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

                             字母                                          概率

                              a1                                            0.2

                              a2                                            0.3

                              a3                                            0.5

解：p(a1=0.2),p(a2=0.3),p(a3=0.5).

      映射：a1=1,a2=2,a3=3.

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

1                                                                 113

     u⁽¹⁾=l⁽⁰⁾+(u⁽⁰⁾-l⁽⁰⁾)F_x(1)=0.2

      l⁽²⁾=l⁽¹⁾+(u⁽¹⁾-l⁽¹⁾)F_x(1)=0.1

11

      u⁽²⁾=l⁽¹⁾+(u⁽¹⁾-l⁽¹⁾)F_x(2)=0.2

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

第三次作业

原文：http://www.cnblogs.com/liuli553400/p/4812533.html