Third time job

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

Available by test instructions:

Character set {a1,a2,a3}, where P (A1) =0.2, P (A2) =0.3, P (A3) =0.5

We can use the formula to determine the upper and lower bounds of the label.

Using the update formula, you get:

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

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

The range of A1A1 for the available sequence is [0,0.2]

The 2nd element of the sequence is A1, with the update formula available:

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

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

The range of A1A1 for the available sequence is [0,0.04]

The 3rd element of the sequence is A3, with the update formula available:

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

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

The range of A1A1A3 for the available sequence is [0.02,0.04]

The 4th element of the sequence is A2, with the update formula available:

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 range of a1a1a3a2 for the available sequence is [0.024,0.03]

The 5th element of the sequence is A3, with the update formula available:

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 range of A1A1A3A2A3 for the available sequence is [0.027,0.03]

The 6th element of the sequence is A1, with the update formula available:

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 tags that can generate sequence a1a1a3a2a3a1 are as follows:

Tx (A1A1A3A2A3A1) = (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.

From table 4-9 we know FX (k) =0, K≤0, FX (1) =0.2, FX (2) =0.5, FX (3) =1, k>3.

Make L (0) =0, u (0) = 1. The sequence of the corresponding elements is x1x2x3x4x5x6x7x8x9x10, and the updated formula is used

X1

L (1) = L (0) + (U (0)-L (0)) *fx (x1-1) =fx (x1-1)

U (1) = L (0) + (U (0)-L (0)) *FX (x1) =fx (x1)

X1=1, the interval is [0,0.2]

x1=2, the interval is [0.2,0.5]

X1=3, the interval is [0.5,1]

Since 0.63215699 is within the interval [0.5,1], the sequence of the first element is 3 and the element is A3

X2

U (2) = L (1) + (U (1)-L (1)) *fx (x2) =0.5+ (1-0.5) *fx (x2) =0.5+0.5fx (x2)

L (2) =l (1) + (U (1)-L (1)) *fx (x2-1) =0.5+ (1-0.5) *fx (x2-1) =0.5+0.5fx (x2-1)

X2=1, the interval is [0.5,0.6]

x2=2, the interval is [0.6,0.75]

X2=3, the interval is [0.75,1]

Since 0.63215699 is within the interval [0.6,0.75], the second element has a sequence of 2 and the element is A2

X3

U (3) = L (2) + (U (2)-L (2)) *fx (x3) =0.6+ (0.75-0.6) *fx (x2) =0.6+0.15fx (x3)

L (3) = L (2) + (U (2)-L (2)) *fx (x3-1) =0.6+ (0.75-0.6) *fx (x2-1) =0.6+0.15fx (x3-1)

X3=1, the interval is [0.6,0.63]

x3=2, the interval is [0.63,0.675]

X3=3, the interval is [0.675,0.75]

Since 0.63215699 is within the interval [0.63,0.675], the third element has a sequence of 2 and the element is A2

X4

U (4) =l (3) + (U (3)-L (3)) *fx (x4) =0.63+ (0.675-0.63) *fx (x4) =0.63+0.045*fx (x4)

L (4) =l (3) + (U (3)-L (3)) *fx (x4-1) =0.63+ (0.675-0.63) *fx (x4-1) =0.63+0.045*fx (x4-1)

X4=1, the interval is [0.63,0.639]

x4=2, the interval is [0.639,0.6525]

X4=3, the interval is [0.6525,0.675]

Since 0.63215699 is within the interval [0.63,0.639], the fourth element has a sequence of 1 and the element is A1

X5

U (5) =l (4) + (U (4)-L (4)) *fx (X5) =0.63+ (0.639-0.63) *fx (X5) =0.63+0.009*fx (X5)

L (5) =l (4) + (U (4)-L (4)) *fx (x5-1) =0.63+ (0.639-0.63) *fx (x5-1) =0.63+0.009*fx (x5-1)

