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。

由题意可得：

字符集{a1,a2,a3} ， 其中p(a₁)=0.2 ,p(a₂)=0.3  ,p(a₃)=0.5

我们可以利用公式确定标签所在的上下限。

利用更新公式，可得：

l⁽¹⁾ =0+(1-0)F_x(0)=0

u⁽¹⁾ =0+(1-0)F_x(1)=0.2 

可得序列a₁a₁的标签所在的区间为[0,0.2)

该序列的第2个元素为a₁，利用更新公式，可得：

l⁽²⁾ =0+(0.2-0)F_x(0)=0

u⁽²⁾ =0+(0.2-0)F_x(1)=0.04

可得序列a₁a₁的标签所在的区间为[0,0.04)

该序列的第3个元素为a₃，利用更新公式，可得：

l⁽³⁾ =0+(0.04-0)F_x(2)=0.02

u⁽³⁾ =0+(0.04-0)F_x(3)=0.04

可得序列a₁a₁a₃的标签所在的区间为[0.02,0.04)

该序列的第4个元素为a₂，利用更新公式，可得：

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

可得序列a₁a₁a₃a₂的标签所在的区间为[0.024,0.03)

该序列的第5个元素为a₃，利用更新公式，可得：

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

可得序列a₁a₁a₃a₂a₃的标签所在的区间为[0.027,0.03)

该序列的第6个元素为a₁，利用更新公式，可得：

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

可以生成序列a₁a₁a₃a₂a₃a₁的标签如下：

T_x(a₁a₁a₃a₂a₃a₁)=(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。

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

令ｌ^（0）=0， u^（0）=1。设玄素对应的序列为x₁x₂x₃x₄x₅x₆x₇x₈x₉x_{10  ,}则利用更新公式

x₁

ｌ^（1）=  ｌ⁽⁰⁾+(u⁽⁰⁾- ｌ⁽⁰⁾)*F_x(x₁-1)=F_x(x₁-1)

u⁽¹⁾= ｌ⁽⁰⁾+(u⁽⁰⁾- ｌ⁽⁰⁾)*F_x(x₁)=F_x(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⁾)*F_x(x₂)=0.5+(1-0.5)*F_x(x₂)=0.5+0.5F_x(x₂)

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

x₂=1时,区间为[0.5,0.6]

x₂=2时,则该区间为[0.6,0.75]

x₂=3时,则该区间为[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⁾)*F_x(x₃)=0.6+(0.75-0.6)*F_x(x₂)=0.6+0.15F_x(x₃)

ｌ^（3）=  ｌ⁽²⁾+(u⁽²⁾- ｌ⁽²⁾)*F_x(x₃-1)=0.6+(0.75-0.6)*F_x(x₂-1)=0.6+0.15F_x(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⁾)*F_x(x₄)=0.63+(0.675-0.63)*F_x(x₄)=0.63+0.045*F_x(x₄)

ｌ⁽⁴⁾=ｌ⁽³⁾+(u⁽³⁾-ｌ⁽³⁾)*F_x(x₄-1)=0.63+(0.675-0.63)*F_x(x₄-1)=0.63+0.045*F_x(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⁾)*F_x(x₅)=0.63+(0.639-0.63)*F_x(x₅)=0.63+0.009*F_x(x₅)

ｌ⁽⁵⁾=ｌ⁽⁴⁾+(u⁽⁴⁾-ｌ⁽⁴⁾)*F_x(x₅-1)=0.63+(0.639-0.63)*F_x(x₅-1)=0.63+0.009*F_x(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⁾)*F_x(x₆)=0.6318+(0.6345-0.6318)*F_x(x₆)=0.6318+0.0027*F_x(x₆)

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

x₆=1,则该区间为[0.6318,0.63234]

x₆=2,则该区间为[0.63234,0.63315]

x₆=3,则该区间为[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⁾)*F_x(x₇)=0.6318+(0.63234-0.6318)*F_x(x₇)=0.6318+0.00054*F_x(x₇)

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

x₇=1,则该区间为[0.6318,0.631908]

x₇=2,则该区间为[0.631908,0.63207]

x₇=3,则该区间为[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⁾)*F_x(x₈)=0.63207+(0.63234-0.63207)*F_x(x₈)=0.63207+0.00027*F_x(x₈)

ｌ⁽⁸⁾=ｌ⁽⁷⁾+(u⁽⁷⁾-ｌ⁽⁷⁾)*F_x(x₈-1)=0.63207+(0.63234-0.63207)*F_x(x₈-1)=0.63207+0.00027*F_x(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⁾)*F_x(x₉)=0.632124+(0.632205-0.632124)*F_x(x₉)=0.632124+（8.1e-5）*F_x(x₉)

ｌ⁽⁹⁾=ｌ⁽⁸⁾+(u⁽⁸⁾-ｌ⁽⁸⁾)*F_x(x₉-1)=0.632124+(0.632205-0.632124)*F_x(x₉-1)=0.632124+（8.1e-5）*F_x(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⁾)*F_x(x₁₀)=0.6321402+(0.6321645-0.6321402)*F_x(x₁₀)=0.6321402+（2.43e-5）*F_x(x₁₀)

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