java - 如何评估 Java 中递归算法的效用？

Question

背景:

我正在尝试学习算法和 java。运行 320x320 的网格，100 次试验的运行速度比非递归 Quick-Union 实现快约 5 倍。但是，在大约 400x400(160,000 个站点)的网格之上，我遇到堆栈溢出错误。

我知道 java 没有针对尾递归进行优化(更不用说非尾递归了)。但是，我认为有时可以选择递归算法而不是非递归版本，因为它可以运行得更快并且同样安全。

请记住，我只是在学习这些东西，我的代码可能不是最佳的。但是，我将其包括在内是为了更好地理解我的问题。

问题

评估何时可以在 Java 应用程序中安全地使用递归算法(假设它比非递归算法运行得更快)的过程是什么？

关于递归与 UNION-FIND 实现的统计

(注意:2x 比率只是之前的当前运行时间除以之前的运行时间)

|-----------|-----------|------------|-------------|-------------|
|     N     | Recursive | Recursive  | Quick-Union | Quick-Union |
|  (sites)  |    time   |  2x Ratio  |    time     |  2x Ratio   |
|===========|===========|============|=============|=============|
|     196   |      35   |            |      42     |             |
|     400   |      25   |    0.71    |      44     |     1.05    |
|     784   |      45   |    1.80    |      46     |     1.05    |
|    1600   |     107   |    2.38    |      86     |     1.87    |
|    3136   |      48   |    0.45    |     113     |     1.31    |
|    6400   |      75   |    1.56    |     303     |     2.68    |
|   12769   |     183   |    2.44    |     858     |     2.83    |
|   25600   |     479   |    2.62    |    2682     |     3.13    |
|   51076   |    1253   |    2.62    |    8521     |     3.18    |
|  102400   |    4730   |    3.77    |   27256     |     3.20    |
|-----------|-----------|------------|-------------|-------------|

递归类

public class PercolateRecur implements Percolation {
  // the site has been opened for percolation but is not connected
  private final int OPEN = 0;
  // the site is not open for percolation (default state)
  private final int BLOCKED = -1;
  // the matrix that will be percolated. Values default to `BLOCKED = -1`
  // two sites that are connected together share the same value.
  private int[][] matrix;
  // the size of the sides of the matrix (1 to n)
  private int size;
  // whether water can flow from top to bottom of the matrix
  private boolean percolated;

  public PercolateRecur(int N) {
    percolated = false;
    size = N;
    initMatrix();
  }

  /**
   * initializes the matrix to default values
   */
  private void initMatrix() {
    matrix = new int[size+1][size+1];
    // open up the top of the matrix
    for (int x = 1; x < size+1; x++)
      matrix[x][0] = x;

    // set all values in matrix to closed
    for (int x = 1; x < size+1; x++)
      for (int y = 1; y < size+1; y++)
        matrix[x][y] = BLOCKED;
  }

  /**
   * indicates site (x,y) is a valid coordinate
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return boolean
   */
  private boolean isValid(int x, int y) {
    return x > 0 && x < size+1 && y > 0 && y < size+1;
  }

  /**
   * returns value of site above (x,y)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return int value
   */
  private int above(int x, int y) {
    if (y <= 0)
      return BLOCKED;
    else
      return matrix[x][y-1];
  }

  /**
   * returns value of site below (x,y)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return int value
   */
  private int below(int x, int y) {
    if (y >= size)
      return BLOCKED;
    else
      return matrix[x][y+1];
  }

  /**
   * returns value of site left of (x,y)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return int value
   */
  private int left(int x, int y) {
    if (x <= 0)
      return BLOCKED;
    return matrix[x-1][y];
  }

  /**
   * returns value of site right of (x,y)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return int value
   */
  private int right(int x, int y) {
    if (x >= size)
      return BLOCKED;
    else
      return matrix[x+1][y];
  }

