Brief solutions to Prelim 2
1.
A binary code of dimension
k
has 2
k
codewords. So the answer is 2.
2.
The multiplicative inverse of
x
3
+
x
+ 1 mod
x
4
+
x
3
+ 1 in
F
2
[
x
] is
x
3
+
x
(by
the Euclidean algorithm, for instance).
3.
Computing powers of 2 mod 17 we find that
H
consists of the 8 elements
±
1
,
±
2
,
±
4
,
±
8.
So the index of
H
is 16
/
8 = 2.
The two cosets are
H
and
3
H
=
{±
3
,
±
6
,
±
5
,
±
7
}
.
4.
(a) Since
G
is in the standard form
G
= (
I

A
), we know that
H
:= (

A
T

I
)
is a parity check matrix:
H
=
2
2
1
1
0
1
0
2
0
1
.
This has no zero column, but it does have two columns that are scalar multiples
of one another. So
d
= 2.
(b) Using the parity check matrix in (a), we can check that the given vector
x
satisfies
H
x
T
6
=
0
. So it is not a codeword. Alternatively, just check directly
that
x
is not a linear combination of the 3 rows of
G
. [This involves a system
of 5 equations in 3 unknowns, which one checks to be inconsistent.]
5.
(a)
φ
(51) = (3

1)(17

1) = 32. The decoding exponent is the multiplicative
inverse of 11 mod 32, which is 3.
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 Spring '08
 BILLERA
 Algebra, Number Theory, Vector Space, Prime number, Hamming Code, Euclidean algorithm

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