2.1 Hamming

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本文介绍了一种生成Hamming码的算法，旨在找到一组给定数量的码字，每个码字具有特定长度，并确保任意两个码字之间的Hamming距离至少为预设值。通过实例解释了如何计算两个十六进制数之间的Hamming距离，并提供了C++实现代码。

Hamming Codes
Rob Kolstad

Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords. The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4):

        0x554 = 0101 0101 0100

        0x234 = 0010 0011 0100

Bit differences: xxx  xx

Since five bits were different, the Hamming distance is 5.

PROGRAM NAME: hamming

INPUT FORMAT

N, B, D on a single line

SAMPLE INPUT (file hamming.in)

16 7 3

OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

SAMPLE OUTPUT (file hamming.out)

0 7 25 30 42 45 51 52 75 76

82 85 97 102 120 127



/*
ID: xdzhbin1
LANG: C++
TASK: hamming
*/

#include <fstream>
#include <vector>
#include <cmath>
using namespace std;

class Hamming
{
private:
	int n,b,d;	//n个数，b位数，d距离
	int dis(int i, int j);	//计算两个整数之间的海明距离

public:
	vector<int> v;	//存放产生的数
	Hamming(int i,int j,int k):	//构造函数
	  n(i),b(j),d(k)
	  {
		  //v.resize(this->n);
	  }
	void calculate(); //产生数
};

//计算两个整数之间的海明距离
int Hamming::dis(int i,int j)
{
	int counter = 0;
	i ^= j;		//异或，并把结果赋给第一个数
	//其对应二进制'1'的个数就是两数之间的海明距离
	while (i != 0)
	{
		if (i%2 == 1)
		{
			counter++;
		}
		i /= 2;
	}
	return counter;
}

void Hamming::calculate()
{
	int start = 0;
	int end = (int)pow(2,b) - 1;
	int cur;
	int i;
	while (v.size() < n)
	{
		if (v.size()==0)
		{
			v.push_back(start);
		}
		for (cur = v[v.size()-1]+1; cur<=end; cur++)//根据向量中最后一个数，产生下一个符合条件的数
		{
			for (i=0; i<v.size(); i++)
			{
				if (dis(v[i],cur)<d) break;
			}
			if (i == v.size())
			{
				v.push_back(cur);
			}
			if (v.size() == n) break;
		}
		if (v.size() == n) break;
		start++;
		v.clear();
	}
}

int main()
{
	ifstream in("hamming.in");
	ofstream out("hamming.out");
	int n,b,d;
	int i;
	in>>n>>b>>d;
	Hamming h(n,b,d);
	h.calculate();
	for (i=0; i<h.v.size(); i++)
	{
		if (i%10!=0)
		{
			out<<" ";
		}
		out<<h.v[i];
		if ((i+1)%10==0)
		{
			out<<endl;
		}
	}
	if (i%10!=0)
	{
		out<<endl;
	}

	return 0;
}