  /**
   * connects (x,y) to open adjacent sites
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   */
  private void connect(int x, int y) {
    if (isFull(x,y))
      return;
    if (above(x,y) > OPEN)
      matrix[x][y] = above(x, y);
    else if (below(x, y) > OPEN)
      matrix[x][y] = below(x, y);
    else if (left(x, y) > OPEN)
      matrix[x][y] = left(x, y);
    else if (right(x, y) > OPEN)
      matrix[x][y] = right(x, y);
    else if (matrix[x][y] == BLOCKED)
      matrix[x][y] = OPEN;
  }

  /**
   * recursively connects open sites in same group as (x,y)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   */
  private void expand(int x, int y) {
    if (!isFull(x, y))
      return;
    if (above(x,y) == OPEN)
      openWith(x,y-1, matrix[x][y]);
    if (below(x,y) == OPEN)
      openWith(x,y+1, matrix[x][y]);
    if (left(x,y) == OPEN)
      openWith(x-1,y, matrix[x][y]);
    if (right(x,y) == OPEN)
      openWith(x+1,y, matrix[x][y]);
  }

  /**
   * opens a site (x,y) on the matrix
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   */
  public void open(int x, int y) {
    if (percolated || !isValid(x, y))
      return;
    connect(x, y);
    expand(x, y);
  }

  /**
   * opens a site with given value
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @param val value of point
   */
  private void openWith(int x, int y, int val) {
    matrix[x][y] = val;
    open(x, y);
  }

  /**
   * Returns whether site (x,y) is open
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return true if not blocked
   */
    public boolean isOpen(int x, int y) {
    return matrix[x][y] > BLOCKED;
  }

  /**
   * Returns whether site (x,y) is full (connected to the top)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return true if is full
   */
  public boolean isFull(int x, int y) {
    return matrix[x][y] > OPEN;
  }

  /**
   * indicates whether site is blocked (not open)
   * @param x x-portion of x/y coordinate
   * @param y y-portion of x/y coordinate
   * @return true if blocked
   */
  public boolean isBlocked(int x, int y) {
    return matrix[x][y] == BLOCKED;
  }

  /**
   * indicates whether water can flow from top to bottom of matrix
   * @return true if matrix is percolated
   */
  public boolean percolates() {
    for (int x = 1; x <= size; x++)
      if (matrix[x][size] > OPEN)
        percolated = true;
    return percolated;
  }

  /**
   * prints the matrix to the command line
   */
  public void print() {
    for (int y = 1; y < size+1; y++) {
      System.out.println();
      for (int x = 1; x < size+1; x++) {
        if (matrix[x][y] == BLOCKED)
          System.out.print("XX ");
        else if (matrix[x][y] < 10)
          System.out.print(matrix[x][y] + "  ");
        else
          System.out.print(matrix[x][y] + " ");
      }
    }
    System.out.println();
  }
}

Answer 1

在 Java 中实现递归算法的问题之一是 Java 平台不执行标准的“尾调用消除”优化。这意味着深度递归需要深度堆栈。由于 Java 线程堆栈不会“增长”，这意味着您很容易受到堆栈溢出的影响。

有两种解决方法:

• 增加线程堆栈大小，方法是在命令行上使用 -Xss 选项，或者在 Thread 中显式提供(更大的)堆栈大小> 构造函数。

• 在您的 Java 代码中显式实现尾调用消除……至少可以说这很丑陋。

在您的情况下，第一个解决方法将为您提供“真正递归”的衡量标准。第二个……好吧，该算法将不再是真正的递归算法，但您可以通过这种方式使递归算法在 Java 中适用于“深层”问题。

注意:您始终可以将 Java 中的递归算法转换为使用“模拟堆栈”数据结构来保存递归状态的等效迭代算法。两个版本的算法复杂度应该相同。也许您也应该尝试这种方法，并将“模拟堆栈”变体作为第三对列包含在您的评估中。