快速排列->数字->排列映射算法

要描述n个元素的排列,您可以看到,对于第一个元素结束的位置,您有n个可能性,所以您可以用一个介于0和n-1之间的数字来描述它。对于下一个元素结束的位置,您还有n-1个可能,所以您可以用一个介于0和n-2之间的数字来描述它。
等等,直到你有N个数字。

以n=5为例,考虑 abcde 到 caebd .

• a ,第一个元素,在第二个位置结束,所以我们为它分配索引 1个 .
• b 最后到达第四个位置,也就是索引3,但它是剩下的第三个,所以我们将其分配给 二 .
• c 在第一个剩余位置结束,这是 零 .
• d 最后一个剩余位置结束,其中(只有两个剩余位置)是 一 .
• e 在唯一剩余的位置结束,索引在 零 .

所以我们有了索引序列 { 1, 2, 0,1, 0 } .

现在您知道了,例如在二进制数中,“xyz”表示z+2y+4x。对于十进制数,
是z+10y+100x,每个数字乘以某个权重,结果相加。当然,权重中的明显模式是,权重是w=b^k,其中b是数字的基数,k是数字的索引。(我总是从右边开始计算数字,从索引0开始计算最右边的数字。同样,当我谈论第一个数字时,我指的是最右边的数字。)

这个 原因 为什么数字的权重遵循这种模式,是指可以用从0到k的数字表示的最高数字必须正好比只能用数字k+1表示的最低数字低1。在二进制中,0111必须小于1000。十进制中,099999必须小于100000。

编码到变量基
后面的数字之间的间距正好是1,这是一个重要的规则。意识到这一点,我们可以用 可变基数 . 每个数字的基数是该数字不同可能性的数量。对于十进制,每个数字有10种可能,对于我们的系统,最右边的数字有1种可能,最左边的数字有N种可能。但是,由于最右边的数字(我们序列中的最后一个数字)总是0,所以我们不考虑它。这意味着我们剩下的基数是2到n。一般来说,K位数的基数是b[k]=k+2。数字k允许的最大值是h[k]=b[k]-1=k+1。

我们关于数字权重w[k]的规则要求h[i]*w[i]的和(其中i从i=0到i=k)等于1*w[k+1]。反复声明,w[k+1]=w[k]+h[k]*w[k]=w[k]*(h[k]+1)。第一个权重w[0]应始终为1。从这里开始,我们有以下值:

k    h[k] w[k]    

0    1    1  
1    2    2    
2    3    6    
3    4    24   
...  ...  ...
n-1  n    n!  

(一般关系w[k-1]=k!很容易通过归纳法证明。)

我们从转换序列中得到的数字将是s[k]*w[k]的和,k从0到n-1。这里s[k]是序列的k'th(最右边,从0开始)元素。以我们的1、2、0、1、0为例,去掉最右边的元素,如前所述: { 1, 2, 0,1 } . 我们的总数是1*1+0*2+2*6+1*24= 三十七 .

注意,如果我们取每个索引的最大位置,我们会得到4、3、2、1、0,然后转换为119。因为我们选择了数字编码中的权重,这样我们就不会跳过任何数字,所以所有数字0到119都是有效的。其中正好有120个,就是n!对于我们的例子中的n=5,精确地表示不同排列的数目。所以您可以看到我们的编码数字完全指定了所有可能的排列。

从可变基解码
解码类似于转换为二进制或十进制。常见的算法是:

int number = 42;
int base = 2;
int[] bits = new int[n];

for (int k = 0; k < bits.Length; k++)
{
    bits[k] = number % base;
    number = number / base;
}

对于我们的可变基数:

int n = 5;
int number = 37;

int[] sequence = new int[n - 1];
int base = 2;

for (int k = 0; k < sequence.Length; k++)
{
    sequence[k] = number % base;
    number = number / base;

    base++; // b[k+1] = b[k] + 1
}

这正确地将我们的37解码回1、2、0、1( sequence 将是 {1, 0, 2, 1} 在这个代码示例中,但是无论如何…只要你的索引合适)。我们只需要在右端添加0(记住最后一个元素的新位置总是只有一个可能性),就可以恢复我们原来的序列1、2、0、1、0。

