使用 LINQ 生成排列
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【中文标题】使用 LINQ 生成排列【英文标题】:Generating Permutations using LINQ 【发布时间】:2011-05-18 04:05:58 【问题描述】:我有一组必须安排的产品。有 P 个产品,每个产品的索引从 1 到 P。每个产品可以安排在 0 到 T 的时间段内。我需要构造满足以下约束的产品时间表的所有排列:
If p1.Index > p2.Index then p1.Schedule >= p2.Schedule.
我正在努力构建迭代器。当产品数量是已知常量时,我知道如何通过 LINQ 执行此操作,但不确定当产品数量是输入参数时如何生成此查询。
理想情况下,我想使用 yield 语法来构造这个迭代器。
public class PotentialSchedule()
public PotentialSchedule(int[] schedulePermutation)
_schedulePermutation = schedulePermutation;
private readonly int[] _schedulePermutation;
private int _numberProducts = ...;
public IEnumerator<PotentialSchedule> GetEnumerator()
int[] permutation = new int[_numberProducts];
//Generate all permutation combinations here -- how?
yield return new PotentialSchedule(permutation);
编辑:_numberProducts = 2 时的示例
public IEnumerable<PotentialSchedule> GetEnumerator()
var query = from p1 in Enumerable.Range(0,T)
from p2 in Enumerable.Range(p2,T)
select new P1 = p1, P2 = p2;
foreach (var result in query)
yield return new PotentialSchedule(new int[] result.P1, result.P2 );
【问题讨论】:
看***.com/questions/774457/… 您意识到这会很快变得非常大,对吗?你介意我问为什么你需要生成所有可能的时间表吗? 是的——我意识到这将很快变得非常大。这是优化程序的一部分。最棘手的部分是扩展正在考虑的可能性。在最坏的情况下,我的问题大小大约是 P 第二个例子的第四行应该是(p1, T),是吗? 请注意,此类项目的数量等于 T 边的 P 维金字塔格子中的节点数。(因此,如果 P 为 3,T 为 5,则为点数一个最长边有五个点的三维金字塔。)一个边上有 15 个点的 20 维金字塔有 大量 个点。那个东西的点数和宇宙中基本粒子的数量差不多,(非常)粗略的近似。 【参考方案1】:如果我理解这个问题:您正在寻找长度为 P 的所有整数序列,其中集合中的每个整数都介于 0 和 T 之间,并且该序列是单调不减。对吗?
使用迭代器块编写这样的程序很简单:
using System;
using System.Collections.Generic;
using System.Linq;
static class Program
static IEnumerable<T> Prepend<T>(T first, IEnumerable<T> rest)
yield return first;
foreach (var item in rest)
yield return item;
static IEnumerable<IEnumerable<int>> M(int p, int t1, int t2)
if (p == 0)
yield return Enumerable.Empty<int>();
else
for (int first = t1; first <= t2; ++first)
foreach (var rest in M(p - 1, first, t2))
yield return Prepend(first, rest);
public static void Main()
foreach (var sequence in M(4, 0, 2))
Console.WriteLine(string.Join(", ", sequence));
这会产生所需的输出:从 0 到 2 绘制的长度为 4 的非递减序列。
0, 0, 0, 0
0, 0, 0, 1
0, 0, 0, 2
0, 0, 1, 1
0, 0, 1, 2
0, 0, 2, 2
0, 1, 1, 1
0, 1, 1, 2
0, 1, 2, 2
0, 2, 2, 2
1, 1, 1, 1
1, 1, 1, 2
1, 1, 2, 2
1, 2, 2, 2
2, 2, 2, 2
请注意,用于连接的多重嵌套迭代器的用法是not very efficient,但谁在乎呢?您已经在生成 指数 数量的序列,因此生成器中存在 多项式 效率低下的事实基本上是无关紧要的。
方法 M 生成长度为 p 的所有单调非递减整数序列,其中整数介于 t1 和 t2 之间。它使用简单的递归递归地执行此操作。基本情况是只有一个长度为零的序列,即空序列。递归情况是,为了计算,比如说 P = 3, t1 = 0, t2 = 2,你计算:
- all sequences starting with 0 followed by sequences of length 2 drawn from 0 to 2.
- all sequences starting with 1 followed by sequences of length 2 drawn from 1 to 2.
