5. Given the probability model shown in table 4-9, the real value label of A1A1A3A2A3A1 is obtained.
Table 4-9 The probabilistic model of Exercise 5, exercise 6 |
Letter Probability |
A1 0.2 |
A2 0.3 |
A3 0.5 |
Solution:
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
by 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:
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:
L (2) =l (1) + (U (1)-L (1)) Fx (0) =0
U (2) =l (1) + (U (1)-L (1)) Fx (1) =0.04
When A3 appears for the third time:
L (3) =l (2) + (U (2)-L (2)) Fx (2) =0.02
U (3) =l (2) + (U (2)-L (2)) Fx (3) =0.04
Fourth time when A2 appears:
L (4) =l (3) + (U (3)-L (3)) Fx (1) =0.024
U (4) =l (3) + (U (3)-L (3)) Fx (2) =0.03
Fifth time when A3 appears:
L (5) =l (4) + (U (4)-L (4)) Fx (2) =0.027
U (5) =l (4) + (U (4)-L (4)) Fx (3) =0.03
Sixth time when A1 appears:
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 serial number with a length of 10 labeled 0.63215699 is decoded.
Solution:
Obtained from table: 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
by 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)
so: Set u (0) =1,L (0) =0
L (1) =0+ (1-0) Fx (x1-1) =fx (x1-1)
U (1) =0+ (1-0) Fx (x1) =fx (x1)
if x1=1, then the interval is [0,0.2]
if x1=2, then the interval is [0.2,0.5]
if x1=3, then the interval is [0.5,1]
1. because 0.63215699 is in [0.5,1] , x1=3
L (2) =0.5+ (1-0.5) Fx (x2-1) =0.5+0.5fx (x2-1)
U (2) =0.5+ (1-0.5) Fx (x2) =0.5+0.5fx (x2)
if x2=1, then the interval is [0.5,0.6]
if x2=2, then the interval is [0.6,0.75]
if x2=3, then the interval is [0.75,1]
2. because 0.63215699 is in [0.6,0.75] , x2=2
L (3) =0.5+ (0.75-0.6) Fx (x3-1) =0.6+0.15fx (x3-1)
U (3) =0.5+ (0.75-0.6) Fx (x3) =0.6+0.15fx (x3)
if x3=1, then the interval is [0.6,0.63]
if x3=2, then the interval is [0.63,0.675]
if x3=3, then the interval is [0.675,0.75]
3. because 0.63215699 is in [0.63,0.675] , x3=2
L (4) =0.63+ (0.675-0.625) Fx (x4-1) =0.63+0.045fx (x4-1)
U (4) =0.63+ (0.675-0.625) Fx (x4) =0.63+0.045fx (x4)
if x4=1, then the interval is [0.63,0.639)
if x4=2, then the interval is [0.639,0.6525]
if x4=3, then the interval is [0.6525,0.675]
4. because 0.63215699 is in [0.63,0.639) , so x4=1
L (5) =0.63+ (0.639-0.63) Fx (x5-1) =0.63+0.009fx (x5-1)
U (5) =0.63+ (0.639-0.63) Fx (X5) =0.63+0.009fx (X5)
if x5=1, then the interval is [0.63,0.6318]
if x5=2, then the interval is [0.6318,0.6345]
if x5=3, then the interval is [0.6345,0.639]
5. because 0.63215699 is in [0.6318,0.6345] , x5=2
L (6) =0.6318+ (0.6345-0.6318) Fx (x6-1) =0.6318+0.0027fx (x6-1)
U (6) =0.6318+ (0.6345-0.6318) Fx (X6) =0.6318+0.0027fx (X6)
if x6=1, then the interval is [0.6318,0.63234]
if x6=2, then the interval is [0.63234,0.63315]
if x6=3, then the interval is [0.63315,0.63345]
6. because 0.63215699 is in [0.6318,0.63234) , so x6=1
L (7) =0.6318+ (0.63234-0.6318) Fx (x7-1) =0.6318+0.00054fx (x7-1)
U (7) =0.6318+ (0.63234-0.6318) Fx (x7) =0.6318+0.00054fx (X7)
if x7=1, then the interval is [0.6318,0.631908]
if x7=2, then the interval is [0.631908,0.63207]
if x7=3, then the interval is [0.63207,0.63234]
7. because 0.63215699 is in [0.63207,0.63234] , x7=3
L (8) =0.63207+ (0.63234-0.63207) Fx (x8-1) =0.63207+0.00027fx (x8-1)
U (8) =0.63207+ (0.63234-0.63207) Fx (x8) =0.63207+0.00027fx (x8)
if x8=1, then the interval is [0.63207,0.632124]
if x8=2, then the interval is [0.632124,0.632205]
if x8=3, then the interval is [0.632205,0.63234]
8. because 0.63215699 is in [0.632124,0.632205] , x8=2
L (9) =0.632124+ (0.632205-0.632124) Fx (x9-1) =0.632124+0.000081fx (x9-1)
U (9) =0.632124+ (0.632205-0.632124) Fx (x9) =0.632124+0.000081fx (x9)
if x9=1, then the interval is [0.632124,0.6321402]
if x9=2, then the interval is [0.6321402,0.6321645]
if x9=3, then the interval is [0.6321645,0.632205]
9. because 0.63215699 is in [0.6321402,0.6321645] , x9=2
L (Ten) =0.6321402+ (0.6321645-0.6321402) Fx (x10-1) =0.6321402+0.00006243fx (x10-1)
U (Ten) =0.6321402+ (0.6321645-0.6321402) Fx (x10) =0.6321402+0.00006243fx (x10)
if x10=1, then the interval is [0.6321402,0.63214506]
if x10=2, then the interval is [0.63214506,0.63215235]
if x10=3, then the interval is [0.63215235.0.6321645]
because 0.63215699 is in [0.63215235,0.6321645] , x10=3
so the sequence is 3221213223 that is a3a2a2a1a2a1a3a2a2a3
Assignment for the fourth time