第三次作业

5.给定如表4-9所示的概率模型，求出序列a1a1a3a2a3a1的实值标签。

                    表4-9  习题5，习题6的概率模型

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          字母                                              概率

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           a1                                                 0.2

           a2                                                 0.3

           a3                                                  0.5

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解：由题可知，映射a1<=>1,a2<=>2,a3<=>3

F_x(k)=0, k≤0, F_x(1)=0.2, F_x(2)=0.5, F_x(3)=1, k>3

下界：  l⁽⁰⁾=0,上界：u⁽⁰⁾=1

该序列的第1个元素为a_1:

           l⁽¹⁾= 0+(1- 0)F_x(0)=0

           u⁽¹⁾=0+(1-0) F_x(1)=0.2

序列的第2个元素为a₁

           l⁽²⁾= 0+(0.2- 0)F_x(0)=0

           u⁽²⁾=0+(0.2-0) F_x(1)=0.04

序列的第3个元素为a₃

           l⁽³⁾= 0+(0.04- 0)F_x(2)=0.02

           u⁽³⁾=0+(0.04-0) F_x(3)=0.04

序列的第4个元素为a2  

         l⁽⁴⁾= 0.02+(0.04- 0.02)F_x(1)=0.024

           u⁽⁴⁾=0.02+(0.04-0.02) F_x(2)=0.03

序列的第5个元素为a3   

           l⁽⁵⁾= 0.024+(0.03- 0.024)F_x(2)=0.027

           u⁽⁵⁾=0.024+(0.03-0.024) F_x(3)=0.03

序列的第6个元素为a1   

           l⁽⁶⁾= 0.027+(0.03- 0.027)F_x(0)=0.027

           u⁽⁶⁾=0.027+(0.03-0.027) F_x(1)=0.0276

该序列a1a1a3a2a3a1的标签所在的区间为[0.027,0.0276)

可以生成序列a₁a₁a₃a₂a₃a₁的标签如下：

T_{x(a1a1a3a2a3a1)}= l⁽⁶⁾+u⁽⁶⁾/2

                       =(0.027+0.0276)/2

                       =0.0273

6、对于表4-9所示的概率模型，对于一个标签为0.63215699的长度为10的序列进行解码。

 

include

 

int main()
{
double tag=0.63215699;
double l[100],u[100];
double t;
l[0]=0;
u[0]=1;
double A0=0.0,A1=0.2,A2=0.5,A3=1.0;
int M[100];
for(int k=1;k<=10;k++)
{
t=(double)(tag-l[k-1])/(u[k-1]-l[k-1]);
if(t>=A0&&t<=A1)
{
u[k]=l[k-1]+(u[k-1]-l[k-1])A1;
l[k]=l[k-1]+(u[k-1]-l[k-1])A0;
M[k]=1;
}
else if(t>A1&&t<=A2)
{
u[k]=l[k-1]+(u[k-1]-l[k-1])A2;
l[k]=l[k-1]+(u[k-1]-l[k-1])A1;
M[k]=2;
}
else if(t>A2&&t<=A3)
{
u[k]=l[k-1]+(u[k-1]-l[k-1])A3;
l[k]=l[k-1]+(u[k-1]-l[k-1])A2;
M[k]=3;
}
printf("%d",M[k]);
}
return 0;
}