- all sequences starting with 2 followed by sequences of length 2 drawn from 2 to 2.
结果就是这样。
或者,您可以在主递归方法中使用查询理解而不是迭代器块:
static IEnumerable<T> Singleton<T>(T first)
yield return first;
static IEnumerable<IEnumerable<int>> M(int p, int t1, int t2)
return p == 0 ?
Singleton(Enumerable.Empty<int>()) :
from first in Enumerable.Range(t1, t2 - t1 + 1)
from rest in M(p - 1, first, t2)
select Prepend(first, rest);
基本上做同样的事情;它只是将循环移动到 SelectMany 方法中。
【讨论】:
我更喜欢查询理解语法以提高可读性——与 yield 实现相比会有什么明显的缺陷吗? @erash:两者的效率都不是特别高或低。这几乎是一个偏好问题。 如果序列元素的范围是从0到k,并且序列的长度是n,那么输出中的序列数由这个公式给出: (n + (k − 1) )! / (k - 1)! /n!通过搜索“重复组合”可以找到许多证明,例如在这篇***文章en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29【参考方案2】:我使用这个库进行组合,发现它运行良好。示例程序有点混乱,但文章解释了使用代码需要什么。
使用 C# 泛型的排列、组合和变体 阿德里安·阿克森 | 2008 年 5 月 23 日 讨论六种主要类型的组合集合,并提供示例和计数公式。扩展了一组基于 C# 泛型的类,用于枚举每个元集合。 从http://www.codeproject.com/KB/recipes/Combinatorics.aspx插入【讨论】:
【参考方案3】:注意:Comparer
请记住,此代码已有几年历史,并且针对 .NET 2.0,因此没有扩展方法等(但修改起来应该很简单)。
它使用Knuth calls "Algorithm L"的算法。它是非递归的、快速的,用于 C++ 标准模板库。
static partial class Permutation
/// <summary>
/// Generates permutations.
/// </summary>
/// <typeparam name="T">Type of items to permute.</typeparam>
/// <param name="items">Array of items. Will not be modified.</param>
/// <param name="comparer">Optional comparer to use.
/// If a <paramref name="comparer"/> is supplied,
/// permutations will be ordered according to the
/// <paramref name="comparer"/>
/// </param>
/// <returns>Permutations of input items.</returns>
public static IEnumerable<IEnumerable<T>> Permute<T>(T[] items, IComparer<T> comparer)
int length = items.Length;
IntPair[] transform = new IntPair[length];
if (comparer == null)
//No comparer. Start with an identity transform.
for (int i = 0; i < length; i++)
transform[i] = new IntPair(i, i);
;
else
//Figure out where we are in the sequence of all permutations
int[] initialorder = new int[length];
for (int i = 0; i < length; i++)
initialorder[i] = i;
Array.Sort(initialorder, delegate(int x, int y)
return comparer.Compare(items[x], items[y]);
);
for (int i = 0; i < length; i++)
transform[i] = new IntPair(initialorder[i], i);
//Handle duplicates
for (int i = 1; i < length; i++)
if (comparer.Compare(
items[transform[i - 1].Second],
items[transform[i].Second]) == 0)
transform[i].First = transform[i - 1].First;
yield return ApplyTransform(items, transform);
while (true)
//Ref: E. W. Dijkstra, A Discipline of Programming, Prentice-Hall, 1997
//Find the largest partition from the back that is in decreasing (non-icreasing) order
int decreasingpart = length - 2;
for (;decreasingpart >= 0 &&
transform[decreasingpart].First >= transform[decreasingpart + 1].First;
--decreasingpart) ;
//The whole sequence is in decreasing order, finished
if (decreasingpart < 0) yield break;
//Find the smallest element in the decreasing partition that is
//greater than (or equal to) the item in front of the decreasing partition
int greater = length - 1;
for (;greater > decreasingpart &&
transform[decreasingpart].First >= transform[greater].First;
greater--) ;
//Swap the two
Swap(ref transform[decreasingpart], ref transform[greater]);
//Reverse the decreasing partition
Array.Reverse(transform, decreasingpart + 1, length - decreasingpart - 1);
yield return ApplyTransform(items, transform);
#region Overloads
public static IEnumerable<IEnumerable<T>> Permute<T>(T[] items)
return Permute(items, null);
public static IEnumerable<IEnumerable<T>> Permute<T>(IEnumerable<T> items, IComparer<T> comparer)
List<T> list = new List<T>(items);
return Permute(list.ToArray(), comparer);
public static IEnumerable<IEnumerable<T>> Permute<T>(IEnumerable<T> items)
