第三次作业

我的答案如下：

由题可知：Fx(0)=0,Fx(1),Fx(2)=0.5,Fx(3)=1 l⁰=0,u⁰=1

由公式

l^k=l^k+(u^k-l^k)Fx(x_k-1)

u^k=l^k+(u^k-l^k)Fx(x_k)

及序列为113231

(1) k=1,input 1

l¹=l⁰+(u⁰-l⁰)Fx(0)=0

u¹=l⁰+(u⁰-l⁰)Fx(1)=0.2

(2) k=2,input 1

l²=l¹+(u¹-l¹)Fx(0)=0

u²=l¹+(u¹-l¹)Fx(1)=0+(1-0)*0.2=0.04

(3) k=3,input 3

l³=l²+(u²-l²)Fx(2)=0+(0.04-0)*0.5=0.02

u³=l²+(u²-l²)Fx(3)=0+(0.04-0)*1=0.04

(4) k=4,input 2

l⁴=l³+(u³-l³)Fx(1)=0.02+(0.04-0.02)*0.2=0.024

u⁴=l³+(u³-l³)Fx(2)=0.02+(0.04-0.02)*0.5=0.03

(5) k=5,input 3

l⁵=l⁴+(u⁴-l⁴)Fx(2)=0.024+(0.03-0.024)*0.5=0.027

u⁵=l⁴+(u⁴-l⁴)Fx(3)=0.024+(0.03-0.024)*1=0.03

(6) k=6,input 1

l⁶=l⁵+(u⁵-l⁵)Fx(0)=0.027+(0.03-0.027)*0=0.027

u⁶=l⁵+(u⁵-l⁵)Fx(1)=0.027+(0.03-0.027)*1=0.03

T_（113231）=( l⁶+u⁶)/2=（0.027+0.03）/2=0.0285

解：由题可知：Fx(0)=0,Fx(1),Fx(2)=0.5,Fx(3)=1 l⁰=0,u⁰=1

（1）k=1

因为t*=(标签-l^k-1)/(u^k-1-l^k-1)=(0.63215699-0)/(1-0)=0.63215699

则Fx(2)=0.5<t*<1=Fx(3) 可得input 3

l¹=l⁰+(u⁰-l⁰)Fx(2)=0+( 1-0 )*0.5=0.5

u¹=l⁰+(u⁰-l⁰)Fx(3)=0+( 1-0 )*1=1

（2）k=2

因为t*=(标签-l^k-1)/(u^k-1-l^k-1)=(0.63215699-0.5)/(1-0.5)=0.26431398

则Fx(1)=0.2<t*<0.5=Fx(2) 可得input 32

l²=l¹+(u¹-l¹)Fx(1)=0.5+( 1- 0.5)*0.2=0.6

u²=l¹+(u¹-l¹)Fx(2)=0.5+( 1-0.5 )*0.5=0.75

（3）k=3

因为t*=(标签-l^k-1)/(u^k-1-l^k-1)=(0.63215699-0.6)/(0.75-0.6)=0.21437993

则Fx(1)=0.2<t*<0.5=Fx(2) 可得input 322

l³=l²+(u²-l²)Fx(1)=0.6+(0.75 -0.6 )*0.2=0.63

u³=l²+(u²-l²)Fx(2)=0.6+(0.75 -0.6 )*0.5=0.675

.........

（9）k=9

因为t*=(标签-l^k-1)/(u^k-1-l^k-1)=(0.63215699-0.632124)/(0.632205-0.632124)=0.000081

则Fx(0)=0<t*<0.2=Fx(1) 可得input 322121321

l⁹=l⁸+(u⁸-l⁸)Fx(0)=0

u⁹=l⁸+(u⁸-l⁸)Fx(1)=0.632124+(0.632205 - 0.632124)*0.2=0.6321402

（10）k=10

因为t*=(标签-l^k-1)/(u^k-1-l^k-1)=(063215699-0.632124)/(0.6321402-0.632124)=0.20364198

则Fx(1)=0.2<t*<0.5=Fx(2) 可得input 3221213212

最后可得序列编码为：a₃a₂a₂a₁a₂a₁a₃a₂a₁a₂