huffman编码——原理与实现

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哈夫曼算法原理

Wikipedia上面说的非常清楚了，这里我就不再赘述，直接贴过来了。

1952年, David A. Huffman提出了一个不同的算法，这个算法能够为不论什么的可能性提供出一个理想的树。香农-范诺编码（Shanno-Fano）是从树的根节点到叶子节点所进行的的编码，哈夫曼编码算法却是从相反的方向，暨从叶子节点到根节点的方向编码的。

为每一个符号建立一个叶子节点，并加上其对应的发生频率
当有一个以上的节点存在时，进行下列循环:
1. 把这些节点作为带权值的二叉树的根节点，左右子树为空
2. 选择两棵根结点权值最小的树作为左右子树构造一棵新的二叉树，且至新的二叉树的根结点的权值为其左右子树上根结点的权值之和。
3. 把权值最小的两个根节点移除
4. 将新的二叉树添�队列中.
最后剩下的节点暨为根节点，此时二叉树已经完毕。

演示样例

Huffman Algorithm

符号	A	B	C	D	E
计数	15	7	6	6	5
概率	0.38461538	0.17948718	0.15384615	0.15384615	0.12820513

在这样的情况下，D，E的最低频率和分配分别为0和1，分组结合概率的0.28205128。如今最低的一双是B和C，所以他们就分配0和1组合结合概率的0.33333333在一起。这使得BC和DE所以0和1的前面加上他们的代码和它们结合的概率最低。然后离开仅仅是一个和BCDE，当中有前缀分别为0和1，然后结合。这使我们与一个单一的节点，我们的算法是完整的。

可得A代码的代码长度是1比特，其余字符是3比特。

字符	A	B	C	D	E
代码	0	100	101	110	111

Entropy：

Pseudo-code

1: begin 2: count frequencies of single characters (source units) 3: output(frequencies using Fibonacci Codes of degree 2) 4: sort them to non-decreasing sequence 5: create a leaf node (character, frequency c, left son = NULL, right son = NULL) 6: of the tree for each character and put nodes into queue F 7: while (|F|>=2) do 8: begin 9: pop the first two nodes (u1, u2) with the lowest 10: frequencies from sorted queue 11: create a node evaluated with sum of the chosen units, 12: successors are chosen units (eps, c(u1)+c(u2), u1, u2) 13: insert new node into queue 14: end 15: node evaluate with way from root to leaf node (left son 1, right son 0) 16: create output from coded intput characters 17: end

哈夫曼算法实现

实现的时候我们用vector<bool>记录每一个char的编码；用map<char,vector<bool>>表示整个字典；

就得到了以下的代码（以下有两个，一对一错）：

先放出来这个错的，以示警戒：

/************************************************************************/
/*File Name: Huffman.cpp
*@Function: Lossless Compression
@Author: Sophia Zhang
@Create Time: 2012-9-26 10:40
@Last Modify: 2012-9-26 11:10
*/
/************************************************************************/#include"iostream"
#include "queue"
#include "map"
#include "string"
#include "iterator"
#include "vector"
#include "algorithm"
using namespace std;#define NChar 8//suppose use at most 8 bits to describe all symbols
#define Nsymbols 1<<NChar//can describe 256 symbols totally (include a-z, A-Z)
typedef vector<bool> Huff_code;//8 bit code of one char
map<char,Huff_code> Huff_Dic;//huffman coding dictionaryclass HTree
{
public :
HTree* left;
HTree* right;
char ch;
int weight;HTree(){left = right = NULL; weight=0;}
HTree(HTree* l,HTree* r,int w,char c){left = l;right = r;weight=w;ch=c;}
~HTree(){delete left; delete right;}
int Getweight(){return weight?weight:left->weight+right->weight;}
bool Isleaf(){return !left && !right; }
bool operator < (const HTree tr) const
{
return tr.weight < weight;
}
};HTree* BuildTree(int *frequency)
{
priority_queue<HTree*> QTree;//1st level add characters
for (int i=0;i<Nsymbols;i++)
{
if(frequency[i])
QTree.push(new HTree(NULL,NULL,frequency[i],(char)i));
}//build
while (QTree.size()>1)
{
HTree* lc = QTree.top();
QTree.pop();
HTree* rc = QTree.top();
QTree.pop();HTree* parent = new HTree(lc,rc,parent->Getweight(),(char)256);
QTree.push(parent);
}
//return tree root
return QTree.top();
}void Huffman_Coding(HTree* root, Huff_code& curcode)
{
if(root->Isleaf())
{
Huff_Dic[root->ch] = curcode;
return;
}
Huff_code& lcode = curcode;
Huff_code& rcode = curcode;
lcode.push_back(false);
rcode.push_back(true);Huffman_Coding(root->left,lcode);
Huffman_Coding(root->right,rcode);
}int main()
{
int freq[Nsymbols] = {0};
char *str = "this is the string need to be compressed";//statistic character frequency
while (*str!='\0')
freq[*str++]++;//build tree
HTree* r = BuildTree(freq);
Huff_code nullcode;
nullcode.clear();
Huffman_Coding(r,nullcode);for(map<char,Huff_code>::iterator it = Huff_Dic.begin(); it != Huff_Dic.end(); it++)
{
cout<<(*it).first<<'\t';
Huff_code vec_code = (*it).second;
for (vector<bool>::iterator vit = vec_code.begin(); vit!=vec_code.end();vit++)
{
cout<<(*vit)<<endl;
}
}
}