return Permute(items, null);
#endregion Overloads
#region Utility
public static IEnumerable<T> ApplyTransform<T>(
T[] items,
IntPair[] transform)
for (int i = 0; i < transform.Length; i++)
yield return items[transform[i].Second];
public static void Swap<T>(ref T x, ref T y)
T tmp = x;
x = y;
y = tmp;
public struct IntPair
public IntPair(int first, int second)
this.First = first;
this.Second = second;
public int First;
public int Second;
#endregion
class Program
static void Main()
int pans = 0;
int[] digits = new int[] 1, 2, 3, 4, 5, 6, 7, 8, 9 ;
Stopwatch sw = new Stopwatch();
sw.Start();
foreach (var p in Permutation.Permute(digits))
pans++;
if (pans == 720) break;
sw.Stop();
Console.WriteLine("0pcs, 1ms", pans, sw.ElapsedMilliseconds);
Console.ReadKey();
【讨论】:
【参考方案4】:-
创建另一个长度为 2^n 的数组,其中 n 是产品数
从 0 到 2^n 以二进制计数,并用每个计数填充数组。例如,如果 n=3
数组将如下所示:
000 001 010 011 100 101 110 111
-
循环遍历二进制数组并在每个数字中找到一个,然后添加具有相同索引的产品:
for each binaryNumber in ar for i = 0 to n-1 if binaryNumber(i) = 1 permunation.add(products(i)) permunations.add(permutation)
示例: 如果 binaryNumber= 001 那么 permunation1 = product1 如果 binaryNumber= 101 那么 permunation1 = product3,product1
【讨论】:
【参考方案5】:这是 C# 7 的简单排列扩展方法(值元组和内部方法)。它源自@AndrasVaas's answer,但只使用了单一级别的惰性(防止由于项目随着时间的推移而发生的错误),失去了IComparer
功能(我不需要它),并且相当短。
public static class PermutationExtensions
/// <summary>
/// Generates permutations.
/// </summary>
/// <typeparam name="T">Type of items to permute.</typeparam>
/// <param name="items">Array of items. Will not be modified.</param>
/// <returns>Permutations of input items.</returns>
public static IEnumerable<T[]> Permute<T>(this T[] items)
T[] ApplyTransform(T[] values, (int First, int Second)[] tx)
var permutation = new T[values.Length];
for (var i = 0; i < tx.Length; i++)
permutation[i] = values[tx[i].Second];
return permutation;
void Swap<U>(ref U x, ref U y)
var tmp = x;
x = y;
y = tmp;
var length = items.Length;
// Build identity transform
var transform = new(int First, int Second)[length];
for (var i = 0; i < length; i++)
transform[i] = (i, i);
yield return ApplyTransform(items, transform);
while (true)
// Ref: E. W. Dijkstra, A Discipline of Programming, Prentice-Hall, 1997
// Find the largest partition from the back that is in decreasing (non-increasing) order
var decreasingpart = length - 2;
while (decreasingpart >= 0 && transform[decreasingpart].First >= transform[decreasingpart + 1].First)
--decreasingpart;
// The whole sequence is in decreasing order, finished
if (decreasingpart < 0)
yield break;
// Find the smallest element in the decreasing partition that is
// greater than (or equal to) the item in front of the decreasing partition
var greater = length - 1;
while (greater > decreasingpart && transform[decreasingpart].First >= transform[greater].First)
greater--;
// Swap the two
Swap(ref transform[decreasingpart], ref transform[greater]);
// Reverse the decreasing partition
Array.Reverse(transform, decreasingpart + 1, length - decreasingpart - 1);
yield return ApplyTransform(items, transform);
【讨论】:
【参考方案6】:今天我偶然发现了这一点,并认为我可以分享我的实现。
对于 N 和 M 之间的所有整数,您必须先创建一个数组:
IEnumerable<int> Range(int n, int m)
for(var i = n; i < m; ++i)
yield return i;
并通过Permutations(Range(1, 10))
运行它:
enum PermutationsOption
None,
SkipEmpty,
SkipNotDistinct
private IEnumerable<IEnumerable<T>> Permutations<T>(IEnumerable<T> elements, PermutationsOption option = PermutationsOption.None, IEqualityComparer<T> equalityComparer = default(IEqualityComparer<T>))
var elementsList = new List<IEnumerable<T>>();
var elementsIndex = 0;
var elementsCount = elements.Count();
var elementsLength = Math.Pow(elementsCount + 1, elementsCount);
