的形式得到合理的回答。如需帮助澄清此问题以便重新打开它， visit the help center 。

已关闭 13 年前。

有没有一种方法可以优化此代码以免耗尽内存？

import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.Stack;

public class TilePuzzle {

    private final static byte ROWS = 4;
    private final static byte COLUMNS = 4;
    private static String SOLUTION = "123456789ABCDEF0";
    private static byte RADIX = 16;

    private char[][] board = new char[ROWS][COLUMNS];
    private byte x; // Row of the space ('0')
    private byte y; // Column of the space ('0') private String representation;
    private boolean change = false; // Has the board changed after the last call to toString?

    private TilePuzzle() {
        this(SOLUTION);
        int times = 1000;
        Random rnd = new Random();
        while(times-- > 0) {
            try {
                move((byte)rnd.nextInt(4));
            }
            catch(RuntimeException e) {
            }
        }
        this.representation = asString();
    }

    public TilePuzzle(String representation) {
        this.representation = representation;
        final byte SIZE = (byte)SOLUTION.length();
        if (representation.length() != SIZE) {
            throw new IllegalArgumentException("The board must have " + SIZE + "numbers.");
        }

        boolean[] used = new boolean[SIZE];
        byte idx = 0;
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                char digit = representation.charAt(idx++);
                byte number = (byte)Character.digit(digit, RADIX);
                if (number < 0 || number >= SIZE) {
                    throw new IllegalArgumentException("The character " + digit + " is not valid.");
                } else if(used[number]) {
                    throw new IllegalArgumentException("The character " + digit + " is repeated.");
                }
                used[number] = true;
                board[i][j] = digit;
                if (digit == '0') {
                    x = i;
                    y = j;
                }
            }
        }
    }

    /**
     * Swap position of the space ('0') with the number that's up to it.
     */
    public void moveUp() {
        try {
            move((byte)(x - 1), y);
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    /**
     * Swap position of the space ('0') with the number that's down to it.
     */
    public void moveDown() {
        try {
            move((byte)(x + 1), y);
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    /**
     * Swap position of the space ('0') with the number that's left to it.
     */
    public void moveLeft() {
        try {
            move(x, (byte)(y - 1));
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    /**
     * Swap position of the space ('0') with the number that's right to it.
     */
    public void moveRight() {
        try {
            move(x, (byte)(y + 1));
        } catch(IllegalArgumentException e) {
            throw new RuntimeException("Move prohibited " + e.getMessage());
        }
    }

    private void move(byte movement) {
        switch(movement) {
        case 0: moveUp(); break;
        case 1: moveRight(); break;
        case 2: moveDown(); break;
        case 3: moveLeft(); break;
        }
    }

    private boolean areValidCoordinates(byte x, byte y) {
        return (x >= 0 && x < ROWS && y >= 0 && y < COLUMNS);
    }

    private void move(byte nx, byte ny) {
        if (!areValidCoordinates(nx, ny)) {
            throw new IllegalArgumentException("(" + nx + ", " + ny + ")");
        }
        board[x][y] = board[nx][ny];
        board[nx][ny] = '0';
        x = nx;
        y = ny;
        change = true;
    }

    public String printableString() {
        StringBuilder sb = new StringBuilder();
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                sb.append(board[i][j] + " ");
            }
            sb.append("\r\n");
        }
        return sb.toString();
    }

    private String asString() {
        StringBuilder sb = new StringBuilder();
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                sb.append(board[i][j]);
            }
        }
        return sb.toString();
    }

    public String toString() {
        if (change) {
            representation = asString();
        }
        return representation;
    }

    private static byte[] whereShouldItBe(char digit) {
        byte idx = (byte)SOLUTION.indexOf(digit);
        return new byte[] { (byte)(idx / ROWS), (byte)(idx % ROWS) };
    }

    private static byte manhattanDistance(byte x, byte y, byte x2, byte y2) {
        byte dx = (byte)Math.abs(x - x2);
        byte dy = (byte)Math.abs(y - y2);
        return (byte)(dx + dy);
    }

    private byte heuristic() {
        byte total = 0;
        for (byte i = 0; i < ROWS; ++i) {
            for (byte j = 0; j < COLUMNS; ++j) {
                char digit = board[i][j];
                byte[] coordenates = whereShouldItBe(digit);
                byte distance = manhattanDistance(i, j, coordenates[0], coordenates[1]);
                total += distance;
            }
        }
        return total;
    }

    private class Node implements Comparable<Node> {
        private String puzzle;
        private byte moves; // Number of moves from original configuration
        private byte value; // The value of the heuristic for this configuration.
        public Node(String puzzle, byte moves, byte value) {
            this.puzzle = puzzle;
            this.moves = moves;
            this.value = value;
        }
        @Override
        public int compareTo(Node o) {
            return (value + moves) - (o.value + o.moves);
        }
    }

    private void print(Map<String, String> antecessor) {
        Stack toPrint = new Stack();
        toPrint.add(SOLUTION);
        String before = antecessor.get(SOLUTION);
        while (!before.equals("")) {
            toPrint.add(before);
            before = antecessor.get(before);
        }
        while (!toPrint.isEmpty()) {
            System.out.println(new TilePuzzle(toPrint.pop()).printableString());
        }
    }

    private byte solve() {
        if(toString().equals(SOLUTION)) {
            return 0;
        }

