小编典典

生成集合的排列(最有效)

c#

我想生成一个集合(一个集合)的所有排列,如下所示:

Collection: 1, 2, 3
Permutations: {1, 2, 3}
              {1, 3, 2}
              {2, 1, 3}
              {2, 3, 1}
              {3, 1, 2}
              {3, 2, 1}

通常,这不是“如何”的问题,而是更多关于如何最有效的问题。另外,我不想生成所有排列并返回它们,而是一次只生成一个排列,并且仅在必要时才继续(很像Iterators,我也曾尝试过,但结果却更少高效)。

我已经测试了许多算法和方法,并提出了这段代码,这是我尝试过的最有效的代码:

public static bool NextPermutation<T>(T[] elements) where T : IComparable<T>
{
    // More efficient to have a variable instead of accessing a property
    var count = elements.Length;

    // Indicates whether this is the last lexicographic permutation
    var done = true;

    // Go through the array from last to first
    for (var i = count - 1; i > 0; i--)
    {
        var curr = elements[i];

        // Check if the current element is less than the one before it
        if (curr.CompareTo(elements[i - 1]) < 0)
        {
            continue;
        }

        // An element bigger than the one before it has been found,
        // so this isn't the last lexicographic permutation.
        done = false;

        // Save the previous (bigger) element in a variable for more efficiency.
        var prev = elements[i - 1];

        // Have a variable to hold the index of the element to swap
        // with the previous element (the to-swap element would be
        // the smallest element that comes after the previous element
        // and is bigger than the previous element), initializing it
        // as the current index of the current item (curr).
        var currIndex = i;

        // Go through the array from the element after the current one to last
        for (var j = i + 1; j < count; j++)
        {
            // Save into variable for more efficiency
            var tmp = elements[j];

            // Check if tmp suits the "next swap" conditions:
            // Smallest, but bigger than the "prev" element
            if (tmp.CompareTo(curr) < 0 && tmp.CompareTo(prev) > 0)
            {
                curr = tmp;
                currIndex = j;
            }
        }

        // Swap the "prev" with the new "curr" (the swap-with element)
        elements[currIndex] = prev;
        elements[i - 1] = curr;

        // Reverse the order of the tail, in order to reset it's lexicographic order
        for (var j = count - 1; j > i; j--, i++)
        {
            var tmp = elements[j];
            elements[j] = elements[i];
            elements[i] = tmp;
        }

        // Break since we have got the next permutation
        // The reason to have all the logic inside the loop is
        // to prevent the need of an extra variable indicating "i" when
        // the next needed swap is found (moving "i" outside the loop is a
        // bad practice, and isn't very readable, so I preferred not doing
        // that as well).
        break;
    }

    // Return whether this has been the last lexicographic permutation.
    return done;
}

它的用法是发送一个元素数组,并返回一个布尔值,指示这是否是最后的字典排列,以及将该数组更改为下一个排列。

用法示例:

var arr = new[] {1, 2, 3};

PrintArray(arr);

while (!NextPermutation(arr))
{
    PrintArray(arr);
}

问题是我对代码的速度不满意。

迭代大小为11的数组的所有排列大约需要4秒钟。尽管可以认为它令人印象深刻,但是因为一组大小11的可能排列数量11!接近4000万。

逻辑上,使用数组12的时间将12!11! * 12,大约要花费12倍的时间,使用数组13的时间将比使用12的时间花费大约13倍的时间,依此类推。

因此,您可以轻松了解如何使用大小为12及更大的数组来进行所有排列,这确实需要很长时间。

而且我有很强的直觉,可以以某种方式节省大量时间(无需切换到C#以外的其他语言-
因为编译器优化确实确实可以很好地进行优化,并且我怀疑我是否可以在Assembly中手动进行优化)。

有谁知道其他方法可以更快地完成此任务?您是否有关于如何使当前算法更快的想法?

请注意,我不想使用外部库或服务来做到这一点-我想拥有代码本身,并且希望它尽可能地高效。


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2020-05-19

共1个答案

小编典典

更新2018-05-28:

有点晚了…

根据最近的测试(更新2018-05-22)

  • 最快的是我的但不是字典顺序
  • 对于最快的词典顺序,Sani Singh Huttunen解决方案似乎是必经之路。

我的机器(millisecs)上发布的10个项目(10!)的性能测试结果:

  • 厄勒:29
  • SimpleVar:95
  • 埃雷兹·罗宾逊(Erez Robinson):156
  • Sani Singh Huttunen:37
  • 彭阳:45047

我的计算机上发布的13个项目(13!)的性能测试结果(秒):

  • 厄勒:48.437
  • SimpleVar:159.869
  • 埃雷兹·罗宾逊(Erez Robinson):327.781
  • Sani Singh Huttunen:64.839

我的解决方案的优势:

  • 堆算法(每个置换单个交换)
  • 没有乘法(就像在网络上看到的一些实现)
  • 内联交换
  • 泛型
  • 没有不安全的代码
  • 就位(内存使用率非常低)
  • 无模(仅比较第一位)

