## Fast permutation -> number -> permutation mapping algorithms

fastest permutation algorithm
recursive permutation algorithm
permutation of a number
permutation of n numbers
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all permutations of an array
permutation of numbers from 1 to n

I have n elements. For the sake of an example, let's say, 7 elements, 1234567. I know there are 7! = 5040 permutations possible of these 7 elements.

I want a fast algorithm comprising two functions:

f(number) maps a number between 0 and 5039 to a unique permutation, and

f'(permutation) maps the permutation back to the number that it was generated from.

I don't care about the correspondence between number and permutation, providing each permutation has its own unique number.

So, for instance, I might have functions where

```f(0) = '1234567'
f'('1234567') = 0
```

The fastest algorithm that comes to mind is to enumerate all permutations and create a lookup table in both directions, so that, once the tables are created, f(0) would be O(1) and f('1234567') would be a lookup on a string. However, this is memory hungry, particularly when n becomes large.

Can anyone propose another algorithm that would work quickly and without the memory disadvantage?

All code variations can be found in https://github.com/egonelbre/exp/blob/master/​permutation/code.go. Let's say you have a permutation of 12  Fast-permutation-entropy Order of ordinal patterns is defined as in [1,3,7,8], i.e. order = n-1 for n defined as in [2]. The values of permutation entropy are normalised by log ( (order+1)!) so that they are from [0,1] as proposed in the original paper [2].

I've found an O(n) algorithm, here's a short explanation 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;
}
```

It shows many fast implementation of permutation algorithm. There is also a contribution by providing a unique way to index permutation  #permutation = math.factorial(n) // math.factorial(n-r) #traditional permutation fast_permutation = 1 for i in range(n, (n-r), -1): fast_permutation = permutation * i Faster way completed in 0.1218 seconds for P(5M, 10K) whereas traditional method completed in 708.82 seconds (11 minutes).

The complexity can be brought down to n*log(n), see section 10.1.1 ("The Lehmer code (inversion table)", p.232ff) of the fxtbook: http://www.jjj.de/fxt/#fxtbook skip to section 10.1.1.1 ("Computation with large arrays" p.235) for the fast method. The (GPLed, C++) code is on the same web page.

Now building up on that, If the string is AB, the number of permutations here consists of putting the character “B” in all the possible positions in each permutation  This handy module makes performing permutation in Perl easy and fast, although perhaps its algorithm is not the fastest on the earth. It supports permutation r of n objects where 0 < r <= n. METHODS new [@list] Returns a permutor object for the given items. next. Returns a list of the items in the next permutation.

Problem solved. However, I am not sure you still need the solution after these years. LOL, I just join this site, so ... Check my Java Permutation Class. You can base on an index to get a symbol permutation, or give a symbol permutation then get the index.

Here is my Premutation Class

```/**
****************************************************************************************************************
* 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
```

and here is my Main Class for showing how to use the class.

```/*
*
* 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
```

Have fun. :)

permutation: Permutation. In Rfast: A Collection of Efficient and Extremely Fast R Functions. Description Usage Arguments Details Value Author(s) See Also  Heap’s algorithm is used to generate all permutations of n objects. The idea is to generate each permutation from the previous permutation by choosing a pair of elements to interchange, without disturbing the other n-2 elements. Following is the illustration of generating all the permutations of n given numbers.

Each element can be in one of seven positions. To describe the position of one element, you would need three bits. That means you can store the position of all the elements in a 32bit value. That's far from being efficient, since this representation would even allow all elements to be in the same position, but I believe the bit-masking should be reasonably fast.

However, with more than 8 positions you'll need something more nifty.

Heap's algorithm is used to generate all permutations of n objects. The idea is to generate each permutation from the previous permutation by choosing a pair of  From that, we can now devise that the number of permutations of a string with a length n equals to the multiplication of: Number of permutations of the string without one character. So permutation(n-1). How many possible positions that character can be positioned in these permutations. Which is equal to n. n* perm(n-1)

METHODS BASED ON EXCHANGES. A natural way to permute an array of elements on a computer is to exchange two of its elements. The fastest permutation. The test is called Fisher's permutation test. Naming it "t-test something" is a misnomer. It is no more t-test than the Wilcoxon rank-sum test (aka Mann-Withney U-test). Because permutations are discrete, "Fisher's permutation test" is actually more closely related to the Wilcoxon test.

Abstract: Given a finite sequence a_1, a_2,\ldots, a_d of d permutations in the symmetric group S_n, and a permutation word k_1k_2\cdots  PLL is the acronym for Permutation of the Last Layer.Permutation of the Last Layer is the last step of many speedsolving methods. In this step, the pieces on the top layer have already been oriented so that the top face has all the same color, and they can now be moved into their solved positions.

Check out all my Algebra 2 Videos and Notes at: http://wowmath.org/Algebra2/​Alg2Notes.html.Duration: 2:10 Posted: Nov 19, 2009 A k-permutation of a multiset M is a sequence of length k of elements of M in which each element appears a number of times less than or equal to its multiplicity in M (an element's repetition number). Circular permutations. Permutations, when considered as arrangements, are sometimes referred to as linearly ordered arrangements. In these

• @IsaacLi, if i am correct, f(4) = {2, 0, 0} = 231. And f'(312) = {1, 1, 0} = 3. For `1234`, f(4) = {0, 2, 0, 0} = 1342. And f'(1423) = {0, 1 1, 0} = 3. This algorithm is really inspiring. I wonder it is the original work from the OP. i have studied and analysed it for a while. And i believe it is correct :)
• How to convert from "our representation" to "common representation", `{1, 2, 0, 1, 0}` --> `{1, 3, 0, 4, 2}`? And vice versa? Is it possible? (by not converting between `{1, 2, 0, 1, 0}` <--> `{C, A, E, B, D}`, which needs O(n^2).) If "our style" and "common style" are not convertible, they are in fact two different separate things, isn't it? Thanks x