参考书《数据压缩导论(第4版)》 Page 100
5、给定如表4-9所示的概率模型,求出序列a1a1a3a2a3a1 的实值标签。
解:
又题意得:求序列a1a1a3a2a3a1 的实值标签相当于求序列113231的标签
则:设定随机变量X(ai)=i,那么P(ai)=P(X=i)
P(a1)=P(X=1)=0.2,P(a2)=P(X=2)=0.3,P(a3)=P(X=3)=0.5
由此可得:FX(1)=P(X=1)=0.2, FX(2)=P(X=1)+P(X=2)=0.2+0.3=0.5, FX(3)=P(X=1)+P(X=2)+P(X=3)=0.2+0.3+0.5=1
l(0)=0,u(0)=1
可根据公式:
l(k)=l(k-1)+(u(k-1)-l(k-1))*Fx(Xk-1)
u(k)=l(k-1)+(u(k-1)-l(k-1))*Fx(Xk)
可得:
1 l(1)=l(0)+(u(0)-l(0))*Fx(0)=0+(1-0)*0=0
u(1)=l(0)+(u(0)-l(0))*Fx(1)=0+(1-0)*0.2=0.2
11 l(2)=l(1)+(u(1)-l(1))*Fx(0)=0+(0.2-0)*0=0
u(2)=l(1)+(u(1)-l(1))*Fx(1)=0+(0.2-0)*0.2=0.04
113 l(3)=l(2)+(u(2)-l(2))*Fx(2)=0+(0.04-0)*0.5=0.02
u(3)=l(2)+(u(2)-l(2))*Fx(3)=0+(0.04-0)*1 =0.04
1132 l(4)=l(3)+(u(3)-l(3))*Fx(1)=0.02+(0.04-0.02)*0.2=0.024
u(4)=l(3)+(u(3)-l(3))*Fx(2)=0.02+(0.04-0.02)*0.5 =0.03
11323 l(5)=l(4)+(u(4)-l(4))*Fx(2)=0.024+(0.03-0.024)*0.5=0.027
u(5)=l(4)+(u(4)-l(4))*Fx(3)=0.024+(0.03-0.024)*1=0.03
113231 l(6)=l(5)+(u(5)-l(5))*Fx(0)=0.027+(0.03-0.027)*0=0.027
u(6)=l(5)+(u(5)-l(5))*Fx(1)=0.027+(0.03-0.027)*0.2=0.0276
因此,该序列的实值标签为:
Tx(113231)= ( u(6) + l(6) )/2
=(0.0276+0.027)/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。
解:此题可编程实现,代码如下
#include<stdio.h>
int main()
{
int a[10];
double t,F[100];
F[0]=0,F[1]=0.2,F[2]=0.5,F[3]=1;
double l[100]={0.0},u[100]={1.0};
double tag=0.63215699;
int n=10,k;
for(k=1;k<=10;k++)
{
t=(tag-l[k-1])/(u[k-1]-l[k-1]);
if(t>F[0]&&t<F[1])
{
l[k]=l[k-1]+(u[k-1]-l[k-1])*F[0];
u[k]=l[k-1]+(u[k-1]-l[k-1])*F[1];
a[k]=1;
}
else
if(t>F[1]&&t<F[2])
{
l[k]=l[k-1]+(u[k-1]-l[k-1])*F[1];
u[k]=l[k-1]+(u[k-1]-l[k-1])*F[2];
a[k]=2;
}
else
{
l[k]=l[k-1]+(u[k-1]-l[k-1])*F[2];
u[k]=l[k-1]+(u[k-1]-l[k-1])*F[3];
a[k]=3;
}
printf("%d
",a[k]);
}
return 0;
}
调试结果如下:
此题还可以通过计算得到:
刚开始时l(0)=0,u(0)=1
则
l(1)=0+(1-0)Fx(x1-1)=Fx(x1-1)
u(1)=0+(1-0)Fx(x1)=Fx(x1)
通过计算,当x1=3时,该区间为[0.5,1),而0.63215699在区间[0.5,1)中,所以x1=3
l(2)=0.5+(1-0.5)Fx(x2-1)=0.5+0.5Fx(x2-1)
u(2)=0.5+(1-0.5)Fx(x2)=0.5+0.5Fx(x2)
通过计算,当x2=2时,该区间为[0.6,0.75),而0.63215699在区间[0.6,0.75)中,所以x2=2
l(3)=0.5+(0.75-0.6)Fx(x3-1)=0.6+0.15Fx(x3-1)
u(3)=0.5+(0.75-0.6)Fx(x3)=0.6+0.15Fx(x3)
通过计算,当x3=2时,该区间为[0.63,0.675),而0.63215699在区间[0.63,0.675)中,所以x3=2
l(4)=0.63+(0.675-0.625)Fx(x4-1)=0.63+0.045Fx(x4-1)
u(4)=0.63+(0.675-0.625)Fx(x4)=0.63+0.045Fx(x4)
通过计算,当x4=1时,该区间为[0.63,0.639),而0.63215699在区间[0.63,0.639)中,所以x4=1
l(5)=0.63+(0.639-0.63)Fx(x5-1)=0.63+0.009Fx(x5-1)
u(5)=0.63+(0.639-0.63)Fx(x5)=0.63+0.009Fx(x5)
通过计算,当x5=2时,该区间为[0.6318,0.6345),而0.63215699在区间[0.6318,0.6345)中,所以x5=2
l(6)=0.6318+(0.6345-0.6318)Fx(x6-1)=0.6318+0.0027Fx(x6-1)
u(6)=0.6318+(0.6345-0.6318)Fx(x6)=0.6318+0.0027Fx(x6)
通过计算,当x6=1时,该区间为[0.6318,0.63234),而0.63215699在区间[0.6318,0.63234)中,所以x6=1
l(7)=0.6318+(0.63234-0.6318)Fx(x7-1)=0.6318+0.00054Fx(x7-1)
u(7)=0.6318+(0.63234-0.6318)Fx(x7)=0.6318+0.00054Fx(x7)
通过计算,当x7=3时,该区间为[0.63207,0.63234),而0.63215699在区间[0.63207,0.63234)中,所以x7=3
l(8)=0.63207+(0.63234-0.63207)Fx(x8-1)=0.63207+0.00027Fx(x8-1)
u(8)=0.63207+(0.63234-0.63207)Fx(x8)=0.63207+0.00027Fx(x8)
通过计算,当x8=2时,该区间为[0.632124,0.632205),而0.63215699在区间[0.632124,0.632205)中,所以x8=2
l(9)=0.632124+(0.632205-0.632124)Fx(x9-1)=0.632124+0.000081Fx(x9-1)
u(9)=0.632124+(0.632205-0.632124)Fx(x9)=0.632124+0.000081Fx(x9)
通过计算,当x9=2时,该区间为[0.6321402,0.6321645),而0.63215699在区间[0.6321402,0.6321645)中,所以x9=2
l(10)=0.6321402+(0.6321645-0.6321402)Fx(x10-1)=0.6321402+0.00006243Fx(x10-1)
u(10)=0.6321402+(0.6321645-0.6321402)Fx(x10)=0.6321402+0.00006243Fx(x10)
通过计算,当x10=3时,该区间为[0.63215235.0.6321645),而0.63215699在区间[0.63215235.0.6321645)中,所以x10=3
所以该序列解码得:
3221213223