生成汉明距离t内的所有位序列
给定一个位向量v
,计算汉明距离为1 v
, 然后 为距离2直至输入参数的位集合t
。
因此对于
011 I should get~~~
111
001
010
~~~ -> 3 choose 1 in number
101
000
110
~~~ -> 3 choose 2
100
~~~ -> 3 choose 3
如何有效地计算呢?向量不一定总是3维,例如可能是6维。这将在我的真实代码中运行很多次,因此也将欢迎一些效率(即使付出更多的内存)。
我的尝试:
#include <iostream>#include <vector>
void print(const std::vector<char>& v, const int idx, const char new_bit)
{
for(size_t i = 0; i < v.size(); ++i)
if(i != idx)
std::cout << (int)v[i] << " ";
else
std::cout << (int)new_bit << " ";
std::cout << std::endl;
}
void find_near_hamming_dist(const std::vector<char>& v, const int t)
{
// if t == 1
for(size_t i = 0; i < v.size(); ++i)
{
print(v, i, v[i] ^ 1);
}
// I would like to produce t == 2
// only after ALL the t == 1 results are reported
/* how to? */
}
int main()
{
std::vector<char> v = {0, 1, 1};
find_near_hamming_dist(v, 1);
return 0;
}
输出:
MacBook-Pro:hammingDist gsamaras$ g++ -Wall -std=c++0x hammingDist.cpp -o hamMacBook-Pro:hammingDist gsamaras$ ./ham
1 1 1
0 0 1
0 1 0
回答:
#include <stdio.h>#include <stdint.h>
#include <string.h>
void magic(char* str, int i, int changesLeft) {
if (changesLeft == 0) {
printf("%s\n", str);
return;
}
if (i < 0) return;
// flip current bit
str[i] = str[i] == '0' ? '1' : '0';
magic(str, i-1, changesLeft-1);
// or don't flip it (flip it again to undo)
str[i] = str[i] == '0' ? '1' : '0';
magic(str, i-1, changesLeft);
}
int main(void) {
char str[] = "011";
printf("%s\n", str);
size_t len = strlen(str);
size_t maxDistance = len;
for (size_t i = 1 ; i <= maxDistance ; ++i) {
printf("Computing for distance %d\n", i);
magic(str, len-1, i);
printf("----------------\n");
}
return 0;
}
输出:
MacBook-Pro:hammingDist gsamaras$ nano kastrinis.cppMacBook-Pro:hammingDist gsamaras$ g++ -Wall kastrinis.cpp
MacBook-Pro:hammingDist gsamaras$ ./a.out
011
Computing for distance 1
010
001
111
----------------
Computing for distance 2
000
110
101
----------------
Computing for distance 3
100
----------------
以上是 生成汉明距离t内的所有位序列 的全部内容, 来源链接: utcz.com/qa/398433.html