C# 通过增加索引和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，目前我使用以下高效内存迭代器块（受python启发）生成它们：

其中，正如您所看到的，组合不是严格按升序和排序的

相反，期望的结果如下：
（具有相同总和的组合顺序并不重要）

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，因为随着

n

的增长，组合的数量变得巨大

我还快速查看了组合格雷码，但我找不到适合这个问题的人

你对如何实现这样的东西有什么想法吗

编辑：

这个问题有一个替代（不幸的是不容易）公式。
给定一组

S

和一个数字

r

，所有可能的和都很难找到，因为它们只是从

S

的前

r

元素和到

S

的最后

r

元素和的所有数字

这就是说，如果对于每个和

T

，我们可以有效地找到所有具有和

T

的组合，我们就解决了原始问题，因为我们只是按升序生成它们

1有效地意味着我不想生成所有组合并丢弃具有不同总和的组合

编辑2:

在@EricLippert建议之后，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);
        }
    }
}

公共静态IEnumerable
GetCombinationsSortedByIndexSum（此IList池，int r）
{
int n=池计数；
如果（r>n）
抛出新ArgumentException（“r不能大于池大小”）；
int minSum=（（r-1）*r）/2；
int maxSum=（n*（n+1））/2-（（n-r-1）*（n-r））/2；
对于（int-sum=minSum；sum-pool[x]）.ToArray（）；
}
}
静态可数
AllMonotIncrSubseqOfLenMWhichSumToN（int-seqFirstElement，int-seqLastElement，int-m，int-n）
{
对于（int i=seqFirstElement；i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);
  }
}       

使用系统；
使用System.Collections.Generic；
使用System.Linq；
班级计划
{
//先决条件：
//*项是一个非负单调递增整数序列
//*n是子序列中的项目数
//*总和是该子序列的期望总和。
//结果:
//原始序列的一系列子序列，其中
//子序列有n个项和给定的和。
静态IEnumerable M（IEnumerable项，整数和，整数n）
{
//让我们从简单的例子开始。如果总和是负数
//那么就没有解决方案了。如果
//子序列为负，则没有解。
如果（总和<0 | | n<0）
屈服断裂；
//如果子序列中的项数为零，则
//唯一可能的解决方案是如果和为零。
如果（n==0）
{
如果（总和=0）
产生返回可枚举的.Empty（）；
屈服断裂；
}
//如果项目数量小于所需数量
//但是，没有解决方案。
if（items.Count（）sum）
屈服断裂；
//可能有一些解决方案涉及第一个元素。
//把它们都找出来。
foreach（M中的var子序列（items.Skip（1），sum-first，n-1））
产生返回新[]{first}.Concat（子序列）；
//可能有一些解决方案不涉及第一个元素。
//把它们都找出来。
foreach（以M为单位的var子序列（项跳过（1），求和，n））
收益率子序列；
}
静态void Main（）
{
int[]x={0,1,2,3,4,5}；
对于（int i=0；i和）
屈服断裂；
//可能有一些解决方案涉及第一个元素。
//把它们都找出来。
foreach（var子序列，单位为M（items.Tail，sum-first，n-1））
收益返回子序列推送（第一）；
//可能有一些解决方案不涉及第一个元素。
//把它们都找出来。
foreach（var子序列，单位为M（items.Tail，sum，n））
收益率子序列；
}
静态void Main（）
{
ImmutableList x=ImmutableList.Empty.Push（5）。
推（4）。推（3）。推（2）。推（1）。推（0）；
为了（int i=0；i为了完整性和清晰性，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;
}

//给定一个元素池将返回所有
//池中长度r组的组合，
//使组合按以下各项之和排序（升序）：
//元素的索引。
//例如pool={A，B，C，D，e}r=3
//返回
//（A，B，C）索引：（0，1，2）和：3
//（A，B，D）索引：（0，1，3）和：4
//（A，B，E）索引：（0，1，4）和：5
//（A，C，D）索引：（0，2，3）和：5
//（A，C，E）索引：（0，2，4）和：6
//（B，C，D）指数：（1，2，3）总和：6
//（A，D，E）索引：（0，3，4）和：7
//（B，C，E）指数：（1，2，4）总和：7
//（B，D，E）指数：（1，3，4）总和：8
//（C，D，E）指数：（2，3，4）总和：9
公共静态IEnumerable
GetCombinationsSortedByIndexSum（此IList池，int r）
{
int n=池计数；
如果（r>n）
抛出新ArgumentException（“r不能大于池大小”）；
int minSum=F（r-1）；
int maxSum=F（n）-F（n-r-1）；
对于（int-sum=minSum；sum-pool[x]）.ToArray（）；
}
}
//给定序列的起始元素和最后一个元素
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);
        }
    }
}

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);
  }
}       

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);
  }
}       

// 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;
}