$e.g \ sequence = { a_1 \ a_1 \ a_2 \ a_1 \ a_3 }$

iv. Solution:

Step 1: Initialization (n=0)

$L (0) =0$

$U (0) = 1$

Step 2: $1^{st}$ Symbol in sequence = $a_1$ = X (n=1)

$L (1) = L (0) + [U (0) – L (0)] F (0) = 0 + [1-0] [0] =0$

$U (1) = L (0) + [U (0) – L (0)] F (a_1) = 0 + [1-0] [0.6] =0.6$

$[0, 0.6]$

Step 3: $2^{nd}$ Symbol in sequence = $a_1$ = X (n=2)

$L (2) = L (1) + [U (1) - L (1)] F (a) = 0 + [0.6-0] [0] =0$

$U (2) = L(1) + [U (1) –L (1)] F (a_1) =0 + [0.6-0] [0.6] =0.36$

$[0, 0.36]$

Step 4: $3^{rd}$ Symbol in sequence = $a_2$=X (n=3)

$L (3) = L (2) + [U (2) -L(2)] F (a_1) = 0 +[ 0.36 -0] 0.6 =0.216$

$U (3) = L (3) + [U (2)- L (2)] F (a_2) = 0+ [0.36 -0] 0.8 = 0.288$

$[0.216 , 0.288]$

Step 5: $4^{th}$ Symbol = $a_1$ = X (n=4)

$L (4) = L(3) + [ U(3) - L(3)] F (0) = 0.216$

$U (4) = L(3) + [U (3) - L(3)] F (a_1) = 0.216 + [0.288-0.216] 0.6 =0.2592$

$[0.216 , 0.2592]$

Step 6: $5^{th}$ Symbol = $a_3$ =X (n=5)

$L(5) = L(4) + [ U (4) – L(4)] F (2) = 0.25056$

$U (5) = L (4) + [U (4) – L(4)] F (a_2) = 0.2592$

Step 7: Tag Generation

Tag = = 0.25488

v. Tag ( Fractional decimal number) represents given input of S symbols.

vi. If number of symbols increases in a sequence then it will just affect to resolution of tag but same tag can be represented in fixed number of bits.

vii. Disadvantage:

• The coded tag is sensitive to position symbol in sequence. If position of symbol changes it will result different value of tag.
• Tag cannot be generated / insert sequence.