huffman编码——原理与实现

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。

1. 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
2. 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.
3. 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 dictionary
 
class 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;
	}
}

Reference：

1. http://www.binaryessence.com/dct/en000042.htm

2. http://zh.wikipedia.org/zh/%E9%A6%99%E5%86%9C-%E8%8C%83%E8%AF%BA%E7%BC%96%E7%A0%81#.E5.93.88.E5.A4.AB.E6.9B.BC.E7.AE.97.E6.B3.95

3. http://www.stringology.org/DataCompression/sh/index_en.html

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