Cantor expansion

The Cantor expansion is a bijection arranged to a natural number. So reversible.

• Cantor expansion: given a number n, and a complete arrangement of n bits, find out the number X of this arrangement

• Inverse Cantor expansion: given a number n, and the number X of this arrangement, find this arrangement

Here X (note that the first arrangement is X = 0, so for convenience, we can directly +1 later.)

An means that in the order from (1,2,3, ... n), it is less than an and is not arranged The number.

Such as 34152,

• a5 = 2 (only 1,2 is less than 3)
• a4 = 2 (3 is ranked, only 1,2 is less than 4)
• a3 = 0 (nothing smaller than 1)
• a2 = 1 (1, 3, 4 are arranged, only 2 less than 5)
• a1 = 0

---

Kang expands code implementation, complexity O (N * N)

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int factorial[20]={1,1,2,6,24,120,720,5040,40320,362880,3628800}; 
int flag[20];
void cantor(int  *num,int n){
	int x=0;
	int tep=n-1;
	for(int i=1;i<=n;i++){//遍历num数组 
		int cot=0;
		for(int j=1;j<num[i];j++)//找到小于num[i]的数字
		{
			if(j<num[i]&&flag[j]==0) cot++;	
		}
		flag[num[i]]=1;
		x+=cot*factorial[tep--]; 
	}
	cout<<x+1; 
}
int main(void)
{
	int num[20]={0,3,4,1,5,2};//求34152在排列中排多少
	int n=5;
	cantor(num,n);//n位 
 } 

The output 34152 is the 62nd permutation

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Reverse expansion

The problem of adding X without adding 1 was discussed earlier. If it is +1, it needs -1; if it is not +1, -1 is not needed.
For example, n = 5, x = 62, the arrangement is 34152, see how to calculate a5, a4 ...

• First minus one: x = 61
• a5 = x / 4! = 61/4! = 2, indicating that there are two smaller than the first, so a5 = 3. x = 61% 4! = 13
• a4 = 13/3! = 2, indicating that the first place is removed, there are two smaller than the second, so a4 = 4. X = 13% 3! = 1
• a3 = 1/2! = 0, it means that the first and second digits are removed, there are 0 less than third digit, so a3 = 1 x = 1% 2! = 1
• a4 = 1/1! = 1, so a4 = 5. x = 1% 1 = 0
• a5 = 0/1 = 0, so a5 = 2. The last number
  is 34152.

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int factorial[20]={1,1,2,6,24,120,720,5040,40320,362880,3628800}; 
int nums[20]={0,1,2,3,4,5,6,7,8,9,10};
vector<int>num(nums+1,nums+11); 
void decantor(int n,int x){
	vector<int>ans;
	for(int i=n;i>=1;i--){
		int a=x/factorial[i-1];
		x=x%factorial[i-1];
		ans.push_back(num[a]);
		num.erase(num.begin()+a);	
	}
	for(auto a:ans) cout<<a;
}
int main(void)
{
	int n=5;
	int x=62;
	decantor(n,x-1);//n位 
 }