第三次作业

参考书《数据压缩导论（第4版）》 Page 100 5， 6

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



求序列a1a1a3a2a3a1 的实值标签就是求序列113231的实值标签

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

令 ｌ（0）=0， u（0）=1

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(2)+(u(1)- ｌ(1))*Fx(0)=0+(1-0)*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

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。

由表4-9可知 Fx(k)=0, k≤0, Fx(1)=0.2, Fx(2)=0.5, Fx(3)=1, k>3.

令ｌ（0）=0， u（0）=1。设玄素对应的序列为x1x2x3x4x5x6x7x8x9x10 ,则利用更新公式

x1

ｌ（1）= ｌ(0)+(u(0)- ｌ(0))*Fx(x1-1)=Fx(x1-1)

u(1)= ｌ(0)+(u(0)- ｌ(0))*Fx(x1)=Fx(x1)

x1=1时,区间为[0,0.2]

x1=2时,区间为[0.2,0.5]

x1=3时,区间为[0.5,1]

因为0.63215699在区间[0.5,1]内，所以，第一个元素的序列为3，元素则为a3

x2

u(2)= ｌ(1)+(u(1)- ｌ(1))*Fx(x2)=0.5+(1-0.5)*Fx(x2)=0.5+0.5Fx(x2)

ｌ(2)=ｌ(1)+(u(1)-ｌ(1))*Fx(x2-1)=0.5+(1-0.5)*Fx(x2-1)=0.5+0.5Fx(x2-1)

x2=1时,区间为[0.5,0.6]

x2=2时,区间为[0.6,0.75]

x2=3时,区间为[0.75,1]

因为 0.63215699在区间[0.6，0.75]内，所以，第二个元素的序列为2，元素则为a2

x3

u(3)= ｌ(2)+(u(2)- ｌ(2))*Fx(x3)=0.6+(0.75-0.6)*Fx(x2)=0.6+0.15Fx(x3)

ｌ（3）= ｌ(2)+(u(2)- ｌ(2))*Fx(x3-1)=0.6+(0.75-0.6)*Fx(x2-1)=0.6+0.15Fx(x3-1)

x3=1时,区间为[0.6,0.63]

x3=2时,区间为[0.63,0.675]

x3=3时,区间为[0.675,0.75]

因为0.63215699在区间[0.63,0.675]内，所以，第三个元素的序列为2，元素则为a2

x4

u(4)=ｌ(3)+(u(3)-ｌ(3))*Fx(x4)=0.63+(0.675-0.63)*Fx(x4)=0.63+0.045*Fx(x4)

ｌ(4)=ｌ(3)+(u(3)-ｌ(3))*Fx(x4-1)=0.63+(0.675-0.63)*Fx(x4-1)=0.63+0.045*Fx(x4-1）

x4=1时,区间为[0.63,0.639]

x4=2时,区间为[0.639，0.6525]

x4=3时,区间为[0.6525,0.675]

因为0.63215699在区间[0.63,0.639]内，所以，第四个元素的序列为1，元素则为a1

x5

u(5)=ｌ(4)+(u(4)-ｌ(4))*Fx(x5)=0.63+(0.639-0.63)*Fx(x5)=0.63+0.009*Fx(x5)

ｌ(5)=ｌ(4)+(u(4)-ｌ(4))*Fx(x5-1)=0.63+(0.639-0.63)*Fx(x5-1)=0.63+0.009*Fx(x5-1）

x5=1时,区间为[0.63,0.6318]

x5=2时,区间为[0.6318,0.6345]

x5=3时,区间为[0.6345,0.639]

因为0.63215699在区间[0.6318,0.6345]内，所以，第五个元素的序列为2，元素则为a2

x6

u(6)=ｌ(5)+(u(5)-ｌ(5))*Fx(x6)=0.6318+(0.6345-0.6318)*Fx(x6)=0.6318+0.0027*Fx(x6)

ｌ(6)=ｌ(5)+(u(5)-ｌ(5))*Fx(x6-1)=0.6318+(0.6345-0.6318)*Fx(x6-1)=0.6318+0.0027*Fx(x6-1)

x6=1时,区间为[0.6318,0.63234]

x6=2时,区间为[0.63234,0.63315]

x6=3时,区间为[0.63315,0.6345]

因为0.63215699在区间[0.6318,0.63234]内，所以，第六个元素的序列为1，元素则为a1

x7

u(7)=ｌ(6)+(u(6)-ｌ(6))*Fx(x7)=0.6318+(0.63234-0.6318)*Fx(x7)=0.6318+0.00054*Fx(x7)

ｌ(7)=ｌ(6)+(u(6)-ｌ(6))*Fx(x7-1)=0.6318+(0.63234-0.6318)*Fx(x7-1)=0.6318+0.00054*Fx(x7-1)

x7=1时,区间为[0.6318,0.631908]

x7=2时,区间为[0.631908,0.63207]

x7=3时,区间为[0.63207,0.63234]

因为 0.63215699在区间[0.63207,0.63234]，所以，第七个元素的序列为3，元素则为a3

x8

u(8)=ｌ(7)+(u(7)-ｌ(7))*Fx(x8)=0.63207+(0.63234-0.63207)*Fx(x8)=0.63207+0.00027*Fx(x8)

ｌ(8)=ｌ(7)+(u(7)-ｌ(7))*Fx(x8-1)=0.63207+(0.63234-0.63207)*Fx(x8-1)=0.63207+0.00027*Fx(x8-1)

x8=1时,区间为[0.63207,0.632124]

x8=2时,区间为[0.632124,0.632205]

x8=3时,区间为[0.632205,0.63234]

因为0.63215699在区间[0.632124,0.632205]内，所以，第八个元素的序列为2，元素则为a2

x9

u(9)=ｌ(8)+(u(8)-ｌ(8))*Fx(x9)=0.632124+(0.632205-0.632124)*Fx(x9)=0.632124+（8.1e-5）*Fx(x9)

ｌ(9)=ｌ(8)+(u(8)-ｌ(8))*Fx(x9-1)=0.632124+(0.632205-0.632124)*Fx(x9-1)=0.632124+（8.1e-5）*Fx(x9-1)

x9=1时,区间为[0.632124,0.6321402]

x9=2时,区间为[0.0.6321402,0.6321645]

x9=3时,区间为[0.6321645,0.63234]

因为0.63215699在区间[0.6321402,0.6321645]内，所以，第九个元素的序列为2，元素则为a2

x10

u(10)=ｌ(9)+(u(9)-ｌ(9))*Fx(x10)=0.6321402+(0.6321645-0.6321402)*Fx(x10)=0.6321402+（2.43e-5）*Fx(x10)

ｌ(10)=ｌ(9)+(u(9)-ｌ(9))*Fx(x10-1)=0.6321402+(0.6321645-0.6321402)*Fx(x10-1)=0.6321402+（2.43e-5）*Fx(x10-1)

x10=1时,区间为[0.6321402,0.63212886]

x10=2时,区间为[0.63212886,0.63215325]

x10=3时,区间为[0.63215325,0.6321645]

因为0.63215699在区间[0.63215235,0.6321645]内，所以，第九个元素的序列为3，元素则为a3

所以根据表4-9所给的概率模型得出标签为0.63215699的长度为10的序列的解码结果为：a3 a2 a2 a1 a2 a1 a3 a2 a2 a3。