通过增加索引之和来生成排序组合的有效方法

对于启发式算法，我需要一个接一个地评估特定集合的组合，直到达到停止标准。

由于它们很多，目前我正在使用以下内存高效迭代器块生成它们（受到 python 的启发）itertools.combinations http://docs.python.org/2/library/itertools.html#itertools.combinations):

public static IEnumerable<T[]> GetCombinations<T>(this IList<T> pool, int r)
{
    int n = pool.Count;
    if (r > n)
        throw new ArgumentException("r cannot be greater than pool size");
    int[] indices = Enumerable.Range(0, r).ToArray();
    yield return indices.Select(idx => pool[idx]).ToArray();
    while (true)
    {
        int i;
        for (i = r - 1; i >= 0; i--)
            if (indices[i] != i + n - r)
                break;
        if (i < 0)
            break;
        indices[i] += 1;
        for (int j = i + 1; j < r; j++)
            indices[j] = indices[j - 1] + 1;
        yield return indices.Select(idx => pool[idx]).ToArray();
    }
}

问题是，为了大大提高启发式的效率，我需要生成按索引之和排序的这些组合（换句话说，我需要首先生成包含该集合的第一个元素的组合）。

e.g.
考虑集合S = {0,1,2,3,4,5}
（为了简单起见，我选择这个集合，因为元素及其索引一致）。
所有可能的组合r=4由给定算法生成的数字是：

(0, 1, 2, 3)  SUM:  6
(0, 1, 2, 4)  SUM:  7
(0, 1, 2, 5)  SUM:  8
(0, 1, 3, 4)  SUM:  8
(0, 1, 3, 5)  SUM:  9
(0, 1, 4, 5)  SUM: 10
(0, 2, 3, 4)  SUM:  9
(0, 2, 3, 5)  SUM: 10
(0, 2, 4, 5)  SUM: 11
(0, 3, 4, 5)  SUM: 12
(1, 2, 3, 4)  SUM: 10
(1, 2, 3, 5)  SUM: 11
(1, 2, 4, 5)  SUM: 12
(1, 3, 4, 5)  SUM: 13
(2, 3, 4, 5)  SUM: 14

如您所见，其中的组合并未严格按升序总和排序。

期望的结果如下：
（具有相同总和的组合的顺序并不重要）

(0, 1, 2, 3)  SUM:  6
(0, 1, 2, 4)  SUM:  7
(0, 1, 2, 5)  SUM:  8
(0, 1, 3, 4)  SUM:  8
(0, 1, 3, 5)  SUM:  9
(0, 2, 3, 4)  SUM:  9
(0, 1, 4, 5)  SUM: 10
(0, 2, 3, 5)  SUM: 10
(1, 2, 3, 4)  SUM: 10
(0, 2, 4, 5)  SUM: 11
(1, 2, 3, 5)  SUM: 11
(0, 3, 4, 5)  SUM: 12
(1, 2, 4, 5)  SUM: 12
(1, 3, 4, 5)  SUM: 13
(2, 3, 4, 5)  SUM: 14

一个简单的解决方案是生成所有组合，然后根据它们的总和对它们进行排序；但这并不是真正有效/可行，因为组合的数量变得巨大n grows.

我还快速浏览了组合格雷码，但找不到适合这个问题的人。

您知道如何实施这样的事情吗？

EDIT :

这个问题有一个替代的（不幸的是并不容易）表述。
给定一组S和一个数字r，所有可能的总和都很容易找到，因为它们只是第一个总和中的所有数字r要点S到最后的总和r要点S.

话虽这么说，如果对于每笔金额T我们可以有效地找到所有具有总和的组合T我们解决了原来的问题，因为我们只是按升序生成它们。

有效意味着我不想生成所有组合并丢弃具有不同总和的组合。

EDIT 2:

根据@EricLippert的建议，我创建了以下代码：

public static IEnumerable<T[]> 
GetCombinationsSortedByIndexSum<T>(this IList<T> pool, int r)
{
    int n = pool.Count;
    if (r > n)
        throw new ArgumentException("r cannot be greater than pool size");
    int minSum = ((r - 1) * r) / 2;
    int maxSum = (n * (n + 1)) / 2 - ((n - r - 1) * (n - r)) / 2;

    for (int sum = minSum; sum <= maxSum; sum++)
    {
        foreach (var indexes in AllMonotIncrSubseqOfLenMWhichSumToN(0, n - 1, r, sum))
            yield return indexes.Select(x => pool[x]).ToArray();
    }
}

static IEnumerable<IEnumerable<int>> 
AllMonotIncrSubseqOfLenMWhichSumToN(int seqFirstElement, int seqLastElement, int m, int n)
{
    for (int i = seqFirstElement; i <= seqLastElement - m + 1; i++)
    {
        if (m == 1)
        {
            if (i == n)
                yield return new int[] { i };
        }
        else
        {
            foreach (var el in AllMonotIncrSubseqOfLenMWhichSumToN(i + 1, seqLastElement, m - 1, n - i))
                yield return new int[] { i }.Concat(el);
        }
    }
}

