Third time job

5. Given the probability model shown in table 4-9, the real value label of the sequence 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: By the problem, map 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.

Include

int main ()
{
Double tag=0.63215699;
Double l[100],u[100];
Double T;
l[0]=0;
U[0]=1;
Double a0=0.0,a1=0.2,a2=0.5,a3=1.0;
int m[100];
for (int k=1;k<=10;k++)
{
T= (Double) (tag-l[k-1])/(u[k-1]-l[k-1]);
if (T>=A0&&T<=A1)
{
u[k]=l[k-1]+ (u[k-1]-l[k-1]) A1;
l[k]=l[k-1]+ (u[k-1]-l[k-1]) A0;
M[k]=1;
}
else if (T>A1&&T<=A2)
{
u[k]=l[k-1]+ (u[k-1]-l[k-1]) A2;
l[k]=l[k-1]+ (u[k-1]-l[k-1]) A1;
m[k]=2;
}
else if (T>A2&&T<=A3)
{
u[k]=l[k-1]+ (u[k-1]-l[k-1]) A3;
l[k]=l[k-1]+ (u[k-1]-l[k-1]) A2;
m[k]=3;
}
printf ("%d", m[k]);
}
return 0;
}

Third time job