Fourth time assignment

5, given the probability model as shown in table 4-9, the real value label of the sequence A1A1A3A2A3A1 is obtained.

Solution: By the know,

P (A1) =0.2, P (A2) =0.3, P (A3) =0.5

FX (0) =0,FX (1) =0.2, FX (2) =0.5, FX (3) =1.0, U (0) =1, L (0) =0

Because X (AI) =i, so x (A1) =1,x (A2) =2,x (A3) =3

by formula, L (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn-1)

U (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn)

When the A1 first appears, there are:

L (1) =l (0) + (U (0)-L (0)) Fx (0) =0

U (1) =l (0) + (U (0)-L (0)) Fx (1) =0.2

When A1 appears for the second time, there are:

L (2) =l (1) + (U (1)-L (1)) Fx (0) =0

U (2) =l (1) + (U (1)-L (1)) Fx (1) =0.04

When the A3 appears for the third time, there are:

L (3) =l (2) + (U (2)-L (2)) Fx (2) =0.02

U (3) =l (2) + (U (2)-L (2)) Fx (3) =0.04

When the A2 appears for the fourth time, there are:

L (4) =l (3) + (U (3)-L (3)) Fx (1) =0.024

U (4) =l (3) + (U (3)-L (3)) Fx (2) =0.03

When the A3 appears for the fifth time, there are:

L (5) =l (4) + (U (4)-L (4)) Fx (2) =0.027

U (5) =l (4) + (U (4)-L (4)) Fx (3) =0.03

When the A1 appears for the sixth time, there are:

L (6) =l (5) + (U (5)-L (5)) Fx (0) =0.027

U (6) =l (5) + (U (5)-L (5)) Fx (1) =0.0276

Therefore, the real value tag of the sequence A1A1A3A2A3A1 is: T (113231) = (L (6) + U (6))/2=0.0273;

6, for the probability model given in table 4-9, a sequence with a length of 10 labeled 0.63215699 is decoded.

Solution: The F (x1) =0.2,f (x2) =0.5,f (x3) =1 can be known by Table 4-9.

First assume L (0) =0,u (0) =1.

1. t*= (0.63215699-0)/(1-0) =0.63215699

FX (2) =0.5<= t*<= FX (3) =1

L (1) = L (0) + (U (0)-L (0)) Fx (2) =0.5

U (1) = L (0) + (U (0)-L (0)) Fx (3) =1

So the first character is A3

2. t*= (0.63215699-0.5)/(1-0.5) =0.26431398

FX (1) =0.2<= t*<= FX (2) =0.5

L (2) = L (1) + (U (1)-L (1)) Fx (1) =0.6

U (2) = L (1) + (U (1)-L (1)) Fx (2) =0.75

So the second character is A2

3. t*= (0.63215699-0.6)/(0.75-0.6) =0.21437993

FX (1) =0.2<= t*<= FX (2) =0.5

L (3) = L (2) + (U (2)-L (2) Fx (1) =0.63

U (3) = L (2) + (U (2)-L (2)) Fx (2) =0.635

So the third character is A2

4. t*= (0.63215699-0.63)/(0.635-0.63) =0.431398

FX (1) =0.2<= t*<= FX (2) =0.5

L (4) = L (3) + (U (3)-L (3)) Fx (1) =0.631

U (4) = L (3) + (U (3)-L (3)) Fx2) =0.6325

So the fourth character is A2

5. t*= (0.63215699-0.631)/(0.6325-0.631) =0.77132667

FX (2) =0.5<= t*<= FX (3) =1

L (5) = L (4) + (U (4)-L (4)) Fx (2) =0.63175

U (5) = L (4) + (U (4)-L (4)) Fx (3) =0.6325

So the fifth character is A3

6. t*= (0.63215699-0.63175)/(0.6325-0.63175) =0.5426533

FX (2) =0.5<= t*<= FX (3) =1

L (6) = L (5) + (U (5)-L (5)) Fx (2) =0.632125

U (6) = L (5) + (U (5)-L (5)) Fx (3) =0.6325

So the sixth character is A3

7. t*= (0.63215699-0.632125)/(0.6325-0.632125) =0.04265333

FX (k) =0<= t*<= FX (1) =0.2

L (7) = L (6) + (U (6)-L (6)) Fx (0) =0.632125

U (7) = L (6) + (U (6)-L (6)) Fx (1) =0.632275

So the seventh character is A1

8. t*= (0.63215699-0.632125)/(0. 632125-0.632275) =0.21326667

FX (1) =0.2<= t*<= FX (2) =0.5

L (8) = L (7) + (U (7)-L (7)) Fx (1) =0.632155

U (8) = L (7) + (U (7)-L (7)) Fx (5) =0.6322

So the eighth character is A2

9. t*= (0.63215699-0.632155)/(0.6322-0.632155) =0.04422222

FX (0) =0<= t*<= FX (1) =0.2

L (9) = L (8) + (U (8)-L (8)) Fx (0) =0.632155

U (9) = L (8) + (U (8)-L (8)) Fx (1) =0.632164

So the nineth character is A1

t*= (0.63215699-0.632155)/(0.632164-0.632155) =0.22111111

FX (1) =0.2<= t*<= FX (2) =0.5

L (Ten) = L (9) + (U (9)-L (9)) Fx (1) =0.6321568

U (Ten) = L (9) + (U (9)-L (9)) Fx (2) =0.6321595

So the tenth character is A2

Therefore, a sequence with a label value of 0.63215699 for a length of 10 decodes the sequence to A3A2A2A2A3A3A1A2A1A2

Fourth time assignment