Third time job

Reference book "Introduction to Data Compression (4th edition)" Page 100 5, 6

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

By the probability model, the mapping 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

The 1th element of the sequence is A1:

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

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

The 2nd element of a sequence is A1

L (2) = 0+ (0.2-0) Fx (0) =0

U (2) =0+ (0.2-0) Fx (1) =0.04

The 3rd element of a sequence is A3

L (3) = 0+ (0.04-0) Fx (2) =0.02

U (3) =0+ (0.04-0) Fx (3) =0.04

The 4th element of a sequence is A2

L (4) = 0.02+ (0.04-0.02) Fx (1) =0.024

U (4) =0.02+ (0.04-0.02) Fx (2) =0.03

The 5th element of a sequence is A3

L (5) = 0.024+ (0.03-0.024) Fx (2) =0.027

U (5) =0.024+ (0.03-0.024) Fx (3) =0.03

The 6th element of a sequence is A1

L (6) = 0.027+ (0.03-0.027) Fx (0) =0.027

U (6) =0.027+ (0.03-0.027) Fx (1) =0.0276

The range of the label for the sequence A1A1A3A2A3A1 is [0.027,0.0276]

The tags that can generate sequence a1a1a3a2a3a1 are as follows:

Tx (A1A1A3A2A3A1) = L (6) +u (6)/2

= (0.027+0.0276)/2

=0.0273

6, for the probability model shown in table 4-9, for a label 0.63215699 of the length of a sequence of 10 decoding.

By the probability model, the mapping 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)

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

FX (0) =0<=t*<=1= FX (3)

Third time job