作业三

Posted 2020-08-09

第1题：

2.利用程序huff_enc和huff_dec进行以下操作.

（a）对Sena、Sensin和Omaha图像进行编码

文件名	压缩前大小	压缩后大小	压缩比
Sena	64kb	57kb	0.89
Sensin	64kb	61kb	0.95
Omaha	64kb	58kb	0.91

 

 

 

 

 

4.一个信源从符号集A={a1,a2,a3,a4,a5}中选择字母，概率为p(a1)=0.15,p(a2)=0.04,p(a3)=0.26,p(a4)=0.05,p(a5)=0.50.

(a)计算这个信源的熵.

解：H(A) = -0.15*log₂0.15-0.04*log₂0.04-0.26*log₂0.26-0.05*log₂0.05-0.50*log₂0.50

               =0.15* 2.737+0.04*4.644+0.26*1.943+0.05*4.322+0.5*1

               =0.411+0.186+0.505+0.216+0.5

               =1.818（bits/symbol）  

  

(b)求这个信源的赫夫曼码.

解： a1:001

       a2:0000

       a3:01

       a4:0001

       a5:1

(c)求(b)中代码的平均长度及其冗余度.

解：平均长度：L=0.15*3+0.04*4+0.26*2+0.05*4+0.5*1

                   =0.45+0.16+0.52+0.2+0.5

                   =1.83（bits/symbol）

       冗余度：L-H(A)=1.83-1.818=0.012（bits/symbol）

 

第2题：

思考：为什么压缩领域中的编码方法总和二叉树联系在一起呢？

       答：我们之前学过前缀编码，为了使用不固定的码长表示单个字符，编码必须符合“前缀编码”的要求，即较短的编码决不能是较长编码的前缀，没有码字是其他码字的前缀。要构造符合这一要求的二进制编码体系，二叉树是最理想的选择，可将码字放在叶子节点上。

第3题：

选做：试将"Shannon-Fano"编程实现.

C++编程：

#include"iostream"
#include "queue"
#include "map"
#include "string"
#include "iterator"
#include "vector"
#include "algorithm"
#include "math.h"
using namespace std;

#define NChar 8
#define Nsymbols 1<<NChar
#define INF 1<<31-1

typedef vector<bool> SF_Code;
map<char,SF_Code> SF_Dic;
int Sumvec[Nsymbols];

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; }
};

bool comp(const HTree* t1, const HTree* t2)
{ return (*t1).weight>(*t2).weight; }

typedef vector<HTree*> TreeVector;
TreeVector TreeArr;

void Optimize_Tree(int a,int b,HTree& root)
{
if(a==b)
{
root = *TreeArr[a-1];
return;
}
else if(b-a==1)
{
root.left = TreeArr[a-1];
root.right=TreeArr[b-1];
return;
}

int x,minn=INF,curdiff;
for(int i=a;i<b;i++)
{
curdiff = Sumvec[i]*2-Sumvec[a-1]-Sumvec[b];
if(abs(curdiff)<minn){
x=i;
minn = abs(curdiff);
}
else break;
}
HTree*lc = new HTree; HTree *rc = new HTree;
root.left = lc; root.right = rc;
Optimize_Tree(a,x,*lc);
Optimize_Tree(x+1,b,*rc);
}

HTree* BuildTree(int* freqency)
{
int i;
for(i=0;i<Nsymbols;i++)
{
if(freqency[i])
TreeArr.push_back(new HTree (NULL,NULL,freqency[i], (char)i));
}
sort(TreeArr.begin(), TreeArr.end(), comp);
memset(Sumvec,0,sizeof(Sumvec));
for(i=1;i<=TreeArr.size();i++)
Sumvec[i] = Sumvec[i-1]+TreeArr[i-1]->weight;
HTree* root = new HTree;
Optimize_Tree(1,TreeArr.size(),*root);
return root;
}

void Generate_Coding(HTree* root, SF_Code& curcode)
{
if(root->Isleaf())
{
SF_Dic[root->ch] = curcode;
return;
}
SF_Code lcode = curcode;
SF_Code rcode = curcode;
lcode.push_back(false);
rcode.push_back(true);
Generate_Coding(root->left,lcode);
Generate_Coding(root->right,rcode);
}

int main()
{
int freq[Nsymbols] = {0};
char *str = "bbbbbbbccccccaaaaaaaaaaaaaaaeeeeedddddd";

//statistic character frequency
while (*str!=‘\0‘) freq[*str++]++;


HTree* r = BuildTree(freq);
SF_Code nullcode;
Generate_Coding(r,nullcode);

for(map<char,SF_Code>::iterator it = SF_Dic.begin(); it != SF_Dic.end(); it++) {
cout<<(*it).first<<‘\t‘;
std::copy(it->second.begin(),it->second.end(),std::ostream_iterator<bool>(cout));
cout<<endl;
}
}