Fourth time assignment

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

Table 4-9

Letters	Probability
A1	0.2
A2	0.3
A3	0.5

Solution:

By graph: 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

∵x (AI) =i,

∴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

The real value label for 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:

Mapping Relationship A1=>1,a2=>2,a3=>3

FX (k) =0, K≤0, FX (1) =0.2, FX (2) =0.5, FX (3) =1, k>3.

Nether: L (0) = 0, upper bound: U (0) =1

L (k) = L (k-1) + (U (k-1)-L (k-1)) Fx (xk-1)

U (k) =l (k-1) + (U (k-1)-L (k-1)) Fx (XK)

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

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

If x1=1, then the interval is [0,0.2], x1=2, the interval is [0.2,0.5], x1=3, then the interval is [0.5,1]

0.63215699 within the range [0.5,1]

∴ The first sequence is A3

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

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

If x2=1, then the interval is [0.5,0.6], x2=2, the interval is [0.6,0.75], x2=3, then the interval is [0.75,1]

0.63215699 within the interval [0.6,0.75] so the second sequence is A2

∴

When x3=2

The interval is [0.63,0.675] 0.63215699 within that interval

When X4=1

The interval is [0.63,0.639] 0.63215699 within that interval

When x5=2

The interval is [0.6318,0.6345] 0.63215699 within that interval

When X6=1

The interval is [0.6318,0.63234] 0.63215699 within that interval

When x7=3

The interval is [0.63207,0.63234] 0.63215699 within that interval

When x8=2

The interval is [0.632124,0.632205] 0.63215699 within that interval

When x9=2

The interval is [0.6321402,0.6321645] 0.63215699 within that interval

When x10=3

The interval is [0.63215235,0.6321645] 0.63215699 within that interval

A sequence of length 10 with a ∴ label of 0.63215699 is A3A2A2A1A2A1A3A2A2A3

Fourth time assignment