11

对于启发式算法，我需要一个接一个地评估某个集合的组合，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（受 python 的启发itertools.combinations）生成它们：

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算法的效率，我需要生成这些组合，这些组合按它们的索引之和排序（换句话说，我需要首先生成包含集合中第一个元素的组合）。

例如
考虑集合S = {0,1,2,3,4,5}
（我选择这个集合是为了简单，因为元素和它们的索引是一致的）。从给定算法生成的
所有可能的数字组合是：r=4

(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和排序。

相反，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/可行的，因为组合的数量随着n增长而变得巨大。

我还快速浏览了组合格雷码，但找不到适合这个问题的人。

你对如何实现这样的事情有想法吗？

编辑 ：

这个问题有一个替代（不幸的是不是更容易）的公式。
给定一个集合S和一个数字r，所有可能的和都很容易找到，因为它们只是从 的第一个r元素之S和到 的最后一个r元素之和的所有数字S。

话虽如此，如果对于每个总和，T我们可以有效地¹找到所有具有总和的组合，T我们就可以解决原始问题，因为我们只是按升序生成它们。

¹ 有效意味着我不想生成所有组合并丢弃总和不同的组合。

编辑2：

在@EricLippert 建议后，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);
        }
    }
}

这很好用（希望是 Eric 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，而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（现在真的很难说这有什么复杂性）。

检查我的答案以查看最终代码。

4

2 回答 2

5

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);
  }
}

输出是您想要的输出。

我没有尝试对此进行优化。分析它并查看大部分时间都花在了哪里会很有趣。

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);
  }
}
于 2013-05-25T18:21:59.373 回答
1

为了完整和清晰起见，我将发布我的最终代码：

// Given a pool of elements returns all the 
// combinations of the groups of lenght r in pool, 
// such that the combinations are ordered (ascending) by the sum of 
// the indexes of the elements.
// e.g. pool = {A,B,C,D,E} r = 3
// returns
// (A, B, C)   indexes: (0, 1, 2)   sum: 3
// (A, B, D)   indexes: (0, 1, 3)   sum: 4
// (A, B, E)   indexes: (0, 1, 4)   sum: 5
// (A, C, D)   indexes: (0, 2, 3)   sum: 5
// (A, C, E)   indexes: (0, 2, 4)   sum: 6
// (B, C, D)   indexes: (1, 2, 3)   sum: 6
// (A, D, E)   indexes: (0, 3, 4)   sum: 7
// (B, C, E)   indexes: (1, 2, 4)   sum: 7
// (B, D, E)   indexes: (1, 3, 4)   sum: 8
// (C, D, E)   indexes: (2, 3, 4)   sum: 9
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 = F(r - 1);
    int maxSum = F(n) - F(n - r - 1);

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


// Given a start element and a last element of a sequence of consecutive integers
// returns all the monotonically increasing subsequences of length "m" having sum "sum"
// e.g. seqFirstElement = 1, seqLastElement = 5, m = 3, sum = 8
//      returns {1,2,5} and {1,3,4}
static IEnumerable<IEnumerable<int>>
AllSubSequencesWithGivenSum(int seqFirstElement, int seqLastElement, int m, int sum)
{
    int lb = sum - F(seqLastElement) + F(seqLastElement - m + 1);
    int ub = sum - F(seqFirstElement + m - 1) + F(seqFirstElement);

    lb = Math.Max(seqFirstElement, lb);
    ub = Math.Min(seqLastElement - m + 1, ub);

    for (int i = lb; i <= ub; i++)
    {
        if (m == 1)
        {
            if (i == sum) // this check shouldn't be necessary anymore since LB/UB should automatically exclude wrong solutions
                yield return new int[] { i };
        }
        else
        {
            foreach (var el in AllSubSequencesWithGivenSum(i + 1, seqLastElement, m - 1, sum - i))
                yield return new int[] { i }.Concat(el);
        }
    }
}

// Formula to compute the sum of the numbers from 0 to n
// e.g. F(4) = 0 + 1 + 2 + 3 + 4 = 10
static int F(int n)
{
    return (n * (n + 1)) / 2;
}
于 2013-05-26T10:25:09.607 回答