X5=1, the interval is [0.63,0.6318]

x5=2, the interval is [0.6318,0.6345]

X5=3, the interval is [0.6345,0.639]

Since 0.63215699 is within the interval [0.6318,0.6345], the Fifth element has a sequence of 2 and the element is A2

X6

U (6) =l (5) + (U (5)-L (5)) *fx (X6) =0.6318+ (0.6345-0.6318) *fx (X6) =0.6318+0.0027*fx (X6)

L (6) =l (5) + (U (5)-L (5)) *fx (x6-1) =0.6318+ (0.6345-0.6318) *fx (x6-1) =0.6318+0.0027*fx (x6-1)

X6=1, the interval is [0.6318,0.63234]

x6=2, the interval is [0.63234,0.63315]

X6=3, the interval is [0.63315,0.6345]

Since 0.63215699 is within the interval [0.6318,0.63234], the sixth element has a sequence of 1 and the element is A1

X7

U (7) =l (6) + (U (6)-L (6)) *fx (X7) =0.6318+ (0.63234-0.6318) *fx (x7) =0.6318+0.00054*fx (X7)

L (7) =l (6) + (U (6)-L (6)) *fx (x7-1) =0.6318+ (0.63234-0.6318) *fx (x7-1) =0.6318+0.00054*fx (x7-1)

X7=1, the interval is [0.6318,0.631908]

x7=2, the interval is [0.631908,0.63207]

X7=3, the interval is [0.63207,0.63234]

Since 0.63215699 is in the interval [0.63207,0.63234], the seventh element has a sequence of 3 and the element is A3

x8

U (8) =l (7) + (U (7)-L (7)) *fx (x8) =0.63207+ (0.63234-0.63207) *fx (x8) =0.63207+0.00027*fx (x8)

L (8) =l (7) + (U (7)-L (7)) *fx (x8-1) =0.63207+ (0.63234-0.63207) *fx (x8-1) =0.63207+0.00027*fx (x8-1)

X8=1, the interval is [0.63207,0.632124]

x8=2, the interval is [0.632124,0.632205]

X8=3, the interval is [0.632205,0.63234]

Since 0.63215699 is within the interval [0.632124,0.632205], the eighth element has a sequence of 2 and the element is A2

X9

U (9) =l (8) + (U (8)-L (8)) *fx (x9) =0.632124+ (0.632205-0.632124) *fx (x9) =0.632124+ (8.1e-5) *fx (x9)

L (9) =l (8) + (U (8)-L (8)) *fx (x9-1) =0.632124+ (0.632205-0.632124) *fx (x9-1) =0.632124+ (8.1e-5) *fx (x9-1)

X9=1, the interval is [0.632124,0.6321402]

x9=2, the interval is [0.0.6321402,0.6321645]

X9=3, the interval is [0.6321645,0.63234]

Since 0.63215699 is within the interval [0.6321402,0.6321645], the nineth element has a sequence of 2 and the element is A2

X10

U (Ten) =l (9) + (U (9)-L (9)) *fx (x10) =0.6321402+ (0.6321645-0.6321402) *fx (x10) =0.6321402+ (2.43e-5) *fx (x10)

L (Ten) =l (9) + (U (9)-L (9)) *fx (x10-1) =0.6321402+ (0.6321645-0.6321402) *fx (x10-1) =0.6321402+ (2.43e-5) *fx (x10-1)

X10=1, the interval is [0.6321402,0.63212886]

x10=2, the interval is [0.63212886,0.63215325]

X10=3, the interval is [0.63215325,0.6321645]

Since 0.63215699 is within the interval [0.63215235,0.6321645], the nineth element has a sequence of 3 and the element is A3

Therefore, according to the probability model given in table 4-9, the decoding result of a sequence with a length of 10 labeled 0.63215699 is: A3 A2 A2 A1 A2 A1 A3 A2 A2.

Third time job