使用索引序列排列列表
您可以使用下面的算法根据特定的索引序列排列列表。不幸的是,这是一个O(n_2)算法。

int n = 5;
int[] sequence = new int[] { 1, 2, 0, 1, 0 };
char[] list = new char[] { 'a', 'b', 'c', 'd', 'e' };
char[] permuted = new char[n];
bool[] set = new bool[n];

for (int i = 0; i < n; i++)
{
    int s = sequence[i];
    int remainingPosition = 0;
    int index;

    // Find the s'th position in the permuted list that has not been set yet.
    for (index = 0; index < n; index++)
    {
        if (!set[index])
        {
            if (remainingPosition == s)
                break;

            remainingPosition++;
        }
    }

    permuted[index] = list[i];
    set[index] = true;
}

排列的共同表示法
通常情况下,您不会像我们所做的那样毫无意义地表示置换,而是只通过应用置换后每个元素的绝对位置来表示。我们的示例1、2、0、1、0用于 ABCDE 到 CAEBD 通常用1、3、0、4、2表示。从0到4的每个索引(或者通常是从0到N-1)在此表示中只出现一次。

以这种形式应用排列很容易:

int[] permutation = new int[] { 1, 3, 0, 4, 2 };

char[] list = new char[] { 'a', 'b', 'c', 'd', 'e' };
char[] permuted = new char[n];

for (int i = 0; i < n; i++)
{
    permuted[permutation[i]] = list[i];
}

倒置非常相似:

for (int i = 0; i < n; i++)
{
    list[i] = permuted[permutation[i]];
}

从我们的表象到共同表象的转换
注意,如果我们使用我们的索引序列对一个列表进行排序,并将其应用于标识排列0、1、2、…、n-1,我们得到 逆 排列,以共同形式表示。( { 2, 0, 4,1, 3 } 在我们的例子中)。

为了得到非倒置的预变形,我们应用了我刚才展示的置换算法:

int[] identity = new int[] { 0, 1, 2, 3, 4 };
int[] inverted = { 2, 0, 4, 1, 3 };
int[] normal = new int[n];

for (int i = 0; i < n; i++)
{
    normal[identity[i]] = list[i];
}

或者,您可以直接应用置换,使用逆置换算法:

char[] list = new char[] { 'a', 'b', 'c', 'd', 'e' };
char[] permuted = new char[n];

int[] inverted = { 2, 0, 4, 1, 3 };

for (int i = 0; i < n; i++)
{
    permuted[i] = list[inverted[i]];
}

注意,所有处理常用形式排列的算法都是O(n),而在我们的形式中应用排列的算法是O(n_2)。如果需要多次应用置换,请首先将其转换为公共表示。

我在O(n)中做了一个算法,你可以在这里得到我的函数: http://antoinecomeau.blogspot.ca/2014/07/mapping-between-permutations-and.html

public static int[] perm(int n, int k)
{
    int i, ind, m=k;
    int[] permuted = new int[n];
    int[] elems = new int[n];

    for(i=0;i<n;i++) elems[i]=i;

    for(i=0;i<n;i++)
    {
            ind=m%(n-i);
            m=m/(n-i);
            permuted[i]=elems[ind];
            elems[ind]=elems[n-i-1];
    }

    return permuted;
}

public static int inv(int[] perm)
{
    int i, k=0, m=1;
    int n=perm.length;
    int[] pos = new int[n];
    int[] elems = new int[n];

    for(i=0;i<n;i++) {pos[i]=i; elems[i]=i;}

    for(i=0;i<n-1;i++)
    {
            k+=m*pos[perm[i]];
            m=m*(n-i);
            pos[elems[n-i-1]]=pos[perm[i]];
            elems[pos[perm[i]]]=elems[n-i-1];
    }

    return k;
}

复杂性可降至n*log(n),见第10.1.1节。 fxtbook的(“Lehmer代码(倒置表)”,第232ff页): http://www.jjj.de/fxt/#fxtbook 对于快速方法,请跳到第10.1.1.1节(“大数组计算”P.235)。 (GPUL,C++)代码在同一个网页上。

每个元素可以位于七个位置中的一个。要描述一个元素的位置,需要三位。这意味着您可以以32位值存储所有元素的位置。这远不是有效的,因为这种表示甚至允许所有元素处于相同的位置,但我相信位屏蔽应该相当快。

