Third time job

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

For:

obtained by test instructions: The real value label of the sequence A1A1A3A2A3A1 is the real value label of the sequence 113231.

Set upper bound U (0) = 1, Nether L (0) = 0, symbol set A={A1,A2,A3}, and P (A1) =0.2,p (A2) =0.3,p (A3) =0.5

Because P (x=i) =p (AI), so P (x=1) =p (a1) = 0.2,

P (x=2) = P (a2) = 0.3,

P (x=3) =p (A3) =0.5.

FX (0) =0, FX (1) =p (A0) + p (A1) =0.2, FX (2) =p (A1) + p (A2) =0.5, FX (3) =p (A1) + p (A2) +p (A3) =1

Again because U (k) =l (k-1) + (U (k-1)-L (k-1)) *fx (XK)

L (k) =l (k-1) + (U (k-1)-L (k-1)) *fx (xk-1)

So the results are as follows:

The first element of the sequence is 1

U (1) =l (0) + (U (0)-L (0)) *fx (1) =0+ (1-0) *0.2=0.2

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

So the range of the label [0,0.2]

The second element of the sequence is 1

U (2) =l (1) + (U (1)-L (1)) *FX (1) =0+ (0.2-0) *0.2=0.04

L (2) =l (1) + (U (1)-L (1)) *fx (0) =0+ (0.2-0) *0=0

So sequence 1 1 label interval [0,0.04]

The third element of the sequence is 3

U (3) =l (2) + (U (2)-L (2)) *FX (3) =0+ (0.04-0) * * =0.04

L (3) =l (2) + (U (2)-L (2)) *fx (2) =0+ (0.04-0) *0.5=0.02

So the range of the sequence label [0.02,0.04)

The fourth element of the sequence is 2

U (4) =l (3) + (U (3)-L (3)) *fx (2) =0.02+ (0.04-0.02) *0.5 =0.03

L (4) =l (3) + (U (3)-L (3)) *fx (1) =0.02+ (0.04-0.02) *0.2=0.024

So sequence 2 3 label interval [0.024,0.03]

The fifth element of the sequence is 3

U (5) =l (4) + (U (4)-L (4)) *fx (3) =0.024+ (0.03-0.024) *1=0.03

L (5) =l (4) + (U (4)-L (4)) *fx (2) =0.024+ (0.03-0.024) *0.5=0.027

So the range of the sequence label [0.027,0.03)

The sixth element of the sequence is 1

U (6) =l (5) + (U (5)-L (5)) *fx (1) =0.027+ (0.03-0.027) *0.2=0.0276

L (6) =l (5) + (U (5)-L (5)) *fx (0) =0.027+ (0.03-0.027) *0=0.027

So the range of the sequence label [0.027,0.0276)

The real value tags for the A1A1A3A2A3A1 series are:

Tx (113231) = (U (6) + L (6))/2

= (0.0276+0.027)/2

=0.0546/2

=0.0273

6. For the probability model given in table 4-9, a sequence with a length of 10 labeled 0.63215699 is decoded.

Solution: The following procedures can be obtained according to test instructions:

#include <stdio.h>

int main ()
{
int m[20];
Double f0=0,f1=0.2,f2=0.5,f3=1.0;
Double tag=0.63215699,t1;
Double l[20],u[20];
l[0]=0;
u[0]=1.0;
for (int j=1;j<=10;j++)
{
T1= (Tag-l[j-1])/(u[j-1]-l[j-1]);
if (T1>F0&&T1<F1)
{
l[j]=l[j-1]+ (u[j-1]-l[j-1]) *f0;
u[j]=l[j-1]+ (u[j-1]-l[j-1]) *f1;
M[j]=1;
}
else if (T1>F1&&T1<F2)
{
l[j]=l[j-1]+ (u[j-1]-l[j-1]) *f1;
u[j]=l[j-1]+ (u[j-1]-l[j-1]) *f2;
m[j]=2;
}
Else
{
l[j]=l[j-1]+ (u[j-1]-l[j-1]) *f2;
u[j]=l[j-1]+ (u[j-1]-l[j-1]) *f3;
m[j]=3;
}
printf ("%d\n", M[j]);
}

return 0;

}

The result of debugging is 3221213223.

Third time job