University of Illinois Fall 2005
ECE 556/CS 577/MATH 579: Solutions to Problem Set 3
1. (a)A(1) = A0+A1+· · · +An= 2k, the total number of codewords.
(b)A(1) = A0−A1+A2− · · · + (−1)nAnis the difference between the number of codewords of even
weight and the number of codewords of odd weight. This is either 2kor 0 depending on whether
the code has vectors of even weight only, or of both odd and even weights.
(c)A0(1) = 0 ·A0+ 1 ·A1+·· · +n·Anis the total number of 1s in all the codewords. This sum is
2k−1nsince each coordinate has equal numbers of 1s and 0s.
2. Let µ1s and ν0s be erased where µ+ν≤2tby assumption. Note that this implies that min{µ, ν} ≤ t.
Then R0has µerrors in it and R1has νerrors in it.
(a) Since at least one of R0and R1has at most terrors in it, both decodings cannot end in deco ding
failure.
(b) A decoding failure might occur when the number of errors is greater than t. But, when max{µ, ν}>
tand one decoder fails, the other decoding is guaranteed to produce the correct codeword because
it is correcting min{µ, ν}< t errors.
(c) When max{µ, ν }> t, a successful decoding (resulting in the wrong codeword) is also a possibility.
This decoding will have corrected 2t+ 1 −max{µ, ν}>min{µ, ν}errors. The other decoding will
correct min{µ, ν}errors and produce the correct codeword. Thus, we can decide on which of the
two codewords produced is the correct one by checking the number of errors corrected in each
decoding.
If max{µ, ν}< t, both decodings produce the same (correct) codeword.
3. (a) There are k(n−k) entries in Pwhich can be chosen to be 0 or 1. Hence, there are 2k(n−k)choices
of P. Now, consider all the codes with minimum weight at least 3. A row of Ghas weight 1 or 2
if the corresponding row of Phas weight 0 or 1. Thus, codes of minimum weight 3 cannot have
rows in Pof weight 0 or 1. Now, the sum of irows of Ghas weight at least i. Thus, if Phad two
identical rows, there would be a codeword of weight 2. Hence, a code has minimum distance at
least 3 if the rows of Pare distinct and have minimum weight at least 2. There are N= (n−k) +1
binary vectors of length n−kof weight less than 2. Thus, we have 2n−k−Nchoices for the first
row of P, 2n−k−N−1 choices for the second, . . . , 2n−k−N−(k−1) = 2n−k−nchoices for the
k-th row of P. Hence there are
k−1
Y
i=0 2n−k−n+i≥2n−k−nk
codes of minimum weight at least 3.
(b) The maximum weight of any row of Gis n−k+1 (when the row of Pis 111. . .1). Such a row is a
codeword of weight n−k+1, and thus, the minimum weight cannot be any larger than n−k+ 1
. . ., in fact, the minimum weight is usually much smaller (for a binary code; Reed-Solomon codes,
which we shall study later, do satisfy the bound with equality).
(c)H= [PT|I] where Iis a (n−k)×(n−k) identity matrix. We write x= [y,z] where yis the
first kbits in xand zthe last n−kbits. Then, HxT=PTyT+zT. Now, if y=0and z6=0,
then HxT=zT6=0and hence xcannot belong to any of the codes in the ensemble. So suppose
that y6=0. The i-th bit in HxTis pi1x1+pi2x2+· · · pikxk+xk+iwhere the pij ’s are entries in
the i-th row of P. Now, remember that xis a fixed vector. We think of this i-th bit in HxTas
a nonzero linear function of the pij’s which change from code to code. This function has value 0
for 2k−1choices of the pij’s and value 1 for the other 2k−1choices. Since there are n−krows of
Pand we have 2k−1choices for each row that force the corresponding syndrome bit to be 0, we
conclude that 2(k−1)(n−k)of the 2k(n−k)codes in the ensemble contain x.
(d) Each nonzero vector of weight i < d can belong to 2(k−1)(n−k)different codes in the ensemble (it
might not belong to any code too). Hence, the total number of codes that contain a nonzero vector
