Question

Can somebody give me advice on how to create a recursive version of GetEnumerator()? The well-known Towers of Hanoi problem may serve as an example that is comparable to the actual problem I have. A simple algorithm to show all moves for a stack of disks of height n is:

void MoveTower0 (int n, Needle start, Needle finish, Needle temp)
{
  if (n > 0)
  {
    MoveTower0 (n - 1, start, temp, finish);
    Console.WriteLine ("Moving disk from {0} to {1}", start, finish);
    MoveTower0 (n - 1, temp, finish, start);
  }
}

我实际上想要做的是建立一个“河内托韦莫夫”级,执行“E”数字,使我能够在所有行动上调取:

foreach (Move m in HanoiTowerMoves) Console.WriteLine (m);

实现“GetE”指标的第一步似乎已经排除了Move Tower参数。可以通过使用一刀子来做到这一点。我还介绍了将参数纳入单一变量的一组变。

class Move
{
  public int N { private set; get; }
  public Needle Start { private set; get; }
  public Needle Finish { private set; get; }
  public Needle Temp { private set; get; }

  public Move (int n, Needle start, Needle finish, Needle temp)
  {
    N = n;
    Start = start;
    Finish = finish;
    Temp = temp;
  }

  public override string ToString ()
  {
    return string.Format ("Moving disk from {0} to {1}", Start, Finish);
  }
}

现在可以改写如下:

void MoveTower1 ()
{
  Move m = varStack.Pop ();

  if (m.N > 0)
  {
    varStack.Push (new Move (m.N - 1, m.Start, m.Temp, m.Finish));
    MoveTower1 ();
    Console.WriteLine (m);
    varStack.Push (new Move (m.N - 1, m.Temp, m.Finish, m.Start));
    MoveTower1 ();
  }
}

这一版本必须如下:

varStack.Push (new Move (n, Needle.A, Needle.B, Needle.Temp));
MoveTower1 ();

走向可操作版本的下一步是执行这一类:

class HanoiTowerMoves : IEnumerable<Move>
{
  Stack<Move> varStack;
  int n; // number of disks

  public HanoiTowerMoves (int n)
  {
    this.n = n;
    varStack = new Stack<Move> ();
  }

  public IEnumerator<Move> GetEnumerator ()
  {
    // ????????????????????????????  }

  // required by the compiler:
  IEnumerator IEnumerable.GetEnumerator ()
  {
    return GetEnumerator ();
  }
}

Now the big question to me is: what does the body of GetEnumerator () look like? Can somebody solve this mystery for me?

下面是I所创立的青少年申请方案法典。

using System;
using System.Collections.Generic;
using System.Collections;

/* Towers of Hanoi
 * ===============
 * Suppose you have a tower of N disks on needle A, which are supposed to end up on needle B.
 * The big picture is to first move the entire stack of the top N-1 disks to the Temp needle,
 * then move the N-th disk to B, then move the Temp stack to B using A as the new Temp needle.
 * This is reflected in the way the recursion is set up.
 */

namespace ConsoleApplication1
{
  static class main
  {
    static void Main (string [] args)
    {
      int n;
      Console.WriteLine ("Towers of Hanoi");

      while (true)
      {
        Console.Write ("
Enter number of disks: ");

        if (!int.TryParse (Console.ReadLine (), out n))
        {
          break;
        }

        HanoiTowerMoves moves = new HanoiTowerMoves (n);
        moves.Run (1); // algorithm version number, see below
      }
    }
  }

  class Move
  {
    public int N { private set; get; }
    public Needle Start { private set; get; }
    public Needle Finish { private set; get; }
    public Needle Temp { private set; get; }

    public Move (int n, Needle start, Needle finish, Needle temp)
    {
      N = n;
      Start = start;
      Finish = finish;
      Temp = temp;
    }

    public override string ToString ()
    {
      return string.Format ("Moving disk from {0} to {1}", Start, Finish);
    }
  }

  enum Needle { A, B, Temp }

  class HanoiTowerMoves : IEnumerable<Move>
  {
    Stack<Move> varStack;
    int n;            // number of disks

    public HanoiTowerMoves (int n)
    {
      this.n = n;
      varStack = new Stack<Move> ();
    }

    public void Run (int version)
    {
      switch (version)
      {
        case 0: // Original version
          MoveTower0 (n, Needle.A, Needle.B, Needle.Temp);
          break;

