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

        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

     1        

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

                u⁽¹⁾= ｌ⁽⁰⁾+(u⁽⁰⁾- ｌ⁽⁰⁾)*F_x(1)=0+(1-0)*0.2=0.2

    11          

               ｌ^（2）=  ｌ⁽²⁾+(u⁽¹⁾- ｌ⁽¹⁾)*F_x(0)=0+(1-0)*0=0

                u⁽²⁾= ｌ⁽¹⁾+(u⁽¹⁾- ｌ⁽¹⁾)*F_x(1)=0+(0.2-0)*0.2=0.04

   113   

                ｌ^（3）=  ｌ⁽²⁾+(u⁽²⁾- ｌ⁽²⁾)*F_x(2)=0+(0.04-0)*0.5=0.02

                u⁽³⁾= ｌ⁽²⁾+(u⁽²⁾- ｌ⁽²⁾)*F_x(3)=0+(0.04-0)*1=0.04

   1132 

                ｌ⁽⁴⁾=ｌ⁽³⁾+(u⁽³⁾-ｌ⁽³⁾)*F_x(1)=0.02+(0.04-0.02)*0.2=0.024 

                u⁽⁴⁾=ｌ⁽³⁾+(u⁽³⁾-ｌ⁽³⁾)*F_x(2)=0.02+(0.04-0.02)*0.5 =0.03

    11323     

                  ｌ⁽⁵⁾=ｌ⁽⁴⁾+(u⁽⁴⁾-ｌ⁽⁴⁾)*F_x(2)=0.024+(0.03-0.024)*0.5=0.027

                  u⁽⁵⁾=ｌ⁽⁴⁾+(u⁽⁴⁾-ｌ⁽⁴⁾)*F_x(3)=0.024+(0.03-0.024)*1=0.03

    113231

            ｌ⁽⁶⁾=ｌ⁽⁵⁾+(u⁽⁵⁾-ｌ⁽⁵⁾)*F_x(0)=0.027+(0.03-0.027)*0=0.027

            u⁽⁶⁾=ｌ⁽⁵⁾+(u⁽⁵⁾-ｌ⁽⁵⁾)*F_x(1)=0.027+(0.03-0.027)*0.2=0.0276

             T_x(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 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,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⁾)*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⁾)*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⁾)*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。