MATH 3360
Solutions for Selected Porblems in Assignment 8
CS 2.12
One can use the spherepacking bound equation to see if the code is perfect.
However, a more intuitive way is this:

C

= 9,
d
(
C
) = 3 (Check!), and umber of possible words = 81. Since
d
(
C
) = 2(1)+1, the sphere has radius 1. Therefore, if
c
= (
a
1
,a
2
,a
3
,a
4
)
∈
C
,
the sphere containing
c
contains the words
{
c,
(
a
1
±
1
,a
2
,a
3
,a
4
)
,
(
a
1
,a
2
±
1
,a
3
,a
4
)
,
(
a
1
,a
2
,a
3
±
1
,a
4
)
,
(
a
1
,a
2
,a
3
,a
4
±
1)
}
which has 9 elements. So the 9 spheres corresponding to the 9 codes covers
9
×
9 = 81 elements, i.e. the code is perfect.
CS 2.19
Note that we want 3 linear dependent columns in the parity check matrix
H
.
No two columns are linearly dependent since (ONLY) in binary code, two
vectors are linearly depdendent iﬀ they are equal or one or more of them are
zero vectors.
13B 6
i)
g
(
x
) is a unit is not equivalent to
g
(1)
,g
(2)
,g
(3)
,g
(0)
6
= 0
,
2.
ii) One can easily check (1+2