但是,有8个以上的职位,你需要更漂亮的东西。

这恰好是内置函数 J :

   A. 1 2 3 4 5 6 7
0
   0 A. 1 2 3 4 5 6 7
1 2 3 4 5 6 7

   ?!7
5011
   5011 A. 1 2 3 4 5 6 7
7 6 4 5 1 3 2
   A. 7 6 4 5 1 3 2
5011

问题解决了。但是,我不确定这些年后你是否仍然需要这个解决方案。哈哈,我刚加入这个网站,所以… 检查我的Java置换类。您可以基于索引获取符号排列,或者给出符号排列,然后获取索引。

这是我的预备班

/**
 ****************************************************************************************************************
 * Copyright 2015 Fred Pang fred@pnode.com
 ****************************************************************************************************************
 * A complete list of Permutation base on an index.
 * Algorithm is invented and implemented by Fred Pang fred@pnode.com
 * Created by Fred Pang on 18/11/2015.
 ****************************************************************************************************************
 * LOL this is my first Java project. Therefore, my code is very much like C/C++. The coding itself is not
 * very professional. but...
 *
 * This Permutation Class can be use to generate a complete list of all different permutation of a set of symbols.
 * nPr will be n!/(n-r)!
 * the user can input       n = the number of items,
 *                          r = the number of slots for the items,
 *                          provided n >= r
 *                          and a string of single character symbols
 *
 * the program will generate all possible permutation for the condition.
 *
 * Say if n = 5, r = 3, and the string is "12345", it will generate sll 60 different permutation of the set
 * of 3 character strings.
 *
 * The algorithm I used is base on a bin slot.
 * Just like a human or simply myself to generate a permutation.
 *
 * if there are 5 symbols to chose from, I'll have 5 bin slot to indicate which symbol is taken.
 *
 * Note that, once the Permutation object is initialized, or after the constructor is called, the permutation
 * table and all entries are defined, including an index.
 *
 * eg. if pass in value is 5 chose 3, and say the symbol string is "12345"
 * then all permutation table is logically defined (not physically to save memory).
 * It will be a table as follows
 *  index  output
 *      0   123
 *      1   124
 *      2   125
 *      3   132
 *      4   134
 *      5   135
 *      6   143
 *      7   145
 *      :     :
 *      58  542
 *      59  543
 *
 * all you need to do is call the "String PermGetString(int iIndex)" or the "int[] PermGetIntArray(int iIndex)"
 * function or method with an increasing iIndex, starting from 0 to getiMaxIndex() - 1. It will return the string
 * or the integer array corresponding to the index.
 *
 * Also notice that in the input string is "12345" of  position 01234, and the output is always in accenting order
 * this is how the permutation is generated.
 *
 * ***************************************************************************************************************
 * ====  W a r n i n g  ====
 * ***************************************************************************************************************
 *
 * There is very limited error checking in this class
 *
 * Especially the  int PermGetIndex(int[] iInputArray)  method
 * if the input integer array contains invalid index, it WILL crash the system
 *
 * the other is the string of symbol pass in when the object is created, not sure what will happen if the
 * string is invalid.
 * ***************************************************************************************************************
 *
 */
public class Permutation
{
    private boolean bGoodToGo = false;      // object status
    private boolean bNoSymbol = true;
    private BinSlot slot;                   // a bin slot of size n (input)
    private int nTotal;                     // n number for permutation
    private int rChose;                     // r position to chose
    private String sSymbol;                 // character string for symbol of each choice
    private String sOutStr;
    private int iMaxIndex;                  // maximum index allowed in the Get index function
    private int[] iOutPosition;             // output array
    private int[] iDivisorArray;            // array to do calculation

    public Permutation(int inCount, int irCount, String symbol)
    {
        if (inCount >= irCount)
        {
            // save all input values passed in
            this.nTotal = inCount;
            this.rChose = irCount;
            this.sSymbol = symbol;

            // some error checking
            if (inCount < irCount || irCount <= 0)
                return;                                 // do nothing will not set the bGoodToGo flag

            if (this.sSymbol.length() >= inCount)
            {
                bNoSymbol = false;
            }

            // allocate output storage
            this.iOutPosition = new int[this.rChose];