        case 1: // No parameters (i.e. argument values passed via stack)
          varStack.Push (new Move (n, Needle.A, Needle.B, Needle.Temp));
          MoveTower1 ();
          break;

        case 2: // Enumeration
          foreach (Move m in this)
          {
            Console.WriteLine (m);
          }

          break;
      }
    }

    void MoveTower0 (int n, Needle start, Needle finish, Needle temp)
    {
      if (n > 0)
      {
        MoveTower0 (n - 1, start, temp, finish);
        Console.WriteLine ("Moving disk from {0} to {1}", start, finish);
        MoveTower0 (n - 1, temp, finish, start);
      }
    }

    void MoveTower1 ()
    {
      Move m = varStack.Pop ();

      if (m.N > 0)
      {
        varStack.Push (new Move (m.N - 1, m.Start, m.Temp, m.Finish));
        MoveTower1 ();
        Console.WriteLine (m);
        varStack.Push (new Move (m.N - 1, m.Temp, m.Finish, m.Start));
        MoveTower1 ();
      }
    }

    public IEnumerator<Move> GetEnumerator ()
    {
      yield break; // ????????????????????????????
    }

    /*
      void MoveTower1 ()
      {
        Move m = varStack.Pop ();

        if (m.N > 0)
        {
          varStack.Push (new Move (m.N - 1, m.Start, m.Temp, m.Finish));
          MoveTower1 ();
          Console.WriteLine (m); ? yield return m;
          varStack.Push (new Move (m.N - 1, m.Temp, m.Finish, m.Start));
          MoveTower1 ();
        }
      }
    */

    // required by the compiler:
    IEnumerator IEnumerable.GetEnumerator ()
    {
      return GetEnumerator ();
    }
  }
}

Answer 1

你的做法非常好,但我认为你对问题的看法有些过分。让我们退步。您有回馈算:

void MoveTowerConsole (int n, Needle start, Needle finish, Needle temp) 
{   
  if (n > 0)   
  {
    MoveTowerConsole (n - 1, start, temp, finish);
    Console.WriteLine ("Moving disk from {0} to {1}", start, finish);
    MoveTowerConsole (n - 1, temp, finish, start);
  } 
}

算法的输出量是青春期产出的bu。附录让我说明这种方法会是什么样子。

首先,我们改名。第二,其返回类型不能无效。它必须是<条码>。缩略语

IEnumerable<string> MoveTower(int n, Needle start, Needle finish, Needle temp) 
{
  if (n > 0)   
  {
    MoveTower(n - 1, start, temp, finish);
    Console.WriteLine ("Moving disk from {0} to {1}", start, finish);
    MoveTower(n - 1, temp, finish, start);
  } 
}

这项权利吗? 页: 1 我们没有归还任何东西,我们仍把事情重新倾倒到ole。 <>strong>,我们祝愿主持人产生什么成果? 我们祝愿主持人:

all the moves necessary for the first recursive step
the current move
all the moves necessary for the second recursive step

因此,我们修改算法,以得出以下结果:

IEnumerable<string> MoveTower(int n, Needle start, Needle finish, Needle temp) 
{
  if (n > 0)   
  {
    foreach(string move in MoveTower(n - 1, start, temp, finish))
        yield return move;
    yield return string.Format("Moving disk from {0} to {1}", start, finish);
    foreach(string move in MoveTower(n - 1, temp, finish, start))
        yield return move;
  } 
}

而我们又做了! 简便。没有必要界定一个整类,把收回算法变成一个补缺的计数人;让汇编者为你工作。

如果你想把这个方法变成一个列举“移除”的方法,那么:

IEnumerable<Move> MoveTower(int n, Needle start, Needle finish, Needle temp) 
{
  if (n > 0)   
  {
    foreach(Move move in MoveTower(n - 1, start, temp, finish))
        yield return move;
    yield return new Move(start, finish);
    foreach(Move move in MoveTower(n - 1, temp, finish, start))
        yield return move;
  } 
}

现在,我将在效率的基础上批评这一法典。通过以这种方式提供补习生,你正在建立一套 n。当你需要下一个项目时,顶级统计员打电话给下一位统计员......打下下至下,深线。因此,现在每个步骤实际上都采取一些步骤来完成。由于这一原因,我倾向于不再次处理这个问题。

Exercise: Rewrite the iterator block above so that it does no recursion at all. Your solution that uses an explicit stack is a step in the right direction, but it still does recursion. Can you adapt it so that no recursion is done?

