康托展开及其逆运算

By | 2014/05/08

前文:

这个东东是我准备进攻一道A*算法的八数码题目时,遇到的。

决定先搞懂这个,再进攻八数码(传说中 不做人生不完整的 题目)。

康托展开是什么?

定义:

X=an*(n-1)!+an-1*(n-2)!+…+ai*(i-1)!+…+a2*1!+a1*0!

ai为整数,并且0<=ai<i(1<=i<=n)

简单点说就是,判断这个数在其各个数字全排列中从小到大排第几位。

比如 132,在1、2、3的全排列中排第2位。

康托展开有啥用呢?

维基百科:n位(0~n-1)全排列后,其康托展开唯一且最大约为n!,因此可以由更小的空间来储存这些排列。由公式可将X逆推出对应的全排列。

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它可以应用于哈希表中空间压缩,

而且在搜索某些类型题时,将VIS数组量压缩。比如:八数码魔板。。

康托展开求法:

比如2143 这个数,求其展开:

从头判断,至尾结束,

① 比 2(第一位数)小的数有多少个->1个就是1,1*3!

② 比 1(第二位数)小的数有多少个->0个0*2!

③ 比 4(第三位数)小的数有多少个->3个就是1,2,3,但是1,2之前已经出现,所以是 1*1!

将所有乘积相加=7

比该数小的数有7个,所以该数排第8的位置。

1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 personal finance assignment 8 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312

4321

用程序来实现就是:

int fac[] cell spy = {1,1,2,6,24,120,720,5040,40320}; //i的阶乘为fac[i] // 康托展开-> 表示数字a是 a的全排列中从小到大排,排第几 // n表示1~n个数 a数组表示数字。 int kangtuo(int n,char a[]) { int i,j,t,sum; sum=0; for( i=0; i<n ;++i) { t=0; for(j=i+1;j<n;++j) if( a[i]>a[j] ) ++t; sum+=t*fac[n-i-1]; } return sum+1; }


康托展开的逆:

康托展开是一个全排列到自然数的双射,可以作为哈希函数。

所以当然也可以求逆运算了。

逆运算的方法:

假设求4位数中第19个位置的数字。

① 19减去1 → 18

② 18 对3!作除法 → 得3余0

③ 0对2!作除法 → 得0余0

④ 0对1!作除法 → 得0余0

据上面的可知:

我们第一位数(最左面的数),比第一位数小的数有3个,显然 第一位数为→ 4

比第二位数小的数字有0个,所以 第二位数为→1

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比第三位数小的数字有0个,因为1已经用过,所以第三位数为→2

第四位数剩下 3

该数字为 4123 (正解)

用代码实现上述步骤为:

int fac[] = {1,1,2,6,24,120,720,5040,40320}; //康托展开的逆运算,{1...n}的全排列,中的第k个数为s[] void reverse_kangtuo(int n,int k,char s[]) { int i, j, t, vst[8]={0}; --k; for (i=0; i<n; i++) { t = k/fac[n-i-1]; for (j=1; j<=n; j++) if (!vst[j]) { if (t == 0) break; --t; } s[i] = '0'+j; vst[j] = 1; k %= fac[n-i-1]; } }



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