            // initialize the bin slot with the right size
            this.slot = new BinSlot(this.nTotal);

            // allocate and initialize divid array
            this.iDivisorArray = new int[this.rChose];

            // calculate default values base on n & r
            this.iMaxIndex = CalPremFormula(this.nTotal, this.rChose);

            int i;
            int j = this.nTotal - 1;
            int k = this.rChose - 1;

            for (i = 0; i < this.rChose; i++)
            {
                this.iDivisorArray[i] = CalPremFormula(j--, k--);
            }
            bGoodToGo = true;       // we are ready to go
        }
    }

    public String PermGetString(int iIndex)
    {
        if (!this.bGoodToGo) return "Error: Object not initialized Correctly";
        if (this.bNoSymbol) return "Error: Invalid symbol string";
        if (!this.PermEvaluate(iIndex)) return "Invalid Index";

        sOutStr = "";
        // convert string back to String output
        for (int i = 0; i < this.rChose; i++)
        {
            String sTempStr = this.sSymbol.substring(this.iOutPosition[i], iOutPosition[i] + 1);
            this.sOutStr = this.sOutStr.concat(sTempStr);
        }
        return this.sOutStr;
    }

    public int[] PermGetIntArray(int iIndex)
    {
        if (!this.bGoodToGo) return null;
        if (!this.PermEvaluate(iIndex)) return null ;
        return this.iOutPosition;
    }

    // given an int array, and get the index back.
    //
    //  ====== W A R N I N G ======
    //
    // there is no error check in the array that pass in
    // if any invalid value in the input array, it can cause system crash or other unexpected result
    //
    // function pass in an int array generated by the PermGetIntArray() method
    // then return the index value.
    //
    // this is the reverse of the PermGetIntArray()
    //
    public int PermGetIndex(int[] iInputArray)
    {
        if (!this.bGoodToGo) return -1;
        return PermDoReverse(iInputArray);
    }


    public int getiMaxIndex() {
    return iMaxIndex;
}

    // function to evaluate nPr = n!/(n-r)!
    public int CalPremFormula(int n, int r)
    {
        int j = n;
        int k = 1;
        for (int i = 0; i < r; i++, j--)
        {
            k *= j;
        }
        return k;
    }


//  PermEvaluate function (method) base on an index input, evaluate the correspond permuted symbol location
//  then output it to the iOutPosition array.
//
//  In the iOutPosition[], each array element corresponding to the symbol location in the input string symbol.
//  from location 0 to length of string - 1.

    private boolean PermEvaluate(int iIndex)
    {
        int iCurrentIndex;
        int iCurrentRemainder;
        int iCurrentValue = iIndex;
        int iCurrentOutSlot;
        int iLoopCount;

        if (iIndex >= iMaxIndex)
            return false;

        this.slot.binReset();               // clear bin content
        iLoopCount = 0;
        do {
            // evaluate the table position
            iCurrentIndex = iCurrentValue / this.iDivisorArray[iLoopCount];
            iCurrentRemainder = iCurrentValue % this.iDivisorArray[iLoopCount];

            iCurrentOutSlot = this.slot.FindFreeBin(iCurrentIndex);     // find an available slot
            if (iCurrentOutSlot >= 0)
                this.iOutPosition[iLoopCount] = iCurrentOutSlot;
            else return false;                                          // fail to find a slot, quit now

            this.slot.setStatus(iCurrentOutSlot);                       // set the slot to be taken
            iCurrentValue = iCurrentRemainder;                          // set new value for current value.
            iLoopCount++;                                               // increase counter
        } while (iLoopCount < this.rChose);

        // the output is ready in iOutPosition[]
        return true;
    }

    //
    // this function is doing the reverse of the permutation
    // the input is a permutation and will find the correspond index value for that entry
    // which is doing the opposit of the PermEvaluate() method
    //
    private int PermDoReverse(int[] iInputArray)
    {
        int iReturnValue = 0;
        int iLoopIndex;
        int iCurrentValue;
        int iBinLocation;

        this.slot.binReset();               // clear bin content

        for (iLoopIndex = 0; iLoopIndex < this.rChose; iLoopIndex++)
        {
            iCurrentValue = iInputArray[iLoopIndex];
            iBinLocation = this.slot.BinCountFree(iCurrentValue);
            this.slot.setStatus(iCurrentValue);                          // set the slot to be taken
            iReturnValue = iReturnValue + iBinLocation * this.iDivisorArray[iLoopIndex];
        }
        return iReturnValue;
    }


