DeCantor Expansion （逆康托展开）

$\text{Background}$

$\text{The }$Listen&Say $\text{Test will be hold on May 11, so I decided to fill my blog }$
$\text{with English words until that day.}$

$\text{Problem}$

$\text{There goes a problem.}$

$\text{You’ve got 2 intergers }N,k\text{. Please calculate the }k\text{th permutation of }\forall k\in [i,n].$

$\text{Solution}$

$\text{It’s easy to know that we can got it by Depth-first Search, }$$\text{but its Time complexity is }O(n!).$

DeCantor Expansion $\text{is a algorithm which can solve problems like these calculating the }k\text{th permutation}$$\text{in }O(n\log n)\text{ with heap optimization.}$


$\text{Let’s explain how it works in a simple example. Set }N=5,k=61,\text{ the answer is }a[].$

$\text{1.Let 61 / 4! = 2 ... 13, it shows that there’re 2 numbers behind }a[1]\text{ are smaller than a[1].}$
$\text{Therefore, }a[1]=3;$

$\text{2.Let 13 / 3! = 2 ... 1, it shows that there’re 2 numbers behind }a[2]\text{ are smaller than a[2].}$
$\text{Therefore, }a[2]=4;$

$\text{3.Let 1 / 2! = 0 ... 1, it shows that there’re 0 number behind }a[3]\text{ are smaller than a[3].}$
$\text{Therefore, }a[3]=1;$

$\text{4.Let 1 / 1! = 1 ... 0, it shows that there’re 1 number behind }a[4]\text{ are smaller than a[4].}$
$\text{Therefore, }a[4]=5;$


Therefore,  a [ 5 ] = 2 , a [ ] = { 3 , 4 , 1 , 5 , 2 } . \text{Therefore, }a[5]=2, a[]=\{3,4,1,5,2\}. Therefore, a[5]=2,a[]={3,4,1,5,2}.

$\text{Summary}$

$$\begin{gathered} \forall i\in [1,n-1],\text{ let }k/(n-1)!\text{, the answer you’ve got is the number of interger } \\ j\in [i+1,n]\text{ which has }a[j]<a[i].\text{ And let }k\text{ equals to the remainder.} \end{gathered}$$

$\text{The End}$

$\text{Reference material}$