我对Heap算法的实现

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;

namespace WpfPermutations
{
    /// <summary>
    /// EO: 2016-04-14
    /// Generator of all permutations of an array of anything.
    /// Base on Heap's Algorithm. See: https://en.wikipedia.org/wiki/Heap%27s_algorithm#cite_note-3
    /// </summary>
    public static class Permutations
    {
        /// <summary>
        /// Heap's algorithm to find all pmermutations. Non recursive, more efficient.
        /// </summary>
        /// <param name="items">Items to permute in each possible ways</param>
        /// <param name="funcExecuteAndTellIfShouldStop"></param>
        /// <returns>Return true if cancelled</returns> 
        public static bool ForAllPermutation<T>(T[] items, Func<T[], bool> funcExecuteAndTellIfShouldStop)
        {
            int countOfItem = items.Length;

            if (countOfItem <= 1)
            {
                return funcExecuteAndTellIfShouldStop(items);
            }

            var indexes = new int[countOfItem];
            for (int i = 0; i < countOfItem; i++)
            {
                indexes[i] = 0;
            }

            if (funcExecuteAndTellIfShouldStop(items))
            {
                return true;
            }

            for (int i = 1; i < countOfItem;)
            {
                if (indexes[i] < i)
                { // On the web there is an implementation with a multiplication which should be less efficient.
                    if ((i & 1) == 1) // if (i % 2 == 1)  ... more efficient ??? At least the same.
                    {
                        Swap(ref items[i], ref items[indexes[i]]);
                    }
                    else
                    {
                        Swap(ref items[i], ref items[0]);
                    }

                    if (funcExecuteAndTellIfShouldStop(items))
                    {
                        return true;
                    }

                    indexes[i]++;
                    i = 1;
                }
                else
                {
                    indexes[i++] = 0;
                }
            }

            return false;
        }

        /// <summary>
        /// This function is to show a linq way but is far less efficient
        /// From: StackOverflow user: Pengyang : http://stackoverflow.com/questions/756055/listing-all-permutations-of-a-string-integer
        /// </summary>
        /// <typeparam name="T"></typeparam>
        /// <param name="list"></param>
        /// <param name="length"></param>
        /// <returns></returns>
        static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> list, int length)
        {
            if (length == 1) return list.Select(t => new T[] { t });

            return GetPermutations(list, length - 1)
                .SelectMany(t => list.Where(e => !t.Contains(e)),
                    (t1, t2) => t1.Concat(new T[] { t2 }));
        }

        /// <summary>
        /// Swap 2 elements of same type
        /// </summary>
        /// <typeparam name="T"></typeparam>
        /// <param name="a"></param>
        /// <param name="b"></param>
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        static void Swap<T>(ref T a, ref T b)
        {
            T temp = a;
            a = b;
            b = temp;
        }

        /// <summary>
        /// Func to show how to call. It does a little test for an array of 4 items.
        /// </summary>
        public static void Test()
        {
            ForAllPermutation("123".ToCharArray(), (vals) =>
            {
                Console.WriteLine(String.Join("", vals));
                return false;
            });

            int[] values = new int[] { 0, 1, 2, 4 };

            Console.WriteLine("Ouellet heap's algorithm implementation");
            ForAllPermutation(values, (vals) =>
            {
                Console.WriteLine(String.Join("", vals));
                return false;
            });

            Console.WriteLine("Linq algorithm");
            foreach (var v in GetPermutations(values, values.Length))
            {
                Console.WriteLine(String.Join("", v));
            }

            // Performance Heap's against Linq version : huge differences
            int count = 0;

            values = new int[10];
            for (int n = 0; n < values.Length; n++)
            {
                values[n] = n;
            }

            Stopwatch stopWatch = new Stopwatch();

            ForAllPermutation(values, (vals) =>
            {
                foreach (var v in vals)
                {
                    count++;
                }
                return false;
            });

            stopWatch.Stop();
            Console.WriteLine($"Ouellet heap's algorithm implementation {count} items in {stopWatch.ElapsedMilliseconds} millisecs");

            count = 0;
            stopWatch.Reset();
            stopWatch.Start();

            foreach (var vals in GetPermutations(values, values.Length))
            {
                foreach (var v in vals)
                {
                    count++;
                }
            }

            stopWatch.Stop();
            Console.WriteLine($"Linq {count} items in {stopWatch.ElapsedMilliseconds} millisecs");
        }
    }
}

这是我的测试代码:

Task.Run(() =>
            {

                int[] values = new int[12];
                for (int n = 0; n < values.Length; n++)
                {
                    values[n] = n;
                }

                // Eric Ouellet Algorithm
                int count = 0;
                var stopwatch = new Stopwatch();
                stopwatch.Reset();
                stopwatch.Start();
                Permutations.ForAllPermutation(values, (vals) =>
                {
                    foreach (var v in vals)
                    {
                        count++;
                    }
                    return false;
                });
                stopwatch.Stop();
                Console.WriteLine($"This {count} items in {stopwatch.ElapsedMilliseconds} millisecs");

                // Simple Plan Algorithm
                count = 0;
                stopwatch.Reset();
                stopwatch.Start();
                PermutationsSimpleVar permutations2 = new PermutationsSimpleVar();
                permutations2.Permutate(1, values.Length, (int[] vals) =>
                {
                    foreach (var v in vals)
                    {
                        count++;
                    }
                });
                stopwatch.Stop();
                Console.WriteLine($"Simple Plan {count} items in {stopwatch.ElapsedMilliseconds} millisecs");

                // ErezRobinson Algorithm
                count = 0;
                stopwatch.Reset();
                stopwatch.Start();
                foreach(var vals in PermutationsErezRobinson.QuickPerm(values))
                {
                    foreach (var v in vals)
                    {
                        count++;
                    }
                };
                stopwatch.Stop();
                Console.WriteLine($"Erez Robinson {count} items in {stopwatch.ElapsedMilliseconds} millisecs");
            });

用法示例:

ForAllPermutation("123".ToCharArray(), (vals) =>
    {
        Console.WriteLine(String.Join("", vals));
        return false;
    });

int[] values = new int[] { 0, 1, 2, 4 };
ForAllPermutation(values, (vals) =>
        {
            Console.WriteLine(String.Join("", vals));
            return false;
        });
2020-05-19