    /*******************************************************************************************************************
     *******************************************************************************************************************
     * Created by Fred on 18/11/2015.   fred@pnode.com
     *
     * *****************************************************************************************************************
     */
    private static class BinSlot
    {
        private int iBinSize;       // size of array
        private short[] eStatus;    // the status array must have length iBinSize

        private BinSlot(int iBinSize)
        {
            this.iBinSize = iBinSize;               // save bin size
            this.eStatus = new short[iBinSize];     // llocate status array
        }

        // reset the bin content. no symbol is in use
        private void binReset()
        {
            // reset the bin's content
            for (int i = 0; i < this.iBinSize; i++) this.eStatus[i] = 0;
        }

        // set the bin position as taken or the number is already used, cannot be use again.
        private void  setStatus(int iIndex) { this.eStatus[iIndex]= 1; }

        //
        // to search for the iIndex th unused symbol
        // this is important to search through the iindex th symbol
        // because this is how the table is setup. (or the remainder means)
        // note: iIndex is the remainder of the calculation
        //
        // for example:
        // in a 5 choose 3 permutation symbols "12345",
        // the index 7 item (count starting from 0) element is "1 4 3"
        // then comes the index 8, 8/12 result 0 -> 0th symbol in symbol string = '1'
        // remainder 8. then 8/3 = 2, now we need to scan the Bin and skip 2 unused bins
        //              current the bin looks 0 1 2 3 4
        //                                    x o o o o     x -> in use; o -> free only 0 is being used
        //                                      s s ^       skipped 2 bins (bin 1 and 2), we get to bin 3
        //                                                  and bin 3 is the bin needed. Thus symbol "4" is pick
        // in 8/3, there is a remainder 2 comes in this function as 2/1 = 2, now we have to pick the empty slot
        // for the new 2.
        // the bin now looks 0 1 2 3 4
        //                   x 0 0 x 0      as bin 3 was used by the last value
        //                     s s   ^      we skip 2 free bins and the next free bin is bin 4
        //                                  therefor the symbol "5" at the symbol array is pick.
        //
        // Thus, for index 8  "1 4 5" is the symbols.
        //
        //
        private int FindFreeBin(int iIndex)
        {
            int j = iIndex;

            if (j < 0 || j > this.iBinSize) return -1;               // invalid index

            for (int i = 0; i < this.iBinSize; i++)
            {
                if (this.eStatus[i] == 0)       // is it used
                {
                    // found an empty slot
                    if (j == 0)                 // this is a free one we want?
                        return i;               // yes, found and return it.
                    else                        // we have to skip this one
                        j--;                    // else, keep looking and count the skipped one
                }
            }
            assert(true);           // something is wrong
            return -1;              // fail to find the bin we wanted
        }

        //
        // this function is to help the PermDoReverse() to find out what is the corresponding
        // value during should be added to the index value.
        //
        // it is doing the opposite of int FindFreeBin(int iIndex) method. You need to know how this
        // FindFreeBin() works before looking into this function.
        //
        private int BinCountFree(int iIndex)
        {
            int iRetVal = 0;
            for (int i = iIndex; i > 0; i--)
            {
                if (this.eStatus[i-1] == 0)       // it is free
                {
                    iRetVal++;
                }
            }
            return iRetVal;
        }
    }
}
// End of file - Permutation.java

这是我的主要课程,展示如何使用这个课程。

/*
 * copyright 2015 Fred Pang
 *
 * This is the main test program for testing the Permutation Class I created.
 * It can be use to demonstrate how to use the Permutation Class and its methods to generate a complete
 * list of a permutation. It also support function to get back the index value as pass in a permutation.
 *
 * As you can see my Java is not very good. :)
 * This is my 1st Java project I created. As I am a C/C++ programmer for years.
 *
 * I still have problem with the Scanner class and the System class.
 * Note that there is only very limited error checking
 *
 *
 */

import java.util.Scanner;

public class Main
{
    private static Scanner scanner = new Scanner(System.in);

    public static void main(String[] args)
    {
        Permutation perm;       // declear the object
        String sOutString = "";
        int nCount;
        int rCount;
        int iMaxIndex;

