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。
答:
Fx(k)=0, k≤0, Fx(1)=0.2; Fx(2)=0.5; Fx(3)=1, k>3.
利用公式确定标签所在的上下限,将u(0)初始化为1,将l(0)初始化为0。
l(1) =0+(1-0)Fx(0)=0
u(1) =0+(1-0)Fx(1)=0.2
标签在区间[0,0.2)中,
l(2) =0+(0.2-0)Fx(0)=0
u(2) =0+(0.2-0)Fx(1)=0.04
标签在区间[0,0.04)中。
l(3) =0+(0.04-0)Fx(2)=0.02
u(3) =0+(0.04-0)Fx(3)=0.04
标签在区间[0.02,0.04)。
l(4) =0.02+(0.04-0.02)Fx(1)=0.024
u(4) =0.02+(0.04-0.02)Fx(2)=0.03
标签的区间[0.024,0.03)。
l(5) =0.024+(0.03-0.024)Fx(2)=0.027
u(5) =0.024+(0.03-0.024)Fx(3)=0.03
标签在区间[0.027,0.03)。
l(6) =0.027+(0.03-0.027)Fx(0)=0.027
u(6) =0.027+(0.03-0.027)Fx(1)=0.0276
Tx(113231)= ( u(6) + l(6) )/2=(0.0276+0.027)/2=0.0273
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。
答:
t*=(0.63215699-0)/(1-0)=0.63215699
Fx(2)≤t*≤Fx(3)
l(1) =l(0) +(u(0) -l(0) )Fx(2)=0+(1-0)*0.5=0.5
u(1) =l(0) +(u(0) -l(0) )Fx(3)=0+(1-0)*1=1
第一个序列为:a3
t*=(0.63215699-0.5)/(1-0.5)=0.264314
Fx(1)≤t*≤Fx(2)
l(2) =l(1) +(u(1) -l(1) )Fx(1)=0.5+(1-0.5)*0.2=0.6
u(2) =l(1) +(u(1) -l(1) )Fx(2)=0.5+(1-0.5)*0.5=0.75
第二个序列为:a2
t*=(0.63215699-0.6)/(0.75-0.6)=0.2143799
Fx(1)≤t*≤Fx(2)
l(3) =l(2) +(u(2) -l(2) )Fx(1)=0.6+(0.75-0.6)*0.2=0.63
u(3) =l(2) +(u(2) -l(2) )Fx(2)=0.6+(0.75-0.6)*0.5=0.675
第三个序列为:a2
t*=(0.63215699-0.63)/(0.675-0.63)=0.04793311
Fx(0)≤t*≤Fx(1)
l(4) =0.63+(0.675-0.63)*0=0.63
u(4) =0.63+(0.675-0.63)*0.2=0.639
第四个序列为:a1
t*=(0.63215699-0.63)/(0.639-0.63)=0.2396656
Fx(1)≤t*≤Fx(2)
l(5) =0.63+(0.639-0.63)*0.2=0.6318
u(5) =0.63+(0.639-0.63)*0.5=0.6345
第五个序列为:a2
t*=(0.63215699-0.6318)/(0.6345-0.6318)=0.1322185
Fx(0)≤t*≤Fx(1)
l(6) =0.6318+(0.6345-0.6318)*0=0.6318
u(6) =0.6318+(0.6345-0.6318)*0.2=0.63234
第六个序列为:a1
t*=(0.63215699-0.6318)/(0.63234-0.6318)=0.6610926
Fx(2)≤t*≤Fx(3)
l(7) =0.6318+(0.63234-0.6318)*0.5=0.63207
u(7) =0.6318+(0.63234-0.6318)*1=0.63234
第七个序列为:a3
t*=(0.63215699-0.63207)/(0.63234-0.63207)=0.3221852
Fx(1)≤t*≤Fx(2)
l(8) =0.63207+(0.63234-0.63207)*0.2=0.632124
u(8) =0.63207+(0.63234-0.63207)*0.5=0.632205
第八个序列为:a2
t*=(0.63215699-0.632124)/(0.632205-0.632124)=0.40728395
Fx(1)≤t*≤Fx(2)
l(9) =0.632124+(0.632205-0.632124)*0.2=0.6321402
u(9) =0.632124+(0.632205-0.632124)*0.5=0.6321645
第九个序列为:a2
t*=(0.63215699-0.6321402)/(0.6321645-0.6321402)=0.6909
Fx(2)≤t*≤Fx(3)
第十个序列为:a3
解码:3221213223