这工作得很好（希望这就是埃里克的意思：P），但我仍然担心递归方法的复杂性。事实上，我们似乎正在为每个总和重新生成所有组合，并丢弃那些总和未达到所需值的组合。

为了降低内部函数的复杂性，我找到了一种通过使用有效上限和下限来限制迭代的方法（现在很难说它的复杂性是多少）。

Check 我的答案 https://stackoverflow.com/a/16758273/316644查看最终代码。

我想到的解决方案是：

using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
  // Preconditions:
  // * items is a sequence of non-negative monotone increasing integers
  // * n is the number of items to be in the subsequence
  // * sum is the desired sum of that subsequence.
  // Result:
  // A sequence of subsequences of the original sequence where each 
  // subsequence has n items and the given sum.
  static IEnumerable<IEnumerable<int>> M(IEnumerable<int> items, int sum, int n)
  {
    // Let's start by taking some easy outs. If the sum is negative
    // then there is no solution. If the number of items in the
    // subsequence is negative then there is no solution.

    if (sum < 0 || n < 0)
      yield break;

    // If the number of items in the subsequence is zero then
    // the only possible solution is if the sum is zero.

    if (n == 0)
    {
      if (sum == 0)
        yield return Enumerable.Empty<int>();
      yield break;
    }

    // If the number of items is less than the required number of 
    // items, there is no solution.

    if (items.Count() < n)
      yield break;

    // We have at least n items in the sequence, and
    // and n is greater than zero, so First() is valid:

    int first = items.First();

    // We need n items from a monotone increasing subsequence
    // that have a particular sum. We might already be too 
    // large to meet that requirement:

    if (n * first > sum)
      yield break;

    // There might be some solutions that involve the first element.
    // Find them all.

    foreach(var subsequence in M(items.Skip(1), sum - first, n - 1))
      yield return new[]{first}.Concat(subsequence);      

    // And there might be some solutions that do not involve the first element.
    // Find them all.

    foreach(var subsequence in M(items.Skip(1), sum, n))
      yield return subsequence;
  }
  static void Main()
  {
    int[] x = {0, 1, 2, 3, 4, 5};
    for (int i = 0; i <= 15; ++i)
      foreach(var seq in M(x, i, 4))
        Console.WriteLine("({0}) SUM {1}", string.Join(",", seq), i);
  }
}       

输出就是你想要的输出。

我没有尝试对此进行优化。对其进行分析并了解大部分时间都花在哪里会很有趣。

更新：只是为了好玩，我编写了一个使用不可变堆栈而不是任意可枚举的版本。享受！

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

abstract class ImmutableList<T> : IEnumerable<T>
{
  public static readonly ImmutableList<T> Empty = new EmptyList();
  private ImmutableList() {}  
  public abstract bool IsEmpty { get; }
  public abstract T Head { get; }
  public abstract ImmutableList<T> Tail { get; }
  public ImmutableList<T> Push(T newHead)
  {
    return new List(newHead, this);
  }  

  private sealed class EmptyList : ImmutableList<T>
  {
    public override bool IsEmpty { get { return true; } }
    public override T Head { get { throw new InvalidOperationException(); } }
    public override ImmutableList<T> Tail { get { throw new InvalidOperationException(); } }
  }
  private sealed class List : ImmutableList<T>
  {
    private readonly T head;
    private readonly ImmutableList<T> tail;
    public override bool IsEmpty { get { return false; } }
    public override T Head { get { return head; } }
    public override ImmutableList<T> Tail { get { return tail; } }
    public List(T head, ImmutableList<T> tail)
    {
      this.head = head;
      this.tail = tail;
    }
  }
  System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
  {
    return this.GetEnumerator();
  }
  public IEnumerator<T> GetEnumerator()
  {
    for (ImmutableList<T> current = this; !current.IsEmpty; current = current.Tail)
      yield return current.Head;
  }
}  

class Program
{
  // Preconditions:
  // * items is a sequence of non-negative monotone increasing integers
  // * n is the number of items to be in the subsequence
  // * sum is the desired sum of that subsequence.
  // Result:
  // A sequence of subsequences of the original sequence where each 
  // subsequence has n items and the given sum.
  static IEnumerable<ImmutableList<int>> M(ImmutableList<int> items, int sum, int n)
  {
    // Let's start by taking some easy outs. If the sum is negative
    // then there is no solution. If the number of items in the
    // subsequence is negative then there is no solution.

    if (sum < 0 || n < 0)
      yield break;

    // If the number of items in the subsequence is zero then
    // the only possible solution is if the sum is zero.
    if (n == 0)
    {
      if (sum == 0)
        yield return ImmutableList<int>.Empty;
      yield break;
    }

    // If the number of items is less than the required number of 
    // items, there is no solution.

    if (items.Count() < n)
      yield break;

    // We have at least n items in the sequence, and
    // and n is greater than zero.
    int first = items.Head;

    // We need n items from a monotone increasing subsequence
    // that have a particular sum. We might already be too 
    // large to meet that requirement:

    if (n * first > sum)
      yield break;

    // There might be some solutions that involve the first element.
    // Find them all.

    foreach(var subsequence in M(items.Tail, sum - first, n - 1))
      yield return subsequence.Push(first);      

    // And there might be some solutions that do not involve the first element.
    // Find them all.
    foreach(var subsequence in M(items.Tail, sum, n))
      yield return subsequence;
  }
  static void Main()
  {
    ImmutableList<int> x = ImmutableList<int>.Empty.Push(5).
                           Push(4).Push(3).Push(2).Push(1).Push(0);
    for (int i = 0; i <= 15; ++i)
      foreach(var seq in M(x, i, 4))
        Console.WriteLine("({0}) SUM {1}", string.Join(",", seq), i);
  }
}