        // Get user input
        System.out.println("Enter n: ");
        nCount = scanner.nextInt();

        System.out.println("Enter r: ");
        rCount = scanner.nextInt();

        System.out.println("Enter Symbol: ");
        sOutString = scanner.next();

        if (sOutString.length() < rCount)
        {
            System.out.println("String too short, default to numbers");
            sOutString = "";
        }

        // create object with user requirement
        perm = new Permutation(nCount, rCount, sOutString);

        // and print the maximum count
        iMaxIndex = perm.getiMaxIndex();
        System.out.println("Max count is:" + iMaxIndex);

        if (!sOutString.isEmpty())
        {
            for (int i = 0; i < iMaxIndex; i++)
            {   // print out the return permutation symbol string
                System.out.println(i + " " + perm.PermGetString(i));
            }
        }
        else
        {
            for (int i = 0; i < iMaxIndex; i++)
            {
                System.out.print(i + " ->");

                // Get the permutation array
                int[] iTemp = perm.PermGetIntArray(i);

                // print out the permutation
                for (int j = 0; j < rCount; j++)
                {
                    System.out.print(' ');
                    System.out.print(iTemp[j]);
                }

                // to verify my PermGetIndex() works. :)
                if (perm.PermGetIndex(iTemp)== i)
                {
                    System.out.println(" .");
                }
                else
                {   // oops something is wrong :(
                    System.out.println(" ***************** F A I L E D *************************");
                    assert(true);
                    break;
                }
            }
        }
    }
}
//
// End of file - Main.java

玩得高兴。:)

可以使用递归算法对排列进行编码。如果n-排列(数字0,..,n-1的某种顺序)是x,…的形式,那么将其编码为x+n*在数字0,n-1-x上用“…”表示的(n-1)-排列的编码。听起来像一口,下面是一些代码:

// perm[0]..perm[n-1] must contain the numbers in {0,..,n-1} in any order.
int permToNumber(int *perm, int n) {
  // base case
  if (n == 1) return 0;

  // fix up perm[1]..perm[n-1] to be a permutation on {0,..,n-2}.
  for (int i = 1; i < n; i++) {
    if (perm[i] > perm[0]) perm[i]--;
  }

  // recursively compute
  return perm[0] + n * permToNumber(perm + 1, n - 1);
}

// number must be >=0, < n!
void numberToPerm(int number, int *perm, int n) {
  if (n == 1) {
    perm[0] = 0;
    return;
  }
  perm[0] = number % n;
  numberToPerm(number / n, perm + 1, n - 1);

  // fix up perm[1] .. perm[n-1]
  for (int i = 1; i < n; i++) {
    if (perm[i] >= perm[0]) perm[i]++;
  }
}

这个算法是O(n^2)。如果有人有O(N)算法,则可获得奖励积分。

多么有趣的问题!

如果所有元素都是数字,那么您可能需要考虑将它们从字符串转换为实际数字。然后,您将能够通过将排列排列顺序排列来对所有排列进行排序,并将它们放置在一个数组中。在那之后,您将可以使用任何一种不同的搜索算法。

我在先前的回答(删除)中很仓促,但我确实有实际的答案。它由类似的概念提供,即 factoradic 和排列有关(我的答案与组合有关,我为这种混淆道歉)。我不喜欢仅仅发布维基百科链接,但我写了一段时间前我做的事情,因为某些原因是不明白的。因此,如果有要求,我可以稍后对此进行扩展。

有一本关于这个的书。对不起,我不记得它的名字了(你很可能会在维基百科上找到它)。 但无论如何,我写了一个枚举系统的python实现: http://kks.cabal.fi/Kombinaattori 其中一些是芬兰语的,但只需复制代码和名称变量…

一个相关的问题是计算逆排列,当只知道排列数组时,它将排列向量恢复到原来的顺序。以下是O(N)代码(PHP):

// Compute the inverse of a permutation
function GetInvPerm($Perm)
    {
    $n=count($Perm);
    $InvPerm=[];
    for ($i=0; $i<$n; ++$i)
        $InvPerm[$Perm[$i]]=$i;
    return $InvPerm;
    } // GetInvPerm

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