我有n个元素.为了举个例子,让我们说,7个元素,1234567.我知道有7个!=这些7个元素可能有5040个排列.
我想要一个包含两个函数的快速算法:
f(number)将0到5039之间的数字映射到唯一的排列,并且
f'(置换)将置换映射回其生成的数字.
我不关心数字和排列之间的对应关系,只要每个排列都有自己唯一的数字.
所以,举个例子,我可能会在哪里有功能
f(0) = '1234567'
f'('1234567') = 0
想到的最快的算法是枚举所有排列并在两个方向上创建查找表,这样,一旦创建表,f(0)将是O(1)并且f('1234567')将是查找字符串.然而,这是内存饥饿,特别是当n变大时.
任何人都可以提出另一种算法,它可以快速工作,没有内存缺点吗？
要描述n个元素的排列,您会看到对于第一个元素结束的位置,您有n种可能性,因此您可以使用0到n-1之间的数字来描述它.对于下一个元素结束的位置,您有n-1个剩余的可能性,因此您可以使用0到n-2之间的数字来描述它.
等等,直到你有n个数字.
作为对于n = 5的示例中,考虑使置换abcde到caebd.
a,第一个元素,最终在第二个位置,所以我们为它指定索引1.
b最终在第四个位置,这将是索引3,但它是剩下的第三个位置,所以我们分配它2.
c结束于第一个剩余的位置,始终为0.
d结束于最后剩余的位置,其中(仅剩下两个位置)为1.
e最终在唯一的剩余位置,索引为0.
所以我们有索引序列{1,2,0,1,0}.
现在您知道例如二进制数,'xyz'表示z + 2y + 4x.对于十进制数,
它是z + 10y + 100x.每个数字乘以一些权重,并将结果相加.权重中的明显模式当然是权重是w = b ^ k,其中b是数字的基数,k是数字的索引.(我将始终计算右边的数字,从最右边的数字开始在索引0处.同样,当我谈到'第一个'数字时,我指的是最右边的数字.)
的理由为什么数字的权重遵循此模式是可以通过从0到k中的数字来表示的最高数目必须正好1比,可以仅通过使用数字K + 1来表示的最低数目更低.在二进制中,0111必须小于1000.在十进制中,099999必须低于100000.
编码到变量基数
后续数字之间的间距恰好为1是重要的规则.意识到这一点,我们可以用变量基数表示我们的索引序列.每个数字的基数是该数字的不同可能性的数量.对于十进制,每个数字有10种可能性,对于我们的系统,最右边的数字有1种可能性,最左边的数字有n种可能性.但由于最右边的数字(我们序列中的最后一个数字)始终为0,我们将其遗漏.这意味着我们留下了2到n的基数.通常,第k个数字将具有基数b [k] = k + 2.数字k允许的最高值是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]之和,其中i从i = 0变为i = k,等于1*w [k + 1].反复陈述,w [k + 1] = w [k] + h [k]*w [k] = w [k]*(h [k] + 1).第一个权重w [0]应该始终为1.从那里开始,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!很容易通过归纳证明.)
我们从转换序列得到的数字将是s [k]*w [k]的总和,其中k从0到n-1.这里s [k]是序列的第k个(最右边,从0开始)元素.举一个例子,取我们的{1,2,0,1,0},如前所述剥去最右边的元素:{ 1,2,0,1 }.我们的总和是1*1 + 0*2 + 2*6 + 1*24 = 37.
请注意,如果我们为每个索引取最大位置,我们将有{4,3,2,1,0},并且转换为119.由于我们的数字编码中的权重被选中以便我们不会跳过任何数字,所有数字0到119都有效.其中正好有120个,这是n!对于我们的例子中的n = 5,恰好是不同排列的数量.因此,您可以看到我们的编码数字完全指定了所有可能的排列.
从基于变量的
解码解码类似于转换为二进制或十进制.常见的算法是这样的:
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{1, 0, 2, 1}在此代码示例中,但无论如何......只要您正确索引).我们只需要在右端添加0(记住最后一个元素总是只有一个可能的新位置)来恢复原始序列{1,2,0,1,0}.
使用索引序列
置换列表您可以使用以下算法根据特定索引序列置换列表.不幸的是,这是一个O(n²)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;
}
排列的常见表示
通常,您不会像我们所做的那样直观地表示排列,而只是表示在应用排列后每个元素的绝对位置.我们的例子中{1,2,0,1,0}为abcde到caebd通常由下式表示的{1,3,0,4,2}.从0到4(或通常,0到n-1)的每个索引在该表示中恰好出现一次.
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]];
}
从我们的表示转换为公共表示
请注意,如果我们使用我们的算法使用我们的索引序列置换列表,并将其应用于身份置换{0,1,2,...,n-1},我们得到逆置换,以常见形式表示.(在我们的例子中{2,0,4,1,3}).
为了得到非倒置的前置,我们应用我刚刚展示的置换算法:
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];
}
或者您可以通过使用逆置换算法直接应用置换:
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,用于处理常见形式的排列的所有算法都是O(n),而在我们的形式中应用置换是O(n²).如果需要多次应用置换,请先将其转换为通用表示.
我在O(n)中创建了一个算法,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
3> 小智..：
复杂性可以降低到n*log(n),参见fxtbook的第10.1.1节("Lehmer代码(反演表)",第2332页):http://www.jjj.de/fxt/
#fxtbook
跳转到快速方法的第10.1.1.1节("使用大型数组进行计算"第235页).(GPLed,C++)代码位于同一网页上.
4> 小智..：
问题解决了。但是，我不确定这些年来您是否仍需要解决方案。大声笑，我只是加入此站点，所以...检查我的Java置换类。您可以基于索引来获取符号排列，或者给出符号排列然后获取索引。
这是我的预变异班
/**
****************************************************************************************************************
* 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
这是我的主班，展示如何使用该班。
/*
* 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
玩得开心。:)
京公网安备 11010802040832号 | 京ICP备19059560号-6 