上面这段代码，我执行出来不正确。在调试的时候发现了一个问题，就是QTree优先队列中的排序出了问题，说来也是，上面的代码中，我重载小于号是对HTree object做的；而实际上我建树时用的是指针，那么优先级队列中元素为指针时该怎么办呢？

那我们将friend bool operator >(Node node1,Node node2)改动为friend bool operator >(Node* node1,Node* node2)，也就是传递的是Node的指针行不行呢?

答案是不能够，由于依据c++primer中重载操作符中讲的“程序猿仅仅能为类类型或枚举类型的操作数定义重载操作符，在把操作符声明为类的成员时，至少有一个类或枚举类型的參数依照值或者引用的方式传递”，也就是说friend
bool operator >(Node* node1,Node* node2)形參中都是指针类型的是不能够的。我们仅仅能再建一个类，用当中的重载（）操作符作为优先队列的比較函数。

就得到了以下正确的代码：

/************************************************************************/
/*File Name: Huffman.cpp
*@Function: Lossless Compression
@Author: Sophia Zhang
@Create Time: 2012-9-26 10:40
@Last Modify: 2012-9-26 12:10
*/
/************************************************************************/#include"iostream"
#include "queue"
#include "map"
#include "string"
#include "iterator"
#include "vector"
#include "algorithm"
using namespace std;#define NChar 8//suppose use 8 bits to describe all symbols
#define Nsymbols 1<<NChar//can describe 256 symbols totally (include a-z, A-Z)
typedef vector<bool> Huff_code;//8 bit code of one char
map<char,Huff_code> Huff_Dic;//huffman coding dictionary/************************************************************************/
/* Tree Class elements:
*2 child trees
*character and frequency of current node
*/
/************************************************************************/
class HTree
{
public :
HTree* left;
HTree* right;
char ch;
int weight;HTree(){left = right = NULL; weight=0;ch ='\0';}
HTree(HTree* l,HTree* r,int w,char c){left = l;right = r;weight=w;ch=c;}
~HTree(){delete left; delete right;}
bool Isleaf(){return !left && !right; }
};/************************************************************************/
/* prepare for pointer sorting*/
/*because we cannot use overloading in class HTree directly*/
/************************************************************************/
class Compare_tree
{
public:
bool operator () (HTree* t1, HTree* t2)
{
return t1->weight> t2->weight;
}
};/************************************************************************/
/* use priority queue to build huffman tree*/
/************************************************************************/
HTree* BuildTree(int *frequency)
{
priority_queue<HTree*,vector<HTree*>,Compare_tree> QTree;//1st level add characters
for (int i=0;i<Nsymbols;i++)
{
if(frequency[i])
QTree.push(new HTree(NULL,NULL,frequency[i],(char)i));
}//build
while (QTree.size()>1)
{
HTree* lc = QTree.top();
QTree.pop();
HTree* rc = QTree.top();
QTree.pop();HTree* parent = new HTree(lc,rc,lc->weight+rc->weight,(char)256);
QTree.push(parent);
}
//return tree root
return QTree.top();
}/************************************************************************/
/* Give Huffman Coding to the Huffman Tree*/
/************************************************************************/
void Huffman_Coding(HTree* root, Huff_code& curcode)
{
if(root->Isleaf())
{
Huff_Dic[root->ch] = curcode;
return;
}
Huff_code lcode = curcode;
Huff_code rcode = curcode;
lcode.push_back(false);
rcode.push_back(true);Huffman_Coding(root->left,lcode);
Huffman_Coding(root->right,rcode);
}int main()
{
int freq[Nsymbols] = {0};
char *str = "this is the string need to be compressed";//statistic character frequency
while (*str!='\0')
freq[*str++]++;//build tree
HTree* r = BuildTree(freq);
Huff_code nullcode;
nullcode.clear();
Huffman_Coding(r,nullcode);for(map<char,Huff_code>::iterator it = Huff_Dic.begin(); it != Huff_Dic.end(); it++)
{
cout<<(*it).first<<'\t';
std::copy(it->second.begin(),it->second.end(),std::ostream_iterator<bool>(cout));
cout<<endl;
}
}