if (option.HasFlag(PermutationsOption.SkipEmpty))
elementsIndex = 1;
if (elements.Count() > 0)
do
var elementStack = new Stack<T>();
for (var i = 0; i < elementsCount; ++i)
var ind = (int)(elementsIndex / Math.Pow(elementsCount + 1, i) % (elementsCount + 1));
if (ind == 0)
continue;
elementStack.Push(elements.ElementAt(ind - 1));
var elementsCopy = elementStack.ToArray() as IEnumerable<T>;
if (option.HasFlag(PermutationsOption.SkipNotDistinct))
elementsCopy = elementsCopy.Distinct();
elementsCopy = elementsCopy.ToArray();
if (elementsList.Any(p => CompareItemEquality(p, elementsCopy, equalityComparer)))
continue;
elementsList.Add(elementsCopy);
while (++elementsIndex < elementsLength);
return elementsList.ToArray();
private bool CompareItemEquality<T>(IEnumerable<T> elements1, IEnumerable<T> elements2, IEqualityComparer<T> equalityComparer = default(IEqualityComparer<T>))
if (equalityComparer == null)
equalityComparer = EqualityComparer<T>.Default;
return (elements2.Count() == elements2.Count()) && (elements2.All(p => elements1.Contains(p, equalityComparer)));
【讨论】:
【参考方案7】:Lippert 先生的答案的输出可以看作是元素在 4 个槽中 0 和 2 之间的所有可能分布。 比如 0 3 1 读作“没有 0,三个 1 和一个 2” 这远没有利珀特先生的回答那么优雅,但至少效率不低
public static void Main()
var distributions = Distributions(4, 3);
PrintSequences(distributions);
/// <summary>
/// Entry point for the other recursive overload
/// </summary>
/// <param name="length">Number of elements in the output</param>
/// <param name="range">Number of distinct values elements can take</param>
/// <returns></returns>
static List<int[]> Distributions(int length, int range)
var distribution = new int[range];
var distributions = new List<int[]>();
Distributions(0, length, distribution, 0, distributions);
distributions.Reverse();
return distributions;
/// <summary>
/// Recursive methode. Not to be called directly, only from other overload
/// </summary>
/// <param name="index">Value of the (possibly) last added element</param>
/// <param name="length">Number of elements in the output</param>
/// <param name="distribution">Distribution among element distinct values</param>
/// <param name="sum">Exit condition of the recursion. Incremented if element added from parent call</param>
/// <param name="distributions">All possible distributions</param>
static void Distributions(int index,
int length,
int[] distribution,
int sum,
List<int[]> distributions)
//Uncomment for exactness check
//System.Diagnostics.Debug.Assert(distribution.Sum() == sum);
if (sum == length)
distributions.Add(distribution.Reverse().ToArray());
for (; index < distribution.Length; index++)
sum -= distribution[index];
distribution[index] = 0;
return;
if (index < distribution.Length)
Distributions(index + 1, length, distribution, sum, distributions);
distribution[index]++;
Distributions(index, length, distribution, ++sum, distributions);
static void PrintSequences(List<int[]> distributions)
for (int i = 0; i < distributions.Count; i++)
for (int j = distributions[i].Length - 1; j >= 0; j--)
for (int k = 0; k < distributions[i][j]; k++)
Console.Write("0:D1 ", distributions[i].Length - 1 - j);
Console.WriteLine();
【讨论】:
【参考方案8】: public static IList<IList<T>> Permutation<T>(ImmutableList<ImmutableList<T>> dimensions)
IList<IList<T>> result = new List<IList<T>>();
Step(ImmutableList.Create<T>(), dimensions, result);
return result;
private static void Step<T>(ImmutableList<T> previous,
ImmutableList<ImmutableList<T>> rest,
IList<IList<T>> result)
if (rest.IsEmpty)
result.Add(previous);
return;
var first = rest[0];
rest = rest.RemoveAt(0);
foreach (var label in first)
Step(previous.Add(label), rest, result);
【讨论】:
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