University of Illinois
Fall 2005
ECE 556:
Solutions to the First MidSemester Exam
1
.
(
a
) Applying the result wt(
x
⊕
y
) = wt(
x
) + wt(
y
)

2
·
wt(
x
∧
y
) to codewords
x
and
y
, we use the
fact that wt(
x
), wt(
y
), and wt(
x
⊕
y
) are multiples of 4 to deduce that wt(
x
∧
y
) is even. Hence
x
⊥
y
for any two codewords,
C ⊆ C
⊥
, and the rate of
C
is
≤
1
2
.
(
b
) Two binary vectors of odd Hamming weight sum to a vector of even Hamming weight. Now, the
minimum Hamming weight of a code is the smallest number of columns of the paritycheck matrix
H
that sum to 0. Since the columns are
distinct
vectors of odd Hamming weight, two columns
cannot sum to zero. The sum of three columns has odd Hamming weight, and hence cannot be
zero. Thus, the minimum Hamming weight must be at least 4.
2
.
(
a
) If
f
(
x
) =
x
3
+
x
2
+2 were reducible over GF(3), it would have at least one linear factor, which must
be
x,
(
x

1)
,
or (
x

2) and correspondingly we would have
f
(0) = 0 or
f
(1) = 0
,
or
f
(2) = 0.
But,
f
(0) =

f
(1) =
f
(2) = 2, and hence
f
(
x
) is irreducible over GF(3).
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 Fall '08
 Milenkovic,O
 Coding theory, 1 m, 2 m, Hamming weight, odd Hamming weight

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