Rapide permutation -> nombre> permutation de cartographie des algorithmes

J'ai trouvé un algorithme O(n), voici une courte explication 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;
}

Chaque élément peut être dans l'un des sept postes. Pour décrire la position d'un élément, vous aurait besoin de trois bits. Cela signifie que vous pouvez stocker la position de tous les éléments en 32 bits valeur. C'est loin d'être efficace, puisque cette représentation serait à même de permettre tous les éléments pour être dans la même position, mais je crois que le bit de masquage doit être assez rapide.

Cependant, avec plus de 8 postes, vous aurez besoin de quelque chose de plus astucieux.

Le problème est résolu. Cependant, je ne suis pas sûr que vous avez encore besoin de la solution, après toutes ces années. LOL, je viens de rejoindre ce site, donc ...
Vérifier mon Java Permutation de Classe. Vous pouvez baser sur un index pour obtenir un symbole de permutation, ou de donner un symbole de permutation puis obtenir l'indice.

Voici mon Premutation Classe

/**
****************************************************************************************************************
* Copyright 2015 Fred Pang [email protected]
****************************************************************************************************************
* A complete list of Permutation base on an index.
* Algorithm is invented and implemented by Fred Pang [email protected]
* 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.   [email protected]
*
* *****************************************************************************************************************
*/
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

et voici ma Classe Principale pour montrer comment utiliser la classe.

/*
* 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

Avoir du plaisir. 🙂

Ce une question intéressante!

Si tous les éléments sont des nombres, vous pourriez envisager de les convertir à partir de chaînes de nombres réels. Alors vous seriez en mesure de trier toutes les permutations en les mettant dans l'ordre, et les placer dans un tableau. Après cela, vous seriez ouvert à l'un des différents algorithmes de recherche là-bas.

Il y a un livre écrit sur ce sujet. Désolé, mais je ne me souviens pas le nom de celui-ci (vous trouverez très probablement à partir de wikipedia).
mais de toute façon j'ai écrit un python de mise en œuvre de ce système de recensement: http://kks.cabal.fi/Kombinaattori
Certains de il est en finnois, mais il suffit de copier le code et le nom des variables ...

J'ai eu cette question exacte et pensé que je pourrais donner mon Python solution. Il est O(n^2).

import copy
def permute(string, num):
''' generates a permutation '''
def build_s(factoradic): # Build string from factoradic in list form
string0 = copy.copy(string)
n = []
for i in range(len(factoradic)):
n.append(string0[factoradic[i]])
del string0[factoradic[i]]
return n
f = len(string)
factoradic = []
while(f != 0): # Generate factoradic number list
factoradic.append(num % f)
num = (num - factoradic[-1])//f
f -= 1
return build_s(factoradic)
s = set()
# Print 120 permutations of this string
for i in range(120):
m = permute(list('abcde'), i)
s.add(''.join(m))
print(len(s)) # Check that we have 120 unique permutations

C'est assez simple; après la génération de la factoradic représentation du nombre, je viens de prendre et de supprimer les caractères de la chaîne. La suppression de la chaîne est pourquoi c'est un O(n^2) solution.

Antoine solution est meilleure pour la performance.