e : $B^3 —\gt B^5$ defined by

e(000) = 00000 e(100) - 10011

e(001) = 00110 0101) - 10101

e(010) = 01001 e(110) = 11010

e(011)= 01111 e(111)= 11100

Decode the following words relative to a maximum likelyhood decoding function.

1. 11001
2. 01010
3. 00111

1. To decode the word 11001 we first arrange the given code as

   $00000 \ \ \ \ \ 00110 \ \ \ \ \ 01001 \ \ \ \ \ 01111 \ \ \ \ \ 10011 \ \ \ \ \ 10101 \ \ \ \ \ 11010 \ \ \ \ \ 11000$

   Then

   $d(11001, 00000)= 3 \ \ \ \ \ \ \ \ \ \ d(11001, 00110)=5 \\ d(11001, 01001)=2 \ \ \ \ \ \ \ \ \ \ d(11001, 01111)=2 \\ d(11001, 10011)=3 \ \ \ \ \ \ \ \ \ \ d(11001, 10101)=2 \\ d(11001, 11010)=4 \ \ \ \ \ \ \ \ \ \ d(11001, 11000)=3$

   We observe that 01001, 01111 and 10101 are equidistance from 11001. But 01001 comes first in our list of codes. Since e(010)=01001, we decode 11001 as 010.

2. To decode the word 01010 we first arrange the given code as

   $00000 \ \ \ \ \ 00110 \ \ \ \ \ 01001 \ \ \ \ \ 01111 \ \ \ \ \ 10011 \ \ \ \ \ 10101 \ \ \ \ \ 11010 \ \ \ \ \ 11000$

   Then

   $d(01010, 00000)= 2 \ \ \ \ \ \ \ \ \ \ d(01010, 00110)=1 \\ d(01010, 01001)=3 \ \ \ \ \ \ \ \ \ \ d(01010, 01111)=3 \\ d(01010, 10011)=4 \ \ \ \ \ \ \ \ \ \ d(01010, 10101)=5 \\ d(01010, 11010)=2 \ \ \ \ \ \ \ \ \ \ d(01010, 11000)=2$

   We observe that 00110 are equidistance from 01010. Since e (001) = 00110, we decode 01010 as 001.

3. To decode the word 00111 we first arrange the given code as

   $00000 \ \ \ \ \ 00110 \ \ \ \ \ 01001 \ \ \ \ \ 01111 \ \ \ \ \ 10011 \ \ \ \ \ 10101 \ \ \ \ \ 11010 \ \ \ \ \ 11000$

   Then

   $d(00111, 00000)= 3 \ \ \ \ \ \ \ \ \ \ d(00111, 00110)=3 \\ d(00111, 01001)=2 \ \ \ \ \ \ \ \ \ \ d(00111, 01111)=3 \\ d(00111, 10011)=2 \ \ \ \ \ \ \ \ \ \ d(00111, 10101)=3 \\ d(00111, 11010)=2 \ \ \ \ \ \ \ \ \ \ d(00111, 11000)=5$

   We observe that 01001, 10011 and 11010 are equidistance from 00111. But 01001 comes first in our list of codes. Since e (010) =01001, we decode 00111 as 010.