如果你打算撰写一个执行<条码>电子计算法”的班子;Move>的班子,你可以直截了当地修改上述代码:

class MoveIterator : IEnumerable<Move>
{
    public IEnumerator<Move> GetEnumerator()
    {
        foreach(Move move in MoveTower(whatever))
            yield return move;
    }

You can use yield return to implement a method that returns an enumerator or an enumerable.

Answer 2

Your non-recursive solution is good -- building a pushdown automaton (a state machine with a stack, essentially) is a standard technique for building an iterative version of a recursive solution. And in fact, this is very similar to how we generate code for iterator and async blocks.

然而,在这种具体情况下,你不需要用一个开关和目前状态撤出一个推倒的汽车顿的重型机械。你可以这样做:

IEnumerable<Move> MoveTowerConsole (int size, Needle start, Needle finish, Needle temp) 
{   
  if (size <= 0) yield break;
  var stack = new Stack<Work>();
  stack.Push(new Work(size, start, finish, temp));
  while(stack.Count > 0)
  {
    var current = stack.Pop();
    if (current.Size == 1) 
      yield return new Move(current.Start, current.Finish);
    else
    {
       // Push the work in the *opposite* order that it needs to be done.
       stack.Push(new Work(current.Size - 1, current.Temp, current.Finish, current.Start));
       stack.Push(new Work(1, current.Start, current.Finish, current.Temp));
       stack.Push(new Work(current.Size - 1, current.Start, current.Temp, current.Finish));

     }
}

You already know exactly what work you need to be doing after the current recursive step, so there s no need to bounce around a switch to put the three bits of work on the stack. Just queue all the work up at once for a given step.

Answer 3

不收复本:

// Non-recursive version -- state engine
//rta.Push (State.Exit);
//parameters.Push (new Move (n, Needle.A, Needle.B, Needle.Temp));
//MoveTower3 ();

enum State { Init, Call1, Call2, Rtrn, Exit }

{  
  ...

  #region Non-recursive version -- state engine
  static void MoveTower3 ()
  {
    State s = State.Init;
    Move m = null;

    while (true)
      switch (s)
      {
        case State.Init:
          m = moveStack.Pop ();
          s = (m.n <= 0) ? State.Rtrn : State.Call1;
          break;
        case State.Call1:
          rta.Push (State.Call2); // where do I want to go after the call is finished
          moveStack.Push (m);    // save state for second call
          moveStack.Push (new Move (m.n-1, m.start, m.temp, m.finish)); // parameters
          s = State.Init;
          break;
        case State.Call2:
          m = moveStack.Pop ();  // restore state from just before first call
          Console.WriteLine (m);
          rta.Push (State.Rtrn);
          moveStack.Push (new Move (m.n-1, m.temp, m.finish, m.start));
          s = State.Init;
          break;
        case State.Rtrn:
          s = rta.Pop ();
          break;
        case State.Exit:
          return;
      }
  }
  #endregion

  #region Enumeration
  static IEnumerable<Move> GetEnumerable (int n)
  {
    Stack<Move> moveStack = new Stack<Move> ();
    Stack<State> rta = new Stack<State> (); //  return addresses 
    rta.Push (State.Exit);
    moveStack.Push (new Move (n, Needle.A, Needle.B, Needle.Temp));
    State s = State.Init;
    Move m = null;

    while (true)
      switch (s)
      {
        case State.Init:
          m = moveStack.Pop ();
          s = (m.n <= 0) ? State.Rtrn : State.Call1;
          break;
        case State.Call1:
          rta.Push (State.Call2); // where do I want to go after the call is finished
          moveStack.Push (m);    // save state for second call
          moveStack.Push (new Move (m.n-1, m.start, m.temp, m.finish)); // parameters
          s = State.Init;
          break;
        case State.Call2:
          m = moveStack.Pop ();  // restore state from just before first call
          yield return m;
          rta.Push (State.Rtrn);
          moveStack.Push (new Move (m.n-1, m.temp, m.finish, m.start));
          s = State.Init;
          break;
        case State.Rtrn:
          s = rta.Pop ();
          break;
        case State.Exit:
          yield break;
      }
  }
  #endregion
}

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