        PriorityQueue<Node> toProcess = new PriorityQueue();
        Node initial = new Node(toString(), (byte)0, heuristic());
        toProcess.add(initial);

        Map<String, String> antecessor = new HashMap<String, String>();
        antecessor.put(toString(), "");

        while(!toProcess.isEmpty()) {
            Node actual = toProcess.poll();
            for (byte i = 0; i < 4; ++i) {
                TilePuzzle t = new TilePuzzle(actual.puzzle);
                try {
                    t.move(i);
                } catch(RuntimeException e) {
                    continue;
                }
                if (t.toString().equals(SOLUTION)) {
                    antecessor.put(SOLUTION, actual.puzzle);
                    print(antecessor);
                    return (byte)(actual.moves + 1);
                } else if (!antecessor.containsKey(t.toString())) {
                    byte v = t.heuristic();
                    Node neighbor = new Node(t.toString(), (byte)(actual.moves + 1), v);
                    toProcess.add(neighbor);
                    antecessor.put(t.toString(), actual.puzzle);
                }
            }
        }
        return -1;
    }

    public static void main(String... args) {
        TilePuzzle puzzle = new TilePuzzle();
        System.out.println(puzzle.solve());
    }
}

Answer 1

问题

根本原因是您正在创建并存储在 toProcess 队列和 antecessor 映射中的大量 String 对象。你为什么要这么做？

看看你的算法。看看您是否真的需要在每个节点中存储 >200 万个节点和 500 万个字符串。

调查

这很难发现，因为程序很复杂。实际上，我什至没有尝试理解所有代码。相反，我使用了 VisualVM – Java 分析器、采样器和 CPU/内存使用监视器。

我启动了它:

并查看了内存使用情况。我注意到的第一件事是(明显的)事实，您正在创建大量对象。

这是应用程序的屏幕截图:

正如您所看到的，使用的内存量是巨大的。在短短 40 秒内，消耗了 2 GB 空间并填满了整个堆。

死胡同

我最初认为问题与 Node 类有关，因为即使它实现了 Comparable，它也没有实现 equals >。所以我提供了方法:

public boolean equals( Object o ) {
    if( o instanceof Node ) {
        Node other = ( Node ) o;
        return this.value == other.value && this.moves == other.moves;
    }
    return false;
}

但这不是问题。

事实证明，实际问题是顶部所述的问题。

解决方法

如前所述，真正的解决方案是重新考虑您的算法。与此同时，无论采取什么其他措施，都只会拖延问题的解决。

但是解决方法可能很有用。一种是重用您生成的字符串。您非常频繁地使用 TilePuzzle.toString() 方法；这最终会经常创建重复的字符串。

由于您要生成字符串排列，因此您可能会在几秒钟内创建许多 12345ABCD 字符串。如果它们是相同的字符串，则创建数百万个具有相同值的实例是没有意义的。

String.intern()方法允许重复使用字符串。医生说:

Returns a canonical representation for the string object.

A pool of strings, initially empty, is maintained privately by the class String.

When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals() method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.

对于常规应用程序，使用 String.intern() 可能不是一个好主意，因为它不允许 GC 回收实例。但在这种情况下，由于您无论如何都在 Map 和 Queue 中保存引用，所以这是有道理的。

因此进行此更改:

public String toString() {
    if (change) {
        representation = asString();
    }
    return representation.intern(); // <-- Use intern
}

基本上解决了内存问题。

这是更改后的屏幕截图:

现在，即使几分钟后，堆使用量也不会达到 100 MB。

额外备注

备注#1

您使用异常来验证运动是否有效，这是可以的；但当你捕获它们时，你只是忽略它们:

try {
    t.move(i);
} catch(RuntimeException e) {
    continue;
}

如果您无论如何都不使用它们，则可以通过首先不创建异常来节省大量计算。否则，您将创建数百万个未使用的异常。

进行此更改:

if (!areValidCoordinates(nx, ny)) {
    // REMOVE THIS LINE:
    // throw new IllegalArgumentException("(" + nx + ", " + ny + ")");

    // ADD THIS LINE:
    return;
}

并使用验证代替:

// REMOVE THESE LINES:
// try {
//     t.move(i);
// } catch(RuntimeException e) {
//     continue;
// }

// ADD THESE LINES:
if(t.isValidMovement(i)){
    t.move(i);
} else {
    continue;
}

备注 #2

您正在为每个新的 TilePuzzle 实例创建一个新的 Random 对象。如果整个程序只使用一个会更好。毕竟，您只使用单个线程。

备注#3

该解决方法解决了堆内存问题，但创建了另一个涉及 PermGen 的问题。我只是增加了 PermGen 的大小，如下所示:

java -Xmx1g -Xms1g -XX:MaxPermSize=1g TilePuzzle

备注#4

输出有时是 49，有时是 50。矩阵打印如下:

1 2 3 4 
5 6 7 8 
9 A B C 
D E 0 F 

1 2 3 4 
5 6 7 8 
9 A B C 
D E F 0 

... 50 次